One Application of Duistermaat-Heckman Measure in Quantum Information Theory

Lin Zhang; Xiaohan Jiang; Bing Xie

doi:10.2478/qic-2025-0033

Full Article

1.

Introduction

Quantum entanglement, a hallmark of quantum mechanics, is a fundamental resource for quantum information processing. Distinguishing entangled states from separable (non-entangled) states is therefore a central problem in quantum information theory. For bipartite systems, particularly the simplest non-trivial case of two-qubits, determining the likelihood of encountering a separable state within the ensemble of all possible states is a question of significant theoretical interest. This is formalized as the separability probability [1,2].

Under the Hilbert-Schmidt measure—a natural metric induced by the Hilbert-Schmidt distance on the space of density matrices [3]—the exact value of the separability probability for two-qubit states has been a subject of extensive numerical investigation and conjecture. Early numerical studies strongly suggested the fraction 8/33 as the exact value (with 29/64 for the real two-qubit case) [4–6]. This conjecture was rigorously confirmed recently by Huong and Khoi [7] (separately, the fraction 29/64 was analytically proved by Lovas and Andai [8]). However, while their result establishes the exact separability probability, the detailed mathematical derivation presented in [7] remains inaccessible to a significant portion of the community.

This paper aims to bridge this gap in accessibility and understanding. Our primary contribution is to provide a detailed, self-contained, and pedagogically clear derivation of the 8/33¹ separability probability for two-qubit states under the Hilbert-Schmidt measure. We achieve this by leveraging a powerful geometric framework centered on the computation of volumes [9] in the relevant state spaces and their substructures. The core of the used approach lies in the systematic computation of Hilbert-Schmidt/symplectic volumes associated with key geometric objects:

(1)
Flag manifold: This is a specific instance constructed by taking the quotient of the unitary group U(N) by its maximal torus.
(2)
A regular adjoint orbit: It is an orbit of a fixed diagonal matrix with distinct diagonal entries under the conjugate action of unitary group U(N). Note that this fixed diagonal matrix can be translated to a regular element in the positive Weyl chamber for the unitary group U(N).
(3)
The space of all qudit states: This is the compact manifold of all density matrices acting on the same underlying space ℂ^N.
(4)
A regular co-adjoint orbit: It is a orbit of a regular element in the positive Weyl chamber under the coadjoint action of the unitary group U(N).

A critical insight underpinning the derivation is the profound connection between the Hilbert-Schmidt volumes of adjoint orbits and the symplectic volumes of their corresponding regular co-adjoint orbits. This connection is rigorously established through the Duistermaat-Heckman (DH) measure [10], a fundamental tool in symplectic geometry and geometric quantization that intrinsically relates these volume measures. The DH measure’s properties are well-documented, with recent generalizations further extending its scope [11,12]. Crucially, its density relative to the Lebesgue measure on the dual Lie algebra is piecewise polynomial, computable via the Boylal-Vergne-Paradan jump formula [13]. Moreover, the DH formalism has proven instrumental in analyzing joint eigenvalue distributions of marginal states in random multipartite quantum systems [14], demonstrating its utility in quantum information theory.

The remainder of this paper is structured as follows: In Section 2, we review necessary background on the Hilbert-Schmidt metric, the geometry of the quantum state space, and separability criteria. We provide details about the computation of the Hilbert-Schmidt volume of the full quantum state space in Section 3 by calculating volume of flag manifolds and (co)adjoint orbit. Section 4 explores the connection to symplectic geometry via co-adjoint orbits and the Duistermaat-Heckman measure, deriving the corresponding symplectic volumes and linking them back to the Hilbert-Schmidt framework. In Section 5, we synthesizes the volume calculations from the previous sections to compute the volume of the separable set and derives the exact separability probability. Finally, in Section 6, we conclude with a discussion of the significance of the result and potential extensions.

2.

Preliminaries and Notations

Let the set of all N × N complex Hermitian matrices be denoted by Herm(ℂ^N). Furthermore, Herm_s(ℂ^N) denote the set of all elements in Herm(ℂ^N) with simple spectrum (i.e., those N × N complex Hermitian matrices with N distinct eigenvalues).

Proposition 1 ([15,16]).

It holds that the subset Herm_s(ℂ^N) is an open and dense of full measure in Herm(ℂ^N). Thatis, the Lebesgue measure of the complement is zero, i.e., the Lebesgue measure of Herm(ℂ^N)\Herm_s(ℂ^N) is zero.

Proof

The technical proof is omitted here.

We will denote by ℋ_N = ℋ₁ ⊗ ℋ₂ the N-dimensional Hilbert space that describes the composite system of two components ℋ₁, ℋ₂ with dim ℋ₁ = n, dim ℋ₂ = m and N = nm. The set D(ℂⁿ ⊗ ℂ^m) ofbipartite states on ℋ_N can be represented by complex N × N positive semi-definite matrices of unit trace. Let 1 $D_{s} (ℂ^{n} \otimes ℂ^{m}) := D (ℂ^{n} \otimes ℂ^{m}) \cap^{} {Herm}_{s} (ℂ^{n} \otimes ℂ^{m}) .$ {{\rm{D}}_{\rm{s}}}({^n} \otimes {^m}): = {\rm{D}}({^n} \otimes {^m})\mathop \cap \nolimits^ {\rm{Her}}{{\rm{m}}_{\rm{s}}}({^n} \otimes {^m}).

From Proposition 1, we see easily that D_s(ℂⁿ ⊗ ℂ^m) is also open and dense of full measure in D(ℂⁿ ⊗ ℂ^m). A bipartite state ρ_AB in D(ℂⁿ ⊗ ℂ^m) is called separable if and only if it can be written as a convex combination of local product states: $ρ_{A B} = \sum_{k = 1}^{K} p_{k} ρ_{A, k} \otimes ρ_{B, k}$ {\rho _{AB}} = \sum\nolimits_{k = 1}^K {{p_k}{\rho _{A,k}} \otimes {\rho _{B,k}}} for some ρ_A,k ∈ D(ℂⁿ), ρ_B,k ∈ D(ℂ^m), and a probability vector (p₁, …, p_K).

A natural parametrization of ρ ∈ D(ℂⁿ ⊗ ℂ^m) makes use of the traceless Hermitian generators of SU(n) and SU( m): 2 $ρ = \frac{1}{n m} (𝟙_{n m} + \sum_{i = 1}^{n^{2} - 1} a_{i} A_{i} \otimes 𝟙_{m} + 𝟙_{n} \otimes \sum_{j = 1}^{m^{2} - 1} b_{j} B_{j} + \sum_{k = 1}^{n^{2} - 1} \sum_{l = 1}^{m^{2} - 1} c_{k l} A_{k} \otimes B_{l})$ \rho = {1 \over {nm}}\left( {{X_{nm}} + \mathop \sum \limits_{i = 1}^{{n^2} - 1} {a_i}{A_i} \otimes {X_m} + {X_n} \otimes \mathop \sum \limits_{j = 1}^{{m^2} - 1} {b_j}{B_j} + \mathop \sum \limits_{k = 1}^{{n^2} - 1} \mathop \sum \limits_{l = 1}^{{m^2} - 1} {c_{kl}}{A_k} \otimes {B_l}} \right) where

all a_i, b_j, c_kl are in ℝ,
A_k and B_l are the Hermitian generators of the SU(n) and SU(m), respectively, chosen to satisfy 〈A_i, A_j〉 = 2δ_ij = 〈B_i, B_j〉,
two reduced states are: $ρ_{ℂ^{n}} = \frac{1}{n} (𝟙_{n} + \sum_{i = 1}^{n^{2} - 1} a_{i} A_{i})$ {\rho _{{^n}}} = {1 \over n}({\diamondsuit _n} + \sum\nolimits_{i = 1}^{{n^2} - 1} {{a_i}{{\bf{A}}_i})} and $ρ_{ℂ^{m}} = \frac{1}{m} (𝟙_{m} + \sum_{j = 1}^{m^{2} - 1} b_{j} B_{j})$ {\rho _{{^m}}} = {1 \over m}({\diamondsuit _m} + \sum\nolimits_{j = 1}^{{m^2} - 1} {{b_j}{{\bf{B}}_j})} ,
The vector $x = (a_{1}, \dots, a_{n^{2} - 1}, b_{1}, \dots, b_{m^{2} - 1}, c_{11}, \dots, c_{n^{2} - 1, m^{2} - 1}) ⊤ \in ℝ^{(n m)^{2} - 1}$ x = {({a_1}, \ldots ,{a_{{n^2} - 1}},{b_1}, \ldots ,{b_{{m^2} - 1}},{c_{11}}, \ldots ,{c_{{n^2} - 1,{m^2} - 1}})^ \top } \in {^{{{(nm)}^2} - 1}} completely determines the state ρ ∈ D(ℂⁿ ⊗ ℂ^m) and vice versa. The positivity of density matrices restricts the possible vectors x to a proper subset D^{(n × m)} of ℝ^{(nm)² – 1}.

According to the parametrization Eq. (2) of bipartite states, the Hilbert-Schmidt distance $d_{HS} (ρ^{'}, ρ) := \sqrt{{〈 ρ^{'} - ρ, ρ^{'} - ρ 〉}_{HS}}$ {d_{{\rm{HS}}}}({\rho ^\prime },\rho ): = \sqrt {{{\left\langle {{\rho ^\prime } - \rho ,{\rho ^\prime } - \rho } \right\rangle }_{{\rm{HS}}}}} (see below Eq. (8)) induces a flat metric on D^{(n × m)} because 3 $d_{HS} (ρ^{'}, ρ) = \frac{1}{n m} {(2 m \sum_{i = 1}^{n^{2} - 1} {(a_{i}^{'} - a_{i})}^{2} + 2 n \sum_{j = 1}^{m^{2} - 1} {(b_{j}^{'} - b_{j})}^{2} + 4 \sum_{k = 1}^{n^{2} - 1} \sum_{l = 1}^{m^{2} - 1} {(c_{k l}^{'} - c_{k l})}^{2})}^{\frac{1}{2}} .$ {d_{{\rm{HS}}}}\left( {{\rho ^\prime },\rho } \right) = {1 \over {nm}}{\left( {2m\sum\limits_{i = 1}^{{n^2} - 1} {{{\left( {a_i^\prime - {a_i}} \right)}^2}} + 2n\sum\limits_{j = 1}^{{m^2} - 1} {{{\left( {b_j^\prime - {b_j}} \right)}^2}} + 4\sum\limits_{k = 1}^{{n^2} - 1} {\sum\limits_{l = 1}^{{m^2} - 1} {{{\left( {c_{kl}^\prime - {c_{kl}}} \right)}^2}} } } \right)^{{1 \over 2}}}.

In particular, for n = m = 2, we get that 4 $d_{HS} (ρ^{'}, ρ) = \frac{1}{2} {(\sum_{i = 1}^{3} {(a_{i}^{'} - a_{i})}^{2} + \sum_{j = 1}^{3} {(b_{j}^{'} - b_{j})}^{2} + \sum_{k = 1}^{3} \sum_{l = 1}^{3} {(c_{k l}^{'} - c_{k l})}^{2})}^{\frac{1}{2}} = \frac{1}{2} d_{Euclid} (x^{'}, x),$ {d_{{\rm{HS}}}}\left( {{\rho ^\prime },\rho } \right) = {1 \over 2}{\left( {\sum\limits_{i = 1}^3 {{{\left( {a_i^\prime - {a_i}} \right)}^2}} + \sum\limits_{j = 1}^3 {{{\left( {b_j^\prime - {b_j}} \right)}^2}} + \sum\limits_{k = 1}^3 {\sum\limits_{l = 1}^3 {{{\left( {c_{kl}^\prime - {c_{kl}}} \right)}^2}} } } \right)^{{1 \over 2}}} = {1 \over 2}{d_{{\rm{Euclid}}}}\left( {{{\bf{x}}^\prime },{\bf{x}}} \right), where $x^{'} = (a^{'}, b^{'}, c_{k l}^{'})$ {x^\prime } = ({a^\prime },{b^\prime },c_{kl}^\prime ) and x = (a, b, c_kl). In such case, up to an overall constant $\frac{1}{2}$ {1 \over 2}, the following mapping is bijective and isometric: 5 $(D (ℂ^{2} \otimes ℂ^{2}), d_{HS}) ≅ (D^{(2 \times 2)}, d_{Euclid}) .$ ({\rm{D}}({^2} \otimes {^2}),{d_{{\rm{HS}}}}) \cong ({D^{(2 \times 2)}},{d_{{\rm{Euclid}}}}).

3.

Riemannian Volumes of Manifolds

Recall that an N-dimensional oriented manifold M with a pseudo-Riemannian metric g has a standard volume form ω, known as the Riemannian volume form, whose expression in an oriented chart (x¹, …, x^N) is given by 6 $ω = \sqrt{\det (g)} d x^{1} \land \dots \land d x^{N},$ \omega = \sqrt {\det (g)} {\rm{d}}{x^1} \wedge \cdots \wedge {\rm{d}}{x^N}, corresponding to the line element (aka, arc length differential or differential of arc length) $d s^{2} = \sum_{i, j = 1}^{N} d x^{i} g_{i j} d x^{j}$ {\rm{d}}{s^2} = \sum\nolimits_{i,j = 1}^N {{\rm{d}}{x^i}{g_{ij}}{\rm{d}}{x^j}} .

If D is a domain of integration on M, then 7 ${vol}_{g} (D) := \int_{D} ω$ {\rm{vo}}{{\rm{l}}_g}(D): = \int_D \omega is called the Riemannian volume of D.

In particular, on D_s(ℂ^N), we have the Hilbert-Schmidt inner product, which is defined by 8 $〈 X, ϒ 〉 := Tr (X^{†} ϒ) .$ \left\langle {X,\Upsilon } \right\rangle : = {\rm{Tr}}({X^\dag }\Upsilon ).

Differentiating this inner product yields a metric on D_s(ℂ^N) which we denote by g_HS. We shall denote by vol_HS the Riemannian volume form associated with g_HS and refer to the volume measured by vol_HS as the HS volume [17].

3.1.

Hilbert-Schmidt Volumes of Flag Manifolds

A complex matrix Z can always be written as Z = X + iϒ, where X = Re(Z) and ϒ = Im(Z) are real matrices.

Denote 9 $[d Z] := [d X] [d ϒ],$ [{\rm{d}}Z]: = [{\rm{d}}X][{\rm{d}}\Upsilon ], where [dX] means the product of independent differentials in X, and [dY] has a similar meaning.

Assume that U = (u_ij) ∈ U(N), where u_ij ∈ ℂ. Let u_kk = |u_kk|e^iθ_k be its polar form, where θ_k ∈ [— π, π).

Then 10 $U = (U T^{- 1}) T, T = diag (e^{i θ_{1}}, \dots, e^{i θ_{N}}) .$ U = (U{T^{ - 1}})T,\quad T = {\rm{diag}}({e^{{\rm{i}}{\theta _1}}}, \ldots ,{e^{{\rm{i}}{\theta _N}}}).

This means that any U ∈ U (N) can be factorized into a product of a unitary matrix with diagonal entries being nonnegative and a diagonal unitary matrix. Thus, we have the following diffeomorphism: 11 $U (N) ≅ U (N) / T^{N} \times T^{N},$ {\rm{U}}(N) \cong {\rm{U}}(N)/{^N} \times {^N}, where $T^{N} := {diag (e^{i θ_{1}}, \dots, e^{i θ_{N}}) : θ_{k} \in [- π, π), k = 1, \dots, N}$ {^N}: = \left\{ {{\mathop{\rm diag}\nolimits} \left( {{e^{{\rm{i}}{\theta _1}}}, \ldots ,{e^{{\rm{i}}{\theta _N}}}} \right):{\theta _k} \in [ - \pi ,\pi ),k = 1, \ldots ,N} \right\}, which is the maximal torus of U(N). For any measurable f over U(N), 12 $\int_{U (N)} f (U) [U^{†} d U] = \int_{U (N) / T^{N}} \int_{T^{N}} f (V T) [V^{†} d V] [T^{†} d T],$ \int_{{\rm{U}}(N)} {f(U)[{U^\dag }{\rm{d}}U]} = \int_{{\rm{U}}(N)/{^N}} {\int_{{^N}} {f(VT)[{V^\dag }{\rm{d}}V][{T^\dag }{\rm{d}}T]} } , where U = VT for V ∈ U(N)/𝕋^N and T ∈ 𝕋^N, and [U^†dU] = [V ^†dV][T^†dT], where [U^†dU] is a leftinvariant matrix-valued differential form, called Maurer-Cartan form. Since v_kk ∈ ℝ_>0, and moreover v_kk is not an independent variable, it follows that 13 $[V^{†} d V] = \prod_{1 ⩽ i < j ⩽ N} Re {(V^{†} d V)}_{i j}) Im {(V^{†} d V)}_{i j}),$ \matrix{ {[{V^\dag }{\rm{d}}V]} & = & {\mathop \prod \limits_{1i < jN} {\mathop{\rm Re}\nolimits} {{({V^\dag }{\rm{d}}V)}_{ij}}){\rm{Im}}{{({V^\dag }{\rm{d}}V)}_{ij}}),} \cr } 14 $[U^{†} d U] = \prod_{k = 1}^{N} Im ({(U^{†} d U)}_{k k}) \times \prod_{1 ⩽ i < j ⩽ N} Re {(U^{†} d U)}_{i j} Im {(U^{†} d U)}_{i j}$ \matrix{ {[{U^\dag }{\rm{d}}U]} & = & {\mathop \prod \limits_{k = 1}^N {\rm{Im}}({{({U^\dag }{\rm{d}}U)}_{kk}}) \times \mathop \prod \limits_{1i < jN} {\mathop{\rm Re}\nolimits} {{({U^\dag }{\rm{d}}U)}_{ij}}{\rm{Im}}{{({U^\dag }{\rm{d}}U)}_{ij}}} \cr }

The Euclid volume of the flag manifold U(N)/𝕋^N is given by [15]: 15 ${vol}_{Euclid} (U (N) / T^{N}) = \int_{U (N) / T^{N}} [V^{†} d V] = \frac{π^{(\begin{matrix} N \\ 2 \end{matrix})}}{\prod_{k = 1}^{N} Γ (k)} .$ {{\mathop{\rm vol}\nolimits} _{{\rm{Euclid }}}}\left( {{\rm{U}}(N)/{^N}} \right) = \int_{{\rm{U}}(N)/{^N}} {\left[ {{{\bf{V}}^\dag }{\rm{d}}{\bf{V}}} \right]} = {{{\pi ^{\left( {\scriptstyle N \atop \scriptstyle 2} \right)}}} \over {\prod\nolimits_{k = 1}^N {\Gamma (k)} }}.

Proposition 2 ([15]).

The Hilbert-Schmidt volume of the flag manifold U(N)/𝕋^N is given by 16 ${vol}_{HS} (U (N) / T^{N}) = 2^{(\begin{matrix} N \\ 2 \end{matrix})} {Vol}_{Euclid} (U (N) / T^{N}) = \frac{{(2 π)}^{(\begin{matrix} N \\ 2 \end{matrix})}}{\prod_{k = 1}^{N} Γ (k)},$ {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{\rm{U}}(N)/{^N}} \right) = {2^{\left( {\scriptstyle N \atop \scriptstyle 2} \right)}}{{\mathop{\rm Vol}\nolimits} _{{\rm{Euclid }}}}\left( {{\rm{U}}(N)/{^N}} \right) = {{{{(2\pi )}^{\left( {\scriptstyle N \atop \scriptstyle 2} \right)}}} \over {\prod\nolimits_{k = 1}^N {\Gamma (k)} }},

Proof

In fact, the line element for the Hilbert-Schmidt inner metric is given by $d s_{HS}^{2} = 〈 d V, d V 〉 = 〈 V^{†} d V, V^{†} d V 〉, V \in U (N) / T^{N} .$ {\rm{d}}s_{{\rm{HS}}}^2 = \left\langle {{\rm{d}}V,{\rm{d}}V} \right\rangle = \left\langle {{V^\dag }{\rm{d}}V,{V^\dag }{\rm{d}}V} \right\rangle ,\quad V \in {\rm{U}}(N)/{^N}.

It is clearly seen that V^†dV is skew-Hermitian and its diagonal part is real. Thus the diagonal part of V^†dV must be vanished, and $\begin{array}{l} {ds}_{HS}^{2} & = & \sum_{k = 1}^{N} {[{(V^{†} d V)}_{k k}]}^{2} + 2 \underset{1 ⩽ i < j ⩽ N}{\sum^{}} ({[Re {(V^{†} d V)}_{i j}]}^{2} + {[Im {(V^{†} d V)}_{i j}]}^{2}) \\ = & 2 \underset{1 ⩽ i < j ⩽ N}{\sum^{}} ({[Re {(V^{†} d V)}_{i j}]}^{2} + {[Im {(V^{†} d V)}_{i j}]}^{2}) . \end{array}$ \matrix{ {{\rm{d}}s_{{\rm{HS}}}^2} \hfill & = \hfill & {\mathop \sum \limits_{k = 1}^N {{[{{({V^\dag }{\rm{d}}V)}_{kk}}]}^2} + 2\mathop \sum \limits_{1i < jN} ({{[{\mathop{\rm Re}\nolimits} {{({V^\dag }{\rm{d}}V)}_{ij}}]}^2} + {{[Im{{({V^\dag }{\rm{d}}V)}_{ij}}]}^2})} \hfill \cr {} \hfill & = \hfill & {2\mathop \sum \limits_{1i < jN} ({{[{\mathop{\rm Re}\nolimits} {{({V^\dag }{\rm{d}}V)}_{ij}}]}^2} + {{[Im{{({V^\dag }{\rm{d}}V)}_{ij}}]}^2}).} \hfill \cr }

The corresponding HS volume element is given by $\begin{array}{l} d V_{HS} & = & \prod_{1 ⩽ i < j ⩽ N} (\sqrt{2} Re {(V^{†} d V)}_{i j} \sqrt{2} Im {(V^{†} d V)}_{i j}) \\ = & 2^{(\begin{matrix} N \\ 2 \end{matrix})} \prod_{1 ⩽ i < j ⩽ N} (Re {(V^{†} d V)}_{i j} Im {(V^{†} d V)}_{i j}) = 2^{(\begin{matrix} N \\ 2 \end{matrix})} [V^{†} d V], \end{array}$ \matrix{ {{\rm{d}}{V_{{\rm{HS}}}}} \hfill & = \hfill & {\mathop \prod \limits_{1i < jN} (\sqrt 2 {\mathop{\rm Re}\nolimits} {{({V^\dag }{\rm{d}}V)}_{ij}}\sqrt 2 {\rm{Im}}{{({V^\dag }{\rm{d}}V)}_{ij}})} \hfill \cr {} \hfill & = \hfill & {{2^{(\matrix{ N \cr 2 \cr } )}}\mathop \prod \limits_{1i < jN} ({\mathop{\rm Re}\nolimits} {{({V^\dag }{\rm{d}}V)}_{ij}}{\rm{Im}}{{({V^\dag }{\rm{d}}V)}_{ij}}) = {2^{(\matrix{ N \cr 2 \cr } )}}[{V^\dag }{\rm{d}}V],} \hfill \cr } which leads to the conclusion $\begin{array}{l} {vol}_{HS} (U (N) / T^{N}) & = & \int_{U (N) / T^{N}} d V_{HS} = 2^{(\begin{matrix} N \\ 2 \end{matrix})} \int_{U (N) / T^{N}} [V^{†} d V] \\ = & 2^{(\begin{matrix} N \\ 2 \end{matrix})} {vol}_{Euclid} (U (N) / T^{N}) \\ = & 2^{(\begin{matrix} N \\ 2 \end{matrix})} \frac{π^{(\begin{matrix} N \\ 2 \end{matrix})}}{\prod_{k = 1}^{N} Γ (k)} = \frac{{(2 π)}^{(\begin{matrix} N \\ 2 \end{matrix})}}{\prod_{k = 1}^{N} Γ (k)} . \end{array}$ \matrix{ {{\rm{vo}}{{\rm{l}}_{{\rm{HS}}}}({\rm{U}}(N)/{^N})} \hfill & = \hfill & {{\smallint _{{\rm{U}}(N)/{^N}}}{\rm{d}}{V_{{\rm{HS}}}} = {2^{(\matrix{ N \cr 2 \cr } )}}{\smallint _{{\rm{U}}(N)/{^N}}}[{V^\dag }{\rm{d}}V]} \hfill \cr {} \hfill & = \hfill & {{2^{(\matrix{ N \cr 2 \cr } )}}{\rm{vo}}{{\rm{l}}_{{\rm{Euclid}}}}({\rm{U}}(N)/{^N})} \hfill \cr {} \hfill & = \hfill & {{2^{(\matrix{ N \cr 2 \cr } )}}{{{\pi ^{(\matrix{ N \cr 2 \cr } )}}} \over {\mathop \prod \nolimits_{k = 1}^N {\rm{\Gamma }}(k)}} = {{{{(2\pi )}^{(\matrix{ N \cr 2 \cr } )}}} \over {\mathop \prod \nolimits_{k = 1}^N {\rm{\Gamma }}(k)}}.} \hfill \cr }

This completes the proof.

3.2.

Hilbert-Schmidt Volumes of Regular Adjoint Orbits

Denote the adjoint orbit 𝒰_x for x = (x₁, …, X_N) ∈ ℝ^N, where x₁ > ⋯ > x_N, as 17 $U_{x} := {U d i a g (x_{1}, \dots, x_{N}) U^{†} : U \in U (N)}$ {{\cal U}_x}: = \{ U{\rm{diag}}({x_1}, \ldots ,{x_N}){U^\dag }:U \in {\rm{U}}(N)\} .

Let $V_{N} (x) := \prod_{1 ⩽ i < j ⩽ N} (x_{i} - x_{j})$ {V_N}(x): = \prod\nolimits_{1i < jN} {({x_i} - {x_j})} . The stablizer group of x is just the flag manifold U(N)/𝕋^N. Apparently, 𝒰_x ≌ {x} × U(N)/𝕋^N, where 𝕋^N = U(1)^×N is N-torus. Suppose that the unitary group U(N) is equipped with Haar probability measure μ_Haar. Consider the following map $\begin{matrix} π : & (U (N), μ_{Haar}) \mapsto (U_{x}, v := π_{*} μ_{Haar}), \\ U \mapsto U diag (x_{1}, \dots, x_{N}) U^{†} . \end{matrix}$ \matrix{ {\pi :} & {({\rm{U}}(N),{\mu _{{\rm{Haar}}}}) \mapsto ({{\cal U}_x},v: = {\pi _ * }{\mu _{{\rm{Haar}}}}),} \cr {} & {U \mapsto U{\rm{diag}}({x_1}, \ldots ,{x_N}){U^\dag }.} \cr }

The push-forward measure of μ_Haar along the map π is v := π_*μ_Haar. called the orbital measure [18]. So the integral along the orbit is given by the following formula $\begin{array}{l} \int_{U_{x}} f d v = 〈 v, f 〉 = 〈 π_{*} μ_{Haar}, f 〉 = 〈 μ_{Haar}, π^{*} f 〉 \\ = \int_{U (N)} f (U diag (x_{1}, \dots, x_{N}) U^{†}) d μ_{Haar} (U) . \end{array}$ \matrix{ {\int_{{{\cal U}_x}} {f{\rm{d}}v = \left\langle {v,f} \right\rangle = \langle {\pi _ * }{\mu _{{\rm{Haar}}}},f\rangle = \langle {\mu _{{\rm{Haar}}}},{\pi ^ * }f\rangle } } \hfill \cr { = \int_{{\rm{U}}(N)} {f(U{\rm{diag}}({x_1}, \ldots ,{x_N}){U^\dag }){\rm{d}}{\mu _{{\rm{Haar}}}}(U).} } \hfill \cr }

The following result [19] is mentioned without proof. Here we provide detailed proof for the reference.

Proposition 3.

The Hilbert-Schmidt volume of the adjoint orbit 𝒰_x ≌ U(N)/𝕋^N is given by 18 ${vol}_{HS} (U_{x}) = V_{N} {(x)}^{2} {vol}_{HS} (U (N) / T^{N}) .$ {\rm{vo}}{{\rm{l}}_{{\rm{HS}}}}({{\cal U}_x}) = {V_N}{(x)^2}{\rm{vo}}{{\rm{l}}_{{\rm{HS}}}}({\rm{U}}(N)/{^N}).

Proof

Indeed, for X ∈ 𝒰_x, i.e., X = Vdiag(x₁, …, x_N)V^†, where V ∈ U(N)/𝕋^N. Then $d X = d V diag (x_{1}, \dots, x_{N}) V^{†} + V diag (x_{1}, \dots, x_{N}) d V^{†},$ {\rm{d}}X = {\rm{d}}V{\rm{diag}}({x_1}, \ldots ,{x_N}){V^\dag } + V{\rm{diag}}({x_1}, \ldots ,{x_N}){\rm{d}}{V^\dag }, which implies that $V^{†} (d X) V = V^{†} d V diag (x_{1}, \dots, x_{N}) + diag (x_{1}, \dots, x_{N}) d V^{†} V .$ {V^\dag }({\rm{d}}X)V = {V^\dag }{\rm{d}}V{\rm{diag}}({x_1}, \ldots ,{x_N}) + {\rm{diag}}({x_1}, \ldots ,{x_N}){\rm{d}}{V^\dag }V.

Due to the fact that V^†V =𝟙_N, we infer that $(d V^{†}) V + V^{†} d V = 0 \Leftrightarrow (d V^{†}) V = - V^{†} d V .$ ({\rm{d}}{V^\dag })V + {V^\dag }{\rm{d}}V = 0 \Leftrightarrow ({\rm{d}}{V^\dag })V = - {V^\dag }{\rm{d}}V.

