Average entropy of a subsystem over a global unitary orbit of a mixed bipartite state

Abstract

We investigate the average entropy of a subsystem within a global unitary orbit of a given mixed bipartite state in the finite-dimensional space. Without working out the closed-form expression of such average entropy for the mixed state case, we provide an analytical lower bound for this average entropy. In deriving this analytical lower bound, we get some useful by-products of independent interest. We also apply these results to estimate average correlation along a global unitary orbit of a given mixed bipartite state. When the notion of von Neumann entropy is replaced by linear entropy, the similar problem can be considered also, and moreover the exact average linear entropy formula is derived for a subsystem over a global unitary orbit of a mixed bipartite state.

Appendices

Appendix 1: The proof of Lemma 2.1

In this paper, we will utilize some notion of matrix integral [4, 5, 22]. The formula in Lemma 2.1 is given firstly. A detailed reasoning is presented here.

The proof of Lemma 2.1

Firstly, we note that

$$\begin{aligned}&\left\langle i_1 \left| UAU^\dagger BUXU^\dagger CUDU^\dagger \right| i'_1 \right\rangle \\&\quad = \sum _{j_1,j'_1} \left\langle i_1 \left| U \right| j_1 \right\rangle \left\langle j_1 \left| AU^\dagger BUXU^\dagger CUD \right| j'_1 \right\rangle \left\langle j'_1 \left| U^\dagger \right| i'_1 \right\rangle \\&\quad =\sum _{i_2,j_1,j_2,i'_2,j'_1,j'_2} U_{i_1j_1}\overline{U}_{i'_1j'_1}\left\langle j_1 \left| A \right| j'_2 \right\rangle \left\langle j'_2 \left| U^\dagger \right| i'_2 \right\rangle \left\langle i'_2 \left| BUXU^\dagger C \right| i_2 \right\rangle \\&\qquad \left\langle i_2 \left| U \right| j_2 \right\rangle \left\langle j_2 \left| D \right| j'_1 \right\rangle \\&\quad =\sum _{i_2,j_1,j_2,i'_2,j'_1,j'_2} U_{i_1j_1}U_{i_2j_2}\overline{U}_{i'_1j'_1}\overline{U}_{i'_2j'_2}\left\langle j_1 \left| A \right| j'_2 \right\rangle \left\langle i'_2 \left| BUXU^\dagger C \right| i_2 \right\rangle \left\langle j_2 \left| D \right| j'_1 \right\rangle \\&\quad =\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} U_{i_1j_1}U_{i_2j_2}\overline{U}_{i'_1j'_1}\overline{U}_{i'_2j'_2}\\&\quad ~~~\times \left\langle j_1 \left| A \right| j'_2 \right\rangle \left\langle i'_2 \left| B \right| i_3 \right\rangle \left\langle i_3 \left| U \right| j_3 \right\rangle \left\langle j_3 \left| X \right| j'_3 \right\rangle \left\langle j'_3 \left| U^\dagger \right| i'_3 \right\rangle \left\langle i'_3 \left| C \right| i_2 \right\rangle \left\langle j_2 \left| D \right| j'_1 \right\rangle \\&\quad =\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} U_{i_1j_1}U_{i_2j_2}U_{i_3j_3}\overline{U}_{i'_1j'_1}\overline{U}_{i'_2j'_2}\overline{U}_{i'_3j'_3}\left\langle j_1 \left| A \right| j'_2 \right\rangle \\&\qquad \left\langle i'_2 \left| B \right| i_3 \right\rangle \left\langle j_3 \left| X \right| j'_3 \right\rangle \left\langle i'_3 \left| C \right| i_2 \right\rangle \left\langle j_2 \left| D \right| j'_1 \right\rangle . \end{aligned}$$

Then we have:

$$\begin{aligned}&\left\langle i_1 \left| \int UAU^\dagger BUXU^\dagger CUDU^\dagger dU \right| i'_1 \right\rangle \\&\quad =\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3}A_{j_1,j'_2} B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \\&\qquad \left( \int U_{i_1j_1}U_{i_2j_2}U_{i_3j_3}\overline{U}_{i'_1j'_1}\overline{U}_{i'_2j'_2}\overline{U}_{i'_3j'_3}dU\right) \\&\quad = \sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2} B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \\&\quad ~~~\times \left( \sum _{\pi ,\sigma \in S_3} \langle i_1 | i'_{\pi (1)}\rangle \langle i_2 | i'_{\pi (2)}\rangle \langle i_3 | i'_{\pi (3)}\rangle \langle j_1 | j'_{\sigma (1)}\rangle \langle j_2 | j'_{\sigma (2)}\rangle \langle j_3 | j'_{\sigma (3)}\rangle \mathrm {Wg}(\sigma \pi ^{-1})\right) \\&\quad = \sum _{\pi ,\sigma \in S_3} \mathrm {Wg}(\sigma \pi ^{-1})\\&\quad ~~~\times \left( \sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1}\right. \\&\quad \left. \langle i_1 | i'_{\pi (1)}\rangle \langle i_2 | i'_{\pi (2)}\rangle \langle i_3 | i'_{\pi (3)}\rangle \langle j_1 | j'_{\sigma (1)}\rangle \langle j_2 | j'_{\sigma (2)}\rangle \langle j_3 | j'_{\sigma (3)}\rangle \right) . \end{aligned}$$

In what follows, we compute this value step by step.

• (1). If $(\pi ,\sigma )=((1),(1))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_1\rangle \langle i_2 | i'_2\rangle \langle i_3 | i'_3\rangle \langle j_1 | j'_1\rangle \langle j_2 | j'_2\rangle \langle j_3 | j'_3\rangle \end{aligned}$$

  (4.4)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( BC\right) \langle i_1 | i'_{1}\rangle . \end{aligned}$$

  (4.5)

• (2). If $(\pi ,\sigma )=((1),(12))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_1\rangle \langle i_2 | i'_2\rangle \langle i_3 | i'_3\rangle \langle j_1 | j'_2\rangle \langle j_2 | j'_1\rangle \langle j_3 | j'_3\rangle \end{aligned}$$

  (4.6)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( BC\right) \langle i_1 | i'_{1}\rangle . \end{aligned}$$

  (4.7)

• (3). If $(\pi ,\sigma )=((1),(13))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_1\rangle \langle i_2 | i'_2\rangle \langle i_3 | i'_3\rangle \langle j_1 | j'_3\rangle \langle j_2 | j'_2\rangle \langle j_3 | j'_1\rangle \end{aligned}$$

  (4.8)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( ADX\right) {{\mathrm{Tr}}}_{}\left( BC\right) \langle i_1 | i'_1\rangle . \end{aligned}$$

  (4.9)

• (4). If $(\pi ,\