This indicates that V^†(dX)V = [V^†dV, diag(x₁, …, x_N)]. Thus $\begin{array}{l} [d X] & = & [V^{†} d X V] = \prod_{i < j} {[V^{†} d V, diag (x_{1}, \dots, x_{N})]}_{i j} \\ = & \prod_{i < j} (x_{j} - x_{i}) Re {(V^{†} d V)}_{i j} (x_{j} - x_{i}) Im {(V^{†} d V)}_{i j} \\ = & V_{N} {(x)}^{2} [V^{†} d V], \end{array}$ \matrix{ {[{\rm{d}}X]} \hfill & = \hfill & {[{V^\dag }{\rm{d}}XV] = \mathop \prod \limits_{i < j} {{[{V^\dag }{\rm{d}}V,diag({x_1}, \ldots ,{x_N})]}_{ij}}} \hfill \cr {} \hfill & = \hfill & {\mathop \prod \limits_{i < j} ({x_j} - {x_i}){\mathop{\rm Re}\nolimits} {{({V^\dag }{\rm{d}}V)}_{ij}}({x_j} - {x_i}){\rm{Im}}{{({V^\dag }{\rm{d}}V)}_{ij}}} \hfill \cr {} \hfill & = \hfill & {{V_N}{{(x)}^2}[{V^\dag }{\rm{d}}V],} \hfill \cr } where $[V^{†} d V] = \prod_{1 ⩽ i < j ⩽ N} Re {(V^{†} d V)}_{i j} Im {(V^{†} d V)}_{i j}$ \left[ {{{\bf{V}}^\dag }{\rm{d}}{\bf{V}}} \right] = \prod\nolimits_{1i < jN} {{\mathop{\rm Re}\nolimits} } {\left( {{{\bf{V}}^\dag }{\rm{d}}{\bf{V}}} \right)_{ij}}{\mathop{\rm Im}\nolimits} {\left( {{{\bf{V}}^\dag }{\rm{d}}{\bf{V}}} \right)_{ij}} is the Euclid volume element on the flag manifold U(N)/𝕋^N. We can derive that [dX] = V_N(x)²[V^†dV]. Now $\begin{array}{l} {vol}_{Euclid} (U_{x}) & = & \int_{U_{x}} [d X] = V_{N} {(x)}^{2} \int_{U (N) / T^{N}} [V^{†} d V] \\ = & V_{N} {(x)}^{2} {vol}_{Euclid} (U (N) / T^{N}) . \end{array}$ \matrix{ {{\rm{vo}}{{\rm{l}}_{{\rm{Euclid}}}}({{\cal U}_x})} \hfill & = \hfill & {\int_{{{\cal U}_x}} {[{\rm{d}}X] = {V_N}{{(x)}^2}} \int_{{\rm{U}}(N)/{^N}} {[{V^\dag }{\rm{d}}V]} } \hfill \cr {} \hfill & = \hfill & {{V_N}{{(x)}^2}{\rm{vo}}{{\rm{l}}_{{\rm{Euclid}}}}({\rm{U}}(N)/{^N}).} \hfill \cr }

The differential of arc length is given by $\begin{array}{l} d s_{HS}^{2} & = & 〈 d X, d X 〉 = 〈 [V^{†} d V, diag (x_{1}, \dots, x_{N})], [V^{†} d V, diag (x_{1}, \dots, x_{N})] 〉 \\ = & \sum_{i, j = 1}^{N} {| {[V^{†} d V, diag (x_{1}, \dots, x_{N})]}_{i j} |}^{2} = \sum_{i, j = 1}^{N} {| (x_{j} - x_{i}) {(V^{†} d V)}_{i j} |}^{2} \\ = & \sum_{i \neq j} {(x_{j} - x_{i})}^{2} {| {(V^{†} d V)}_{i j} |}^{2} = 2 \sum_{1 ⩽ i < j ⩽ N} {(x_{j} - x_{i})}^{2} {| {(V^{†} d V)}_{i j} |}^{2} \\ = & \sum_{1 ⩽ i < j ⩽ N} {(\sqrt{2} (x_{j} - x_{i}) Re {(V^{†} d V)}_{i j})}^{2} + \sum_{1 ⩽ i < j ⩽ N} {(\sqrt{2} (x_{j} - x_{i}) Im {(V^{†} d V)}_{i j})}^{2} . \end{array}$ \matrix{ {{\rm{d}}s_{{\rm{HS}}}^{\rm{2}}} \hfill & = \hfill & {\left\langle {{\rm{d}}X,{\rm{d}}X} \right\rangle = \langle [{V^\dag }{\rm{d}}V,{\rm{diag}}({x_1}, \ldots ,{x_N})],[{V^\dag }{\rm{d}}V,{\rm{diag}}({x_1}, \ldots ,{x_N})]\rangle } \hfill \cr {} \hfill & = \hfill & {\mathop \sum \limits_{i,j = 1}^N {{\left| {{{[{V^\dag }{\rm{d}}V,{\rm{diag}}({x_1}, \ldots ,{x_N})]}_{ij}}} \right|}^2} = \mathop \sum \limits_{i,j = 1}^N {{\left| {({x_j} - {x_i}){{({V^\dag }{\rm{d}}V)}_{ij}}} \right|}^2}} \hfill \cr {} \hfill & = \hfill & {\mathop \sum \limits_{i \ne j} {{({x_j} - {x_i})}^2}{{\left| {{{({V^\dag }{\rm{d}}V)}_{ij}}} \right|}^2} = 2\mathop \sum \limits_{1i < jN} {{({x_j} - {x_i})}^2}{{\left| {{{({V^\dag }{\rm{d}}V)}_{ij}}} \right|}^2}} \hfill \cr {} \hfill & = \hfill & {\mathop \sum \limits_{1i < jN} {{(\sqrt 2 ({x_j} - {x_i}){\mathop{\rm Re}\nolimits} {{({V^\dag }{\rm{d}}V)}_{ij}})}^2} + \mathop \sum \limits_{1i < jN} {{(\sqrt 2 ({x_j} - {x_i}){\rm{Im}}{{({V^\dag }{\rm{d}}V)}_{ij}})}^2}.} \hfill \cr }

The corresponding HS volume is given by $d V_{HS} = 2^{(\begin{matrix} N \\ 2 \end{matrix})} V_{N} (x)^{2} [V^{†} d V]$ {\rm{d}}{V_{{\rm{HS}}}} = {2^{(\matrix{ N \cr 2 \cr } )}}{V_N}{(x)^2}[{{\bf{V}}^\dag }{\rm{d}}{\bf{V}}]. Thus $\begin{array}{l} {vol}_{HS} (U_{x}) = 2^{(\begin{matrix} N \\ 2 \end{matrix})} {vol}_{Euclid} (U_{x}) = 2^{(\begin{matrix} N \\ 2 \end{matrix})} V_{N} {(x)}^{2} {vol}_{Euclid} (U (N) / T^{N}) \\ = 2^{(\begin{matrix} N \\ 2 \end{matrix})} V_{N} {(x)}^{2} \frac{π^{(N)} \begin{matrix} N \end{matrix})}{\prod_{k = 1}^{N} Γ (k)} = V_{N} {(x)}^{2} \frac{{(2 π)}^{(\begin{matrix} N \\ 2 \end{matrix})}}{\prod_{k = 1}^{N} Γ (k)}, \end{array}$ \matrix{ {{\rm{vo}}{{\rm{l}}_{{\rm{HS}}}}({{\cal U}_x}) = {2^{(\matrix{ N \cr 2 \cr } )}}{\rm{vo}}{{\rm{l}}_{{\rm{Euclid}}}}({{\cal U}_x}) = {2^{(\matrix{ N \cr 2 \cr } )}}{V_N}{{(x)}^2}{\rm{vo}}{{\rm{l}}_{{\rm{Euclid}}}}({\rm{U}}(N)/{^N})} \hfill \cr { = {2^{(\matrix{ N \cr 2 \cr } )}}{V_N}{{(x)}^2}{{{\pi ^{(N)}}\matrix{ N \cr } )} \over {\mathop \prod \nolimits_{k = 1}^N {\rm{\Gamma }}(k)}} = {V_N}{{(x)}^2}{{{{(2\pi )}^{(\matrix{ N \cr 2 \cr } )}}} \over {\mathop \prod \nolimits_{k = 1}^N {\rm{\Gamma }}(k)}},} \hfill \cr } implying that the HS volume of adjoint orbit 𝒰_x ≌ U(N)/𝕋^N is given by the following formula: ${vol}_{HS} (U_{x}) = V_{N} (x)^{2} \frac{{(2 π)}^{(\begin{matrix} N \\ 2 \end{matrix})}}{\prod_{k = 1}^{N} Γ (k)} = V_{N} (x)^{2} {vol}_{HS} (U (N) / T^{N}) .$ {\rm{vo}}{{\rm{l}}_{{\rm{HS}}}}({{\cal U}_x}) = {V_N}{(x)^2}{{{{(2\pi )}^{(\matrix{ N \cr 2 \cr } )}}} \over {\mathop \prod \nolimits_{k = 1}^N {\rm{\Gamma }}(k)}} = {V_N}{(x)^2}{\rm{vo}}{{\rm{l}}_{{\rm{HS}}}}({\rm{U}}(N)/{^N}).

We are done.

3.3.

Hilbert-Schmidt Volumes of Quantum State Spaces

Let 19 $C_{N} := {(x_{1}, \dots, x_{N}) \in ℝ^{N} : x_{1} > \dots > x_{N}},$ {{\cal C}_N}: = \left\{ {\left( {{x_1}, \ldots ,{x_N}} \right) \in {^N}:{x_1} > \cdots > {x_N}} \right\}, 20 $Δ_{N - 1} := {(x_{1}, \dots, x_{N}) \in ℝ_{⩾ 0}^{N} : \sum_{k = 1}^{N} x_{k} = 1} .$ {\Delta _{N - 1}}: = \left\{ {\left( {{x_1}, \ldots ,{x_N}} \right) \in _{0}^N:\sum\limits_{k = 1}^N {{x_k}} = 1} \right\}.

The natural projection U(N) → U(N)/𝕋^N, where 𝕋^N = {diag(e^iθ₁, …, e^iθ_N) : θ₁, …, θ_N ∈ ℝ} such that: $U \mapsto [U] \equiv U T^{N} \in U (N) / T^{N} .$ {\bf{U}} \mapsto [{\bf{U}}] \equiv {\bf{U}}{^N} \in {\rm{U}}(N)/{^N}.

Define the map τ : Herm_s(ℂ^N) → 𝒞_N × U(N)/𝕋^N.

Theorem 1 ([16]).

It holds that the above map τ is a diffeomorphism: 21 ${Herm}_{s} (ℂ^{N}) \overset{τ}{≅} C_{N} \times (U (N) / T^{N}) .$ {{\mathop{\rm Herm}\nolimits} _{\rm{s}}}\left( {{^N}} \right)\mathop \cong \limits^\tau {{\cal C}_N} \times \left( {{\rm{U}}(N)/{^N}} \right).

Moreover, the map τ induces the following two diffeomorphisms: 22 $D_{s} (ℂ^{N}) ≅ (Δ_{N - 1} \cap C_{N}) \times (U (N) / T^{N}),$ {{\rm{D}}_{\rm{s}}}\left( {{^N}} \right) \cong \left( {{\Delta _{N - 1}} \cap {{\cal C}_N}} \right) \times \left( {{\rm{U}}(N)/{^N}} \right), 23 $U_{x} ≅ {x} \times (U (N) / T^{N}) ≅ U (N) / T^{N} .$ {{\cal U}_x} \cong \{ x\} \times \left( {{\rm{U}}(N)/{^N}} \right) \cong {\rm{U}}(N)/{^N}.

Proof

The proof is omitted here.

Although the HS volume of the set of all density matrices acting on ℂ^N is derived already by Życzkowski, we still include the proof here for completeness.

Theorem 2 ([3]).

It holds that 24 ${vol}_{HS} (D (ℂ^{N})) = \sqrt{N} (2 π)^{(\begin{matrix} N \\ 2 \end{matrix})} \frac{\prod_{k = 1}^{N} Γ (k)}{Γ (N^{2})} .$ {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{\rm{D}}\left( {{^N}} \right)} \right) = \sqrt N {(2\pi )^{\left( {\scriptstyle N \atop \scriptstyle 2} \right)}}{{\prod\nolimits_{k = 1}^N \Gamma (k)} \over {\Gamma \left( {{N^2}} \right)}}.

Proof

The infinitesimal distance takes a particularly simple form $d s_{HS}^{2} = 〈 d ρ, d ρ 〉_{HS}$ {\rm{d}}s_{{\rm{HS}}}^2 = {\langle {\rm{d}}\rho ,{\rm{d}}\rho \rangle _{{\rm{HS}}}} valid for any dimension N. Making use of the diagonal form ρ = UΛU^†, we may write $d ρ = V (d Λ + [V^{†} d V, Λ]) V^{†},$ {\rm{d}}\rho = V\left( {{\rm{d}}\Lambda + \left[ {{V^\dag }{\rm{d}}V,\Lambda } \right]} \right){V^\dag }, where V ∈ U(N)/𝕋^N. Thus the differential of arc length can be rewritten as $d s_{HS}^{2} = \sum_{j = 1}^{N} d λ_{j}^{2} + 2 \sum_{1 ⩽ i < j ⩽ N}^{N} {(λ_{i} - λ_{j})}^{2} {| 〈 i | V^{†} d V | j 〉 |}^{2} .$ {\rm{d}}s_{{\rm{HS}}}^2 = \sum\limits_{j = 1}^N {{\rm{d}}} \lambda _j^2 + 2\sum\limits_{1i < jN}^N {{{\left( {{\lambda _i} - {\lambda _j}} \right)}^2}} {\left| {\left\langle {i\left| {{V^\dag }{\rm{d}}V} \right|j} \right\rangle } \right|^2}.

Apparently, $\sum_{j = 1}^{N} d λ_{j} = 0$ \sum\nolimits_{j = 1}^N {\rm{d}} {\lambda _j} = 0 since $\sum_{j = 1}^{N} λ_{j} = 1$ \sum\nolimits_{j = 1}^N {{\lambda _j} = 1} . Thus $d s_{HS}^{2} = \sum_{i, j = 1}^{N - 1} d λ_{i} g_{i j} d λ_{j} + 2 \sum_{1 ⩽ i < j ⩽ N}^{N} {(λ_{i} - λ_{j})}^{2} {| 〈 i | V^{†} d V | j 〉 |}^{2} .$ {\rm{d}}s_{{\rm{HS}}}^2 = \sum\limits_{i,j = 1}^{N - 1} {{\rm{d}}{\lambda _i}{g_{ij}}{\rm{d}}{\lambda _j} + 2} \sum\limits_{1i < jN}^N {{{\left( {{\lambda _i} - {\lambda _j}} \right)}^2}} {\left| {\left\langle {i\left| {{V^\dag }{\rm{d}}V} \right|j} \right\rangle } \right|^2}.

The HS volume element is given by $d V_{HS} = \sqrt{N} 2^{(\begin{matrix} N \\ 2 \end{matrix})} V_{N} (λ)^{2} \prod_{k = 1}^{N - 1} d λ_{k} \prod_{1 ⩽ i < j ⩽ N} Re {(V^{†} d V)}_{i j} Im {(V^{†} d V)}_{i j} .$ {\rm{d}}{V_{{\rm{HS}}}} = \sqrt N {2^{\left( {\scriptstyle N \atop \scriptstyle 2} \right)}}{V_N}{(\lambda )^2}\prod\limits_{k = 1}^{N - 1} {{\rm{d}}} {\lambda _k}\prod\limits_{1i < jN} {{\mathop{\rm Re}\nolimits} } {\left( {{V^\dag }{\rm{d}}V} \right)_{ij}}{\mathop{\rm Im}\nolimits} {\left( {{V^\dag }{\rm{d}}V} \right)_{ij}}.

The corresponding volume element gains a factor $\sqrt{\det (g)} = \sqrt{N}$ \sqrt {\det (g)} = \sqrt N , where g = (g_ij)_{(N–1) × (N–1)} is the metric in the (N – 1)-dimensional simplex △_{N 1} of eigenvalues. Note that g = 𝟙_{N 1} + |e〉〈e| for (N – 1)-dimensional vector e := (1, …, 1)^⊤. Therefore $\begin{array}{l} {vol}_{HS} (D (ℂ^{N})) & = & \int_{D (ℂ^{N})} d V_{HS} = \int_{Δ_{N - 1} \cap C_{N}} V_{N} {(λ)}^{2} \sqrt{N} \prod_{k = 1}^{N - 1} d λ_{k} \times \int_{U (N) / T^{N}} 2^{(\begin{matrix} N \\ 2 \end{matrix})} [V^{†} d V] \\ = & {vol}_{HS} (U (N) / T^{N}) \int_{Δ_{N - 1} \cap C_{N}} V_{N} {(λ)}^{2} \sqrt{N} \prod_{k = 1}^{N - 1} d λ_{k}, \end{array}$ \matrix{ {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{\rm{D}}\left( {{^N}} \right)} \right)} \hfill & = \hfill & {\int_{{\rm{D}}\left( {{^N}} \right)} {\rm{d}} {V_{{\rm{HS}}}} = \int_{{\Delta _{N - 1}} \cap {{\cal C}_N}} {{V_N}} {{(\lambda )}^2}\sqrt N \prod\limits_{k = 1}^{N - 1} {{\rm{d}}} {\lambda _k} \times \int_{{\rm{U}}(N)/{^N}} {{2^{\left( {\scriptstyle N \atop \scriptstyle 2} \right)}}} \left[ {{V^\dag }{\rm{d}}V} \right]} \hfill \cr {} \hfill & = \hfill & {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{\rm{U}}(N)/{^N}} \right)\int_{{\Delta _{N - 1}} \cap {{\cal C}_N}} {{V_N}} {{(\lambda )}^2}\sqrt N \prod\limits_{k = 1}^{N - 1} {{\rm{d}}} {\lambda _k},} \hfill \cr } where $\int_{Δ_{N - 1} \cap C_{N}} V_{N} {(λ)}^{2} \prod_{k = 1}^{N - 1} d λ_{k} = \frac{1}{N!} \int_{ℝ_{⩾ 0}^{N}} δ (1 - \sum_{k = 1}^{N} λ_{k}) V_{N} {(λ)}^{2} \prod_{k = 1}^{N} d λ_{k} = \frac{{(\prod_{k = 1}^{N} Γ (k))}^{2}}{Γ (N^{2})} .$ \int_{{\Delta _{N - 1}} \cap {{\cal C}_N}} {{V_N}} {(\lambda )^2}\prod\limits_{k = 1}^{N - 1} {{\rm{d}}} {\lambda _k} = {1 \over {N!}}\int_{_{0}^N} \delta \left( {1 - \sum\limits_{k = 1}^N {{\lambda _k}} } \right){V_N}{(\lambda )^2}\prod\limits_{k = 1}^N {{\rm{d}}} {\lambda _k} = {{{{\left( {\prod\limits_{k = 1}^N \Gamma (k)} \right)}^2}} \over {\Gamma \left( {{N^2}} \right)}}.

Substituting the last result into the above, we get the desired result.

Remark 1.

We can partition the set D(ℂ^N) of all density matrices acting on ℂ^N according to the spectrum Λ = (λ₁, …, λ_N) ∈ △_N–1 ⋂ 𝒞_N: 25 $D (ℂ^{N}) = \underset{Λ \in Δ_{N - 1} \cap C_{N}}{⊎} U_{Λ} .$ {\rm{D}}\left( {{^N}} \right) = \mathop \uplus \limits_{\Lambda \in {\Delta _{N - 1}} \cap {{\cal C}_N}} {{\cal U}_\Lambda }.

From this observation and previous discussion, we get that $\begin{array}{l} {vol}_{HS} (D (ℂ^{N})) = \sqrt{N} \frac{{(\prod_{k = 1}^{N} Γ (k))}^{2}}{Γ (N^{2})} \cdot {vol}_{HS} (U (N) / T^{N}) \\ = \sqrt{N} \int_{Δ_{N - 1} \cap C_{N}} V_{N} {(Λ)}^{2} [d Λ] \cdot {vol}_{HS} (U (N) / T^{N}) \\ = \sqrt{N} \int_{Δ_{N - 1} \cap C_{N}} {vol}_{HS} (U_{Λ}) [d Λ] . \end{array}$ \matrix{ {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{\rm{D}}\left( {{{\rm{C}}^N}} \right)} \right) = \sqrt N {{{{\left( {\prod\nolimits_{k = 1}^N \Gamma (k)} \right)}^2}} \over {\Gamma \left( {{N^2}} \right)}}\cdot{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{\rm{U}}(N)/{^N}} \right)} \hfill \cr { = \sqrt N \int_{{\Delta _{N - 1}} \cap {{\cal C}_N}} {{V_N}} {{(\Lambda )}^2}[{\rm{d}}\Lambda ]\cdot{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{\rm{U}}(N)/{^N}} \right)} \hfill \cr { = \sqrt N \int_{{\Delta _{N - 1}} \cap {{\cal C}_N}} {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}} \left( {{{\cal U}_\Lambda }} \right)[{\rm{d}}\Lambda ].} \hfill \cr }

Based on it, we infer that every adjoint orbit 𝒰_Λ is subject to the distrubtion 26 $d m (Λ) = \sqrt{N} δ (1 - Tr (Λ)) [d Λ] .$ {\rm{d}}m(\Lambda ) = \sqrt N \delta (1 - {\mathop{\rm Tr}\nolimits} (\Lambda ))[{\rm{d}}\Lambda ].

It holds that 27 ${vol}_{HS} (D (ℂ^{N})) = \int_{C_{N} \cap Δ_{N - 1}} {vol}_{HS} (U_{Λ}) d m (Λ) .$ {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{\rm{D}}\left( {{^N}} \right)} \right) = \int_{{{\cal C}_N} \cap {\Delta _{N - 1}}} {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}} \left( {{{\cal U}_\Lambda }} \right){\rm{d}}m(\Lambda ).

4.

Symplectic Volumes of Regular Co-Adjoint Orbits

Consider a compact connected Lie group K with its Lie algebra . Let T be the maximal torus of K with its Lie algebra t, namely, the Cartan subalgebra. Choose an K-invariant inner product on . By restricting such K-invariant inner product on t, we can identify t*, the dual space of t, with t.

Let R⁺ be the set of all positive roots of K. Then R⁺ determines the positive Weyl chamber t_⩾0 via the way 28 $t_{⩾ 0} = {X \in {it}_{⩾ 0} : α (X) ⩾ 0, \forall α \in R^{+}} .$ {{\rm{t}}_{0}} = \left\{ {{\bf{X}} \in {\rm{i}}{{\rm{t}}_{0}}:\alpha ({\bf{X}})\;\;0,\forall \alpha \in {{\bf{R}}^ + }} \right\}.

Analogously, we also identity $t_{⩾ 0}^{*}$ t_{0}^* with t_⩾0.

Consider an element $\hat{λ} \in {it}_{> 0}^{*}$ \hat \lambda \in {\rm{it}}_{ > 0}^*, the interior of positive Weyl chamber ${it}_{⩾ 0}^{*}$ t_{0}^*. Let 29 $O_{\hat{λ}} := K \cdot \hat{λ} = {{Ad}_{g}^{*} \hat{λ} : g \in K}$ {{\cal O}_{\hat \lambda }}: = K\cdot\hat \lambda = \left\{ {{\mathop{\rm Ad}\nolimits} _g^*\hat \lambda :g \in K} \right\} be the co-adjoint orbit, identified by co-adjoint action of K on ${it}_{> 0}^{*} ≅ {it}_{> 0}$ {\rm{it}}_{ > 0}^* \cong {\rm{i}}{{\rm{t}}_{ > 0}}. There is a K-equivariantly diffeomorphism $O_{\hat{λ}} ≅ K / T .$ {{\cal O}_{\hat \lambda }} \cong K/T.

Moreover, the coadjoint orbit $O_{\hat{λ}}$ {{\cal O}_{\hat \lambda }} carry the so-called standard symplectic form, i.e., Kirillov-Kostant-Souriau form ω_KKS [20]. Now we get a symplectic form $(O_{\hat{λ}}, ω_{KKS})$ \left( {{{\cal O}_{\hat \lambda }},{\omega _{{\rm{KKS}}}}} \right). The top degree of ω_kks will identify a volume form $\frac{ω_{KKS}^{d}}{d!}$ {{\omega _{{\rm{KKS}}}^d} \over {d!}}, where $d := \frac{1}{2} \dim (O_{λ}) = \frac{1}{2} \dim (K / T)$ d: = {1 \over 2}\dim ({{\cal O}_\lambda }) = {1 \over 2}\dim (K/T).

Definition 1.

The Liouville measure on $O_{\hat{λ}}$ {{\cal O}_{\hat \lambda }} is defined by 30 $μ_{O_{\hat{λ}}} (B) := \int_{B} \frac{ω_{KKS}^{d}}{d!},$ {\mu _{{{\cal O}_{\hat \lambda }}}}(B): = \int_B {{{\omega _{{\rm{KKS}}}^d} \over {d!}}} , where B is a Borel subset of $O_{\hat{λ}}$ {{\cal O}_{\hat \lambda }}. The symplectic volume of $O_{\hat{λ}}$ {{\cal O}_{\hat \lambda }} _{is just} ${vol}_{symp} (O_{\hat{λ}}) = μ_{O_{\hat{λ}}} (O_{\hat{λ}})$ {{\mathop{\rm vol}\nolimits} _{{\mathop{\rm symp}\nolimits} }}\left( {{{\cal O}_{\hat \lambda }}} \right) = {\mu _{{{\cal O}_{\hat \lambda }}}}\left( {{{\cal O}_{\hat \lambda }}} \right).

Proposition 4 ([21]).

Let T be a maximal torus of a compact connected Lie group K. Let $\hat{λ} \in {it}_{> 0}^{*}$ \hat \lambda \in {\rm{it}}_{ > 0}^* for which $O_{\hat{λ}}$ {{\cal O}_{\hat \lambda }} is the coadjoint orbit through $\hat{λ}$ {\hat \lambda }. The symplectic volume of $O_{\hat{λ}}$ {{\cal O}_{\hat \lambda }}, with respect to the KKS form ω_KKS, _{is given by} 31 ${vol}_{symp} (O_{\hat{λ}}) = (\prod_{α \in R^{+}} 〈 \hat{λ}, α 〉) (\prod_{α \in R^{+}} \frac{2 π}{〈 ρ, α 〉}),$ {{\mathop{\rm vol}\nolimits} _{{\rm{symp}}}}\left( {{{\cal O}_{\hat \lambda }}} \right) = \left( {\prod\limits_{\alpha \in {R^ + }} {\langle \hat \lambda ,\alpha \rangle } } \right)\left( {\prod\limits_{\alpha \in {R^ + }} {{{2\pi } \over {\langle \rho ,\alpha \rangle }}} } \right), where $ρ := \frac{1}{2} \sum_{α \in R^{+}} α$ \rho : = {1 \over 2}\sum\nolimits_{\alpha \in {R^ + }} \alpha .

Remark 2.

According to Harish-Chandra’s volume formula [22], 32 ${vol}_{HS} (K / T) = \prod_{α \in R^{+}} \frac{2 π}{〈 ρ, α 〉} .$ {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}(K/T) = \prod\limits_{\alpha \in {R^ + }} {{{2\pi } \over {\langle \rho ,\alpha \rangle }}} .

Based on this formula, we get that 33 ${vol}_{symp} (O_{\hat{λ}}) = (\prod_{α \in R^{+}} 〈 \hat{λ}, α 〉) {vol}_{HS} (K / T) .$ {{\mathop{\rm vol}\nolimits} _{{\rm{symp }}}}\left( {{{\cal O}_{\hat \lambda }}} \right) = \left( {\prod\limits_{\alpha \in {R^ + }} {\langle \hat \lambda ,\alpha \rangle } } \right){{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}(K/T).

Example 1.

Let K = U(N). The maximal torus of U(N) is T = 𝕋^N, and its Lie algebra t such that it^* ≌ it ≌ ℝ^N. Moreover ${it}_{> 0}^{*} ≅ C_{N}$ {\rm{it}}_{ > 0}^* \cong {{\cal C}_N}. Then the set of all positive roots R⁺, where 34 $R^{+} = {α_{i} - α_{j} :, 1 ⩽ i < j ⩽ N},$ {R^ + } = \left\{ {{\alpha _i} - {\alpha _j}:,1\;\;i < j\;\;N} \right\}, where α_k is a linear functional taking the k-th position’s imaginary part. Moreover $d = | R^{+} | = (\begin{matrix} N \\ 2 \end{matrix})$ d = \left| {{R^ + }} \right| = \left( \matrix{ N \cr 2 \cr} \right). The symplectic volume of the coadjoint orbit $O_{\hat{λ}}$ {{\cal O}_{\hat \lambda }} is 35 ${vol}_{symp} (O_{\hat{λ}}) = V_{N} (\hat{λ}) {vol}_{HS} (U (N) / T^{N}) .$ {{\mathop{\rm vol}\nolimits} _{{\rm{symp }}}}\left( {{{\cal O}_{\hat \lambda }}} \right) = {V_N}(\hat \lambda ){{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{\rm{U}}(N)/{^N}} \right).

When we consider both HS volume and symplectic volume for the same manifold $O_{\hat{λ}}$ {{\cal O}_{\hat \lambda }}, we will get the following relationship between such volumes.

Proposition 5.

Let $\hat{λ} \in {it}_{> 0}^{*} ≅ {it}_{> 0}$ \hat \lambda \in {\rm{it}}_{ > 0}^* \cong {\rm{i}}{{\rm{t}}_{ > 0}} for the unitary group U(N). The HS volume and the symplectic volume of $O_{\hat{λ}}$ {{\cal O}_{\hat \lambda }} are connected via the formula: 36 ${vol}_{HS} (O_{\hat{λ}}) = V_{N} (\hat{λ}) \times {vol}_{symp} (O_{\hat{λ}}) .$ {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{{\cal O}_{\hat \lambda }}} \right) = {V_N}(\hat \lambda ) \times {{\mathop{\rm vol}\nolimits} _{{\rm{symp }}}}\left( {{{\cal O}_{\hat \lambda }}} \right).

Proof

According to Proposition 3, $\begin{array}{l} {vol}_{HS} (O_{\hat{λ}}) & = & V_{N} {(\hat{λ})}^{2} {vol}_{HS} (U (N) / T^{N}) = V_{N} (\hat{λ}) [V_{N} (\hat{λ}) {vol}_{HS} (U (N) / T^{N})] \\ = & V_{N} (\hat{λ}) {vol}_{Symp} (O_{\hat{λ}}), \end{array}$ \matrix{ {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\cal O}_{\hat \lambda }}} \right)} \hfill & = \hfill & {{V_N}{{(\hat \lambda )}^2}{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{\rm{U}}(N)/{^N}} \right) = {V_N}(\hat \lambda )\left[ {{V_N}(\hat \lambda ){{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{\rm{U}}(N)/{^N}} \right)} \right]} \hfill \cr {} \hfill & = \hfill & {{V_N}(\hat \lambda ){{{\mathop{\rm vol}\nolimits} }_{{\rm{Symp}}}}\left( {{{\cal O}_{\hat \lambda }}} \right),} \hfill \cr } where the last equality is obtained from Eq. (35).

5.

Duistermaat-Heckman Measure

The Duistermaat-Heckman (DH) Theorem, introduced in [10], revolutionized symplectic geometry and mathematical physics by uncovering profound connections between Hamiltonian dynamics, localization, and topology. Its key conclusions — including the exact stationary phase approximation, the convexity of images of moment map, the polynomiality of reduced symplectic volumes, relations to equivariant cohomology and localization [23], non-Abelian generalizations, applications in Physics, and combinatorial and algebraic consequences — have each had significant impact.

Central to these developments is the powerful Duistermaat-Heckman (DH) measure [24]. This measure provides a crucial tool across symplectic geometry, representation theory, and mathematical physics. Its strength lies in transforming complex global geometric problems into manageable local computations concentrated at fixed points, thereby effectively bridging symplectic geometry with representation theory, combinatorics, and physics.

Definition 2 (Push-forward of a measure).

Given measurable spaces (X, ℱ) and (ϒ, 𝒢), a measurable mapping Φ : X → ϒ and a measure μ : ℱ → [0, +∞), the push-forward of μ is defined to be the measure Φ_∗μ : 𝒢 → [0, +∞) given by $(Φ_{*} μ) (B) := μ (Φ^{- 1} (B)) f o r B \in G .$ \left( {{\Phi _*}\mu } \right)(B): = \mu \left( {{\Phi ^{ - 1}}(B)} \right)\quad {\rm{ }}for{\rm{ }}\quad B \in {\cal G}.

Proposition 6 (Change of variables formula).

A measurable function f on ϒ is integrable with respect to the pushforward measure Φ_∗μ if and only if the composition Φ∗ f = f ∘ Φ is integrable with respect to the measure μ. In that case, the integrals coincide $\int_{ϒ} f d (Φ_{*} μ) = \int_{X} f \circ Φ d μ = \int_{X} Φ^{*} f d μ .$ \int_\Upsilon f {\rm{d}}\left( {{\Phi _*}\mu } \right) = \int_X f \circ \Phi {\rm{d}}\mu = \int_X {{\Phi ^*}} f{\rm{d}}\mu .

In the sense of distribution, 〈Φ_{∗μ, f}〉 = 〈μ, Φ*f).

Definition 3 (Liouville measure).

Let (M, ω) be a symplectic manifold of 2n dimension. A Borel set in M is a set generated from compact subsets of M under countable union and complementation. Given a Borel set B in M, the Liouville measure of B is defined as $μ (B) = \int_{B} \frac{ω^{n}}{n!} .$ \mu (B) = \int_B {{{{\omega ^n}} \over {n!}}} .

The symplectic volume of M is defined by ${vol}_{symp} (M) = μ (M) = \int_{M} \frac{ω^{n}}{n!} .$ {{\mathop{\rm vol}\nolimits} _{{\rm{symp }}}}({\bf{M}}) = \mu ({\bf{M}}) = \int_{\bf{M}} {{{{\omega ^n}} \over {n!}}} .

In order to define Duistermaat-Heckman measure, we need the following notion of moment map (or momentum map) is a fundamental concept in symplectic geometry and mathematical physics, providing a bridge between symplectic group actions and conserved quantities. It generalizes the idea of conserved momenta in Hamiltonian mechanics to more abstract geometric settings.

Definition 4 (Moment map).

Given a smooth manifold M equipped with a symplectic form ω (a closed, non-degenerate 2-form). A Lie group K acts on M via symplectomorphisms (symmetry transformations preserving ω). The so-called moment map is a map satisfying:

(i)
For every , the function Φ^X(m) := 〈Φ(m), X〉 is a Hamiltonian function for the vector field X_M on M generated by X: $d Φ^{X} = ι_{X_{M}} ω .$ {\rm{d}}{\Phi ^X} = {\iota _{{X_M}}}\omega .
(ii)
Φ is K-equivariant with respect to the coadjoint action Ad* on : $Φ (g \cdot m) = ({Ad}_{g}^{*} Φ) (m), \forall g \in K .$ \Phi (g \cdot m) = \left( {{\mathop{\rm Ad}\nolimits} _g^*\Phi } \right)(m),\quad \forall g \in K.

In such a case, this action of K on M is called Hamiltonian action. At this time, (M, K, ω) is called Hamiltonian K-manifold.

Definition 5 (Duistermaat-Heckman measure).

Let (M, K, ω) (here ω is a symplectic form) be a compact, connected Hamiltonian K-manifold of dimension 2n and a choice of moment map .

(i)
The non-Abelian Duistermaat-Heckman measure ${DH}_{M}^{K}$ {\rm{DH}}_M^K{\rm{DH}}_M^K = {1 \over {{p_K}}}{\left( {{\tau _K}} \right)_*}{\Phi _*}\mu is defined as follows: 37 ${DH}_{M}^{K} = \frac{1}{p_{K}} {(τ_{K})}_{*} Φ_{*} μ$ {\tau _K}\left( {{{\cal O}_{\hat \lambda }}} \right) = \{ \hat \lambda \} where is defined as $τ_{K} (O_{\hat{λ}}) = {\hat{λ}}$ \hat \lambda \in {\rm{it}}_{0}^* for $\hat{λ} \in {it}_{⩾ 0}^{*}$ {p_K}(\hat \lambda ) = {{\mathop{\rm vol}\nolimits} _{{\rm{symp }}}}\left( {{{\cal O}_{\hat \lambda }}} \right) and $p_{K} (\hat{λ}) = {vol}_{symp} (O_{\hat{λ}})$ {\rm{DH}}_M^T = {\tau _*}{\Phi _*}\mu ,, which is known from Eq. (31).
(ii)
The Abelian Duistermaat-Heckman measure is defined as: 38 ${DH}_{M}^{T} = τ_{*} Φ_{*} μ,$ {\rm{it}}_{ > 0}^* where is the projection dual to the inclusion map .

The computation of the density of the pushforward measure with respect to the Lebesgue measure on the positive Weyl chamber is governed by the Duistermaat-Heckman (DH) theorem [10] and its generalizations. But in a special case, for instance, where the underlying symplectic manifold is just a coadjoint orbit through a regular element in ${it}_{> 0}^{*}$ {{\cal O}_{\hat \lambda }}, there is an explicit formula for computing the density. In the next subsection, we describe it explicitly.

5.1.

Harish-Chandra Formula and Derivative Principle

The following result gives the Fourier transform formula of Abelian Duistermaat-Heckman measure for the coadjoint orbit $O_{\hat{λ}}$ {\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^T. By this, we can identify the analytical expression of Abelian Duistermaat-Heckman measure.

Theorem 3 (Harish-Chandra, [22]).

The Fourier transform of abelian Duistermaat-Heckman ${DH}_{O_{\hat{λ}}}^{T}$ \left\langle {{\rm{DH}}_{{\cal O}_{{{\hat \lambda }^\prime }}^T}^T{e^{{\rm{i}}\langle - ,X\rangle }}} \right\rangle = \sum\limits_{w \in W} {{{( - 1)}^{l(w)}}} {e^{{\rm{i}}\langle w\hat \lambda ,X\rangle }}\prod\limits_{\alpha > 0} {{1 \over {{\rm{i}}\langle \alpha ,X\rangle }}} . measure is given by 39 $〈 {DH}_{O_{\hat{λ}}}^{T}, e^{i 〈 -, X 〉} 〉 = \sum_{w \in W} {(- 1)}^{l (w)} e^{i 〈 w \hat{λ}, X 〉} \prod_{α > 0} \frac{1}{i 〈 α, X 〉} .$ {\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^T = \sum\limits_{w \in W} {{{( - 1)}^{l(w)}}} {\delta _{w\hat \lambda }} \star {H_{ - {\alpha _1}}} \star \cdots \star {H_{ - {\alpha _R}}}. for every X in the Lie algebra of T which is not orthogonal to any root. That is, 40 ${DH}_{O_{\hat{λ}}}^{T} = \sum_{w \in W} {(- 1)}^{l (w)} δ_{w \hat{λ}} ⋆ H_{- α_{1}} ⋆ \dots ⋆ H_{- α_{R}} .$ {\left. {{\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^K} \right|_{{\rm{it}}_{0}^*}} = {\left. {\left( {\prod\limits_{\alpha > 0} {{\partial _{ - \alpha }}} } \right){\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^T} \right|_{{\rm{it}}_{0}^*}}. Here l(w) is the length of the Weyl group element w ∈ W, and δ_α for the Dirac measure at α; ★ means the convolution.

Subsequently, the so-called derivative principle is provided in which the non-Abelian Duistermaat-Heckman measure is obtained from Abelian Duistermaat-Heckman measure.

Theorem 4 (Derivative principle, [14]).

It holds that ${{DH}_{O_{\hat{λ}}}^{K} |}_{{it}_{⩾ 0}^{*}} = {(\prod_{α > 0} \partial_{- α}) {DH}_{O_{\hat{λ}}}^{T} |}_{{it}_{⩾ 0}^{*}} .$ {\tilde T}

Consider more generally the action of $\tilde{T}$ {{\cal O}_{\hat \lambda }} on a co-adjoint K-orbit $O_{\hat{λ}}$ \tilde T \to T \subset K induced by a group homomorphism $\tilde{T} \to T \subset K$ {\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^{\tilde T} = {\pi _*}{\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^T. Note that it follows directly that ${DH}_{O_{\hat{λ}}}^{\tilde{T}} = π_{*} {DH}_{O_{\hat{λ}}}^{T}$ {\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^{\tilde T} = {\pi _*}{\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^T where $π : {it}^{*} \to i {\tilde{t}}^{*}$ \pi :{\rm{i}}{{\rm{t}}^*} \to {\rm{i}}{\widetilde {\rm{t}}^*} is the restriction map from the dual Lie algebra t* of T to that ${\tilde{t}}^{*}$ {\widetilde ^*} of $\tilde{T}$ {\tilde T}. Thus we get that ${DH}_{O_{\hat{λ}}}^{\tilde{T}} = \sum_{w \in W} {(- 1)}^{l (w)} δ_{π (w \hat{λ})} ⋆ H_{- π (α_{1})} ⋆ \dots ⋆ H_{- π (α_{R})} .$ {\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^{\tilde T} = \sum\limits_{w \in W} {{{( - 1)}^{l(w)}}} {\delta _{\pi (w\hat \lambda )}} \star {H_{ - \pi \left( {{\alpha _1}} \right)}} \star \cdots \star {H_{ - \pi \left( {{\alpha _R}} \right)}}.

In what follows, we will now use the non-Abelian Heckman algorithm to treat the case of random two-qubit states with fixed, non-degenerate global eigenvalue spectrum. That is, we consider the action of $\tilde{K} = SU (2) \times SU (2)$ \tilde K = {\rm{SU}}(2) \times {\rm{SU}}(2) on a coadjoint K = SU (4)-orbit through a point $\hat{λ}$ {\hat \lambda } in the interior of positive Weyl chamber of K.

Let $\tilde{T}$ {\tilde T} be the maximal torus of $\tilde{K}$ {\tilde K}.
Symmetric group W = S₄ is the Weyl group of K = SU(4).
${DH}_{O_{\hat{λ}}}^{\tilde{T}} = \sum_{w \in S_{4}} sign (w) δ_{π (w \hat{λ})} ⋆ H_{- π (α_{1})} ⋆ \dots ⋆ H_{- π (α_{6})}$ {\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^{\tilde T} = \sum\nolimits_{w \in {S_4}} {{\mathop{\rm sign}\nolimits} } (w){\delta _{\pi (w\hat \lambda )}} \star {H_{ - \pi \left( {{\alpha _1}} \right)}} \star \cdots \star {H_{ - \pi \left( {{\alpha _6}} \right)}}, where {α_k : k = 1, …, 6} is the set of positive roots of K = SU(4).
The dual Lie algebra ${\tilde{t}}^{*}$ {\widetilde ^*} of $\tilde{T}$ {\tilde T} can be identified with it $i {\tilde{t}}^{*} ≅ ℝ^{2}$ {\rm{i}}{\widetilde {\rm{t}}^*} \cong {^2}.
The maps π is given by $π ({\hat{λ}}_{1}, \dots, {\hat{λ}}_{4}) = 2 i ({\hat{λ}}_{1} + {\hat{λ}}_{2}, {\hat{λ}}_{1} + {\hat{λ}}_{3}) .$ \pi \left( {{{\hat \lambda }_1}, \ldots ,{{\hat \lambda }_4}} \right) = 2{\rm{i}}\left( {{{\hat \lambda }_1} + {{\hat \lambda }_2},{{\hat \lambda }_1} + {{\hat \lambda }_3}} \right).
One computes readily that the –π(α_k) are precisely the weights ${(- 2, 2), {(- 2, 0)}_{(2)}, (- 2, - 2), {(0, - 2)}_{(2)}} .$ \left\{ {( - 2,2),{{( - 2,0)}_{(2)}},( - 2, - 2),{{(0, - 2)}_{(2)}}} \right\}.

In particular, two negative roots of $\tilde{K}$ {\tilde K} are contained in this list (each of them is in fact contained twice).

The following result appears in [14] without proof. We provide a detailed proof for completeness.

Proposition 7.

The non-Abelian Duistermaat-Heckman measure for the action of $\tilde{K} = SU (2) \times SU (2)$ \tilde K = {\rm{SU}}(2) \times {\rm{SU}}(2) on a coad-joint K-orbit $O_{\hat{λ}} = K \cdot \hat{λ}$ {{\cal O}_{\hat \lambda }} = K\cdot\hat \lambda with $\hat{λ}$ {\hat \lambda }, where K = SU(4), in the interior of positive Weyl chamber of T is given by 41 ${{DH}_{O_{\hat{λ}}}^{\tilde{K}} |}_{i {\tilde{t}}_{⩾ 0}^{*}} = {(\sum_{w \in S_{4}} sign (w) δ_{π (w λ)}) ⋆ H_{(- 2, 2)} ⋆ H_{(- 2, 0)} ⋆ H_{(- 2, - 2)} ⋆ H_{(0, - 2)} |}_{i {\tilde{t}}_{⩾ 0}^{*}},$ {\left. {{\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^{\tilde K}} \right|_{{\rm{i}}\widetilde _{0}^*}} = {\left. {\left( {\sum\limits_{w \in {S_4}} {{\mathop{\rm sign}\nolimits} } (w){\delta _{\pi (w\lambda )}}} \right) \star {H_{( - 2,2)}} \star {H_{( - 2,0)}} \star {H_{( - 2, - 2)}} \star {H_{(0, - 2)}}} \right|_{{\rm{i}}\widetilde _{0}^*}}, where ${\tilde{it}}_{⩾ 0}^{*} ≅ ℝ_{⩾ 0}^{2}$ \widetilde {\rm{t}}_{0}^* \cong _{0}^2 is the positive Weyl chamber of $\tilde{T}$ {\tilde T}, the maximal torus of $\tilde{K}$ {\tilde K}.

Proof

Recall that the Weyl group of K = SU(4) is the symmetric group W = S₄, with (–1)^l(w) equal to the signum of a permutation w ∈ S₄. Then ${DH}_{O_{\hat{λ}}}^{\tilde{T}} = π_{*} {DH}_{O_{\hat{λ}}}^{T} = \sum_{w \in S_{4}} sign (w) δ_{π (w \hat{λ})} ⋆ H_{- π (α_{1})} ⋆ \dots ⋆ H_{- π (α_{6})},$ {\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^{\tilde T} = {\pi _*}{\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^T = \sum\limits_{w \in {S_4}} {{\mathop{\rm sign}\nolimits} } (w){\delta _{\pi (w\hat \lambda )}} \star {H_{ - \pi \left( {{\alpha _1}} \right)}} \star \cdots \star {H_{ - \pi \left( {{\alpha _6}} \right)}}, where T is is the maximal torus of K = SU(4), and $\tilde{T}$ {\tilde T} is the maximal torus of $\tilde{K} = SU (2) \times SU (2)$ \tilde K = {\mathop{\rm SU}\nolimits} (2) \times {\mathop{\rm SU}\nolimits} (2), and {α_k : k = 1, … ,6} are the positive roots of K = SU(4), i.e., $\begin{array}{l} α_{1} = (1, - 1, 0, 0), α_{2} = (1, 0, - 1, 0), α_{3} = (1, 0, 0, - 1), \\ α_{4} = (0, 1, - 1, 0), α_{5} = (0, 1, 0, - 1), α_{6} = (0, 0, 1, - 1) . \end{array}$ \matrix{ {{\alpha _1} = (1, - 1,0,0),{\alpha _2} = (1,0, - 1,0),{\alpha _3} = (1,0,0, - 1),} \hfill \cr {{\alpha _4} = (0,1, - 1,0),{\alpha _5} = (0,1,0, - 1),{\alpha _6} = (0,0,1, - 1).} \hfill \cr }

Since ${\hat{λ}}_{1} > \dots > {\hat{λ}}_{4}$ {{\hat \lambda }_1} > \cdots > {{\hat \lambda }_4}, it follows that this π maps each $\hat{λ}$ {\hat \lambda } to $2 i ({\hat{λ}}_{1} + {\hat{λ}}_{2}, {\hat{λ}}_{1} + {\hat{λ}}_{3})$ 2{\rm{i}}\left( {{{\hat \lambda }_1} + {{\hat \lambda }_2},{{\hat \lambda }_1} + {{\hat \lambda }_3}} \right), where $2 ({\hat{λ}}_{1} + {\hat{λ}}_{2})$ 2\left( {{{\hat \lambda }_1} + {{\hat \lambda }_2}} \right) and $2 ({\hat{λ}}_{1} + {\hat{λ}}_{3})$ 2\left( {{{\hat \lambda }_1} + {{\hat \lambda }_3}} \right) are the maximal eigenvalues of two reduced density matrices determined by a global density matrix Λ, where $Λ = diag ({\hat{λ}}_{1} + \frac{1}{4}, \dots, {\hat{λ}}_{4} + \frac{1}{4})$ \Lambda = {\mathop{\rm diag}\nolimits} \left( {{{\hat \lambda }_1} + {1 \over 4}, \ldots ,{{\hat \lambda }_4} + {1 \over 4}} \right). One computes readily that the –π(α_k) are precisely the weights ${(- 2, 2), {(- 2, 0)}_{(2)}, (- 2, - 2), {(0, - 2)}_{(2)}} .$ \left\{ {( - 2,2),{{( - 2,0)}_{(2)}},( - 2, - 2),{{(0, - 2)}_{(2)}}} \right\}.

Indeed, $\begin{array}{l} - π (α_{1}) = - π (α_{6}) = (0, - 2), - π (α_{3}) = (- 2, - 2), \\ - π (α_{2}) = - π (α_{5}) = (- 2, 0), - π (α_{4}) = (- 2, 2) . \end{array}$ \matrix{ { - \pi \left( {{\alpha _1}} \right) = - \pi \left( {{\alpha _6}} \right) = (0, - 2), - \pi \left( {{\alpha _3}} \right) = ( - 2, - 2),} \hfill \cr {\;\; - \pi \left( {{\alpha _2}} \right) = - \pi \left( {{\alpha _5}} \right) = ( - 2,0), - \pi \left( {{\alpha _4}} \right) = ( - 2,2).} \hfill \cr }

That is, ${- π (α_{k}) : k \in [6]} = {(- 2, 2), {(- 2, 0)}_{(2)}, (- 2, - 2), {(0, - 2)}_{(2)}} .$ \left\{ { - \pi \left( {{\alpha _k}} \right):k \in [6]} \right\} = \left\{ {( - 2,2),{{( - 2,0)}_{(2)}},( - 2, - 2),{{(0, - 2)}_{(2)}}} \right\}.

In particular, the two negative roots {(–2,0), (0, – 2)} of $\tilde{K} = SU (2) \times SU (2)$ \tilde K = {\rm{SU}}(2) \times {\rm{SU}}(2) are contained in this list t (each of them is in fact contained twice). From the above discussion, we see that ${DH}_{O_{\hat{λ}}}^{\tilde{T}} = (\sum_{w \in S_{4}} sign (w) δ_{π (w \hat{λ})}) ⋆ H_{(0, - 2)} ⋆ H_{(- 2, 0)} ⋆ H_{(- 2, - 2)} ⋆ H_{(- 2, 2)} ⋆ H_{(- 2, 0)} ⋆ H_{(0, - 2)} .$ {\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^{\tilde T} = \left( {\sum\limits_{w \in {S_4}} {{\mathop{\rm sign}\nolimits} } (w){\delta _{\pi (w\hat \lambda )}}} \right) \star {H_{(0, - 2)}} \star {H_{( - 2,0)}} \star {H_{( - 2, - 2)}} \star {H_{( - 2,2)}} \star {H_{( - 2,0)}} \star {H_{(0, - 2)}}.

Therefore, we arrived at the following formula: $\begin{array}{l} {{DH}_{O_{\hat{λ}}}^{\tilde{K}} |}_{{i \tilde{t}}_{⩾ 0}^{*}} = {(\prod_{α > 0} \partial_{- α}) {DH}_{O_{\hat{λ}}}^{\tilde{T}} |}_{i {\tilde{t}}_{⩾ 0}^{*}} \\ = {(\sum_{w \in S_{4}} sign (w) δ_{π (w \hat{λ})}) ⋆ H_{(- 2, - 2)} ⋆ H_{(- 2, 2)} ⋆ H_{(- 2, 0)} ⋆ H_{(0, - 2)} |}_{{i \tilde{t}}_{⩾ 0}^{*}} . \end{array}$ \matrix{ {{{\left. {{\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^{\tilde K}} \right|}_{\widetilde {\rm{i}}_{0}^*}} = {{\left. {\left( {\prod\limits_{\alpha > 0} {{\partial _{ - \alpha }}} } \right){\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^{\tilde T}} \right|}_{{\rm{i}}\widetilde {\rm{t}}_{0}^*}}} \hfill \cr { = {{\left. {\left( {\sum\limits_{w \in {S_4}} {{\mathop{\rm sign}\nolimits} } (w){\delta _{\pi (w\hat \lambda )}}} \right) \star {H_{( - 2, - 2)}} \star {H_{( - 2,2)}} \star {H_{( - 2,0)}} \star {H_{(0, - 2)}}} \right|}_{\widetilde {\rm{i}}_{0}^*}}.} \hfill \cr }

We are done.

We have already known the fact that a density matrix of a generic two-qubit state is represented by the 2 × 2 block matrix $ρ_{12} = (\begin{matrix} A & C \\ C^{†} & B \end{matrix}),$ {\rho _{12}} = \left( {\matrix{ {\bf{A}} & {\bf{C}} \cr {{{\bf{C}}^\dag }} & {\bf{B}} \cr } } \right), where A, B are 2 × 2 positive semi-definite complex matrices. Now we present the two reduced marginal states, respectively: $ρ_{1} = (\begin{matrix} Tr (A) & Tr (C) \\ Tr (C^{†}) & Tr (B) \end{matrix}), ρ_{2} = A + B .$ {\rho _1} = \left( {\matrix{ {{\mathop{\rm Tr}\nolimits} ({\bf{A}})} & {{\mathop{\rm Tr}\nolimits} ({\bf{C}})} \cr {{\mathop{\rm Tr}\nolimits} \left( {{{\bf{C}}^\dag }} \right)} & {{\mathop{\rm Tr}\nolimits} ({\bf{B}})} \cr } } \right),\quad {\rho _2} = {\bf{A}} + {\bf{B}}.

From this, we see that for ρ₁₂ = Λ with Λ₁ ⩾ ⋯ ⩾ Λ₄ ⩾ 0 and $\sum_{j = 1}^{4} Λ_{j} = 1$ \sum\limits_{j = 1}^4 {{\Lambda _j}} = 1, it can be rewritten as $ρ - \frac{𝟙_{4}}{4} = diag ({\hat{λ}}_{1}, \dots, {\hat{λ}}_{4})$ \rho - {{{_4}} \over 4} = {\mathop{\rm diag}\nolimits} \left( {{{\hat \lambda }_1}, \ldots ,{{\hat \lambda }_4}} \right), where ${\hat{λ}}_{1} ⩾ \dots ⩾ {\hat{λ}}_{4}$ {{\hat \lambda }_1}\;\; \cdots \;\;{{\hat \lambda }_4} and $\sum_{j = 1}^{4} {\hat{λ}}_{j} = 0$ \sum\nolimits_{j = 1}^4 {{{\hat \lambda }_j} = 0} . Thus $\hat{λ}$ {\hat \lambda } is in the positive Weyl chamber t_⩾0 of T.

Now the map ρ ↦ (ρ₁, ρ₂), where ρ₁ = Tr₂(ρ) and ρ₂ = Tr₁(ρ), can be equivalently represented as $\begin{array}{l} π (\hat{λ}) & = & ((\begin{matrix} {\hat{λ}}_{1} + {\hat{λ}}_{2} & 0 \\ 0 & {\hat{λ}}_{3} + {\hat{λ}}_{4} \end{matrix}), (\begin{matrix} {\hat{λ}}_{1} + {\hat{λ}}_{3} & 0 \\ 0 & {\hat{λ}}_{2} + {\hat{λ}}_{4} \end{matrix})) \\ \overset{def}{=} & (2 ({\hat{λ}}_{1} + {\hat{λ}}_{2}), 2 ({\hat{λ}}_{1} + {\hat{λ}}_{3})), \end{array}$ \matrix{ {\pi (\hat \lambda )} \hfill & = \hfill & {\left( {\left( {\matrix{ {{{\hat \lambda }_1} + {{\hat \lambda }_2}} & 0 \cr 0 & {{{\hat \lambda }_3} + {{\hat \lambda }_4}} \cr } } \right),\left( {\matrix{ {{{\hat \lambda }_1} + {{\hat \lambda }_3}} & 0 \cr 0 & {{{\hat \lambda }_2} + {{\hat \lambda }_4}} \cr } } \right)} \right)} \hfill \cr {} \hfill & {\mathop = \limits^{{\rm{ def }}} } \hfill & {\left( {2\left( {{{\hat \lambda }_1} + {{\hat \lambda }_2}} \right),2\left( {{{\hat \lambda }_1} + {{\hat \lambda }_3}} \right)} \right),} \hfill \cr } where $(\begin{matrix} {\hat{λ}}_{1} + {\hat{λ}}_{2} & 0 \\ 0 & {\hat{λ}}_{3} + {\hat{λ}}_{4} \end{matrix}) \overset{def}{=} {\hat{λ}}_{1} + {\hat{λ}}_{2} - ({\hat{λ}}_{3} + {\hat{λ}}_{4}) = 2 ({\hat{λ}}_{1} + {\hat{λ}}_{2})$ \left( {\matrix{ {{{\hat \lambda }_1} + {{\hat \lambda }_2}} & 0 \cr 0 & {{{\hat \lambda }_3} + {{\hat \lambda }_4}} \cr } } \right)\mathop = \limits^{{\rm{ def }}} {{\hat \lambda }_1} + {{\hat \lambda }_2} - \left( {{{\hat \lambda }_3} + {{\hat \lambda }_4}} \right) = 2\left( {{{\hat \lambda }_1} + {{\hat \lambda }_2}} \right) and $(\begin{matrix} {\hat{λ}}_{1} + {\hat{λ}}_{3} & 0 \\ 0 & {\hat{λ}}_{2} + {\hat{λ}}_{4} \end{matrix}) \overset{def}{=} {\hat{λ}}_{1} + {\hat{λ}}_{3} - ({\hat{λ}}_{2} + {\hat{λ}}_{4}) = 2 ({\hat{λ}}_{1} + {\hat{λ}}_{3}) .$ \left( {\matrix{ {{{\hat \lambda }_1} + {{\hat \lambda }_3}} & 0 \cr 0 & {{{\hat \lambda }_2} + {{\hat \lambda }_4}} \cr } } \right)\mathop = \limits^{{\rm{ def }}} {{\hat \lambda }_1} + {{\hat \lambda }_3} - \left( {{{\hat \lambda }_2} + {{\hat \lambda }_4}} \right) = 2\left( {{{\hat \lambda }_1} + {{\hat \lambda }_3}} \right).

Denote h = diag(1, –1). Then $diag ({\hat{λ}}_{1} + {\hat{λ}}_{2}, {\hat{λ}}_{3} + {\hat{λ}}_{4}) = ({\hat{λ}}_{1} + {\hat{λ}}_{2}) \cdot h$ {\mathop{\rm diag}\nolimits} \left( {{{\hat \lambda }_1} + {{\hat \lambda }_2},{{\hat \lambda }_3} + {{\hat \lambda }_4}} \right) = \left( {{{\hat \lambda }_1} + {{\hat \lambda }_2}} \right)\cdot{\bf{h}} and $diag ({\hat{λ}}_{1} + {\hat{λ}}_{3}, {\hat{λ}}_{2} + {\hat{λ}}_{4}) = ({\hat{λ}}_{1} + {\hat{λ}}_{3}) \cdot h$ {\mathop{\rm diag}\nolimits} \left( {{{\hat \lambda }_1} + {{\hat \lambda }_3},{{\hat \lambda }_2} + {{\hat \lambda }_4}} \right) = \left( {{{\hat \lambda }_1} + {{\hat \lambda }_3}} \right)\cdot{\bf{h}}. Since ${\hat{λ}}_{1} + {\hat{λ}}_{3} ⩾ {\hat{λ}}_{2} + {\hat{λ}}_{4}$ {{\hat \lambda }_1} + {{\hat \lambda }_3}\;\;{{\hat \lambda }_2} + {{\hat \lambda }_4} and $({\hat{λ}}_{1} + {\hat{λ}}_{3}) + ({\hat{λ}}_{2} + {\hat{λ}}_{4}) = 0$ \left( {{{\hat \lambda }_1} + {{\hat \lambda }_3}} \right) + \left( {{{\hat \lambda }_2} + {{\hat \lambda }_4}} \right) = 0, thus ${\hat{λ}}_{1} + {\hat{λ}}_{3} ⩾ 0$ {{\hat \lambda }_1} + {{\hat \lambda }_3}\;\;0, a fortiori ${\hat{λ}}_{1} + {\hat{λ}}_{2} ⩾ {\hat{λ}}_{1} + {\hat{λ}}_{3} ⩾ 0$ {{\hat \lambda }_1} + {{\hat \lambda }_2}\;\;{{\hat \lambda }_1} + {{\hat \lambda }_3}\;\;0. Note that ${\hat{λ}}_{1} + {\hat{λ}}_{4}$ {{\hat \lambda }_1} + {{\hat \lambda }_4} is not non-negative in general.

5.2.

Boysal-Vergne-Paradan Jump Formula

In 2009, Bosyal and Vergne published a paper [13] in which they investigated the push-forward of Lebesgue measures on the cone $ℝ_{⩾ 0}^{N}$ _{0}^N along a linear map. Later in his PhD thesis, Walter found that the density of Abelian Duistermaat-Heckman measure with respect to the Lebesgue measure on the affine hull of the Abelian moment polytope is given by the volume of a parametrized polytope [25]. In view of this result, Boysal and Vergne’s result is connected with Abelian Duistermaat-Heckman measure, and can be adapted to calculate the density. To this end, we need firstly introduce some notions involved in question.

Definition 6 ([13]).

Let e ∈ V be a primitive vector. It defines a hyperplane in V* (the dual space of V): $W = {α \in V^{*} : 〈 α, e 〉 = 0} \overset{def}{=} e^{⊥} .$ W = \left\{ {\alpha \in {V^*}:\langle \alpha ,{\bf{e}}\rangle = 0} \right\}\mathop = \limits^{{\rm{ def }}} {{\bf{e}}^ \bot }.

Let P be a polynomial function on V* and let Ψ be a sequence of vectors not belong to W, i.e., Ψ ⋂ W = ∅. We define, for α ∈ V*, 12 $Pol (P, Ψ, e) (α) = {Res}_{z = 0} [{(P (\partial_{x}) \frac{e^{〈 α, x + z e 〉}}{\prod_{ψ \in Ψ} 〈 ψ, x + z e 〉})}_{x = 0}] .$ {\mathop{\rm Pol}\nolimits} (P,\Psi ,{\bf{e}})(\alpha ) = {{\mathop{\rm Res}\nolimits} _{z = 0}}\left[ {{{\left( {P\left( {{\partial _x}} \right){{{e^{\langle \alpha ,x + z{\bf{e}}\rangle }}} \over {\prod\nolimits_{\psi \in \Psi } {\langle \psi ,{\bf{x}} + z{\bf{e}}\rangle } }}} \right)}_{x = 0}}} \right].

Remark 3.

Note that the function Pol(P, Ψ, e) depends only on the restriction p of P to W. If p is a polynomial function on W = e^⊥. We define $Pol (p, Ψ, e) := Pol (P, Ψ, e),$ {\mathop{\rm Pol}\nolimits} (p,\Psi ,{\bf{e}}): = {\mathop{\rm Pol}\nolimits} (P,\Psi ,{\bf{e}}), where P is any polynomial on V* extending p.

Definition 7 (Wall).

Let Ψ = [ψ₁, …, ψ_N] be a sequence of non-zero, not necessarily distinct, linear forms on V, i.e., ψ_k ∈ V*, lying in an open half-space. If Ψ spans the whole space V* with dim(V) = n, then a wall of Ψ is a (real) hyperplane generated by n – 1 linearly independent elements of Ψ.

Note that V* is separated into two open half-spaces by the wall W := {φ ∈ V* : 〈φ, e〉 = 0} in V*. Let $V_{\pm}^{*}$ V_ \pm ^* denote the corresponding open half-spaces, that is, $V_{+}^{*} := {ϕ \in V^{*} : 〈 ϕ, e 〉 > 0} and V_{-}^{*} := {ϕ \in V^{*} : 〈 ϕ, e 〉 < 0} .$ V_ + ^*: = \left\{ {\phi \in {V^*}:\langle \phi ,{\bf{e}}\rangle > 0} \right\}{\rm{ and }}V_ - ^*: = \left\{ {\phi \in {V^*}:\langle \phi ,{\bf{e}}\rangle < 0} \right\}.

Let $C_{1} \subset V_{+}^{*}$ {C_1} \subset V_ + ^* and $C_{2} \subset V_{-}^{*}$ {C_2} \subset V_ - ^* be two chambers on two sides of W and adjacent. We choose the measures dx on V and dϕ on V*, respectively, and choose Lebesgue measure dt on ℝ^N. We also choose the measure dw on W such that dϕ = dwdt with t = 〈ϕ, e〉 for φ ∈ V*. Based on this, we can write Ψ as $Ψ = [Ψ_{0}, Ψ^{+}, Ψ^{-}],$ \Psi = \left[ {{\Psi _0},{\Psi ^ + },{\Psi ^ - }} \right], where $Ψ^{\pm} = Ψ \cap V_{\pm}^{*}$ {\Psi ^ \pm } = \Psi \cap V_ \pm ^*, and Ψ₀ = Ψ ⋂ W.

Theorem 5 (Boysal-Vergne-Paradan jump formula, [13]).

Let W be a wall, determined by a vector e ∈ V, Ψ be a sequence of vectors spanning the whole space V*. Denote Ψ₀ := Ψ ⋂ W. Let v₁₂ = v(Ψ₀, dw, C₁₂) be the polynomial function on W associated to the chamber C₁₂ of Ψ₀. Then, if 〈C₁, e〉 > 0 and V₁₂ is any extension of v₁₂ on V*, 43 $v (Ψ, d x, C_{1}) - v (Ψ, d x, C_{2}) = Pol (v_{12}, Ψ \ Ψ_{0}, e) .$ v\left( {\Psi ,{\rm{d}}x,{C_1}} \right) - v\left( {\Psi ,{\rm{d}}x,{C_2}} \right) = {\mathop{\rm Pol}\nolimits} \left( {{v_{12}},\Psi \backslash {\Psi _0},{\bf{e}}} \right).

Although the following result is obtained in [14], the details of proof is not provided. For reader’s convenience, we present it here using Boysal-Vergne-Paradan jump formula.

Proposition 8.

The measure H_(–2,2) ★ H_(–2,0) ★ H_(–2,–2) ★ H_(0,–2) has Lebesgue density $p (r, s) = {\begin{array}{l} p_{1} (r, s) \equiv \frac{{(r + s)}^{2}}{64}, & i f (r, s) \in C_{1} = {(r, s) : 0 ⩽ s ⩽ - r}; \\ p_{2} (r, s) \equiv \frac{r^{2} + 2 r s - s^{2}}{64}, & i f (r, s) \in C_{2} = {(r, s) : r ⩽ s ⩽ 0}; \\ p_{3} (r, s) \equiv \frac{r^{2}}{32}, & i f (r, s) \in C_{3} = {(r, s) : s ⩽ r ⩽ 0}, \\ p_{0} (r, s) \equiv 0, & i f (r, s) \in C_{0} = ℝ^{2} \ (C_{1} \cup C_{2} \cup C_{3}) . \end{array}$ p(r,s) = \left\{ {\matrix{ {{p_1}(r,s) \equiv {{{{(r + s)}^2}} \over {64}},} \hfill & {{\rm{ }}if{\rm{ }}(r,s) \in {C_1} = \{ (r,s):0\;\;s\; - r\} ;} \hfill \cr {{p_2}(r,s) \equiv {{{r^2} + 2rs - {s^2}} \over {64}},} \hfill & {{\rm{ }}if{\rm{ }}(r,s) \in {C_2} = \{ (r,s):r\;\;s\;\;0\} ;} \hfill \cr {{p_3}(r,s) \equiv {{{r^2}} \over {32}},} \hfill & {{\rm{ }}if{\rm{ }}(r,s) \in {C_3} = \{ (r,s):s\;\;r\;\;0\} ,} \hfill \cr {{p_0}(r,s) \equiv 0,} \hfill & {{\rm{ }}if{\rm{ }}(r,s) \in {C_0} = {^2}\backslash \left( {{C_1} \cup {C_2} \cup {C_3}} \right).} \hfill \cr } } \right.

The density function p(r, s) (its graph has already appeared in [14]) and its support that is decomposed into three chammbers can be visualized in the following Figure 1.

Proof

The measure H_(–2,2) ★ H_(–2,0) ★ H_(–2,–2) ★ H_(0,–2) is, in fact, the non-Abelian Duistermaat-Heckman measure that is on the closures of the regular chambers containing the vertex (0,0) given by the above convolution: $δ_{(0, 0)} ⋆ H_{(- 2, 2)} ⋆ H_{(- 2, 0)} ⋆ H_{(- 2, - 2)} ⋆ H_{(0, - 2)} .$ {\delta _{(0,0)}} \star {H_{( - 2,2)}} \star {H_{( - 2,0)}} \star {H_{( - 2, - 2)}} \star {H_{(0, - 2)}}.

Then its density is denote by p(r,s). Denote O(0, 0),P₁(–2, 2),P₂(–2, 0),P₃(–2, –2),P₄(0, –2). Then $| O P_{1} P_{2} | := C_{1}, | O P_{2} P_{3} | := C_{2}, | O P_{3} P_{4} | := C_{3} .$ \left| {O{P_1}{P_2}} \right|: = {C_1},\quad \left| {O{P_2}{P_3}} \right|: = {C_2},\quad \left| {O{P_3}{P_4}} \right|: = {C_3}.

Denote ℝ² – C₁ – C₂ – C₃ := C₀. Thus $ℝ^{2} = C_{0} \cup C_{1} \cup C_{2} \cup C_{3} .$ {^2} = {C_0} \cup {C_1} \cup {C_2} \cup {C_3}.

(i)
Clearly p ≡ 0 on C₀.
- The wall W₀₁ separating C₀ and C₁ is given by the equation: r + s = 0. Its normal vector ξ = (–1, –1). Just only one weights ω₁ = (–2,2) lies on the linear hyperplane spanned by W₀₁ (other weights are outside of W₀₁: ω₂ = (–2,0), ω₃ = (–2, –2), ω₄ = (0, –2)).
- Consider the push-forward of Lebesgue measure on $ℝ_{⩾ 0}^{1}$ _{0}^1 along the linear map P_W₀₁ : u ↦ uω₁. Its density with respect to dw is given by a single homogeneous polynomial on the wall W₀₁.
- Denote by p_W₀₁ any polynomial function extending it to all of the dual Lie algebra of $\tilde{T}$ {\tilde T}.
- Clearly $p_{W_{01}} = \frac{1}{2}$ {p_{{W_{01}}}} = {1 \over 2}. Indeed, $\begin{array}{l} p_{W_{01}}^{- 1} = d λ (ω_{1}, ξ / ‖ ξ ‖^{2}) = | \det (\begin{matrix} - 2 & 2 \\ - 1 / 2 & - 1 / 2 \end{matrix}) | = 2 \\ \Leftrightarrow p_{W_{01}} = \frac{1}{2} . \end{array}$ \matrix{ {p_{{W_{01}}}^{ - 1} = {\rm{d}}\lambda \left( {{\omega _1},\xi /\xi {^2}} \right) = \left| {\det \left( {\matrix{ { - 2} & 2 \cr { - 1/2} & { - 1/2} \cr } } \right)} \right| = 2} \hfill \cr { \Leftrightarrow {p_{{W_{01}}}} = {1 \over 2}.} \hfill \cr }

Note that, during the proof, we abuse notation by using identical symbols for a differential form and its induced measure. Hence μ = (r, s) $p (μ) = \frac{1}{2} {Res}_{z = 0} {(\frac{e^{〈 μ, x + z ξ 〉}}{\prod_{k = 2}^{4} 〈 ω_{k}, x + z ξ 〉})}_{x = 0},$ p(\mu ) = {1 \over 2}{{\mathop{\rm Res}\nolimits} _{z = 0}}{\left( {{{{e^{\langle \mu ,x + z\xi \rangle }}} \over {\prod\nolimits_{k = 2}^4 {\left\langle {{\omega _k},x + z\xi } \right\rangle } }}} \right)_{x = {\bf{0}}}}, where $\frac{e^{〈 μ, x + z ξ 〉}}{\prod_{k = 2}^{4} 〈 ω_{k}, x + z ξ 〉} = \frac{\exp (r (x_{1} - z) + s (x_{2} - z))}{[- 2 (x_{1} - z)] [[- 2 (x_{1} - z) - 2 (x_{2} - z)]] [- 2 (x_{2} - z)]} .$ {{{e^{\langle \mu ,x + z\xi \rangle }}} \over {\prod\nolimits_{k = 2}^4 {\left\langle {{\omega _k},x + z\xi } \right\rangle } }} = {{\exp \left( {r\left( {{x_1} - z} \right) + s\left( {{x_2} - z} \right)} \right)} \over {\left[ { - 2\left( {{x_1} - z} \right)} \right]\left[ {\left[ { - 2\left( {{x_1} - z} \right) - 2\left( {{x_2} - z} \right)} \right]} \right]\left[ { - 2\left( {{x_2} - z} \right)} \right]}}.

Thus $\begin{array}{l} p (μ) & = & \frac{1}{2} \times (- \frac{1}{8}) {Res}_{z = 0} {(\frac{\exp (r (x_{1} - z) + s (x_{2} - z))}{(x_{1} - z) (x_{1} + x_{2} - 2 z) (x_{2} - z)})}_{x = 0} \\ = & \frac{1}{32} {Res}_{z = 0} (\frac{e^{- (r + s) z}}{z^{3}}) = \frac{1}{64} {(r + s)}^{2} . \end{array}$ \matrix{ {p(\mu )} \hfill & = \hfill & {{1 \over 2} \times \left( { - {1 \over 8}} \right){{{\mathop{\rm Res}\nolimits} }_{z = 0}}{{\left( {{{\exp \left( {r\left( {{x_1} - z} \right) + s\left( {{x_2} - z} \right)} \right)} \over {\left( {{x_1} - z} \right)\left( {{x_1} + {x_2} - 2z} \right)\left( {{x_2} - z} \right)}}} \right)}_{x = 0}}} \hfill \cr {} \hfill & = \hfill & {{1 \over {32}}{{{\mathop{\rm Res}\nolimits} }_{z = 0}}\left( {{{{e^{ - (r + s)z}}} \over {{z^3}}}} \right) = {1 \over {64}}{{(r + s)}^2}.} \hfill \cr }

(ii)
The wall W₁₂ separating C₁ and C₂ is given by the equation: s = 0. Its normal vector ξ = (0, –1). Just only one weights ω₁ = (–2,0) lies on the linear hyperplane spanned by W₁₂ (other weights are outside of W₁₂: ω₂ = (–2,2), ω₃ = (–2, – 2), ω₄ = (0, – 2)).
- Consider the push-forward of Lebesgue measure on $ℝ_{⩾ 0}^{1}$ _{0}^1 along the linear map P_W₁₂ : uω₁. Its density with respect to dw is given by a single homogeneous polynomial on the wall W₁₂.
- Denote by p_W₁₂ any polynomial function extending it to all of the dual Lie algebra of $\tilde{T}$ {\tilde T}. Clearly $p_{W_{12}} = \frac{1}{2}$ {p_{{W_{12}}}} = {1 \over 2}. Indeed, $p_{W_{12}}^{- 1} = d λ (ω_{1}, ξ / ‖ ξ ‖^{2}) = | \det (\begin{matrix} - 2 & 0 \\ 0 & - 1 \end{matrix}) | = 2 \Leftrightarrow p_{W_{12}} = \frac{1}{2} .$ p_{{W_{12}}}^{ - 1} = {\rm{d}}\lambda \left( {{\omega _1},\xi /\xi {^2}} \right) = \left| {\det \left( {\matrix{ { - 2} & 0 \cr 0 & { - 1} \cr } } \right)} \right| = 2 \Leftrightarrow {p_{{W_{12}}}} = {1 \over 2}.

Hence $p (μ) - \frac{1}{64} {(r + s)}^{2} = \frac{1}{2} {Res}_{z = 0} {(\frac{e^{〈 μ, x + z ξ 〉}}{\prod_{k = 2}^{4} 〈 ω_{k}, x + z ξ 〉})}_{x = 0},$ p(\mu ) - {1 \over {64}}{(r + s)^2} = {1 \over 2}{{\mathop{\rm Res}\nolimits} _{z = 0}}{\left( {{{{e^{\left\langle {\mu ,x + z\xi } \right\rangle }}} \over {\prod\nolimits_{k = 2}^4 {\left\langle {{\omega _k},x + z\xi } \right\rangle } }}} \right)_{x = 0}}, where $\frac{e^{〈 μ, x + z ξ 〉}}{\prod_{k = 2}^{4} 〈 ω_{k}, x + z ξ 〉} = \frac{e^{r x_{1} + s (x_{2} - z)}}{[- 2 x_{1} + 2 (x_{2} - z)] [- 2 x_{1} - 2 (x_{2} - z)] [- 2 (x_{2} - z)]} .$ {{{e^{\langle \mu ,x + z\xi \rangle }}} \over {\prod\nolimits_{k = 2}^4 {\left\langle {{\omega _k},x + z\xi } \right\rangle } }} = {{{e^{r{x_1} + s\left( {{x_2} - z} \right)}}} \over {\left[ { - 2{x_1} + 2\left( {{x_2} - z} \right)} \right]\left[ { - 2{x_1} - 2\left( {{x_2} - z} \right)} \right]\left[ { - 2\left( {{x_2} - z} \right)} \right]}}.

Thus $\begin{array}{l} p (μ) - \frac{{(r + s)}^{2}}{64} \\ = \frac{1}{2} \times - \frac{1}{8} {Res}_{z = 0} {(\frac{e^{r x_{1} + s (x_{2} - z)}}{(x_{1} - x_{2} + z) (x_{1} + x_{2} - z) (x_{2} - z)})}_{x = 0} \\ = - \frac{1}{16} {Res}_{z = 0} (\frac{e^{- s z}}{z^{3}}) = - \frac{s^{2}}{32} . \end{array}$ \matrix{ {p(\mu ) - {{{{(r + s)}^2}} \over {64}}} \hfill \cr { = {1 \over 2} \times - {1 \over 8}{{{\mathop{\rm Res}\nolimits} }_{z = 0}}{{\left( {{{{e^{r{x_1} + s\left( {{x_2} - z} \right)}}} \over {\left( {{x_1} - {x_2} + z} \right)\left( {{x_1} + {x_2} - z} \right)\left( {{x_2} - z} \right)}}} \right)}_{x = 0}}} \hfill \cr { = - {1 \over {16}}{{{\mathop{\rm Res}\nolimits} }_{z = 0}}\left( {{{{e^{ - sz}}} \over {{z^3}}}} \right) = - {{{s^2}} \over {32}}.} \hfill \cr }

Therefore, on the chamber C₂ $p (μ) = p (r, s) = \frac{{(r + s)}^{2}}{64} - \frac{s^{2}}{32} = \frac{r^{2} + 2 r s - s^{2}}{64} .$ p(\mu ) = p(r,s) = {{{{(r + s)}^2}} \over {64}} - {{{s^2}} \over {32}} = {{{r^2} + 2rs - {s^2}} \over {64}}.

(iii)
The wall W₂₃ separating C₂ and C₃ is given by the equation: r – s = 0. Its normal vector ξ = (1, –1). Just only one weights ω₁ = (–2, –2) lies on the linear hyperplane spanned by W₂₃ (other weights are outside of W₂₃: ω₂ = (–2,2), ω₃ = (–2, 0), ω₄ = (0, –2)).
- Consider the push-forward of Lebesgue measure on $ℝ_{⩾ 0}^{1}$ _{0}^1 along the linear map P_W₂₃ : u ↦ uω₁. Its density with respect to dw is given by a single homogeneous polynomial on the wall W₂₃.
- Denote by p_W₂₃ any polynomial function extending it to all of the dual Lie algebra of $\tilde{T}$ {\tilde T}. Clearly $p_{W_{23}} = \frac{1}{2}$ {p_{{W_{23}}}} = {1 \over 2}.

Indeed,

$p_{W_{23}}^{- 1} = d λ (ω_{1}, ξ / ‖ ξ ‖^{2}) = | \det (\begin{matrix} - 2 & - 2 \\ 1 / 2 & - 1 / 2 \end{matrix}) | = 2 \Leftrightarrow p_{W_{23}} = \frac{1}{2} .$ p_{{W_{23}}}^{ - 1} = {\rm{d}}\lambda \left( {{\omega _1},\xi /\xi {^2}} \right) = \left| {\det \left( {\matrix{ { - 2} & { - 2} \cr {1/2} & { - 1/2} \cr } } \right)} \right| = 2 \Leftrightarrow {p_{{W_{23}}}} = {1 \over 2}.

Hence $p (μ) - \frac{r^{2} + 2 r s - s^{2}}{64} = \frac{1}{2} {Res}_{z = 0} {(\frac{e^{〈 μ, x + z ξ 〉}}{\prod_{k = 2}^{4} 〈 ω_{k}, x + z ξ 〉})}_{x = 0},$ p(\mu ) - {{{r^2} + 2rs - {s^2}} \over {64}} = {1 \over 2}{{\mathop{\rm Res}\nolimits} _{z = 0}}{\left( {{{{e^{\langle \mu ,x + z\xi \rangle }}} \over {\prod\nolimits_{k = 2}^4 {\left\langle {{\omega _k},x + z\xi } \right\rangle } }}} \right)_{x = 0}}, where $\frac{e^{〈 μ, x + z ξ 〉}}{\prod_{k = 2}^{4} 〈 ω_{k}, x + z ξ 〉} = \frac{e^{r (x_{1} + z) + s (x_{2} - z)}}{[- 2 (x_{1} + z) + 2 (x_{2} - z)] [- 2 (x_{1} + z)] [- 2 (x_{2} - z)]} .$ {{{e^{\langle \mu ,x + z\xi \rangle }}} \over {\prod\nolimits_{k = 2}^4 {\left\langle {{\omega _k},x + z\xi } \right\rangle } }} = {{{e^{r\left( {{x_1} + z} \right) + s\left( {{x_2} - z} \right)}}} \over {\left[ { - 2\left( {{x_1} + z} \right) + 2\left( {{x_2} - z} \right)} \right]\left[ { - 2\left( {{x_1} + z} \right)} \right]\left[ { - 2\left( {{x_2} - z} \right)} \right]}}.

Thus $\begin{array}{l} p (μ) - \frac{r^{2} + 2 r s - s^{2}}{64} \\ = \frac{1}{2} \times - \frac{1}{8} {Res}_{z = 0} {(\frac{e^{r (x_{1} + z) + s (x_{2} - z)}}{(x_{1} - x_{2} + 2 z) (x_{1} + z) (x_{2} - z)})}_{x = 0} \\ = \frac{1}{32} {Res}_{z = 0} (\frac{e^{(r - s) z}}{z^{3}}) = \frac{{(r - s)}^{2}}{64} . \end{array}$ \matrix{ {p(\mu ) - {{{r^2} + 2rs - {s^2}} \over {64}}} \hfill \cr { = {1 \over 2} \times - {1 \over 8}{{{\mathop{\rm Res}\nolimits} }_{z = 0}}{{\left( {{{{e^{r\left( {{x_1} + z} \right) + s\left( {{x_2} - z} \right)}}} \over {\left( {{x_1} - {x_2} + 2z} \right)\left( {{x_1} + z} \right)\left( {{x_2} - z} \right)}}} \right)}_{x = 0}}} \hfill \cr { = {1 \over {32}}{{{\mathop{\rm Res}\nolimits} }_{z = 0}}\left( {{{{e^{(r - s)z}}} \over {{z^3}}}} \right) = {{{{(r - s)}^2}} \over {64}}.} \hfill \cr }

Therefore, on the chamber C₃ $p (μ) = p (r, s) = \frac{r^{2} + 2 r s - s^{2}}{64} + \frac{{(r - s)}^{2}}{64} = \frac{r^{2}}{32} .$ p(\mu ) = p(r,s) = {{{r^2} + 2rs - {s^2}} \over {64}} + {{{{(r - s)}^2}} \over {64}} = {{{r^2}} \over {32}}.

This completes the proof.

6.

Applications of Duistermaat-Heckman Measure

For $λ = (λ_{1}, λ_{2}, λ_{3}, λ_{4}) \in ℝ_{⩾ 0}^{3}$ \lambda = \left( {{\lambda _1},{\lambda _2},{\lambda _3},{\lambda _4}} \right) \in _{0}^3 with λ₁ > λ₂ > λ₃ > λ₄ and $\sum_{k = 1}^{4} λ_{k} = 1$ \sum\nolimits_{k = 1}^4 {{\lambda _k}} = 1, denote the SU(4)-adjoint orbit of Λ = diag(λ₁, λ₂, λ₃, λ₄) by $U_{Λ} := {U Λ U^{†} : U \in SU (4)} .$ {{\cal U}_\Lambda }: = \left\{ {{\bf{U}}\Lambda {{\bf{U}}^\dag }:{\bf{U}} \in {\mathop{\rm SU}\nolimits} (4)} \right\}.

Let $\hat{Λ} := Λ - \frac{𝟙_{4}}{4} = diag ({\hat{λ}}_{1}, {\hat{λ}}_{2}, {\hat{λ}}_{3}, {\hat{λ}}_{4}) .$ \hat \Lambda : = \Lambda - {{{_4}} \over 4} = {\mathop{\rm diag}\nolimits} \left( {{{\hat \lambda }_1},{{\hat \lambda }_2},{{\hat \lambda }_3},{{\hat \lambda }_4}} \right). where ${\hat{λ}}_{k} = λ_{k} - \frac{1}{4}$ {{\hat \lambda }_k} = {\lambda _k} - {1 \over 4}. Denote by $\hat{λ} = ({\hat{λ}}_{1}, {\hat{λ}}_{2}, {\hat{λ}}_{3}, {\hat{λ}}_{4})$ \hat \lambda = \left( {{{\hat \lambda }_1},{{\hat \lambda }_2},{{\hat \lambda }_3},{{\hat \lambda }_4}} \right). Apparently ${\hat{λ}}_{1} > {\hat{λ}}_{2} > {\hat{λ}}_{3} > {\hat{λ}}_{4}$ {{\hat \lambda }_1} > {{\hat \lambda }_2} > {{\hat \lambda }_3} > {{\hat \lambda }_4} and $\sum_{k = 0}^{4} {\hat{λ}}_{k} = 0$ \sum\nolimits_{k = 0}^4 {{{\hat \lambda }_k}} = 0. It is easily seen that $\hat{Λ} \in i h,$ \hat \Lambda \in , where t is the Cartan subalgebra of 𝔰𝔲(4), the Lie algebra of SU(4). By chosing Ad-invariant inner product on 𝔱, we can identify 𝔱* with t. Let $\tilde{t}$ \widetilde be the Lie algebra of maximal torus $\tilde{T}$ {\tilde T} of $\tilde{K} = SU (2) \times SU (2)$ \tilde K = {\mathop{\rm SU}\nolimits} (2) \times {\mathop{\rm SU}\nolimits} (2).

Denote 44 $c_{3} = 2 ({\hat{λ}}_{1} + {\hat{λ}}_{2}), c_{2} = 2 ({\hat{λ}}_{1} + {\hat{λ}}_{3}), c_{1} = 2 | {\hat{λ}}_{1} + {\hat{λ}}_{4} | .$ {c_3} = 2\left( {{{\hat \lambda }_1} + {{\hat \lambda }_2}} \right),\quad {c_2} = 2\left( {{{\hat \lambda }_1} + {{\hat \lambda }_3}} \right),\quad {c_1} = 2\left| {{{\hat \lambda }_1} + {{\hat \lambda }_4}} \right|.

From the above definition, we infer that 45 $c_{3} > c_{2} > c_{1} > 0 .$ {c_{\rm{3}}}\;{\rm{ > }}\;{c_{\rm{2}}}\;{\rm{ > }}\;{c_{\rm{1}}}\;{\rm{ > }}\;{\rm{0}}{\rm{.}}

Proposition 9.

Consider the Lie group $\tilde{K} = SU (2) \times SU (2)$ \tilde K = {\rm{SU}}(2) \times {\rm{SU}}(2) with its maximal torus $\tilde{T}$ {\tilde T} whose Lie algebra is given by $\tilde{t}$ \widetilde . The positive Weyl chamber $i {\tilde{t}}_{⩾ 0}^{*}$ {\rm{i}}\widetilde {\rm{t}}_{0}^* can be identified with $ℝ_{⩾ 0}^{2}$ _{0}^2.

Proof.

The Lie group SU (2) has a rank one Lie algebra, 𝔰𝔲(2), with Cartan subalgebra 𝔥 spanned by H = |0〉〈0| – |1〉〈1|. The root system is of a single positive root α satisfying α(H) = 2. The positive Weyl chamber is a subset of the Cartan subalgebra 𝔥 defined by the condition that all positive roots take nonnegative values. Any element h ∈ i𝔥 can be expressed as h = zH for some z ∈ ℝ. Then α(h) = zα(H) = 2z ⩾ 0. Geometrically, 𝔥 ≌ ℝ. Thus, the positive Weyl chamber for SU(2) is the set of all nonnegative multipliers of H, i.e., $h_{⩾ 0} = {z H : z ⩾ 0} ≅ ℝ_{⩾ 0} .$ {_{0}} = \{ z{\bf{H}}:z\;\;0\} \cong {_{0}}.

The Lie group $\tilde{K} = SU (2) \times SU (2)$ \tilde K = {\rm{SU}}(2) \times {\rm{SU}}(2) has rank two. Its Lie algebra is , and the Cartan subalgebra $\tilde{t}$ \tilde K = {\rm{SU}}(2) \times {\rm{SU}}(2) is spanned by the generators H₁ from the first su(2) factor and H₂ from the second 𝔰𝔲(2) factor. Using the standard representation, the generators are $H_{1} = H \otimes | 0 〉〈 0 | and H_{2} = H \otimes | 1 〉〈 1 | .$ \widetilde {\rm{t}}

The root system of $\tilde{K} = SU (2) \times SU (2)$ {H_1} = H \otimes |0\rangle \langle 0|{\rm{ and }}{H_2} = H \otimes |1\rangle \langle 1|. is the disjoint union of the root systems of each SU(2) factor. The positive roots are α₁ for the first factor and α₂ for the second factor, defined $α_{1} (H_{1}) = 2, α_{1} (H_{2}) = 0, α_{2} (H_{1}) = 0, α_{2} (H_{2}) = 2.$ \tilde K = {\rm{SU}}(2) \times {\rm{SU}}(2)

The positive Weyl chamber is the subset of the Cartan subalgebra where all positive roots take non-negative values. For an element $h = z_{1} H_{1} + z_{2} H_{2} \in i \tilde{t}$ h = {z_1}{{\bf{H}}_1} + {z_2}{{\bf{H}}_2} \in \widetilde , the conditions are $\begin{array}{l} α_{1} (h) = α_{1} (z_{1} H_{2} + z_{2} H_{2}) = 2 z_{1} + 0 z_{2} = 2 z_{1} ⩾ 0, \\ α_{2} (h) = α_{2} (z_{1} H_{2} + z_{2} H_{2}) = 0 z_{1} + 2 z_{2} = 2 z_{2} ⩾ 0. \end{array}$ \matrix{ {{\alpha _1}(h) = {\alpha _1}\left( {{z_1}{{\bf{H}}_2} + {z_2}{{\bf{H}}_2}} \right) = 2{z_1} + 0{z_2} = 2{z_1}\;\;0,} \hfill \cr {{\alpha _2}(h) = {\alpha _2}\left( {{z_1}{{\bf{H}}_2} + {z_2}{{\bf{H}}_2}} \right) = 0{z_1} + 2{z_2} = 2{z_2}\;\;0.} \hfill \cr }

Thus, the positive Weyl chamber is identified with $i {\tilde{t}}_{⩾ 0} = {z_{1} H_{1} + z_{2} H_{2} : z_{1} ⩾ 0, z_{2} ⩾ 0} ≅ ℝ_{⩾ 0}^{2} .$ {\rm{i}}{\widetilde {\rm{t}}_{0}} = \left\{ {{z_1}{{\bf{H}}_1} + {z_2}{{\bf{H}}_2}:{z_1}\;\;0,{z_2}\;\;0} \right\} \cong _{0}^2.

We are done.

With $i {\tilde{t}}_{⩾ 0}^{*} ≅ ℝ_{⩾ 0}^{2}$ \widetilde {\rm{t}}_{0}^* \cong _{0}^2, we have already seen that 46 ${{DH}_{O_{\hat{λ}}}^{\tilde{K}} |}_{i {\tilde{t}}_{⩾ 0}^{*}} = {\sum_{w \in S_{4}} sign (w) δ_{π (w \hat{λ})} ⋆ H_{(- 2, 0)} ⋆ H_{(- 2, - 2)} ⋆ H_{(0, - 2)} ⋆ H_{(- 2, 2)} |}_{{\tilde{i}}_{⩾ 0}^{*}}$ {\left. {{\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^{\tilde K}} \right|_{{\rm{i}}\widetilde _{0}^*}} = {\left. {\sum\limits_{w \in {S_4}} {{\mathop{\rm sign}\nolimits} } (w){\delta _{\pi (w\hat \lambda )}} \star {H_{( - 2,0)}} \star {H_{( - 2, - 2)}} \star {H_{(0, - 2)}} \star {H_{( - 2,2)}}} \right|_{\widetilde _{0}^*}} is supported on $i {\tilde{t}}_{⩾ 0}^{*}$ {\rm{i}}\widetilde {\rm{t}}_{0}^*, which amounts to be also supported on $ℝ_{⩾ 0}^{2}$ _{0}^2. We can write it in the following form:

Proposition 10.

For $\tilde{K} = SU (2) \times SU (2)$ \tilde K = {\rm{SU}}(2) \times {\rm{SU}}(2), it holds that 47 ${{DH}_{O_{\hat{λ}}}^{\tilde{K}} |}_{ℝ_{⩾ 0}^{2}} = {\sum_{w \in S_{4}} sign (w) δ_{π (w \hat{λ})} ⋆ H_{(- 2, 0)} ⋆ H_{(- 2, - 2)} ⋆ H_{(0, - 2)} ⋆ H_{(- 2, 2)} |}_{ℝ_{⩾ 0}^{2}}$ {\left. {{\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^{\tilde K}} \right|_{_{0}^2}} = {\left. {\sum\limits_{w \in {S_4}} {{\mathop{\rm sign}\nolimits} } (w){\delta _{\pi (w\hat \lambda )}} \star {H_{( - 2,0)}} \star {H_{( - 2, - 2)}} \star {H_{(0, - 2)}} \star {H_{( - 2,2)}}} \right|_{_{0}^2}} with its density function being given by 48 $\begin{array}{l} {\sum_{w \in S_{4}} sign (w) δ_{π (w \hat{λ})} ⋆ p (x, y) |}_{ℝ_{⩾ 0}^{2}} \\ = {(δ_{(c_{2}, \pm c_{1})} - δ_{(c_{1}, c_{2})} + δ_{(c_{1}, c_{3})} - δ_{(c_{3}, \pm c_{1})} + δ_{(c_{3}, \pm c_{2})} - δ_{(c_{2}, c_{3})}) ⋆ p (x, y) |}_{ℝ_{⩾ 0}^{2}}, \end{array}$ \matrix{ {{{\left. {\sum\limits_{w \in {S_4}} {{\mathop{\rm sign}\nolimits} } (w){\delta _{\pi (w\hat \lambda )}} \star p(x,y)} \right|}_{_{0}^2}}} \hfill \cr { = {{\left. {\left( {{\delta _{\left( {{c_2}, \pm {c_1}} \right)}} - {\delta _{\left( {{c_1},{c_2}} \right)}} + {\delta _{\left( {{c_1},{c_3}} \right)}} - {\delta _{\left( {{c_3}, \pm {c_1}} \right)}} + {\delta _{\left( {{c_3}, \pm {c_2}} \right)}} - {\delta _{\left( {{c_2},{c_3}} \right)}}} \right) \star p(x,y)} \right|}_{_{0}^2}},} \hfill \cr } where the expression of p(x, y) is from Proposition 8.

Proof

Sketch of proof can be given for Eq. (48): Indeed, we have already known the support of p(x, y) is $\cup_{k = 1}^{3} C_{k}$ \cup _{k = 1}^3{C_k}. Then the support of $(δ_{π (w \hat{λ})} ⋆ p) (x, y)$ \left( {{\delta _{\pi (w\hat \lambda )}} \star p} \right)(x,y) is the shifted version of $\cup_{k = 1}^{3} C_{k}$ \cup _{k = 1}^3{C_k} to the point $π (w \hat{λ})$ \pi (w\hat \lambda ). However, points whose corresponding shifted supports do not intersect with $ℝ_{> 0}^{2}$ _{ > 0}^2 do not contribute to the density. In other words, by calculation, we find that it remains only 9 points can contribute to the density.

Remark 4.

Moreover, the non-Abelian moment polytope for the action of $\tilde{K} = SU (2) \times SU (2)$ \tilde K = {\rm{SU}}(2) \times {\rm{SU}}(2) on a generic coadjoint K = SU (4)-orbit can be identified as 49 $Δ_{\tilde{K}} (O_{\hat{λ}}) := {(x, y) \in {[0, c_{3}]}^{2} : x + y ⩽ c_{2} + c_{3}, | x - y | ⩽ c_{3} - c_{1}},$ {\Delta _{\tilde K}}\left( {{{\cal O}_{\hat \lambda }}} \right): = \left\{ {(x,y) \in {{\left[ {0,{c_3}} \right]}^2}:x + y\;\;{c_2} + {c_3},|x - y|\;{c_3} - {c_1}} \right\}, where c_k’s are from Eq. (44). This is just the support of the density in Eq. (48). Based on the constraint Eq. (49), we can re-derive the compatibility solution for two-qubit quantum marginal problem. Let me describe it below: Let ρ_AB ∈ D(ℂ² ⊗ ℂ²) with two reduced states (ρ_A, ρ_B), where the eigenvalues of ρ_AB are ordered non-increasingly as λ₁ ⩾ ⋯ ⩾ λ₄ ⩾ 0, and the minimal eigenvalue of ρ_X(X = A, B) is denoted by λ_min. Via ${\hat{λ}}_{j} = λ_{j} - \frac{1}{4}$ {{\hat \lambda }_j} = {\lambda _j} - {1 \over 4} for j ∈ {1,2,3,4} and x = 1 – 2λ_min(ρ₁) and y = 1 – 2λ_min(ρ₂), we see that the condition x ∈ [0, c₃] can be rewritten as 50 $\min {λ_{\min} (ρ_{A}), λ_{\min} (ρ_{B})} ⩾ λ_{3} (ρ_{A B}) + λ_{4} (ρ_{A B});$ \min \left\{ {{\lambda _{\min }}\left( {{\rho _A}} \right),{\lambda _{\min }}\left( {{\rho _B}} \right)} \right\}\;\;{\lambda _3}\left( {{\rho _{AB}}} \right) + {\lambda _4}\left( {{\rho _{AB}}} \right); and the condition x + y ⩽ c₂ + c₃ can be rewritten as 51 $λ_{\min} (ρ_{A}) + λ_{\min} (ρ_{B}) ⩾ λ_{2} (ρ_{A B}) + λ_{3} (ρ_{A B}) + 2 λ_{4} (ρ_{A B});$ {\lambda _{\min }}\left( {{\rho _A}} \right) + {\lambda _{\min }}\left( {{\rho _B}} \right)\;\;{\lambda _2}\left( {{\rho _{AB}}} \right) + {\lambda _3}\left( {{\rho _{AB}}} \right) + 2{\lambda _4}\left( {{\rho _{AB}}} \right); and the condition |x – y| ⩽ c₃ – c₁ can be rewritten as 52 $| λ_{\min} (ρ_{A}) - λ_{\min} (ρ_{B}) | ⩾ \min {λ_{1} (ρ_{A B}) - λ_{3} (ρ_{A B}), λ_{2} (ρ_{A B}) - λ_{4} (ρ_{A B})} .$ \left| {{\lambda _{\min }}\left( {{\rho _A}} \right) - {\lambda _{\min }}\left( {{\rho _B}} \right)} \right|\;\;\min \left\{ {{\lambda _1}\left( {{\rho _{AB}}} \right) - {\lambda _3}\left( {{\rho _{AB}}} \right),{\lambda _2}\left( {{\rho _{AB}}} \right) - {\lambda _4}\left( {{\rho _{AB}}} \right)} \right\}.

In summary, for any given 2-tuple (ρ_A, ρ_B) of qubit states, there exists a global two-qubit states ρ_AB such that ρ_A = Tr_A(ρ_AB) and ρ_B = Tr_A(ρ_AB) if and only if the three compatibility conditions Eqs. (50)–(52) hold. This result is obtained already by Bravyi [26].

Consider the pushforward measure $F_{*} (μ_{O_{\hat{λ}}})$ {F_*}\left( {{\mu _{{{\cal O}_{\hat \lambda }}}}} \right) of Liouville measure $μ_{O_{\hat{λ}}}$ {\mu _{{{\cal O}_{\hat \lambda }}}} along the map $F : O_{\hat{λ}} \to i 𝔥_{⩾ 0}^{*} ≅ ℝ_{⩾ 0}$ F:{{\cal O}_{\hat \lambda }} \to i_{0}^* \cong {_{0}}, where $𝔥_{⩾ 0}^{*}$ _{0}^* is the dual Cartan subalgebra of 𝔰𝔲(2) and F = q₁ ∘ π ∘ Φ, 53 $F : O_{\hat{λ}} (\subset k^{*}) \overset{Φ}{↪} i k^{*} \overset{π}{\to} i {\tilde{t}}_{⩾ 0}^{*} = i 𝔥_{⩾ 0}^{*} \oplus i h_{⩾ 0}^{*} \overset{q_{1}}{\to} i h_{⩾ 0}^{*} .$ F:{{\cal O}_{\hat \lambda }}\left( { \subset {^*}} \right)\mathop \limits^\Phi {\rm{i}}{^*}\buildrel \pi \over \longrightarrow {\rm{i}}\widetilde _{0}^* = {\rm{i}}_{0}^* \oplus {\rm{i}}_{0}^*\buildrel {{q_1}} \over \longrightarrow {\rm{i}}_{0}^*.

Proposition 11.

The pushforward measure $F_{*} μ_{O_{\hat{λ}}}$ {F_*}{\mu _{{{\cal O}_{\hat \lambda }}}} is just 54 $F_{*} (μ_{O_{\hat{λ}}}) = {(q_{1})}_{*} π_{*} Φ_{*} (μ_{O_{\hat{λ}}}) = {(q_{1})}_{*} (p_{\tilde{K}} {DH}_{O_{\hat{λ}}}^{\tilde{K}})$ {F_*}\left( {{\mu _{{{\cal O}_{\hat \lambda }}}}} \right) = {\left( {{q_1}} \right)_*}{\pi _*}{\Phi _*}\left( {{\mu _{{{\cal O}_{\hat \lambda }}}}} \right) = {\left( {{q_1}} \right)_*}\left( {{p_{\tilde K}}{\rm{DH}}_{{{\cal O}_{\hat \lambda }}}^{\tilde K}} \right) whose density function is given by $(2 π)^{6} ℐ (x ∣ \hat{λ})$ {(2\pi )^6}{\cal I}(x\mid \hat \lambda ), where $ℐ (x ∣ \hat{λ}) := \int_{ℝ_{⩾ 0}^{2}} x y [(δ_{(c_{2}, \pm c_{1})} - δ_{(c_{1}, c_{2})} + δ_{(c_{1}, c_{3})} - δ_{(c_{3}, \pm c_{1})} + δ_{(c_{3}, \pm c_{2})} - δ_{(c_{2}, c_{3})}) * p] (x, y) d x d y,$ {\cal I}(x\mid \hat \lambda ): = \int_{_{0}^2} x y\left[ {\left( {{\delta _{\left( {{c_2}, \pm {c_1}} \right)}} - {\delta _{\left( {{c_1},{c_2}} \right)}} + {\delta _{\left( {{c_1},{c_3}} \right)}} - {\delta _{\left( {{c_3}, \pm {c_1}} \right)}} + {\delta _{\left( {{c_3}, \pm {c_2}} \right)}} - {\delta _{\left( {{c_2},{c_3}} \right)}}} \right)*p} \right](x,y){\rm{d}}x{\rm{d}}y, where the expression of p(x, y) is from Proposition 8. Moreover, it holds that 55 $ℐ (x ∣ \hat{λ}) = ℐ_{1} (x ∣ \hat{λ}) + ℐ_{2} (x ∣ \hat{λ}) + ℐ_{3} (x ∣ \hat{λ}),$ {\cal I}(x\mid \hat \lambda ) = {{\cal I}_1}(x\mid \hat \lambda ) + {{\cal I}_2}(x\mid \hat \lambda ) + {{\cal I}_3}(x\mid \hat \lambda ), where 56 $ℐ_{1} (x ∣ \hat{λ}) = \frac{c_{3}^{2} - c_{2}^{2}}{64} x {(x - c_{1})}^{2} χ_{[0, c_{1}]} (x),$ {{\cal I}_1}(x\mid \hat \lambda ) = {{c_3^2 - c_2^2} \over {64}}x{\left( {x - {c_1}} \right)^2}{\chi _{\left[ {0,{c_1}} \right]}}(x), 57 $ℐ_{2} (x ∣ \hat{λ}) = \frac{c_{1}^{2} - c_{3}^{2}}{64} x {(x - c_{2})}^{2} χ_{[0, c_{2}]} (x),$ {{\cal I}_2}(x\mid \hat \lambda ) = {{c_1^2 - c_3^2} \over {64}}x{\left( {x - {c_2}} \right)^2}{\chi _{\left[ {0,{c_2}} \right]}}(x), 58 $ℐ_{3} (x ∣ \hat{λ}) = \frac{c_{2}^{2} - c_{1}^{2}}{64} x {(x - c_{3})}^{2} χ_{[0, c_{3}]} (x),$ {{\cal I}_3}(x\mid \hat \lambda ) = {{c_2^2 - c_1^2} \over {64}}x{\left( {x - {c_3}} \right)^2}{\chi _{\left[ {0,{c_3}} \right]}}(x), where c_k’s are from Eq. (44), and χ_A(x) is the indicator of a set A.

Proof

Note that $p_{\tilde{K}} (\frac{x}{2} h, \frac{y}{2} h) = p_{SU (2)} (\frac{x}{2} h) p_{SU (2)} (\frac{y}{2} h) = x y$ {p_{\tilde K}}\left( {{x \over 2}{\bf{h}},{y \over 2}{\bf{h}}} \right) = {p_{{\rm{SU}}(2)}}\left( {{x \over 2}{\bf{h}}} \right){p_{{\rm{SU}}(2)}}\left( {{y \over 2}{\bf{h}}} \right) = xy We consider the following integral $\begin{array}{l} ℐ (x ∣ \hat{λ}) := \int_{ℝ_{⩾ 0}^{2}} x y [(δ_{(c_{2}, \pm c_{1})} - δ_{(c_{1}, c_{2})} + δ_{(c_{1}, c_{3})} - δ_{(c_{3}, \pm c_{1})} + δ_{(c_{3}, \pm c_{2})} - δ_{(c_{2}, c_{3})}) ⋆ p] (x, y) d x d y \\ = \int_{ℝ_{⩾ 0}^{2}} x y [(δ_{(c_{1}, c_{3})} - δ_{(c_{1}, c_{2})}) ⋆ p] (x, y) d x d y \\ + \int_{ℝ_{⩾ 0}^{2}} x y [(- δ_{(c_{2}, c_{3})} + δ_{(c_{2}, c_{1})} + δ_{(c_{2}, - c_{1})}) ⋆ p] (x, y) d x d y \\ + \int_{ℝ_{⩾ 0}^{2}} x y [(δ_{(c_{3}, c_{2})} + δ_{(c_{3}, - c_{2})} - δ_{(c_{3}, c_{1})} - δ_{(c_{3}, - c_{1})}) ⋆ p] (x, y) d x d y \end{array}$ \matrix{ {{\cal I}(x\mid \hat \lambda ): = \int_{_{0}^2} x y\left[ {\left( {{\delta _{\left( {{c_2}, \pm {c_1}} \right)}} - {\delta _{\left( {{c_1},{c_2}} \right)}} + {\delta _{\left( {{c_1},{c_3}} \right)}} - {\delta _{\left( {{c_3}, \pm {c_1}} \right)}} + {\delta _{\left( {{c_3}, \pm {c_2}} \right)}} - {\delta _{\left( {{c_2},{c_3}} \right)}}} \right) \star p} \right](x,y){\rm{d}}x{\rm{d}}y} \hfill \cr { = \int_{_{0}^2} x y\left[ {\left( {{\delta _{\left( {{c_1},{c_3}} \right)}} - {\delta _{\left( {{c_1},{c_2}} \right)}}} \right) \star p} \right](x,y){\rm{d}}x{\rm{d}}y} \hfill \cr {\;\; + \int_{_{0}^2} x y\left[ {\left( { - {\delta _{\left( {{c_2},{c_3}} \right)}} + {\delta _{\left( {{c_2},{c_1}} \right)}} + {\delta _{\left( {{c_2}, - {c_1}} \right)}}} \right) \star p} \right](x,y){\rm{d}}x{\rm{d}}y} \hfill \cr {\;\; + \int_{_{0}^2} x y\left[ {\left( {{\delta _{\left( {{c_3},{c_2}} \right)}} + {\delta _{\left( {{c_3}, - {c_2}} \right)}} - {\delta _{\left( {{c_3},{c_1}} \right)}} - {\delta _{\left( {{c_3}, - {c_1}} \right)}}} \right) \star p} \right](x,y){\rm{d}}x{\rm{d}}y} \hfill \cr } which is the density of the Duistermaat-Heckman measure with respect to the Lebesgue measure on the positive Weyl chamber (which is identified with $ℝ_{⩾ 0}^{2}$ _{0}^2) of SU(2) × SU(2). Denote by 59 $ℐ_{1} (x ∣ \hat{λ}) = \int_{ℝ_{⩾ 0}^{2}} x y [(δ_{(c_{1}, c_{3})} - δ_{(c_{1}, c_{2})}) ⋆ p] (x, y) d x d y,$ {{\cal I}_1}(x\mid \hat \lambda ) = \int_{_{0}^2} x y\left[ {\left( {{\delta _{\left( {{c_1},{c_3}} \right)}} - {\delta _{\left( {{c_1},{c_2}} \right)}}} \right) \star p} \right](x,y){\rm{d}}x{\rm{d}}y, 60 $ℐ_{2} (x ∣ \hat{λ}) = \int_{ℝ_{⩾ 0}^{2}} x y [(- δ_{(c_{2}, c_{3})} + δ_{(c_{2}, c_{1})} + δ_{(c_{2}, - c_{1})}) ⋆ p] (x, y) d x d y,$ {{\cal I}_2}(x\mid \hat \lambda ) = \int_{_{0}^2} x y\left[ {\left( { - {\delta _{\left( {{c_2},{c_3}} \right)}} + {\delta _{\left( {{c_2},{c_1}} \right)}} + {\delta _{\left( {{c_2}, - {c_1}} \right)}}} \right) \star p} \right](x,y){\rm{d}}x{\rm{d}}y, 61 $ℐ_{3} (x ∣ \hat{λ}) = \int_{ℝ_{⩾ 0}^{2}} x y [(δ_{(c_{3}, c_{2})} + δ_{(c_{3}, - c_{2})} - δ_{(c_{3}, c_{1})} - δ_{(c_{3}, - c_{1})}) ⋆ p] (x, y) d x d y .$ {{\cal I}_3}(x\mid \hat \lambda ) = \int_{_{0}^2} x y\left[ {\left( {{\delta _{\left( {{c_3},{c_2}} \right)}} + {\delta _{\left( {{c_3}, - {c_2}} \right)}} - {\delta _{\left( {{c_3},{c_1}} \right)}} - {\delta _{\left( {{c_3}, - {c_1}} \right)}}} \right) \star p} \right](x,y){\rm{d}}x{\rm{d}}y.

Note that $c_{1} = 2 | {\hat{λ}}_{1} + {\hat{λ}}_{4} | = 2 | {\hat{λ}}_{2} + {\hat{λ}}_{3} |$ {c_1} = 2\left| {{{\hat \lambda }_1} + {{\hat \lambda }_4}} \right| = 2\left| {{{\hat \lambda }_2} + {{\hat \lambda }_3}} \right| because ${\hat{λ}}_{1} + {\hat{λ}}_{4} = - ({\hat{λ}}_{2} + {\hat{λ}}_{3})$ {{\hat \lambda }_1} + {{\hat \lambda }_4} = - \left( {{{\hat \lambda }_2} + {{\hat \lambda }_3}} \right). Thus $c_{1} = \max {{\hat{λ}}_{1} + {\hat{λ}}_{4}, {\hat{λ}}_{2} + {\hat{λ}}_{3}} .$ {c_1} = \max \left\{ {{{\hat \lambda }_1} + {{\hat \lambda }_4},{{\hat \lambda }_2} + {{\hat \lambda }_3}} \right\}.

(a)
The range [0, c₃] of parameter x can be decomposed into two parts [0, c₃] = [0, c₁] ∪ [c₁, c₃]. In order to calculate ℐ₁, note that ℐ₁ ≡ 0 if x ∈ [c₁, c₃], it suffices to calculate it for x ∈ [0, c₁]. Thus $\begin{array}{l} ℐ_{1} (x ∣ \hat{λ}) & = & \int_{0}^{x + c_{3} - c_{1}} x y \cdot p_{3} (x - c_{1}, y - c_{3}) d y + \int_{x + c_{3} - c_{1}}^{c_{3}} x y \cdot p_{2} (x - c_{1}, y - c_{3}) d y \\ + \int_{c_{3}}^{- x + c_{3} + c_{1}} x y \cdot p_{1} (x - c_{1}, y - c_{3}) d y - \int_{0}^{x + c_{2} - c_{1}} x y \cdot p_{3} (x - c_{1}, y - c_{2}) d y \\ - \int_{x + c_{2} - c_{1}}^{c_{2}} x y \cdot p_{2} (x - c_{1}, y - c_{2}) d y - \int_{c_{2}}^{- x + c_{1} + c_{2}} x y \cdot p_{1} (x - c_{1}, y - c_{2}) d y \\ = & \frac{c_{3}^{2} - c_{2}^{2}}{64} x {(x - c_{1})}^{2}, \end{array}$ \matrix{ {{{\cal I}_1}(x\mid \hat \lambda )} \hfill & = \hfill & {\int_0^{x + {c_3} - {c_1}} x y\cdot{p_3}\left( {x - {c_1},y - {c_3}} \right){\rm{d}}y + \int_{x + {c_3} - {c_1}}^{{c_3}} x y \cdot {p_2}\left( {x - {c_1},y - {c_3}} \right){\rm{d}}y} \hfill \cr {} \hfill & {} \hfill & { + \int_{{c_3}}^{ - x + {c_3} + {c_1}} x y\cdot{p_1}\left( {x - {c_1},y - {c_3}} \right){\rm{d}}y - \int_0^{x + {c_2} - {c_1}} x y \cdot {p_3}\left( {x - {c_1},y - {c_2}} \right){\rm{d}}y} \hfill \cr {} \hfill & {} \hfill & { - \int_{x + {c_2} - {c_1}}^{{c_2}} x y\cdot{p_2}\left( {x - {c_1},y - {c_2}} \right){\rm{d}}y - \int_{{c_2}}^{ - x + {c_1} + {c_2}} x y \cdot {p_1}\left( {x - {c_1},y - {c_2}} \right){\rm{d}}y} \hfill \cr {} \hfill & = \hfill & {{{c_3^2 - c_2^2} \over {64}}x{{\left( {x - {c_1}} \right)}^2},} \hfill \cr } which leads to 62 $I_{1} (x ∣ \hat{λ}) = {\begin{array}{l} \frac{c_{3}^{2} - c_{2}^{2}}{64} x {(x - c_{1})}^{2}, & if x \in [0, c_{1}], \\ 0, & if x \in [c_{1}, c_{3}] . \end{array}$ {{\cal I}_1}(x\mid \hat \lambda ) = \left\{ {\matrix{ {{{c_3^2 - c_2^2} \over {64}}x{{\left( {x - {c_1}} \right)}^2},} \hfill & {{\rm{ if }}x \in \left[ {0,{c_1}} \right],} \hfill \cr {0,} \hfill & {{\rm{ if }}x \in \left[ {{c_1},{c_3}} \right].} \hfill \cr } } \right.

With the help of the following Figure 2, the above calculation is easily obtained.
(b)
The range [0, c₃] of parameter x can be decomposed into three parts: [0, c₃] = [0, c₂ – c₁] ∪ [c₂ – c₁, c2] ∪ [c₂, c₃]. Note that I₂ ≡ 0 when x ∈ [c₂, c₃]. It suffices to calculate I₂ on [0, c₂ – c₁] ∩ [c₂ – c₁, c₂]. See the following Figure 3.
- (b1)
  For x ∈ [0, c₂ – c₁], we see that $\begin{array}{l} I_{2} (x ∣ \hat{λ}) & = & - \int_{0}^{x + c_{3} - c_{2}} x y p_{3} (x - c_{2}, y - c_{3}) d y - \int_{x + c_{3} - c_{2}}^{c_{3}} x y p_{2} (x - c_{2}, y - c_{3}) d y \\ - \int_{c_{3}}^{- x + c_{3} + c_{2}} x y p_{1} (x - c_{2}, y - c_{3}) d y + \int_{0}^{c_{1}} x y p_{2} (x - c_{2}, y - c_{1}) d y \\ + \int_{c_{1}}^{- x + c_{2} + c_{1}} x y p_{3} (x - c_{2}, y - c_{1}) d y + \int_{0}^{- x + c_{2} - c_{1}} x y p_{3} (x - c_{2}, y + c_{1}) d y \\ = & \frac{c_{1}^{2} - c_{3}^{2}}{64} x {(x - c_{2})}^{2} \end{array}$ \matrix{ {{{\cal I}_2}(x\mid \hat \lambda )} \hfill & = \hfill & { - \int_0^{x + {c_3} - {c_2}} x y{p_3}\left( {x - {c_2},y - {c_3}} \right){\rm{d}}y - \int_{x + {c_3} - {c_2}}^{{c_3}} x y{p_2}\left( {x - {c_2},y - {c_3}} \right){\rm{d}}y} \hfill \cr {} \hfill & {} \hfill & { - \int_{{c_3}}^{ - x + {c_3} + {c_2}} x y{p_1}\left( {x - {c_2},y - {c_3}} \right){\rm{d}}y + \int_0^{{c_1}} x y{p_2}\left( {x - {c_2},y - {c_1}} \right){\rm{d}}y} \hfill \cr {} \hfill & {} \hfill & { + \int_{{c_1}}^{ - x + {c_2} + {c_1}} x y{p_3}\left( {x - {c_2},y - {c_1}} \right){\rm{d}}y + \int_0^{ - x + {c_2} - {c_1}} x y{p_3}\left( {x - {c_2},y + {c_1}} \right){\rm{d}}y} \hfill \cr {} \hfill & = \hfill & {{{c_1^2 - c_3^2} \over {64}}x{{\left( {x - {c_2}} \right)}^2}} \hfill \cr }
- (b2)
  For x ∈ [c₂ – c₁, c₂], $\begin{array}{l} I_{2} (x ∣ \hat{λ}) & = & - \int_{0}^{x + c_{3} - c_{2}} x y p_{3} (x - c_{2}, y - c_{3}) d y - \int_{x + c_{3} - c_{2}}^{c_{3}} x y p_{2} (x - c_{2}, y - c_{3}) d y \\ - \int_{c_{3}}^{- x + c_{3} + c_{2}} x y p_{1} (x - c_{2}, y - c_{3}) d y + \int_{0}^{x + c_{1} - c_{2}} x y p_{3} (x - c_{2}, y - c_{1}) d y \\ + \int_{x + c_{1} - c_{2}}^{c_{1}} x y p_{2} (x - c_{2}, y - c_{1}) d y + \int_{c_{1}}^{- x + c_{2} + - c_{1}} x y p_{1} (x - c_{2}, y - c_{1}) d y \\ = & \frac{c_{1}^{2} - c_{3}^{2}}{64} x {(x - c_{2})}^{2} \end{array}$ \matrix{ {{{\cal I}_2}(x\mid \hat \lambda )} \hfill & = \hfill & { - \int_0^{x + {c_3} - {c_2}} x y{p_3}\left( {x - {c_2},y - {c_3}} \right){\rm{d}}y - \int_{x + {c_3} - {c_2}}^{{c_3}} x y{p_2}\left( {x - {c_2},y - {c_3}} \right){\rm{d}}y} \hfill \cr {} \hfill & {} \hfill & { - \int_{{c_3}}^{ - x + {c_3} + {c_2}} x y{p_1}\left( {x - {c_2},y - {c_3}} \right){\rm{d}}y + \int_0^{x + {c_1} - {c_2}} x y{p_3}\left( {x - {c_2},y - {c_1}} \right){\rm{d}}y} \hfill \cr {} \hfill & {} \hfill & { + \int_{x + {c_1} - {c_2}}^{{c_1}} x y{p_2}\left( {x - {c_2},y - {c_1}} \right){\rm{d}}y + \int_{{c_1}}^{ - x + {c_2} + - {c_1}} x y{p_1}\left( {x - {c_2},y - {c_1}} \right){\rm{d}}y} \hfill \cr {} \hfill & = \hfill & {{{c_1^2 - c_3^2} \over {64}}x{{\left( {x - {c_2}} \right)}^2}} \hfill \cr }

This leads to 63 $I_{2} (x ∣ \hat{λ}) = {\begin{array}{l} \frac{c_{1}^{2} - c_{3}^{2}}{64} x {(x - c_{2})}^{2}, & if x \in [0, c_{2}], \\ 0, & if x \in [c_{2}, c_{3}] . \end{array}$ {{\cal I}_2}(x\mid \hat \lambda ) = \left\{ {\matrix{ {{{c_1^2 - c_3^2} \over {64}}x{{\left( {x - {c_2}} \right)}^2},} \hfill & {{\rm{ if }}x \in \left[ {0,{c_2}} \right],} \hfill \cr {0,} \hfill & {{\rm{ if }}x \in \left[ {{c_2},{c_3}} \right].} \hfill \cr } } \right.

(c)
The range [0, c₃] of parameter x can be decomposed into three parts: [0, c₃] = [0, c₃ – c₂] ∪ [c₃ – c₂, c₃ – c₁] ∪ [c₃ – c₁, c₃]. It suffices to calculate I₂ on [0, c₂ – c₁] ∪ [c₂ – c₁, c₂]. See the following Figure 4.
- (c1)
  For x ∈ [0, c₃ – c₂], we see that $\begin{array}{l} I_{3} (x ∣ \hat{λ}) & = & = \int_{0}^{c_{2}} x y p_{2} (x - c_{3}, y - c_{2}) d x d y + \int_{c_{2}}^{- x + c_{3} + c_{2}} x y p_{1} (x - c_{3}, y - c_{2}) d x d y \\ + \int_{0}^{- x + c_{3} - c_{2}} x y p_{1} (x - c_{3}, y + c_{2}) d x d y - \int_{0}^{c_{1}} x y p_{2} (x - c_{3}, y - c_{1}) d x d y \\ - \int_{c_{1}}^{- x + c_{3} + c_{1}} x y p_{1} (x - c_{3}, y - c_{1}) d x d y - \int_{0}^{- x + c_{3} - c_{1}} x y p_{1} (x - c_{3}, y + c_{1}) d x d y \\ = & \frac{c_{2}^{2} - c_{1}^{2}}{64} x {(x - c_{3})}^{2} \end{array}$ \matrix{ {{{\cal I}_3}(x\mid \hat \lambda )} \hfill & = \hfill & { = \int_0^{{c_2}} x y{p_2}\left( {x - {c_3},y - {c_2}} \right){\rm{d}}x{\rm{d}}y + \int_{{c_2}}^{ - x + {c_3} + {c_2}} x y{p_1}\left( {x - {c_3},y - {c_2}} \right){\rm{d}}x{\rm{d}}y} \hfill \cr {} \hfill & {} \hfill & { + \int_0^{ - x + {c_3} - {c_2}} x y{p_1}\left( {x - {c_3},y + {c_2}} \right){\rm{d}}x{\rm{d}}y - \int_0^{{c_1}} x y{p_2}\left( {x - {c_3},y - {c_1}} \right){\rm{d}}x{\rm{d}}y} \hfill \cr {} \hfill & {} \hfill & { - \int_{{c_1}}^{ - x + {c_3} + {c_1}} x y{p_1}\left( {x - {c_3},y - {c_1}} \right){\rm{d}}x{\rm{d}}y - \int_0^{ - x + {c_3} - {c_1}} x y{p_1}\left( {x - {c_3},y + {c_1}} \right){\rm{d}}x{\rm{d}}y} \hfill \cr {} \hfill & = \hfill & {{{c_2^2 - c_1^2} \over {64}}x{{\left( {x - {c_3}} \right)}^2}} \hfill \cr }
- (c2)
  For x ∈ [c₃ – c₂, c₃ – c₁], $\begin{array}{l} I_{3} (x ∣ \hat{λ}) & = & \int_{0}^{x + c_{2} - c_{3}} x y p_{3} (x - c_{3}, y - c_{2}) d x d y + \int_{x + c_{2} - c_{3}}^{c_{2}} x y p_{2} (x - c_{3}, y - c_{2}) d x d y \\ + \int_{c_{2}}^{- x + c_{3} + c_{2}} x y p_{1} (x - c_{3}, y - c_{2}) d x d y - \int_{0}^{c_{1}} x y p_{2} (x - c_{3}, y - c_{1}) d x d y \\ - \int_{c_{1}}^{- x + c_{3} + c_{1}} x y p_{1} (x - c_{3}, y - c_{1}) d x d y - \int_{0}^{- x + c_{3} - c_{1}} x y p_{1} (x - c_{3}, y + c_{1}) d x d y \\ = & \frac{c_{2}^{2} - c_{1}^{2}}{64} x {(x - c_{3})}^{2} \end{array}$ \matrix{ {{{\cal I}_3}(x\mid \hat \lambda )} \hfill & = \hfill & {\int_0^{x + {c_2} - {c_3}} x y{p_3}\left( {x - {c_3},y - {c_2}} \right){\rm{d}}x{\rm{d}}y + \int_{x + {c_2} - {c_3}}^{{c_2}} x y{p_2}\left( {x - {c_3},y - {c_2}} \right){\rm{d}}x{\rm{d}}y} \hfill \cr {} \hfill & {} \hfill & { + \int_{{c_2}}^{ - x + {c_3} + {c_2}} x y{p_1}\left( {x - {c_3},y - {c_2}} \right){\rm{d}}x{\rm{d}}y - \int_0^{{c_1}} x y{p_2}\left( {x - {c_3},y - {c_1}} \right){\rm{d}}x{\rm{d}}y} \hfill \cr {} \hfill & {} \hfill & { - \int_{{c_1}}^{ - x + {c_3} + {c_1}} x y{p_1}\left( {x - {c_3},y - {c_1}} \right){\rm{d}}x{\rm{d}}y - \int_0^{ - x + {c_3} - {c_1}} x y{p_1}\left( {x - {c_3},y + {c_1}} \right){\rm{d}}x{\rm{d}}y} \hfill \cr {} \hfill & = \hfill & {{{c_2^2 - c_1^2} \over {64}}x{{\left( {x - {c_3}} \right)}^2}} \hfill \cr }
- (c3)
  For x ∈ [c₃ – c₁, c₃], $\begin{array}{l} I_{3} (x ∣ \hat{λ}) & = & \int_{0}^{x + c_{2} - c_{3}} x y p_{3} (x - c_{3}, y - c_{2}) d x d y + \int_{x + c_{2} - c_{3}}^{c_{2}} x y p_{2} (x - c_{3}, y - c_{2}) d x d y \\ + \int_{c_{2}}^{- x + c_{2} + c_{3}} x y p_{1} (x - c_{3}, y - c_{2}) d x d y - \int_{0}^{x + c_{1} - c_{3}} x y p_{3} (x - c_{3}, y - c_{1}) d x d y \\ - \int_{x + c_{1} - c_{3}}^{c_{1}} x y p_{2} (x - c_{3}, y - c_{1}) d x d y - \int_{c_{1}}^{- x + c_{3} + c_{1}} x y p_{1} (x - c_{3}, y - c_{1}) d x d y \\ = & \frac{c_{2}^{2} - c_{1}^{2}}{64} x {(x - c_{3})}^{2} \end{array}$ \matrix{ {{{\cal I}_3}(x\mid \hat \lambda )} \hfill & = \hfill & {\int_0^{x + {c_2} - {c_3}} x y{p_3}\left( {x - {c_3},y - {c_2}} \right){\rm{d}}x{\rm{d}}y + \int_{x + {c_2} - {c_3}}^{{c_2}} x y{p_2}\left( {x - {c_3},y - {c_2}} \right){\rm{d}}x{\rm{d}}y} \hfill \cr {} \hfill & {} \hfill & { + \int_{{c_2}}^{ - x + {c_2} + {c_3}} x y{p_1}\left( {x - {c_3},y - {c_2}} \right){\rm{d}}x{\rm{d}}y - \int_0^{x + {c_1} - {c_3}} x y{p_3}\left( {x - {c_3},y - {c_1}} \right){\rm{d}}x{\rm{d}}y} \hfill \cr {} \hfill & {} \hfill & { - \int_{x + {c_1} - {c_3}}^{{c_1}} x y{p_2}\left( {x - {c_3},y - {c_1}} \right){\rm{d}}x{\rm{d}}y - \int_{{c_1}}^{ - x + {c_3} + {c_1}} x y{p_1}\left( {x - {c_3},y - {c_1}} \right){\rm{d}}x{\rm{d}}y} \hfill \cr {} \hfill & = \hfill & {{{c_2^2 - c_1^2} \over {64}}x{{\left( {x - {c_3}} \right)}^2}} \hfill \cr }

These leads to 64 $I_{3} (x ∣ \hat{λ}) = \frac{c_{2}^{2} - c_{1}^{2}}{64} x {(x - c_{3})}^{2}, (x \in [0, c_{3}]) .$ {{\cal I}_3}(x\mid \hat \lambda ) = {{c_2^2 - c_1^2} \over {64}}x{\left( {x - {c_3}} \right)^2},\quad \left( {x \in \left[ {0,{c_3}} \right]} \right).

This completes the proof.

6.1.

Separability Probability in the Conditioned State Space

For a given qudit state η ∈ D(ℂⁿ), Milz and Strunz considered the conditioned state space [27] 65 $D^{η} (ℂ^{n} \otimes ℂ^{m}) := {ρ \in D (ℂ^{n} \otimes ℂ^{m}) : {Tr}_{ℂ^{m}} (ρ) = η \in D (ℂ^{n})}$ {{\rm{D}}^\eta }\left( {{^n} \otimes {^m}} \right): = \left\{ {\rho \in {\rm{D}}\left( {{^n} \otimes {^m}} \right):{{{\mathop{\rm Tr}\nolimits} }_{{^m}}}(\rho ) = \eta \in {\rm{D}}\left( {{^n}} \right)} \right\} and the set of separable states in D^η(ℂⁿ ⊗ ℂ^m), denoted by $D_{sep}^{η} (ℂ^{n} \otimes ℂ^{m})$ {\rm{D}}_{{\rm{sep }}}^\eta \left( {{^n} \otimes {^m}} \right). They studied the separability probability in the conditioned state space 66 $P_{sep}^{(n \times m)} (η) := \frac{{vol}_{HS} (D_{sep}^{η} (ℂ^{n} \otimes ℂ^{m}))}{{vol}_{HS} (D^{η} (ℂ^{n} \otimes ℂ^{m}))} .$ P_{{\rm{sep}}}^{(n \times m)}(\eta ): = {{{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{\rm{D}}_{{\rm{sep}}}^\eta \left( {{^n} \otimes {^m}} \right)} \right)} \over {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\rm{D}}^\eta }\left( {{^n} \otimes {^m}} \right)} \right)}}.

All further considerations will be simplified by the observation that both the Hilbert-Schmidt measure in the set D^η(ℂⁿ ⊗ ℂ^m) and the separability of a state ρ ∈ D^η(ℂⁿ ⊗ ℂ^m) are invariant under a transformation U ⊗ 𝟙_m, where U ∈ SU(n). Indeed, 67 $D^{U η U^{+}} (ℂ^{n} \otimes ℂ^{m}) = (U \otimes 𝟙_{m}) D^{η} (ℂ^{n} \otimes ℂ^{m}) {(U \otimes 𝟙_{m})}^{†} .$ {{\rm{D}}^{{\bf{U}}\eta {{\bf{U}}^ + }}}\left( {{^n} \otimes {^m}} \right) = \left( {{\bf{U}} \otimes {_m}} \right){{\rm{D}}^\eta }\left( {{^n} \otimes {^m}} \right){\left( {{\bf{U}} \otimes {_m}} \right)^\dag }.

For n = 2, the qubit state η ∈ D(ℂ²) can be represented by Bloch vector $η = \frac{1}{2} (𝟙_{2} + a \cdot σ)$ \eta = {1 \over 2}\left( {{1_2} + a \cdot \sigma } \right). We denote by η ≈ a simply. Thus we make the following identifications: $\begin{matrix} D^{a} (ℂ^{2} \otimes ℂ^{m}) = D^{η} (ℂ^{2} \otimes ℂ^{m}), \\ D_{sep}^{a} (ℂ^{2} \otimes ℂ^{m}) = D_{sep}^{η} (ℂ^{2} \otimes ℂ^{m}), \\ P_{sep}^{(2 \times m)} (a) = P_{sep}^{(2 \times m)} (η) . \end{matrix}$ \matrix{ {{{\rm{D}}^a}\left( {{^2} \otimes {^m}} \right) = {{\rm{D}}^\eta }\left( {{^2} \otimes {^m}} \right),} \cr {{\rm{D}}_{{\rm{sep }}}^a\left( {{^2} \otimes {^m}} \right) = {\rm{D}}_{{\rm{sep }}}^\eta \left( {{^2} \otimes {^m}} \right),} \cr {P_{{\rm{sep }}}^{(2 \times m)}(a) = P_{{\rm{sep }}}^{(2 \times m)}(\eta ).} \cr }

Thus Eq. (66), can be represented as 68 $P_{sep}^{(2 \times m)} (a) := \frac{{vol}_{HS} (D_{sep}^{a} (ℂ^{2} \otimes ℂ^{m}))}{{vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{m}))} .$ P_{{\rm{sep}}}^{(2 \times m)}(a): = {{{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{\rm{D}}_{{\rm{sep}}}^a\left( {{^2} \otimes {^m}} \right)} \right)} \over {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\rm{D}}^a}\left( {{^2} \otimes {^m}} \right)} \right)}}.

We have already known the fact that SU(2) is the double cover of SO(3) from Lie Theory, which leads to the following $U η U^{†} = \frac{1}{2} (𝟙_{2} + (O a) \cdot σ)$ {\bf{U}}\eta {{\bf{U}}^\dag } = {1 \over 2}\left( {{_2} + ({\bf{Oa}})\cdot{\bf{\sigma }}} \right.), where both U ∈ SU(2) and O = (o_ij) ∈ SO(3) are connected via $o_{i j} = \frac{1}{2} Tr (σ_{i} U σ_{j} U^{†})$ {o_{ij}} = {1 \over 2}{\mathop{\rm Tr}\nolimits} \left( {{\sigma _i}{\bf{U}}{\sigma _j}{{\bf{U}}^\dag }} \right).

Proposition 12.

It holds that 69 $P_{sep}^{(2 \times m)} (O a) = P_{sep}^{(2 \times m)} (a)$ P_{{\rm{sep}}}^{(2 \times m)}({\bf{Oa}}) = P_{{\rm{sep}}}^{(2 \times m)}({\bf{a}}) for any O ∈ SO(3) and a ∈ ℝ³ with |a| ⩽ 1.

Proof

In fact, it is easily seen that $\begin{matrix} {vol}_{HS} ((U \otimes 𝟙_{m}) D^{η} (ℂ^{n} \otimes ℂ^{m}) {(U \otimes 𝟙_{m})}^{†}) = vol {HS}^{n} (D^{η} (ℂ^{n} \otimes ℂ^{m})) \\ {vol}_{HS} ((U \otimes 𝟙_{m}) D_{sep}^{η} (ℂ^{n} \otimes ℂ^{m}) {(U \otimes 𝟙_{m})}^{†}) = vol HS (D_{sep}^{η} (ℂ^{n} \otimes ℂ^{m})) \end{matrix}$ \matrix{ {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {\left( {{\bf{U}} \otimes {_m}} \right){{\rm{D}}^\eta }\left( {{^n} \otimes {^m}} \right){{\left( {{\bf{U}} \otimes {_m}} \right)}^\dag }} \right) = {\mathop{\rm vol}\nolimits} {{{\mathop{\rm HS}\nolimits} }^n}\left( {{{\rm{D}}^\eta }\left( {{^n} \otimes {^m}} \right)} \right)} \cr {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {\left( {{\bf{U}} \otimes {_m}} \right){\rm{D}}_{{\rm{sep}}}^\eta \left( {{^n} \otimes {^m}} \right){{\left( {{\bf{U}} \otimes {_m}} \right)}^\dag }} \right) = {\mathop{\rm vol}\nolimits} {\rm{HS}}\left( {{\rm{D}}_{{\rm{sep}}}^\eta \left( {{^n} \otimes {^m}} \right)} \right)} \cr } holds for all U ∈ SU(2), which means that $P_{sep}^{(2 \times m)} (U η U^{†}) = P_{sep}^{(2 \times m)} (η) P_{sep}^{(2 \times m)} (η)$ P_{{\rm{sep }}}^{(2 \times m)}\left( {{\bf{U}}\eta {{\bf{U}}^\dag }} \right) = P_{{\rm{sep }}}^{(2 \times m)}(\eta )P_{{\rm{sep }}}^{(2 \times m)}(\eta ). The desired result is obtained by translating it into the Bloch vector form.

Let a = |a| ∈ [0,1]. From the above Proposition 12, we see that $P_{sep}^{(2 \times m)} (a)$ P_{{\rm{sep }}}^{(2 \times m)}({\bf{a}}) is constant on the sphere S_a := {a ∈ ℝ³: |a| = a}. In view of this reason, we let a = (0,0, a) and 70 $P_{sep}^{(2 \times m)} (a) := \frac{{vol}_{HS} (D_{sep}^{a} (ℂ^{2} \otimes ℂ^{m}))}{{vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{m}))} .$ P_{{\rm{sep}}}^{(2 \times m)}({\bf{a}}): = {{{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{\rm{D}}_{{\rm{sep}}}^a\left( {{^2} \otimes {^m}} \right)} \right)} \over {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\rm{D}}^a}\left( {{^2} \otimes {^m}} \right)} \right)}}. where 71 $D^{a} (ℂ^{2} \otimes ℂ^{m}) := {ρ \in D (ℂ^{2} \otimes ℂ^{m}) : {Tr}_{ℂ^{m}} (ρ) = \frac{1}{2} (𝟙_{2} + a σ_{3})} .$ {{\rm{D}}^a}\left( {{^2} \otimes {^m}} \right): = \left\{ {\rho \in {\rm{D}}\left( {{^2} \otimes {^m}} \right):{{{\mathop{\rm Tr}\nolimits} }_{{^m}}}(\rho ) = {1 \over 2}\left( {{_2} + a{\sigma _3}} \right)} \right\}.

Denote the unit ball by B₁ := {a ∈ ℝ³ : |a| ⩽ 1}. Then B₁ = ∪_a∈[0,1]S_a, which leads to the following identifications: 72 $D (ℂ^{2} \otimes ℂ^{2}) = \underset{η \in D (ℂ^{2})}{\cup} D^{η} (ℂ^{2} \otimes ℂ^{2}) = \underset{a \in B_{1}}{\cup} D^{a} (ℂ^{2} \otimes ℂ^{2})$ {\rm{D}}\left( {{^2} \otimes {^2}} \right) = \bigcup\limits_{\eta \in {\rm{D}}\left( {{^2}} \right)} {{{\rm{D}}^\eta }} \left( {{^2} \otimes {^2}} \right) = \bigcup\limits_{a \in {B_1}} {{{\rm{D}}^a}} \left( {{^2} \otimes {^2}} \right) 73 $= \underset{a \in [0, 1]}{\cup} \underset{a \in S_{a}}{\cup} D^{a} (ℂ^{2} \otimes ℂ^{2}) .$ = \bigcup\limits_{a \in [0,1]} {\bigcup\limits_{a \in {S_a}} {{{\rm{D}}^a}} } \left( {{^2} \otimes {^2}} \right)

Thus we get that

Proposition 13.

It holds that 74 ${vol}_{HS} (D (ℂ^{2} \otimes ℂ^{2})) = \frac{1}{8} \int_{B_{1}} {vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{2})) [d a]$ {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{\rm{D}}\left( {{^2} \otimes {^2}} \right)} \right) = {1 \over 8}\int_{{B_1}} {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}} \left( {{{\rm{D}}^a}\left( {{^2} \otimes {^2}} \right)} \right)[{\rm{d}}a] 75 $= \frac{π}{2} \int_{0}^{1} a^{2} {vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{2})) d a .$ = {\pi \over 2}\int_0^1 {{a^2}} {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{{\rm{D}}^a}\left( {{^2} \otimes {^2}} \right)} \right){\rm{d}}a.

Proof

Recall that $d_{HS} (ρ^{'}, ρ) = \frac{1}{2} d_{Euclid} (x^{'}, x)$ {d_{{\rm{HS}}}}\left( {{\rho ^\prime },\rho } \right) = {1 \over 2}{d_{{\rm{Euclid }}}}\left( {{{\bf{x}}^\prime },{\bf{x}}} \right), where $x^{'} = (a^{'}, b^{'}, c_{k l}^{'})$ {{\bf{x}}^\prime } = \left( {{{\bf{a}}^\prime },{{\bf{b}}^\prime },c_{kl}^\prime } \right) and x = (a, b, c_kl). Thus three parameters in a determine Euclid volume [da], and thus contribute a scaling factor ${(\frac{1}{2})}^{3}$ {\left( {{1 \over 2}} \right)^3} multiple of Euclid volume [da], i.e., in the HS volume element for D(ℂ² ⊗ ℂ²). Therefore we get that $\begin{array}{l} {vol}_{HS} (D (ℂ^{2} \otimes ℂ^{2})) = \int_{B_{1}} {vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{2})) (2^{- 3} [d a]) \\ = \frac{1}{8} \int_{0}^{1} d a \int_{S_{a}} {vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{2})) [d a] \\ = \frac{1}{8} \int_{0}^{1} d a \int δ (a - | a |) {vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{2})) [d a] \\ = \frac{1}{8} \int_{0}^{1} d a {vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{2})) \int δ (a - | a |) [d a] \\ = \frac{1}{8} \int_{0}^{1} (4 π a^{2}) {vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{2})) d a \\ \frac{π}{2} \int_{0}^{1} a^{2} {vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{2})) d a . \end{array}$ \matrix{ {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{\rm{D}}\left( {{^2} \otimes {^2}} \right)} \right) = \int_{{B_1}} {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}} \left( {{{\rm{D}}^a}\left( {{^2} \otimes {^2}} \right)} \right)\left( {{2^{ - 3}}[{\rm{d}}{\bf{a}}]} \right)} \hfill \cr { = {1 \over 8}\int_0^1 {{\rm{d}}} a\int_{{S_a}} {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}} \left( {{{\rm{D}}^a}\left( {{^2} \otimes {^2}} \right)} \right)[{\rm{d}}{\bf{a}}]} \hfill \cr { = {1 \over 8}\int_0^1 {{\rm{d}}} a\int \delta (a - |{\bf{a}}|){{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\rm{D}}^a}\left( {{^2} \otimes {^2}} \right)} \right)[{\rm{d}}{\bf{a}}]} \hfill \cr { = {1 \over 8}\int_0^1 {{\rm{d}}} a{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\rm{D}}^a}\left( {{^2} \otimes {^2}} \right)} \right)\int \delta (a - |{\bf{a}}|)[{\rm{d}}{\bf{a}}]} \hfill \cr { = {1 \over 8}\int_0^1 {\left( {4\pi {a^2}} \right)} {{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\rm{D}}^a}\left( {{^2} \otimes {^2}} \right)} \right){\rm{d}}a} \hfill \cr {{\pi \over 2}\int_0^1 {{a^2}} {{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\rm{D}}^a}\left( {{^2} \otimes {^2}} \right)} \right){\rm{d}}a.} \hfill \cr }

We have done it.

Next, we calculate the HS volume vol_HS(D^a(ℂ² ⊗ ℂ²)). Indeed,

Proposition 14 ([8]).

It holds that 76 ${vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{2})) = {vol}_{HS} (D^{0} (ℂ^{2} \otimes ℂ^{2})) {(1 - a^{2})}^{6},$ {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{{\rm{D}}^a}\left( {{^2} \otimes {^2}} \right)} \right) = {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{{\rm{D}}^0}\left( {{^2} \otimes {^2}} \right)} \right){\left( {1 - {a^2}} \right)^6}, where a ∈ [0,1). Moreover, 77 ${vol}_{HS} (D^{0} (ℂ^{2} \otimes ℂ^{2})) = \frac{π^{5}}{9676800} .$ {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{{\rm{D}}^0}\left( {{^2} \otimes {^2}} \right)} \right) = {{{\pi ^5}} \over {9676800}}.

Proof

This expression in Eq. (76) is conjectured by Milz and Strunz [27], and proved by Lovas and Andai [8]. From the above formula, we get that 78 ${vol}_{HS} (D (ℂ^{2} \otimes ℂ^{2})) = {vol}_{HS} (D^{0} (ℂ^{2} \otimes ℂ^{2})) \times \frac{π}{2} \int_{0}^{1} a^{2} {(1 - a^{2})}^{6} d a$ {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{\rm{D}}\left( {{^2} \otimes {^2}} \right)} \right) = {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{{\rm{D}}^0}\left( {{^2} \otimes {^2}} \right)} \right) \times {\pi \over 2}\int_0^1 {{a^2}} {\left( {1 - {a^2}} \right)^6}{\rm{d}}a 79 $= \frac{2^{9} π}{45045} {vol}_{HS} (D^{0} (ℂ^{2} \otimes ℂ^{2})) .$ = {{{2^9}\pi } \over {45045}}{{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{{\rm{D}}^0}\left( {{^2} \otimes {^2}} \right)} \right).

Note that, from Theorem 2, we can see that 80 ${vol}_{HS} (D (ℂ^{2} \otimes ℂ^{2})) = {vol}_{HS} (D (ℂ^{4})) = \sqrt{4} {(2 π)}^{(\begin{matrix} 4 \\ 2 \end{matrix})} \frac{\prod_{k = 1}^{4} Γ (k)}{Γ (4^{2})}$ {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{\rm{D}}\left( {{^2} \otimes {^2}} \right)} \right) = {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{\rm{D}}\left( {{^4}} \right)} \right) = \sqrt 4 {(2\pi )^{\left( {\scriptstyle 4 \atop \scriptstyle 2} \right)}}{{\prod\nolimits_{k = 1}^4 \Gamma (k)} \over {\Gamma \left( {{4^2}} \right)}} 81 $= \frac{2 {(2 π)}^{6} (2!) (3!)}{15!} .$ = {{2{{(2\pi )}^6}(2!)(3!)} \over {15!}}.

By combining the above formulas, we can derive that ${vol}_{HS} (D^{0} (ℂ^{2} \otimes ℂ^{2})) = \frac{π^{5}}{9676800}$ {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{{\rm{D}}^0}\left( {{^2} \otimes {^2}} \right)} \right) = {{{\pi ^5}} \over {9676800}}.

The following result was conjectured in [27], and proven in [8], which is described as a theorem without proof:

Theorem 6 ([8]).

It holds that $P_{sep}^{(2 \times 2)} (a)$ P_{{\rm{sep}}}^{(2 \times 2)}\left( a \right) is constant for all a ∈ [0,1).

Proof

Please see the proof in [8, Corollary 2].

6.2.

Separability Probability of Two-Qubit States

The separability probability is given by 82 $P_{sep}^{(n \times m)} := \frac{{vol}_{HS} (D_{sep} (ℂ^{n} \otimes ℂ^{m}))}{{vol}_{HS} (D (ℂ^{n} \otimes ℂ^{m}))} .$ P_{{\rm{sep}}}^{(n \times m)}: = {{{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\rm{D}}_{{\rm{sep}}}}\left( {{^n} \otimes {^m}} \right)} \right)} \over {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{\rm{D}}\left( {{^n} \otimes {^m}} \right)} \right)}}.

Here D_sep(ℂⁿ ⊗ ℂ^m) is the set of separable states in D(ℂⁿ ⊗ ℂ^m). In what follows, we focus on the case where n = m = 2.

For ρ ∈ D(ℂ² ⊗ ℂ²), we can view ρ as a 4 × 4 matrix acting on ℂ⁴ and then its partial transpose w.r.t. 1st subsystem, are respectively: 83 $ρ = (\begin{array}{l} ρ_{11} & ρ_{12} & ρ_{13} & ρ_{14} \\ ρ_{21} & ρ_{22} & ρ_{23} & ρ_{24} \\ ρ_{31} & ρ_{32} & ρ_{33} & ρ_{34} \\ ρ_{41} & ρ_{42} & ρ_{43} & ρ_{44} \end{array}) and ρ^{Γ} = (\begin{matrix} ρ_{11} & ρ_{21} & ρ_{13} & ρ_{23} \\ ρ_{12} & ρ_{22} & ρ_{14} & ρ_{24} \\ ρ_{31} & ρ_{41} & ρ_{33} & ρ_{43} \\ ρ_{32} & ρ_{42} & ρ_{34} & ρ_{44} \end{matrix}) .$ \rho = \left( {\matrix{ {{\rho _{11}}} \hfill & {{\rho _{12}}} \hfill & {{\rho _{13}}} \hfill & {{\rho _{14}}} \hfill \cr {{\rho _{21}}} \hfill & {{\rho _{22}}} \hfill & {{\rho _{23}}} \hfill & {{\rho _{24}}} \hfill \cr {{\rho _{31}}} \hfill & {{\rho _{32}}} \hfill & {{\rho _{33}}} \hfill & {{\rho _{34}}} \hfill \cr {{\rho _{41}}} \hfill & {{\rho _{42}}} \hfill & {{\rho _{43}}} \hfill & {{\rho _{44}}} \hfill \cr } } \right){\rm{ and }}{\rho ^\Gamma } = \left( {\matrix{ {{\rho _{11}}} & {{\rho _{21}}} & {{\rho _{13}}} & {{\rho _{23}}} \cr {{\rho _{12}}} & {{\rho _{22}}} & {{\rho _{14}}} & {{\rho _{24}}} \cr {{\rho _{31}}} & {{\rho _{41}}} & {{\rho _{33}}} & {{\rho _{43}}} \cr {{\rho _{32}}} & {{\rho _{42}}} & {{\rho _{34}}} & {{\rho _{44}}} \cr } } \right).

Let E_n(i, j) be the elementary matrix, obtained by changing the i-th and j-th rows/columns of identity matrix 𝟙_n. Now 84 $E_{4} (2, 3) ρ E_{4} (2, 3) = (\begin{array}{l} ρ_{11} & ρ_{13} & ρ_{12} & ρ_{14} \\ ρ_{31} & ρ_{33} & ρ_{32} & ρ_{34} \\ ρ_{21} & ρ_{23} & ρ_{22} & ρ_{24} \\ ρ_{41} & ρ_{43} & ρ_{42} & ρ_{44} \end{array}) = (\begin{matrix} A & B \\ C & D \end{matrix})$ {E_4}(2,3)\rho {E_4}(2,3) = \left( {\matrix{ {{\rho _{11}}} \hfill & {{\rho _{13}}} \hfill & {{\rho _{12}}} \hfill & {{\rho _{14}}} \hfill \cr {{\rho _{31}}} \hfill & {{\rho _{33}}} \hfill & {{\rho _{32}}} \hfill & {{\rho _{34}}} \hfill \cr {{\rho _{21}}} \hfill & {{\rho _{23}}} \hfill & {{\rho _{22}}} \hfill & {{\rho _{24}}} \hfill \cr {{\rho _{41}}} \hfill & {{\rho _{43}}} \hfill & {{\rho _{42}}} \hfill & {{\rho _{44}}} \hfill \cr } } \right) = \left( {\matrix{ A \hfill & B \hfill \cr C \hfill & D \hfill \cr } } \right) and 85 $E_{4} (2, 3) ρ^{Γ} E_{4} (2, 3) = (\begin{matrix} ρ_{11} & ρ_{13} & ρ_{21} & ρ_{23} \\ ρ_{31} & ρ_{33} & ρ_{41} & ρ_{43} \\ ρ_{12} & ρ_{14} & ρ_{22} & ρ_{24} \\ ρ_{32} & ρ_{34} & ρ_{42} & ρ_{44} \end{matrix}) = (\begin{matrix} A & B \\ C & D \end{matrix}) .$ {E_4}(2,3){\rho ^\Gamma }{E_4}(2,3) = \left( {\matrix{ {{\rho _{11}}} & {{\rho _{13}}} & {{\rho _{21}}} & {{\rho _{23}}} \cr {{\rho _{31}}} & {{\rho _{33}}} & {{\rho _{41}}} & {{\rho _{43}}} \cr {{\rho _{12}}} & {{\rho _{14}}} & {{\rho _{22}}} & {{\rho _{24}}} \cr {{\rho _{32}}} & {{\rho _{34}}} & {{\rho _{42}}} & {{\rho _{44}}} \cr } } \right) = \left( {\matrix{ A & C \cr B & D \cr } } \right).

Denote 86 $D_{ss} (ℂ^{2} \otimes ℂ^{2}) := {D (ℂ^{2} \otimes ℂ^{2}) : λ_{\max} (ρ) ⩽ \frac{1}{2}} .$ {{\rm{D}}_{{\rm{ss}}}}\left( {{^2} \otimes {^2}} \right): = \left\{ {{\rm{D}}\left( {{^2} \otimes {^2}} \right):{\lambda _{\max }}(\rho )\;\;{1 \over 2}} \right\}.

Proposition 15 ([7]).

Let ρ ∈ D⁰(ℂ² ⊗ ℂ²). It holds that ρ is separable if and only if ρ ∈ D_ss(ℂ² ⊗ ℂ²). In other words, 87 $D_{sep}^{0} (ℂ^{2} \otimes ℂ^{2}) = D^{0} (ℂ^{2} \otimes ℂ^{2}) \cap D_{ss} (ℂ^{2} \otimes ℂ^{2}) .$ {\rm{D}}_{{\rm{sep }}}^0\left( {{^2} \otimes {^2}} \right) = {{\rm{D}}^0}\left( {{^2} \otimes {^2}} \right) \cap {{\rm{D}}_{{\rm{ss}}}}\left( {{^2} \otimes {^2}} \right).

Proof

Note that the characteristic polynomial of ρ is given by f_ρ(λ) = det(ρ – λ𝟙₄). Then $\begin{array}{l} f_{ρ} (λ) = \det (E_{4} (2, 3) ρ E_{4} (2, 3) - λ 𝟙_{4}) = \det (\begin{matrix} A - λ 𝟙_{2} & B \\ C & D - λ 𝟙_{2} \end{matrix}) \\ = \det (A - λ 𝟙_{2}) \det ((D - λ 𝟙_{2}) - C {(A - λ 𝟙_{2})}^{- 1} B) . \end{array}$ \matrix{ {{f_\rho }(\lambda ) = \det \left( {{{\bf{E}}_4}(2,3)\rho {{\bf{E}}_4}(2,3) - \lambda {_4}} \right) = \det \left( {\matrix{ {{\bf{A}} - \lambda {_2}} & {\bf{B}} \cr {\bf{C}} & {{\bf{D}} - \lambda {_2}} \cr } } \right)} \hfill \cr { = \det \left( {{\bf{A}} - \lambda {_2}} \right)\det \left( {\left( {{\bf{D}} - \lambda {_2}} \right) - {\bf{C}}{{\left( {{\bf{A}} - \lambda {_2}} \right)}^{ - 1}}{\bf{B}}} \right).} \hfill \cr }

Similarly, the characteristic polynomial of ρ^Γ is given by $\begin{array}{l} f_{ρ^{Γ}} (λ) = \det (E_{4} (2, 3) ρ^{Γ} E_{4} (2, 3) - λ 𝟙_{4}) = \det (\begin{matrix} A - λ 𝟙_{2} & C \\ B & D - λ 𝟙_{2} \end{matrix}) \\ = \det (A - λ 𝟙_{2}) \det ((D - λ 𝟙_{2}) - B {(A - λ 𝟙_{2})}^{- 1} C) \\ = \det (D - λ 𝟙_{2}) \det ((A - λ 𝟙_{2}) - C {(D - λ 𝟙_{2})}^{- 1} B) . \end{array}$ \matrix{ {{f_{{\rho ^\Gamma }}}(\lambda ) = \det \left( {{{\bf{E}}_4}(2,3){\rho ^\Gamma }{{\bf{E}}_4}(2,3) - \lambda {_4}} \right) = \det \left( {\matrix{ {{\bf{A}} - \lambda {_2}} & {\bf{C}} \cr {\bf{B}} & {{\bf{D}} - \lambda {_2}} \cr } } \right)} \hfill \cr { = \det \left( {{\bf{A}} - \lambda {_2}} \right)\det \left( {\left( {{\bf{D}} - \lambda {_2}} \right) - {\bf{B}}{{\left( {{\bf{A}} - \lambda {_2}} \right)}^{ - 1}}{\bf{C}}} \right)} \hfill \cr { = \det \left( {{\bf{D}} - \lambda {_2}} \right)\det \left( {\left( {{\bf{A}} - \lambda {_2}} \right) - {\bf{C}}{{\left( {{\bf{D}} - \lambda {_2}} \right)}^{ - 1}}{\bf{B}}} \right).} \hfill \cr }

Due to the fact that ρ ∈ D⁰(ℂ² ⊗ ℂ²), we see that ${Tr}_{2} (ρ) = \frac{1}{2} 𝟙_{2}$ {{\mathop{\rm Tr}\nolimits} _2}(\rho ) = {1 \over 2}{_2}, which is equivalent to $A + D = \frac{1}{2} 𝟙_{2}$ {\bf{A}} + {\bf{D}} = {1 \over 2}{_2}. Now we have shown that 88 $f_{ρ^{Γ}} (λ) = \det (D - λ 𝟙_{2}) \det ((A - λ 𝟙_{2}) - C {(D - λ 𝟙_{2})}^{- 1} B) .$ {f_{{\rho ^\Gamma }}}(\lambda ) = \det \left( {{\bf{D}} - \lambda {_2}} \right)\det \left( {\left( {{\bf{A}} - \lambda {_2}} \right) - {\bf{C}}{{\left( {{\bf{D}} - \lambda {_2}} \right)}^{ - 1}}{\bf{B}}} \right).

Thus replacing the 2-tuple (λ, D) by $(\frac{1}{2} - λ, \frac{1}{2} 𝟙_{2} - A)$ \left( {{1 \over 2} - \lambda ,{1 \over 2}{_2} - {\bf{A}}} \right) in the first factor of Eq. (88), we get that the first factor det(D – λ𝟙₂) is transformed into $\det ((\frac{1}{2} 𝟙_{2} - A) - (\frac{1}{2} - λ) 𝟙_{2}) = \det (λ 𝟙_{2} - A) = \det (A - λ 𝟙_{2}) .$ \det \left( {\left( {{1 \over 2}{_2} - {\bf{A}}} \right) - \left( {{1 \over 2} - \lambda } \right){_2}} \right) = \det \left( {\lambda {_2} - {\bf{A}}} \right) = \det \left( {{\bf{A}} - \lambda {_2}} \right).

Thus replacing λ by $\frac{1}{2} - λ$ {1 \over 2} - \lambda and using $A + D = \frac{1}{2} 𝟙_{2}$ {\bf{A}} + {\bf{D}} = {1 \over 2}{_2} in the second factor of Eq. (88), we get that the second factor det((A – λ𝟙₂) – C(D – λ𝟙₂)⁻¹ B) is transformed into $\det (((\frac{1}{2} 𝟙_{2} - D) - (\frac{1}{2} - λ) 𝟙_{2}) - C {((\frac{1}{2} 𝟙_{2} - A) - (\frac{1}{2} - λ) 𝟙_{2})}^{- 1} B) .$ \det \left( {\left( {\left( {{1 \over 2}{_2} - {\bf{D}}} \right) - \left( {{1 \over 2} - \lambda } \right){_2}} \right) - {\bf{C}}{{\left( {\left( {{1 \over 2}{_2} - {\bf{A}}} \right) - \left( {{1 \over 2} - \lambda } \right){_2}} \right)}^{ - 1}}{\bf{B}}} \right).

That is, the last expression is reduced to the following: $\det ((- D + λ 𝟙_{2}) - C {(- A + λ 𝟙_{2})}^{- 1} B) = \det ((D - λ 𝟙_{2}) - C {(A - λ 𝟙_{2})}^{- 1} B) .$ \det \left( {\left( { - {\bf{D}} + \lambda {_2}} \right) - {\bf{C}}{{\left( { - {\bf{A}} + \lambda {_2}} \right)}^{ - 1}}{\bf{B}}} \right) = \det \left( {\left( {{\bf{D}} - \lambda {_2}} \right) - {\bf{C}}{{\left( {{\bf{A}} - \lambda {_2}} \right)}^{ - 1}}{\bf{B}}} \right).

Thus $f_{ρ^{Γ}} (\frac{1}{2} - λ) = \det (A - λ 𝟙_{2}) \det ((D - λ 𝟙_{2}) - C {(A - λ 𝟙_{2})}^{- 1} B) = f_{ρ} (λ) .$ {f_{{\rho ^\Gamma }}}\left( {{1 \over 2} - \lambda } \right) = \det \left( {{\bf{A}} - \lambda {_2}} \right)\det \left( {\left( {{\bf{D}} - \lambda {_2}} \right) - {\bf{C}}{{\left( {{\bf{A}} - \lambda {_2}} \right)}^{ - 1}}{\bf{B}}} \right) = {f_\rho }(\lambda ).

In summary, we get that $f_{ρ^{Γ}} (\frac{1}{2} - λ) = f_{ρ} (λ)$ {f_{{\rho ^\Gamma }}}\left( {{1 \over 2} - \lambda } \right) = {f_\rho }(\lambda ). With these preparations, we can now present the proof below.

(⇒)
Now suppose that $ρ \in D_{sep}^{0} (ℂ^{2} \otimes ℂ^{2})$ \rho \in {\rm{D}}_{{\rm{sep }}}^0\left( {{^2} \otimes {^2}} \right). That is, ρ ∈ D⁰(ℂ² ⊗ ℂ²) is separable. Choose any eigenvalue t ⩾ 0 of the separable state ρ, i.e., f_ρ(t) = 0, then $f_{ρ^{Γ}} (\frac{1}{2} - t) = f_{ρ} (t) = 0$ {f_{{\rho ^\Gamma }}}\left( {{1 \over 2} - t} \right) = {f_\rho }(t) = 0, namely, $\frac{1}{2} - t$ {1 \over 2} - t satisfying the eigen-equation f_ρ^γ(λ) = 0, thus $\frac{1}{2} - t ⩾ 0$ {1 \over 2} - t0 is an eigenvalue of ρ^Γ ⩾ 0 by Peres-Horodecki criterion. Therefore $t ⩽ \frac{1}{2}$ t{1 \over 2}, or ρ ∈ D_ss(ℂ² ⊗ ℂ²).
(⇐)
If ρ ∈ D⁰(ℂ² ⊗ ℂ²) ∩ D_ss(ℂ² ⊗ ℂ²), i.e., any generic eigenvalue t of ρ satisfying $0 ⩽ t ⩽ \frac{1}{2}$ 0t{1 \over 2}, and moreover $f_{ρ^{Γ}} (\frac{1}{2} - t) = f_{ρ} (t) = 0$ {f_{{\rho ^\Gamma }}}\left( {{1 \over 2} - t} \right) = {f_\rho }(t) = 0, which means that $\frac{1}{2} - t ⩾ 0$ {1 \over 2} - t0 is an eigenvalue of ρ^Γ. All eigenvalues of ρ^Γ are nonnegative, that is, ρ^Γ ⩾ 0. By Peres-Horodecki criterion, ρ is separable.

This completes the proof.

Denote 89 $f (a) := {vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{2}) \cap D_{ss} (ℂ^{2} \otimes ℂ^{2})),$ f(a): = {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{{\rm{D}}^a}\left( {{^2} \otimes {^2}} \right) \cap {{\rm{D}}_{{\rm{ss}}}}\left( {{^2} \otimes {^2}} \right)} \right), where a ∈ [0,1). Since the partial trace is a linear map, the function f(a) is continuous. Therefore, we can find $f (0) = \lim_{a \to 0^{+}} f (a)$ f(0) = {\lim _{a \to {0^ + }}}f(a). Clearly $f (0) = {vol}_{HS} (D_{sep}^{0} (ℂ^{2} \otimes ℂ^{2}))$ f(0) = {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{\rm{D}}_{{\rm{sep }}}^0\left( {{^2} \otimes {^2}} \right)} \right) by Proposition 15. So we firstly calculate f(a) in a small neighborhood of 0. Denote 90 $C_{4} = {(x_{1}, x_{2}, x_{3}, x_{4}) \in ℝ^{4} : x_{1} > x_{2} > x_{3} > x_{4}},$ {{\cal C}_4} = \left\{ {\left( {{x_1},{x_2},{x_3},{x_4}} \right) \in {^4}:{x_1} > {x_2} > {x_3} > {x_4}} \right\}, 91 $\hat{C} = {[- \frac{1}{4}, \frac{1}{4}]}^{4} \cap C_{4} .$ \hat C = {\left[ { - {1 \over 4},{1 \over 4}} \right]^4} \cap {{\cal C}_4}.

In fact, we have the following result:

Proposition 16 ([7]).

It holds that 92 $f (a) = \frac{π^{5}}{319334400} (1 - a)^{9} (33 a^{3} + 162 a^{2} + 72 a + 8), a \in (0, \frac{1}{3}) .$ f(a) = {{{\pi ^5}} \over {319334400}}{(1 - a)^9}\left( {33{a^3} + 162{a^2} + 72a + 8} \right),\quad a \in \left( {0,{1 \over 3}} \right).

Proof

For all t ∈ (0,1), denote B_t := {a ∈ ℝ³ : |a| ⩾ t}. Denote $D^{B_{t}} (ℂ^{2} \otimes ℂ^{2}) := \underset{a \in B_{t}}{\cup} D^{a} (ℂ^{2} \otimes ℂ^{2})$ {{\rm{D}}^{{B_t}}}\left( {{^2} \otimes {^2}} \right): = \bigcup\limits_{a \in {B_t}} {{{\rm{D}}^a}} \left( {{^2} \otimes {^2}} \right)

Note that $\begin{array}{l} {vol}_{HS} (D^{B_{t}} (ℂ^{2} \otimes ℂ^{2}) \cap D_{ss} (ℂ^{2} \otimes ℂ^{2})) = {vol}_{HS} (\underset{a \in B_{t}}{\cup} D^{a} (ℂ^{2} \otimes ℂ^{2}) \cap D_{ss} (ℂ^{2} \otimes ℂ^{2})) \\ = \int_{0}^{t} d a {vol}_{HS} (\underset{a \in S_{a}}{\cup} D^{a} (ℂ^{2} \otimes ℂ^{2}) \cap D_{sS} (ℂ^{2} \otimes ℂ^{2})) \\ = \int_{0}^{t} d a [\frac{1}{8} \int_{S_{a}} [d a] {vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{2}) \cap D_{ss} (ℂ^{2} \otimes ℂ^{2}))] \\ = \int_{0}^{t} d a [\frac{1}{8} \int [d a] δ (a - | a |) {vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{2}) \cap D_{ss} (ℂ^{2} \otimes ℂ^{2}))] \\ = \frac{π}{8} \int_{0}^{t} d a {vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{2}) \cap D_{ss} (ℂ^{2} \otimes ℂ^{2})) \int δ (a - | a |) [d a] . \end{array}$ \matrix{ {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\rm{D}}^{{B_t}}}\left( {{^2} \otimes {^2}} \right)\bigcap {{{\rm{D}}_{{\rm{ss}}}}} \left( {{^2} \otimes {^2}} \right)} \right) = {{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {\bigcup\limits_{a \in {B_t}} {{{\rm{D}}^a}} \left( {{^2} \otimes {^2}} \right)\bigcap {{{\rm{D}}_{{\rm{ss}}}}} \left( {{^2} \otimes {^2}} \right)} \right)} \hfill \cr { = \int_0^t {{\rm{d}}} a{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {\bigcup\limits_{{\bf{a}} \in {S_a}} {{{\rm{D}}^a}} \left( {{^2} \otimes {^2}} \right)\bigcap {{{\rm{D}}_{{\rm{sS}}}}} \left( {{^2} \otimes {^2}} \right)} \right)} \hfill \cr { = \int_0^t {{\rm{d}}} a\left[ {{1 \over 8}{\mkern 1mu} \int_{{S_a}} {[{\rm{d}}a]{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\rm{D}}^a}\left( {{^2} \otimes {^2}} \right)\bigcap {{{\rm{D}}_{{\rm{ss}}}}} \left( {{^2} \otimes {^2}} \right)} \right)} } \right]} \hfill \cr { = \int_0^t {{\rm{d}}} a\left[ {{1 \over 8}\int {[{\rm{d}}a]\delta (a - |a|){{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\rm{D}}^a}\left( {{^2} \otimes {^2}} \right)\bigcap {{{\rm{D}}_{{\rm{ss}}}}} \left( {{^2} \otimes {^2}} \right)} \right)} } \right]} \hfill \cr { = {\pi \over 8}\int_0^t {{\rm{d}}} a{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\rm{D}}^a}\left( {{^2} \otimes {^2}} \right)\bigcap {{{\rm{D}}_{{\rm{ss}}}}} \left( {{^2} \otimes {^2}} \right)} \right)\int \delta (a - |a|)[{\rm{d}}a].} \hfill \cr }

Since $f (a) = {vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{2}) \cap D_{SS} (ℂ^{2} \otimes ℂ^{2}))$ f(a) = {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{{\rm{D}}^a}\left( {{^2} \otimes {^2}} \right) \cap {{\rm{D}}_{{\rm{SS}}}}\left( {{^2} \otimes {^2}} \right)} \right), it follows that 93 $\frac{π}{2} \int_{0}^{t} a^{2} f (a) d a = {vol}_{HS} (D^{B_{t}} (ℂ^{2} \otimes ℂ^{2}) \cap D_{ss} (ℂ^{2} \otimes ℂ^{2})) .$ {\pi \over 2}\int_0^t {{a^2}} f(a){\rm{d}}a = {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{{\rm{D}}^{{B_t}}}\left( {{^2} \otimes {^2}} \right)\bigcap {{{\rm{D}}_{{\rm{ss}}}}} \left( {{^2} \otimes {^2}} \right)} \right).

Now we partition the set D_ss(ℂ² ⊗ ℂ²) by adjoint SU(4)-orbits as 94 $D_{ss} (ℂ^{2} \otimes ℂ^{2}) = \underset{λ \in {[0, \frac{1}{2}]}^{4} \cap C_{4}}{\cup} U_{λ},$ {{\rm{D}}_{{\rm{ss}}}}\left( {{^2} \otimes {^2}} \right) = \bigcup\limits_{\lambda \in {{\left[ {0,{1 \over 2}} \right]}^4} \cap {{\cal C}_4}} {{{\cal U}_\lambda }} , where 𝒰_λ := {UΛU^† : U ∈ SU(4)} for Λ = diag(λ₁, …, λ₄), obtained from λ = (λ₁, …, λ₄). Itis easily seen that the adjoint orbit 𝒰_λ can be identified with the co-adjoint orbit $O_{\hat{λ}}$ {{\cal O}_{\hat \lambda }}, where $\hat{λ} = λ - (\frac{1}{4}, \dots, \frac{1}{4})$ \hat \lambda = \lambda - \left( {{1 \over 4}, \ldots ,{1 \over 4}} \right). That is, $U_{λ} ≅ O_{\hat{λ}},$ {{\cal U}_\lambda } \cong {{\cal O}_{\hat \lambda }}, where $λ \in {[0, \frac{1}{2}]}^{4} \cap C_{4} iff \hat{λ} \in {[- \frac{1}{4}, \frac{1}{4}]}^{4} \cap C_{4}$ \lambda \in {\left[ {0,{1 \over 2}} \right]^4} \cap {{\cal C}_4}\;{\rm{iff}}\;\hat \lambda \in {\left[ { - {1 \over 4},{1 \over 4}} \right]^4} \cap {{\cal C}_4}. Under the above identification, $X = ρ - \frac{1}{4} 𝟙_{4}$ {\bf{X}} = \rho - {1 \over 4}{_4}, we get that $\begin{array}{l} D^{B_{t}} (ℂ^{2} \otimes ℂ^{2}) = {ρ \in D (ℂ^{2} \otimes ℂ^{2}) : {Tr}_{2} (ρ) = \frac{1}{2} (𝟙_{2} + a \cdot σ), | a | ⩽ t} \\ ≅ {X \in Herm (ℂ^{4}) : {Tr}_{2} (X) = \frac{a}{2} \cdot σ, | a | ⩽ t} \\ = {X \in Herm (ℂ^{4}) : Tr (X) = 0, | λ_{\pm} ({Tr}_{2} (X)) | ⩽ \frac{t}{2}} := ε_{t} . \end{array}$ \matrix{ {{{\rm{D}}^{{B_t}}}\left( {{^2} \otimes {^2}} \right) = \left\{ {\rho \in {\rm{D}}\left( {{^2} \otimes {^2}} \right):{{{\mathop{\rm Tr}\nolimits} }_2}(\rho ) = {1 \over 2}\left( {{_2} + {\bf{a}}\cdot{\bf{\sigma }}} \right),|{\bf{a}}|t} \right\}} \hfill \cr { \cong \left\{ {{\bf{X}} \in {\mathop{\rm Herm}\nolimits} \left( {{^4}} \right):{{{\mathop{\rm Tr}\nolimits} }_2}({\bf{X}}) = {{\bf{a}} \over 2}\cdot{\bf{\sigma }},|{\bf{a}}|t} \right\}} \hfill \cr { = \left\{ {{\bf{X}} \in {\mathop{\rm Herm}\nolimits} \left( {{^4}} \right):{\mathop{\rm Tr}\nolimits} ({\bf{X}}) = 0,\left| {{\lambda _ \pm }\left( {{{{\mathop{\rm Tr}\nolimits} }_2}({\bf{X}})} \right)} \right|{t \over 2}} \right\}: = {{\cal E}_t}.} \hfill \cr }

That is, the set D^B_t(ℂ² ⊗ ℂ²) is translated into the set ε_t. Thus the HS volume is invariant under the translation and $\begin{array}{l} {vol}_{HS} (D^{B_{t}} (ℂ^{2} \otimes ℂ^{2}) \cap D_{Ss} (ℂ^{2} \otimes ℂ^{2})) = {vol}_{HS} (\underset{λ \in {[0, \frac{1}{2}]}^{4} \cap C_{4}}{\cup} D^{B_{t}} (ℂ^{2} \otimes ℂ^{2}) \cap U_{λ}) \\ = {vol}_{HS} (\underset{\hat{λ} \in \hat{C}}{\cup} ε_{t} \cap O_{\hat{λ}}) = \int_{\hat{λ} \in \hat{C}} {vol}_{HS} (O_{\hat{λ}} \cap ε_{t}) d m (\hat{λ}), \end{array}$ \matrix{ {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\rm{D}}^{{B_t}}}\left( {{^2} \otimes {^2}} \right)\bigcap {{{\rm{D}}_{{\rm{Ss}}}}} \left( {{^2} \otimes {^2}} \right)} \right) = {{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {\bigcup\limits_{\lambda \in {{\left[ {0,{1 \over 2}} \right]}^4} \cap {{\cal C}_4}} {{{\rm{D}}^{{B_t}}}} \left( {{^2} \otimes {^2}} \right)\bigcap {{{\cal U}_\lambda }} } \right)} \hfill \cr { = {{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {\bigcup\limits_{\hat \lambda \in \hat C} {{{\cal E}_t}} \bigcap {{{\cal O}_{\hat \lambda }}} } \right) = \int_{\hat \lambda \in \hat C} {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}} \left( {{{\cal O}_{\hat \lambda }}\bigcap {{{\cal E}_t}} } \right){\rm{d}}m(\hat \lambda ),} \hfill \cr } where we used the facts in Eq. (26) and Eq. (27). Up to now, we have already obtained that 95 $\frac{π}{2} \int_{0}^{t} a^{2} f (a) d a = \int_{\hat{λ} \in \hat{C}} {vol}_{HS} (O_{\hat{λ}} \cap ε_{t}) d m (\hat{λ})$ {\pi \over 2}\int_0^t {{a^2}} f(a){\rm{d}}a = \int_{\hat \lambda \in \hat C} {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}} \left( {{{\cal O}_{\hat \lambda }}\bigcap {{{\cal E}_t}} } \right){\rm{d}}m(\hat \lambda ) 96 $= \int_{\hat{λ} \in \hat{C}} V_{4} (\hat{λ}) {vol}_{symp} (O_{\hat{λ}} \cap ε_{t}) d m (\hat{λ}),$ = \int_{\hat \lambda \in \hat C} {{V_4}} (\hat \lambda ){{\mathop{\rm vol}\nolimits} _{{\rm{symp }}}}\left( {{{\cal O}_{\hat \lambda }}\bigcap {{{\cal E}_t}} } \right){\rm{d}}m(\hat \lambda ), where we used the relationship (36) between HS volume of adjoint orbit and symplectic volume of co-adjoint orbit in the last equality. Consider the map F in Eq. (53), $F : (O_{\hat{λ}}, μ_{O_{\hat{λ}}}) \to (ℝ_{> 0}, d a),$ F:\left( {{{\cal O}_{\hat \lambda }},{\mu _{{{\cal O}_{\hat \lambda }}}}} \right) \to \left( {{_{ > 0}},{\rm{d}}a} \right), where F(X) = λ +(Tr₂(X)) – λ_–(Tr₂(X)) = 2λ₊(Tr₂(X)) for any $X \in O_{\hat{λ}}$ {\bf{X}} \in {{\cal O}_{\hat \lambda }}, i.e., twice the positive eigenvalue of Tr₂(X). By the definition of F, we see that $F^{- 1} ([0, t]) = O_{\hat{λ}} \cap ε_{t},$ {F^{ - 1}}([0,t]) = {{\cal O}_{\hat \lambda }}\bigcap {{{\cal E}_t}} , implying that 97 ${vol}_{symp} (ε_{t} \cap O_{\hat{λ}}) = {vol}_{symp} (F^{- 1} ([0, t])) = μ_{O_{\hat{λ}}} (F^{- 1} ([0, t]))$ {{\mathop{\rm vol}\nolimits} _{{\rm{symp }}}}\left( {{{\cal E}_t} \cap {{\cal O}_{\hat \lambda }}} \right) = {{\mathop{\rm vol}\nolimits} _{{\rm{symp }}}}\left( {{F^{ - 1}}([0,t])} \right) = {\mu _{{{\cal O}_{\hat \lambda }}}}\left( {{F^{ - 1}}([0,t])} \right) 98 $= (F_{*} μ_{O_{\hat{λ}}}) ([0, t]) = \int_{0}^{t} \frac{d F_{*} μ_{O_{\hat{λ}}}}{d a} d a,$ = \left( {{F_*}{\mu _{{{\cal O}_{\hat \lambda }}}}} \right)([0,t]) = \int_0^t {{{{\rm{d}}{F_*}{\mu _{{{\cal O}_{\hat \lambda }}}}} \over {{\rm{d}}a}}} {\rm{d}}a, where $\frac{d F_{*} μ_{O_{\hat{λ}}}}{d a} := (2 π)^{6} I (a ∣ \hat{λ})$ {{{\rm{d}}{F_*}{\mu _{{{\cal O}_{\hat \lambda }}}}} \over {{\rm{d}}a}}: = {(2\pi )^6}{\cal I}(a\mid \hat \lambda ) is the Radon-Nikodym derivative of push-forward measure $F_{*} μ_{O_{\hat{λ}}}$ {F_*}{\mu _{{{\cal O}_{\hat \lambda }}}} with respect to Lebesgue measure on the ray ℝ_>0. Moreover, the analytical expression of such Radon-Nikodym derivative $\frac{d F_{*} μ O_{\hat{λ}}}{d a} = (2 π)^{6} I (a ∣ \hat{λ})$ {{{\rm{d}}{F_*}\mu {{\cal O}_{\hat \lambda }}} \over {{\rm{d}}a}} = {(2\pi )^6}{\cal I}(a\mid \hat \lambda ) is given by Eq. (55). Once again, we have obtained that $\begin{array}{l} \frac{π}{2} \int_{0}^{t} a^{2} f (a) d a = \int_{\hat{λ} \in \hat{C}} V_{4} (\hat{λ}) {vol}_{symp} (ε_{t} \cap O_{\hat{λ}}) d m (\hat{λ}) \\ = \int_{\hat{λ} \in \hat{C}} V_{4} (\hat{λ}) (\int_{0}^{t} \frac{d F_{*} μ_{\hat{O}}}{d a} d a) d m (\hat{λ}) = \int_{\hat{λ} \in \hat{C}} V_{4} (\hat{λ}) (\int_{0}^{t} {(2 π)}^{6} I (a ∣ \hat{λ}) d a) d m (\hat{λ}) \\ = {(2 π)}^{6} \int_{0}^{t} (\int_{\hat{λ} \in \hat{C}} V_{4} (\hat{λ}) I (a ∣ \hat{λ}) d m (\hat{λ})) d a, \end{array}$ \matrix{ {{\pi \over 2}\int_0^t {{a^2}} f(a){\rm{d}}a = \int_{\hat \lambda \in \hat C} {{V_4}} (\hat \lambda ){{{\mathop{\rm vol}\nolimits} }_{{\rm{symp }}}}\left( {{{\cal E}_t}\bigcap {{{\cal O}_{\hat \lambda }}} } \right){\rm{d}}m(\hat \lambda )} \hfill \cr { = \int_{\hat \lambda \in \hat C} {{V_4}} (\hat \lambda )\left( {\int_0^t {{{{\rm{d}}{F_*}{\mu _{\widehat {\cal O}}}} \over {{\rm{d}}a}}} {\rm{d}}a} \right){\rm{d}}m(\hat \lambda ) = \int_{\hat \lambda \in \hat C} {{V_4}} (\hat \lambda )\left( {\int_0^t {{{(2\pi )}^6}} {\cal I}(a\mid \hat \lambda ){\rm{d}}a} \right){\rm{d}}m(\hat \lambda )} \hfill \cr { = {{(2\pi )}^6}\int_0^t {\left( {\int_{\hat \lambda \in \hat C} {{V_4}} (\hat \lambda ){\cal I}(a\mid \hat \lambda ){\rm{d}}m(\hat \lambda )} \right)} {\rm{d}}a,} \hfill \cr } implying that 99 $\frac{π}{2} \int_{0}^{t} a^{2} f (a) d a = (2 π)^{6} \int_{0}^{t} (\int_{\hat{λ} \in \hat{C}} V_{4} (\hat{λ}) I (a ∣ \hat{λ}) d m (\hat{λ})) d a .$ {\pi \over 2}\int_0^t {{a^2}} f(a){\rm{d}}a = {(2\pi )^6}\int_0^t {\left( {\int_{\hat \lambda \in \hat C} {{V_4}} (\hat \lambda ){\cal I}(a\mid \hat \lambda ){\rm{d}}m(\hat \lambda )} \right)} {\rm{d}}a.

By taking the derivative with respect to t, and then replacing t by a, we get that 100 $\frac{π}{2} a^{2} f (a) = (2 π)^{6} \int_{\hat{λ} \in \hat{C}} V_{4} (\hat{λ}) I (a ∣ \hat{λ}) d m (\hat{λ}) \Rightarrow$ {\pi \over 2}{a^2}f(a) = {(2\pi )^6}\int_{\hat \lambda \in \hat C} {{V_4}} (\hat \lambda ){\cal I}(a\mid \hat \lambda ){\rm{d}}m(\hat \lambda ) \Rightarrow 101 $f (a) = \frac{2}{π} a^{- 2} (2 π)^{6} \int_{\hat{λ} \in \hat{C}} V_{4} (\hat{λ}) I (a ∣ \hat{λ}) d m (\hat{λ}),$ f(a) = {2 \over \pi }{a^{ - 2}}{(2\pi )^6}\int_{\hat \lambda \in \hat C} {{V_4}} (\hat \lambda ){\cal I}(a\mid \hat \lambda ){\rm{d}}m(\hat \lambda ), where $a \in (0, \frac{1}{3})$ a \in \left( {0,{1 \over 3}} \right). Now we compute the following integral 102 $\int_{\hat{λ} \in \hat{C}} V_{4} (\hat{λ}) I (a ∣ \hat{λ}) d m (\hat{λ}) = 2 \int_{{[- \frac{1}{4}, \frac{1}{4}]}^{4} \cap C_{4}} \prod_{1 ⩽ i < j ⩽ 4} ({\hat{λ}}_{i} - {\hat{λ}}_{j}) I (a ∣ \hat{λ}) δ (\sum_{k = 1}^{4} {\hat{λ}}_{k}) \prod_{l = 1}^{4} d {\hat{λ}}_{l}$ \int_{\hat \lambda \in \hat C} {{V_4}} (\hat \lambda ){\cal I}(a\mid \hat \lambda ){\rm{d}}m(\hat \lambda ) = 2\int_{{{\left[ { - {1 \over 4},{1 \over 4}} \right]}^4} \cap {{\cal C}_4}} {\prod\limits_{1i < j4} {\left( {{{\hat \lambda }_i} - {{\hat \lambda }_j}} \right)} } {\cal I}(a\mid \hat \lambda )\delta \left( {\sum\limits_{k = 1}^4 {{{\hat \lambda }_k}} } \right)\prod\limits_{l = 1}^4 {{\rm{d}}} {{\hat \lambda }_l}

Let the change of variables be given below: 103 ${\begin{array}{l} {\hat{λ}}_{1} = t_{1} + \frac{t_{2}}{2} + \frac{t_{3}}{3} + \frac{t_{4}}{4} - \frac{1}{4}, \\ {\hat{λ}}_{2} = \frac{t_{2}}{2} + \frac{t_{3}}{3} + \frac{t_{4}}{4} - \frac{1}{4}, \\ {\hat{λ}}_{3} = \frac{t_{3}}{3} + \frac{t_{4}}{4} - \frac{1}{4}, \\ {\hat{λ}}_{4} = \frac{t_{4}}{4} - \frac{1}{4} . \end{array}$ \left\{ {\matrix{ {{{\hat \lambda }_1} = {t_1} + {{{t_2}} \over 2} + {{{t_3}} \over 3} + {{{t_4}} \over 4} - {1 \over 4},} \hfill \cr {{{\hat \lambda }_2} = {{{t_2}} \over 2} + {{{t_3}} \over 3} + {{{t_4}} \over 4} - {1 \over 4},} \hfill \cr {{{\hat \lambda }_3} = {{{t_3}} \over 3} + {{{t_4}} \over 4} - {1 \over 4},} \hfill \cr {{{\hat \lambda }_4} = {{{t_4}} \over 4} - {1 \over 4}.} \hfill \cr } } \right.

Its Jacobian is given by $J = \det (\frac{\partial ({\hat{λ}}_{1}, {\hat{λ}}_{2}, {\hat{λ}}_{3})}{\partial (t_{1}, t_{2}, t_{3})}) = | \begin{matrix} \frac{3}{4} & \frac{1}{4} & \frac{1}{12} \\ - \frac{1}{4} & \frac{1}{4} & \frac{1}{12} \\ - \frac{1}{4} & - \frac{1}{4} & \frac{1}{12} \end{matrix} | = \frac{1}{4!}$ J = \det \left( {{{\partial \left( {{{\hat \lambda }_1},{{\hat \lambda }_2},{{\hat \lambda }_3}} \right)} \over {\partial \left( {{t_1},{t_2},{t_3}} \right)}}} \right) = \left| {\matrix{ {{3 \over 4}} & {{1 \over 4}} & {{1 \over {12}}} \cr { - {1 \over 4}} & {{1 \over 4}} & {{1 \over {12}}} \cr { - {1 \over 4}} & { - {1 \over 4}} & {{1 \over {12}}} \cr } } \right| = {1 \over {4!}}

We note that

(i)
t_k ⩾ 0(k = 1, … ,4) due to the fact that ${\hat{λ}}_{1} > {\hat{λ}}_{2} > {\hat{λ}}_{3} > {\hat{λ}}_{4}$ {{\hat \lambda }_1} > {{\hat \lambda }_2} > {{\hat \lambda }_3} > {{\hat \lambda }_4}.
(ii)
t₁ + t₂ + t₃ ⩽ 1 because t₄ ⩾ 0 and $\sum_{k = 1}^{4} t_{k} = 1$ \sum\nolimits_{k = 1}^4 {{t_k}} = 1 due to the fact that $\sum_{k = 1}^{4} {\hat{λ}}_{4} = 0$ \sum\nolimits_{k = 1}^4 {{{\hat \lambda }_4}} = 0.
(iii)
$3 t_{1} + t_{2} + \frac{t_{3}}{3} ⩽ 1$ 3{t_1} + {t_2} + {{{t_3}} \over 3}1 because $t_{1} + \frac{t_{2}}{2} + \frac{t_{3}}{3} + \frac{t_{4}}{4} - \frac{1}{4} ⩽ \frac{1}{4}$ {t_1} + {{{t_2}} \over 2} + {{{t_3}} \over 3} + {{{t_4}} \over 4} - {1 \over 4}{1 \over 4} and $\sum_{k = 1}^{4} t_{k} = 1$ \sum\nolimits_{k = 1}^4 {{t_k}} = 1.

In summary, $\hat{C} = {[- \frac{1}{4}, \frac{1}{4}]}^{4} \cap C_{4}$ \widehat {\rm{C}} = {\left[ { - {1 \over 4},{1 \over 4}} \right]^4} \cap {{\cal C}_4} is transformed into the following form: 104 $R = {(t_{1}, t_{2}, t_{3}) \in ℝ_{⩾ 0}^{3} : t_{1} + t_{2} + t_{3} ⩽ 1, 3 t_{1} + t_{2} + \frac{t_{3}}{3} ⩽ 1} .$ R = \left\{ {\left( {{t_1},{t_2},{t_3}} \right) \in _{0}^3:{t_1} + {t_2} + {t_3}1,3{t_1} + {t_2} + {{{t_3}} \over 3}1} \right\}.

According to such change of variables, we get that 105 ${\begin{array}{l} c_{3} = 2 ({\hat{λ}}_{1} + {\hat{λ}}_{2}) = t_{1} + t_{2} + \frac{t_{3}}{3} \\ c_{2} = 2 ({\hat{λ}}_{1} + {\hat{λ}}_{3}) = t_{1} + \frac{t_{3}}{3} \\ c_{1} = 2 | {\hat{λ}}_{1} + {\hat{λ}}_{4} | = | t_{1} - \frac{t_{3}}{3} | \\ V_{4} (\hat{λ}) = \frac{t_{1} t_{2} t_{3} (2 t_{1} + t_{2}) (3 t_{2} + 2 t_{3}) (6 t_{1} + 3 t_{2} + 2 t_{3})}{432} = : V (t_{1}, t_{2}, t_{3}) \end{array}$ \left\{ {\matrix{ {{c_3} = 2\left( {{{\hat \lambda }_1} + {{\hat \lambda }_2}} \right) = {t_1} + {t_2} + {{{t_3}} \over 3}} \hfill \cr {{c_2} = 2\left( {{{\hat \lambda }_1} + {{\hat \lambda }_3}} \right) = {t_1} + {{{t_3}} \over 3}} \hfill \cr {{c_1} = 2\left| {{{\hat \lambda }_1} + {{\hat \lambda }_4}} \right| = \left| {{t_1} - {{{t_3}} \over 3}} \right|} \hfill \cr {{V_4}(\hat \lambda ) = {{{t_1}{t_2}{t_3}\left( {2{t_1} + {t_2}} \right)\left( {3{t_2} + 2{t_3}} \right)\left( {6{t_1} + 3{t_2} + 2{t_3}} \right)} \over {432}} = :V\left( {{t_1},{t_2},{t_3}} \right)} \hfill \cr } } \right.

Therefore we get that via change of variables 106 ${\tilde{I}}_{1} (x ∣ t) = \frac{t_{2} (t_{1} + \frac{t_{2}}{2} + \frac{t_{3}}{3})}{32} {(| t_{1} - \frac{t_{3}}{3} | - x)}^{2} x, | t_{1} - \frac{t_{3}}{3} | ⩾ x$ {\widetilde {\cal I}_1}(x\mid {\bf{t}}) = {{{t_2}\left( {{t_1} + {{{t_2}} \over 2} + {{{t_3}} \over 3}} \right)} \over {32}}{\left( {\left| {{t_1} - {{{t_3}} \over 3}} \right| - x} \right)^2}x,\quad \left| {{t_1} - {{{t_3}} \over 3}} \right|x 107 ${\tilde{I}}_{2} (x ∣ t) = \frac{(t_{1} + \frac{t_{2}}{2}) (\frac{t_{2}}{2} + \frac{t_{3}}{3})}{16} {(t_{1} + \frac{t_{3}}{3} - x)}^{2} x, t_{1} + \frac{t_{3}}{3} ⩾ x$ {\widetilde {\cal I}_2}(x\mid {\bf{t}}) = {{\left( {{t_1} + {{{t_2}} \over 2}} \right)\left( {{{{t_2}} \over 2} + {{{t_3}} \over 3}} \right)} \over {16}}{\left( {{t_1} + {{{t_3}} \over 3} - x} \right)^2}x,\quad {t_1} + {{{t_3}} \over 3}x 108 ${\tilde{I}}_{3} (x ∣ t) = \frac{t_{1} t_{3}}{48} {(t_{1} + t_{2} + \frac{t_{3}}{3} - x)}^{2} x, t_{1} + t_{2} + \frac{t_{3}}{3} ⩾ x$ {\widetilde {\cal I}_3}(x\mid {\bf{t}}) = {{{t_1}{t_3}} \over {48}}{\left( {{t_1} + {t_2} + {{{t_3}} \over 3} - x} \right)^2}x,\quad {t_1} + {t_2} + {{{t_3}} \over 3}x

With these preparations, we can transform the integral in Eq. (102) into the following form: $\begin{array}{l} \int_{\hat{λ} \in \hat{C}} V_{4} (\hat{λ}) I (x ∣ \hat{λ}) d m (\hat{λ}) & = & \frac{2}{4!} \int_{R} V (t_{1}, t_{2}, t_{3}) ({\tilde{I}}_{1} (x ∣ t) + {\tilde{I}}_{2} (x ∣ t) + {\tilde{I}}_{3} (x ∣ t)) d t_{1} d t_{2} d t_{3} \\ = & \frac{2}{4!} \int_{R} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{1} (x ∣ t) d t_{1} d t_{2} d t_{3} \\ + \frac{2}{4!} \int_{R} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{2} (x ∣ t) d t_{1} d t_{2} d t_{3} \\ + \frac{2}{4!} \int_{R} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{3} (x ∣ t) d t_{1} d t_{2} d t_{3} \\ = & M_{1} + M_{2} + M_{3} . \end{array}$ \matrix{ {\int_{\hat \lambda \in \hat C} {{V_4}} (\hat \lambda ){\cal I}(x\mid \hat \lambda ){\rm{d}}m(\hat \lambda )} \hfill & = \hfill & {{2 \over {4!}}\int_R V \left( {{t_1},{t_2},{t_3}} \right)\left( {{{\widetilde {\cal I}}_1}(x\mid {\bf{t}}) + {{\widetilde {\cal I}}_2}(x\mid {\bf{t}}) + {{\widetilde {\cal I}}_3}(x\mid {\bf{t}})} \right){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3}} \hfill \cr {} \hfill & = \hfill & {{2 \over {4!}}\int_R V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_1}(x\mid {\bf{t}}){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3}} \hfill \cr {} \hfill & {} \hfill & { + {2 \over {4!}}\int_R V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_2}(x\mid {\bf{t}}){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3}} \hfill \cr {} \hfill & {} \hfill & { + {2 \over {4!}}\int_R V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_3}(x\mid {\bf{t}}){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3}} \hfill \cr {} \hfill & = \hfill & {{M_1} + {M_2} + {M_3}.} \hfill \cr }

Note that the following three planes ${t_{1} + t_{2} + t_{3} = 1, 3 t_{1} + t_{2} + \frac{t_{3}}{3} = 1, t_{1} - \frac{t_{3}}{3} = 0}$ \left\{ {{t_1} + {t_2} + {t_3} = 1,3{t_1} + {t_2} + {{{t_3}} \over 3} = 1,{t_1} - {{{t_3}} \over 3} = 0} \right\} intersects at the line segment in $ℝ_{⩾ 0}^{3}$ _{0}^3 109 ${\begin{array}{l} t_{1} = t \\ t_{2} = 1 - 4 t \\ t_{3} = 3 t \end{array} (t \in [0, \frac{1}{4}]) .$ \left\{ {\matrix{ {{t_1} = t} \hfill \cr {{t_2} = 1 - 4t} \hfill \cr {{t_3} = 3t} \hfill \cr } \quad \left( {t \in \left[ {0,{1 \over 4}} \right]} \right)} \right..
The following two planes ${3 t_{1} + t_{2} + \frac{t_{3}}{3} = 1, t_{1} - \frac{t_{3}}{3} = x}$ \left\{ {3{t_1} + {t_2} + {{{t_3}} \over 3} = 1,{t_1} - {{{t_3}} \over 3} = x} \right\} intersects at the line segment in $ℝ_{⩾ 0}^{3}$ _{0}^3: 110 ${\begin{array}{l} t_{1} = t \\ t_{2} = 1 + x - 4 t \\ t_{3} = - 3 x + 3 t \end{array} (t \in [x, \frac{1 + x}{4}]) .$ \left\{ {\matrix{ {{t_1} = t} \hfill \cr {{t_2} = 1 + x - 4t} \hfill \cr {{t_3} = - 3x + 3t} \hfill \cr } \quad \left( {t \in \left[ {x,{{1 + x} \over 4}} \right]} \right)} \right..
The following two planes ${t_{1} + t_{2} + t_{3} = 1, t_{1} - \frac{t_{3}}{3} = - x}$ \left\{ {{t_1} + {t_2} + {t_3} = 1,{t_1} - {{{t_3}} \over 3} = - x} \right\} intersects at the line segment in $ℝ_{⩾ 0}^{3}$ _{0}^3: 111 ${\begin{array}{l} t_{1} = t \\ t_{2} = 1 - 3 x - 4 t \\ t_{3} = 3 x + 3 t \end{array} (t \in [0, \frac{1 - 3 x}{4}]) .$ \left\{ {\matrix{ {{t_1} = t} \hfill \cr {{t_2} = 1 - 3x - 4t} \hfill \cr {{t_3} = 3x + 3t} \hfill \cr } \quad \left( {t \in \left[ {0,{{1 - 3x} \over 4}} \right]} \right)} \right..

In summary, the plane $| t_{1} - \frac{t_{3}}{3} | = x$ \left| {{t_1} - {{{t_3}} \over 3}} \right| = x intersects with R iff $x \in [0, \frac{1}{3}]$ x \in \left[ {0,{1 \over 3}} \right]. For m ⩾ 2, denote $Δ_{m} := {(p_{1}, \dots, p_{m}) \in ℝ_{⩾ 0}^{m} : p_{1} + \dots + p_{m} ⩽ 1} .$ {\Delta _m}: = \left\{ {\left( {{p_1}, \ldots ,{p_m}} \right) \in _{0}^m:{p_1} + \cdots + {p_m}1} \right\}.

Now for $x \in [0, \frac{1}{3}]$ x \in \left[ {0,{1 \over 3}} \right], we get that

(1)
For the domain of the first integral for a given x, we get that $\begin{array}{l} R_{1} (x) & := & {(t_{1}, t_{2}, t_{3}) \in ℝ_{⩾ 0}^{3} : t_{1} + t_{2} + t_{3} ⩽ 1, 3 t_{1} + t_{2} + \frac{t_{3}}{3} ⩽ 1, | t_{1} - \frac{t_{3}}{3} | ⩾ x} \\ = & R_{1, a} (x) \cup R_{1, b} (x), \end{array}$ \matrix{ {{R_1}(x)} \hfill & {: = } \hfill & {\left\{ {\left( {{t_1},{t_2},{t_3}} \right) \in _{0}^3:{t_1} + {t_2} + {t_3}1,3{t_1} + {t_2} + {{{t_3}} \over 3}1,\left| {{t_1} - {{{t_3}} \over 3}} \right|x} \right\}} \hfill \cr {} \hfill & = \hfill & {{R_{1,a}}(x) \cup {R_{1,b}}(x),} \hfill \cr } where $\begin{array}{l} R_{1, a} (x) & := & {(t_{1}, t_{2}, t_{3}) \in ℝ_{⩾ 0}^{3} : t_{1} + t_{2} + t_{3} ⩽ 1, 3 t_{1} + t_{2} + \frac{t_{3}}{3} ⩽ 1, t_{1} - \frac{t_{3}}{3} ⩾ x} \\ = & {(t_{1}, t_{2}, t_{3}) \in ℝ_{⩾ 0}^{3} : 0 ⩽ t_{3} ⩽ \frac{3 (1 - 3 x)}{4}, 0 ⩽ t_{2} ⩽ 1 - 3 x - \frac{4 t_{3}}{3}, x + \frac{t_{3}}{3} ⩽ t_{1} ⩽ \frac{1}{3} - \frac{t_{2}}{3} - \frac{t_{3}}{9}} \\ R_{1, b} (x) & := & {(t_{1}, t_{2}, t_{3}) \in ℝ_{⩾ 0}^{3} : t_{1} + t_{2} + t_{3} ⩽ 1, 3 t_{1} + t_{2} + \frac{t_{3}}{3} ⩽ 1, t_{1} - \frac{t_{3}}{3} ⩽ - x} \\ = & {(t_{1}, t_{2}, t_{3}) \in ℝ_{⩾ 0}^{3} : 0 ⩽ t_{2} ⩽ 1 - 3 x, 0 ⩽ t_{1} ⩽ \frac{1 - 3 x - t_{2}}{4}, 3 (x + t_{1}) ⩽ t_{3} ⩽ 1 - t_{1} - t_{2}} . \end{array}$ \matrix{ {{R_{1,a}}(x)} \hfill & {: = } \hfill & {\left\{ {\left( {{t_1},{t_2},{t_3}} \right) \in _{0}^3:{t_1} + {t_2} + {t_3}1,3{t_1} + {t_2} + {{{t_3}} \over 3}1,{t_1} - {{{t_3}} \over 3}x} \right\}} \hfill \cr {} \hfill & = \hfill & {\left\{ {\left( {{t_1},{t_2},{t_3}} \right) \in _{0}^3:0{t_3}{{3(1 - 3x)} \over 4},0{t_2}1 - 3x - {{4{t_3}} \over 3},x + {{{t_3}} \over 3}{t_1}{1 \over 3} - {{{t_2}} \over 3} - {{{t_3}} \over 9}} \right\}} \hfill \cr {{R_{1,b}}(x)} \hfill & {: = } \hfill & {\left\{ {\left( {{t_1},{t_2},{t_3}} \right) \in _{0}^3:{t_1} + {t_2} + {t_3}1,3{t_1} + {t_2} + {{{t_3}} \over 3}1,{t_1} - {{{t_3}} \over 3} - x} \right\}} \hfill \cr {} \hfill & = \hfill & {\left\{ {\left( {{t_1},{t_2},{t_3}} \right) \in _{0}^3:0{t_2}1 - 3x,0{t_1}{{1 - 3x - {t_2}} \over 4},3\left( {x + {t_1}} \right){t_3}1 - {t_1} - {t_2}} \right\}.} \hfill \cr }

In such case, we can evaluate M₁ as follows: For $x \in (0, \frac{1}{3})$ x \in \left( {0,{1 \over 3}} \right), $\begin{array}{l} M_{1} & = & \frac{2}{4!} \int_{R} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{1} (x ∣ t) d t_{1} d t_{2} d t_{3} = \frac{2}{4!} \int_{R_{1} (x)} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{1} (x ∣ t) d t_{1} d t_{2} d t_{3} \\ = & \frac{2}{4!} \int_{R_{1, a} (x)} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{1} (x ∣ t) d t_{1} d t_{2} d t_{3} + \frac{2}{4!} \int_{R_{1, b} (x)} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{1} (x ∣ t) d t_{1} d t_{2} d t_{3} \\ = & \frac{x {(3 x - 1)}^{9} (567 x^{4} - 3564 x^{3} + 5526 x^{2} - 3152 x - 1345)}{774741019852800} \\ + \frac{x {(3 x - 1)}^{9} (567 x^{4} - 3564 x^{3} + 5526 x^{2} - 3152 x - 1345)}{774741019852800} \\ = & \frac{x {(3 x - 1)}^{9} (567 x^{4} - 3564 x^{3} + 5526 x^{2} - 3152 x - 1345)}{387370509926400} \end{array}$ \matrix{ {{M_1}} \hfill & = \hfill & {{2 \over {4!}}\int_R V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_1}(x\mid t){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3} = {2 \over {4!}}\int_{{R_1}(x)} V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_1}(x\mid t){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3}} \hfill \cr {} \hfill & = \hfill & {{2 \over {4!}}\int_{{R_{1,a}}(x)} V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_1}(x\mid t){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3} + {2 \over {4!}}\int_{{R_{1,b}}(x)} V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_1}(x\mid t){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3}} \hfill \cr {} \hfill & = \hfill & {{{x{{(3x - 1)}^9}\left( {567{x^4} - 3564{x^3} + 5526{x^2} - 3152x - 1345} \right)} \over {774741019852800}}} \hfill \cr {} \hfill & {} \hfill & { + {{x{{(3x - 1)}^9}\left( {567{x^4} - 3564{x^3} + 5526{x^2} - 3152x - 1345} \right)} \over {774741019852800}}} \hfill \cr {} \hfill & = \hfill & {{{x{{(3x - 1)}^9}\left( {567{x^4} - 3564{x^3} + 5526{x^2} - 3152x - 1345} \right)} \over {387370509926400}}} \hfill \cr } thus $M_{1} = \frac{x {(3 x - 1)}^{9} (567 x^{4} - 3564 x^{3} + 5526 x^{2} - 3152 x - 1345)}{387370509926400} .$ {M_1} = {{x{{(3x - 1)}^9}\left( {567{x^4} - 3564{x^3} + 5526{x^2} - 3152x - 1345} \right)} \over {387370509926400}}.
(2)
For the domain of the second integral for a given x, we get that $\begin{array}{l} R_{2} (x) & := & {(t_{1}, t_{2}, t_{3}) \in ℝ_{⩾ 0}^{3} : t_{1} + t_{2} + t_{3} ⩽ 1, 3 t_{1} + t_{2} + \frac{t_{3}}{3} ⩽ 1, t_{1} + \frac{t_{3}}{3} ⩾ x} \\ = & (Δ_{3} \cup R_{2, a b} (x)) \ (R_{2, a} (x) \cup R_{2, b} (x)), \end{array}$ \matrix{ {{R_2}(x)} \hfill & {: = } \hfill & {\left\{ {\left( {{t_1},{t_2},{t_3}} \right) \in _{0}^3:{t_1} + {t_2} + {t_3}1,3{t_1} + {t_2} + {{{t_3}} \over 3}1,{t_1} + {{{t_3}} \over 3}x} \right\}} \hfill \cr {} \hfill & = \hfill & {\left( {{\Delta _3} \cup {R_{2,ab}}(x)} \right)\backslash \left( {{R_{2,a}}(x) \cup {R_{2,b}}(x)} \right)} \hfill \cr } where ${\begin{array}{l} R_{2, a} (x) & := & {(t_{1}, t_{2}, t_{3}) \in ℝ_{⩾ 0}^{3} : t_{1} + t_{2} + t_{3} ⩽ 1, 3 t_{1} + t_{2} + \frac{t_{3}}{3} > 1} \\ R_{2, b} (x) & := & {(t_{1}, t_{2}, t_{3}) \in ℝ_{⩾ 0}^{3} : t_{1} + t_{2} + t_{3} ⩽ 1, t_{1} + \frac{t_{3}}{3} < x} \\ R_{2, a b} (x) & := & R_{2, a} (x) \cap R_{2, b} (x) . \end{array}$ \left\{ {\matrix{ {{R_{2,a}}(x)} \hfill & {: = } \hfill & {\left\{ {\left( {{t_1},{t_2},{t_3}} \right) \in _{0}^3:{t_1} + {t_2} + {t_3}1,3{t_1} + {t_2} + {{{t_3}} \over 3} > 1} \right\}} \hfill \cr {{R_{2,b}}(x)} \hfill & {: = } \hfill & {\left\{ {\left( {{t_1},{t_2},{t_3}} \right) \in _{0}^3:{t_1} + {t_2} + {t_3}1,{t_1} + {{{t_3}} \over 3} < x} \right\}} \hfill \cr {{R_{2,ab}}(x)} \hfill & {: = } \hfill & {{R_{2,a}}(x) \cap {R_{2,b}}(x).} \hfill \cr } } \right.

Note that $\begin{array}{l} R_{2, a} (x) & = & {(t_{1}, t_{2}, t_{3}) \in ℝ^{3} : 0 ⩽ t_{3} < \frac{3}{4}, 0 ⩽ t_{2} < 1 - \frac{4 t_{3}}{3}, \frac{1}{3} - \frac{t_{2}}{3} - \frac{t_{3}}{9} < t_{1} ⩽ 1 - t_{2} - t_{3}}, \\ R_{2, b} (x) & = & {(t_{1}, t_{2}, t_{3}) \in ℝ^{3} : 0 ⩽ t_{3} < 3 x, 0 ⩽ t_{1} < x - \frac{t_{3}}{3}, 0 ⩽ t_{2} ⩽ 1 - t_{1} - t_{3}}, \\ R_{2, a b} (x) & = & {(t_{1}, t_{2}, t_{3}) \in ℝ^{3} : 0 ⩽ t_{3} < \frac{3 x}{2}, \frac{t_{3}}{3} < t_{1} < x - \frac{t_{3}}{3}, 1 - 3 t_{1} - \frac{t_{3}}{3} < t_{2} ⩽ 1 - t_{1} - t_{3}} . \end{array}$ \matrix{ {{R_{2,a}}(x)} \hfill & = \hfill & {\left\{ {\left( {{t_1},{t_2},{t_3}} \right) \in {^3}:0{t_3} < {3 \over 4},0{t_2} < 1 - {{4{t_3}} \over 3},{1 \over 3} - {{{t_2}} \over 3} - {{{t_3}} \over 9} < {t_1}1 - {t_2} - {t_3}} \right\},} \hfill \cr {{R_{2,b}}(x)} \hfill & = \hfill & {\left\{ {\left( {{t_1},{t_2},{t_3}} \right) \in {^3}:0{t_3} < 3x,0{t_1} < x - {{{t_3}} \over 3},0{t_2}1 - {t_1} - {t_3}} \right\},} \hfill \cr {{R_{2,ab}}(x)} \hfill & = \hfill & {\left\{ {\left( {{t_1},{t_2},{t_3}} \right) \in {^3}:0{t_3} < {{3x} \over 2},{{{t_3}} \over 3} < {t_1} < x - {{{t_3}} \over 3},1 - 3{t_1} - {{{t_3}} \over 3} < {t_2}1 - {t_1} - {t_3}} \right\}.} \hfill \cr }

In such case, we can evaluate M₂ as follows: For $x \in (0, \frac{1}{3})$ x \in \left( {0,{1 \over 3}} \right), $\begin{array}{l} M_{2} & = & \frac{2}{4!} \int_{R} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{2} (x ∣ t) d t_{1} d t_{2} d t_{3} = \frac{2}{4!} \int_{R_{2} (x)} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{2} (x ∣ t) d t_{1} d t_{2} d t_{3} \\ = & \frac{2}{4!} \int_{Δ_{3}} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{2} (x ∣ t) d t_{1} d t_{2} d t_{3} + \frac{2}{4!} \int_{R_{2, a b} (x)} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{2} (x ∣ t) d t_{1} d t_{2} d t_{3} \\ - \frac{2}{4!} \int_{R_{2, a} (x)} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{2} (x ∣ t) d t_{1} d t_{2} d t_{3} - \frac{2}{4!} \int_{R_{2, b} (x)} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{2} (x ∣ t) d t_{1} d t_{2} d t_{3} \\ = & - \frac{x (725985 x^{2} - 560326 x + 123355)}{96842627481600} \\ - \frac{x^{8} (2894 x^{6} - 24570 x^{5} + 81900 x^{4} - 150150 x^{3} + 160875 x^{2} - 96525 x + 25740)}{177124147200} \\ + \frac{x (637485030 x^{2} - 525965687 x + 120181250)}{99166850541158400} \\ - \frac{x^{7} (1572 x^{7} - 17550 x^{6} + 62712 x^{5} - 120835 x^{4} + 141570 x^{3} - 106821 x^{2} + 51480 x - 12870)}{132843110400}, \end{array}$ \matrix{ {{M_2}} \hfill & = \hfill & {{2 \over {4!}}\int_R V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_2}(x\mid {\bf{t}}){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3} = {2 \over {4!}}\int_{{R_2}(x)} V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_2}(x\mid {\bf{t}}){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3}} \hfill \cr {} \hfill & = \hfill & {{2 \over {4!}}\int_{{\Delta _3}} V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_2}(x\mid {\bf{t}}){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3} + {2 \over {4!}}\int_{{R_{2,ab}}(x)} V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_2}(x\mid {\bf{t}}){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3}} \hfill \cr {} \hfill & {} \hfill & { - {2 \over {4!}}\int_{{R_{2,a}}(x)} V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_2}(x\mid {\bf{t}}){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3} - {2 \over {4!}}\int_{{R_{2,b}}(x)} V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_2}(x\mid {\bf{t}}){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3}} \hfill \cr {} \hfill & = \hfill & { - {{x\left( {725985{x^2} - 560326x + 123355} \right)} \over {96842627481600}}} \hfill \cr {} \hfill & {} \hfill & { - {{{x^8}\left( {2894{x^6} - 24570{x^5} + 81900{x^4} - 150150{x^3} + 160875{x^2} - 96525x + 25740} \right)} \over {177124147200}}} \hfill \cr {} \hfill & {} \hfill & { + {{x\left( {637485030{x^2} - 525965687x + 120181250} \right)} \over {99166850541158400}}} \hfill \cr {} \hfill & {} \hfill & { - {{{x^7}\left( {1572{x^7} - 17550{x^6} + 62712{x^5} - 120835{x^4} + 141570{x^3} - 106821{x^2} + 51480x - 12870} \right)} \over {132843110400}},} \hfill \cr } implying that $\begin{array}{l} M_{2} & = & - \frac{499 x^{14}}{17712414720} + \frac{41 x^{13}}{151388160} - \frac{1061 x^{12}}{1135411200} + \frac{653 x^{11}}{371589120} - \frac{163 x^{10}}{82575360} + \frac{557 x^{9}}{412876800} \\ - \frac{11 x^{8}}{20643840} + \frac{x^{7}}{10321920} - \frac{90533 x^{3}}{84757991915520} + \frac{3677549 x^{2}}{7628219272396800} - \frac{613427 x}{9916685054115840} . \end{array}$ \matrix{ {{M_2}} \hfill & = \hfill & { - {{499{x^{14}}} \over {17712414720}} + {{41{x^{13}}} \over {151388160}} - {{1061{x^{12}}} \over {1135411200}} + {{653{x^{11}}} \over {371589120}} - {{163{x^{10}}} \over {82575360}} + {{557{x^9}} \over {412876800}}} \hfill \cr {} \hfill & {} \hfill & { - {{11{x^8}} \over {20643840}} + {{{x^7}} \over {10321920}} - {{90533{x^3}} \over {84757991915520}} + {{3677549{x^2}} \over {7628219272396800}} - {{613427x} \over {9916685054115840}}.} \hfill \cr }
(3)
For the domain of the third integral for a given x, we get that $\begin{array}{l} R_{3} (x) & := & {(t_{1}, t_{2}, t_{3}) \in ℝ_{⩾ 0}^{3} : t_{1} + t_{2} + t_{3} ⩽ 1, 3 t_{1} + t_{2} + \frac{t_{3}}{3} ⩽ 1, t_{1} + t_{2} + \frac{t_{3}}{3} ⩾ x} \\ = & (Δ_{3} \cup R_{3, a b} (x)) \ (R_{3, a} (x) \cup R_{3, b} (x)), \end{array}$ \matrix{ {{R_3}(x)} \hfill & {: = } \hfill & {\left\{ {\left( {{t_1},{t_2},{t_3}} \right) \in _{0}^3:{t_1} + {t_2} + {t_3}1,3{t_1} + {t_2} + {{{t_3}} \over 3}1,{t_1} + {t_2} + {{{t_3}} \over 3}x} \right\}} \hfill \cr {} \hfill & = \hfill & {\left( {{\Delta _3} \cup {R_{3,ab}}(x)} \right)\backslash \left( {{R_{3,a}}(x) \cup {R_{3,b}}(x)} \right),} \hfill \cr } where ${\begin{array}{l} R_{3, a} (x) & := & {(t_{1}, t_{2}, t_{3}) \in ℝ_{⩾ 0}^{3} : t_{1} + t_{2} + t_{3} ⩽ 1, 3 t_{1} + t_{2} + \frac{t_{3}}{3} > 1}, \\ R_{3, b} (x) & := & {(t_{1}, t_{2}, t_{3}) \in ℝ_{⩾ 0}^{3} : t_{1} + t_{2} + \frac{t_{3}}{3} < x}, \\ R_{3, a b} (x) & := & R_{3, a} (x) \cap R_{3, b} (x), \end{array}$ \left\{ {\matrix{ {{R_{3,a}}(x)} \hfill & {: = } \hfill & {\left\{ {\left( {{t_1},{t_2},{t_3}} \right) \in _{0}^3:{t_1} + {t_2} + {t_3}1,3{t_1} + {t_2} + {{{t_3}} \over 3} > 1} \right\},} \hfill \cr {{R_{3,b}}(x)} \hfill & {: = } \hfill & {\left\{ {\left( {{t_1},{t_2},{t_3}} \right) \in _{0}^3:{t_1} + {t_2} + {{{t_3}} \over 3} < x} \right\},} \hfill \cr {{R_{3,ab}}(x)} \hfill & {: = } \hfill & {{R_{3,a}}(x) \cap {R_{3,b}}(x),} \hfill \cr } } \right.

Note that $\begin{array}{l} R_{3, a} (x) & = & {(t_{1}, t_{2}, t_{3}) \in ℝ^{3} : 0 ⩽ t_{3} < \frac{3}{4}, 0 ⩽ t_{2} < 1 - \frac{4 t_{3}}{3}, \frac{1}{3} - \frac{t_{2}}{3} - \frac{t_{3}}{9} < t_{1} ⩽ 1 - t_{2} - t_{3}}, \\ R_{3, b} (x) & = & {(t_{1}, t_{2}, t_{3}) \in ℝ^{3} : 0 ⩽ t_{3} < 3 x, 0 ⩽ t_{2} < x - \frac{t_{3}}{3}, 0 ⩽ t_{1} ⩽ x - t_{2} - \frac{t_{3}}{3}}, \\ R_{3, a b} (x) & = & R_{3, a} (x) \cap R_{3, b} (x) = \emptyset . \end{array}$ \matrix{ {{R_{3,a}}(x)} \hfill & = \hfill & {\left\{ {\left( {{t_1},{t_2},{t_3}} \right) \in {^3}:0{t_3} < {3 \over 4},0{t_2} < 1 - {{4{t_3}} \over 3},{1 \over 3} - {{{t_2}} \over 3} - {{{t_3}} \over 9} < {t_1}1 - {t_2} - {t_3}} \right\},} \hfill \cr {{R_{3,b}}(x)} \hfill & = \hfill & {\left\{ {\left( {{t_1},{t_2},{t_3}} \right) \in {^3}:0{t_3} < 3x,0{t_2} < x - {{{t_3}} \over 3},0{t_1}x - {t_2} - {{{t_3}} \over 3}} \right\},} \hfill \cr {{R_{3,ab}}(x)} \hfill & = \hfill & {{R_{3,a}}(x) \cap {R_{3,b}}(x) = \emptyset .} \hfill \cr }

In such case, we can evaluate M₃ as follows: For $x \in (0, \frac{1}{3})$ x \in \left( {0,{1 \over 3}} \right), $\begin{array}{l} M_{3} & = & \frac{2}{4!} \int_{R} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{3} (x ∣ t) d t_{1} d t_{2} d t_{3} = \frac{2}{4!} \int_{R_{3} (x)} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{2} (x ∣ t) d t_{1} d t_{2} d t_{3} \\ = & \frac{2}{4!} \int_{Δ_{3}} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{3} (x ∣ t) d t_{1} d t_{2} d t_{3} \\ - \frac{2}{4!} \int_{R_{3, a} (x)} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{3} (x ∣ t) d t_{1} d t_{2} d t_{3} - \frac{2}{4!} \int_{R_{3, b} (x)} V (t_{1}, t_{2}, t_{3}) {\tilde{I}}_{3} (x ∣ t) d t_{1} d t_{2} d t_{3} \\ = & \frac{x (2340 x^{2} - 3341 x + 1220)}{1513166054400} - \frac{x (135789030 x^{2} - 199046263 x + 74163970)}{99166850541158400} - \frac{x^{14}}{691891200}, \end{array}$ \matrix{ {{M_3}} \hfill & = \hfill & {{2 \over {4!}}\int_R V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_3}(x\mid {\bf{t}}){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3} = {2 \over {4!}}\int_{{R_3}(x)} V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_2}(x\mid {\bf{t}}){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3}} \hfill \cr {} \hfill & = \hfill & {{2 \over {4!}}\int_{{\Delta _3}} V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_3}(x\mid {\bf{t}}){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3}} \hfill \cr {} \hfill & {} \hfill & { - {2 \over {4!}}\int_{{R_{3,a}}(x)} V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_3}(x\mid {\bf{t}}){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3} - {2 \over {4!}}\int_{{R_{3,b}}(x)} V \left( {{t_1},{t_2},{t_3}} \right){{\widetilde {\cal I}}_3}(x\mid {\bf{t}}){\rm{d}}{t_1}{\rm{d}}{t_2}{\rm{d}}{t_3}} \hfill \cr {} \hfill & = \hfill & {{{x\left( {2340{x^2} - 3341x + 1220} \right)} \over {1513166054400}} - {{x\left( {135789030{x^2} - 199046263x + 74163970} \right)} \over {99166850541158400}} - {{{x^{14}}} \over {691891200}},} \hfill \cr } implying that $M_{3} = - \frac{x^{14}}{691891200} + \frac{15013 x^{3}}{84757991915520} - \frac{1531501 x^{2}}{7628219272396800} + \frac{115799 x}{1983337010823168} .$ {M_3} = - {{{x^{14}}} \over {691891200}} + {{15013{x^3}} \over {84757991915520}} - {{1531501{x^2}} \over {7628219272396800}} + {{115799x} \over {1983337010823168}}.

In summary, 112 $M_{1} + M_{2} + M_{3} = \frac{{(1 - x)}^{9} x^{2} (33 x^{3} + 162 x^{2} + 72 x + 8)}{40874803200} .$ {M_1} + {M_2} + {M_3} = {{{{(1 - x)}^9}{x^2}\left( {33{x^3} + 162{x^2} + 72x + 8} \right)} \over {40874803200}}.

This indicates, together with Eq. (101), that 113 $f (a) = \frac{π^{5}}{319334400} (1 - a)^{9} (33 a^{3} + 162 a^{2} + 72 a + 8),$ f(a) = {{{\pi ^5}} \over {319334400}}{(1 - a)^9}\left( {33{a^3} + 162{a^2} + 72a + 8} \right), where $a \in (0, \frac{1}{3})$ a \in \left( {0,{1 \over 3}} \right).

Remark 5.

We see that the parameter x is restricted to the open interval $(0, \frac{1}{3})$ \left( {0,{1 \over 3}} \right) due to fact that the plane $| t_{1} - \frac{t_{3}}{3} | = x$ \left| {{t_1} - {{{t_3}} \over 3}} \right| = x intersects with R iff $x \in [0, \frac{1}{3}]$ x \in \left[ {0,{1 \over 3}} \right]. Note that the integrations involved are performed by the mathematical software Mathematica.

In the recent paper [7], Huong and Khoi published a proof of the exact separability probability $\frac{8}{33}$ {8 \over {33}} for the two-qubit system, resolving a long-standing conjecture originally proposed by Slater. Their result demonstrates that within the Hilbert-Schmidt measure, the proportion of separable states among all two-qubit density matrices is precisely $\frac{8}{33}$ {8 \over {33}} This breakthrough provides a definitive answer to a fundamental question in quantum information theory concerning the geometric prevalence of entanglement in simple quantum systems, achieved through a novel combination of advanced geometric probability techniques and symmetry arguments applied to the structure of the two-qubit state space.

Theorem 7.

The separability probability of all two-qubit states is given by 114 $P_{sep}^{(2 \times 2)} = \frac{{vol}_{HS} (D_{sep} (ℂ^{2} \otimes ℂ^{2}))}{{vol}_{HS} (D (ℂ^{2} \otimes ℂ^{2}))} = \frac{8}{33} .$ P_{{\rm{sep}}}^{(2 \times 2)} = {{{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\rm{D}}_{{\rm{sep}}}}\left( {{^2} \otimes {^2}} \right)} \right)} \over {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{\rm{D}}\left( {{^2} \otimes {^2}} \right)} \right)}} = {8 \over {33}}.

Proof

From Theorem 6, we have seen that $P_{sep}^{(2 \times 2)} (a)$ P_{{\rm{sep }}}^{(2 \times 2)}(a) is independent of a ∈ [0,1), which means that $P_{sep}^{(2 \times 2)} (a) = P_{sep}^{(2 \times 2)} (0)$ P_{{\rm{sep }}}^{(2 \times 2)}(a) = P_{{\rm{sep }}}^{(2 \times 2)}(0) for all a ∈ [0,1). Then 115 ${vol}_{HS} (D_{sep}^{a} (ℂ^{2} \otimes ℂ^{2})) = P_{sep}^{(2 \times 2)} (0) {vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{2})) .$ {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{\rm{D}}_{{\rm{sep}}}^a\left( {{^2} \otimes {^2}} \right)} \right) = P_{{\rm{sep}}}^{(2 \times 2)}(0){{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{{\rm{D}}^a}\left( {{^2} \otimes {^2}} \right)} \right).

By multiplying $\frac{π}{2} a^{2}$ {\pi \over 2}{a^2} on both sides above, and taking the integration over [0,1), we see that $\begin{array}{l} \frac{π}{2} \int_{0}^{1} a^{2} {vol}_{HS} (D_{sep}^{a} (ℂ^{2} \otimes ℂ^{2})) d a & = & P_{sep}^{(2 \times 2)} (0) \frac{π}{2} \int_{0}^{1} a^{2} {vol}_{HS} (D^{a} (ℂ^{2} \otimes ℂ^{2})) d a \\ = & P_{sep}^{(2 \times 2)} (0) {vol}_{HS} (D (ℂ^{2} \otimes ℂ^{2})), \end{array}$ \matrix{ {{\pi \over 2}\int_0^1 {{a^2}} {{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{\rm{D}}_{{\rm{sep}}}^a\left( {{^2} \otimes {^2}} \right)} \right){\rm{d}}a} \hfill & = \hfill & {P_{{\rm{sep}}}^{(2 \times 2)}(0){\pi \over 2}\int_0^1 {{a^2}} {{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\rm{D}}^a}\left( {{^2} \otimes {^2}} \right)} \right){\rm{d}}a} \hfill \cr {} \hfill & = \hfill & {P_{{\rm{sep}}}^{(2 \times 2)}(0){{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{\rm{D}}\left( {{^2} \otimes {^2}} \right)} \right),} \hfill \cr } where we used the result in Proposition 13. Analogously, $\frac{π}{2} \int_{0}^{1} a^{2} {vol}_{HS} (D_{sep}^{a} (ℂ^{2} \otimes ℂ^{2})) d a = {vol}_{HS} (D_{sep} (ℂ^{2} \otimes ℂ^{2})) .$ {\pi \over 2}\int_0^1 {{a^2}} {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{\rm{D}}_{{\rm{sep}}}^a\left( {{^2} \otimes {^2}} \right)} \right){\rm{d}}a = {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{{\rm{D}}_{{\rm{sep}}}}\left( {{^2} \otimes {^2}} \right)} \right).

These formulas indicate that $P_{sep}^{(2 \times 2)} = P_{sep}^{(2 \times 2)} (0)$ P_{{\rm{sep }}}^{(2 \times 2)} = P_{{\rm{sep }}}^{(2 \times 2)}(0). With Eq. (115), it suffices to calculate 116 $P_{sep}^{(2 \times 2)} (0) = \frac{{vol}_{HS} (D_{sep}^{0} (ℂ^{2} \otimes ℂ^{2}))}{{vol}_{HS} (D^{0} (ℂ^{2} \otimes ℂ^{2}))} .$ P_{{\rm{sep}}}^{(2 \times 2)}(0) = {{{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{\rm{D}}_{{\rm{sep}}}^0\left( {{^2} \otimes {^2}} \right)} \right)} \over {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\rm{D}}^0}\left( {{^2} \otimes {^2}} \right)} \right)}}.

To this end, it suffices to calculate vol_HS $(D_{sep}^{0} (ℂ^{2} \otimes ℂ^{2}))$ {\rm{vo}}{{\rm{l}}_{{\rm{HS}}}}\left( {{\rm{D}}_{{\rm{sep }}}^0\left( {{^2} \otimes {^2}} \right)} \right) since we have already know the denominator ${vol}_{HS} (D^{0} (ℂ^{2} \otimes ℂ^{2})) = \frac{π^{5}}{9676800}$ {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{{\rm{D}}^0}\left( {{^2} \otimes {^2}} \right)} \right) = {{{\pi ^5}} \over {9676800}} from Eq. (77). Clearly, from Proposition 16, $f (0) = {vol}_{HS} (D_{sep}^{0} (ℂ^{2} \otimes ℂ^{2})) = \frac{π^{5}}{39916800},$ f(0) = {{\mathop{\rm vol}\nolimits} _{{\rm{HS}}}}\left( {{\rm{D}}_{{\rm{sep}}}^0\left( {{^2} \otimes {^2}} \right)} \right) = {{{\pi ^5}} \over {39916800}}, implying that 117 $P_{sep}^{(2 \times 2)} (0) = \frac{{vol}_{HS} (D_{sep}^{0} (ℂ^{2} \otimes ℂ^{2}))}{{vol}_{HS} (D^{0} (ℂ^{2} \otimes ℂ^{2}))} = \frac{\frac{π^{5}}{39916800}}{\frac{π^{5}}{9676800}} = \frac{8}{33} .$ P_{{\rm{sep}}}^{(2 \times 2)}(0) = {{{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{\rm{D}}_{{\rm{sep}}}^0\left( {{^2} \otimes {^2}} \right)} \right)} \over {{{{\mathop{\rm vol}\nolimits} }_{{\rm{HS}}}}\left( {{{\rm{D}}^0}\left( {{^2} \otimes {^2}} \right)} \right)}} = {{{{{\pi ^5}} \over {39916800}}} \over {{{{\pi ^5}} \over {9676800}}}} = {8 \over {33}}.

This completes the proof.

7.

Concluding Remarks

This study establishes a comprehensive geometric derivation of the exact separability probability 8/33 for two-qubit states under the Hilbert-Schmidt measure — a rigorously confirmed result previously inaccessible. By leveraging the intrinsic connection between Hilbert-Schmidt geometry and symplectic geometry formalized through the Duistermaat-Heckman (DH) measure, and explicitly computing the Hilbert-Schmidt volume of the full state space, volumes of critical substructures (flag manifolds and regular adjoint orbits), and the symplectic volumes of corresponding regular co-adjoint orbits, we integrate these perspectives to isolate the volume of separable states and rigorously confirm the 8/33 ratio.

This work achieves pedagogical clarity through a self-contained derivation of a fundamental quantum information constant and conceptual synthesis by unifying Hilbert-Schmidt geometry, symplectic mechanics, and representation-theoretic tools within a coherent probabilistic framework. Beyond providing an alternative pathway to this constant, our results elucidate the geometric structure governing the transition from classical correlation to quantum entanglement. Future work may extend this framework to higher-dimensional systems, alternative metrics (e.g., Bures), and generalized entanglement witnesses, underscoring its potential as a universal paradigm for probing the classical-quantum correlation boundary.

One Application of Duistermaat-Heckman Measure in Quantum Information Theory

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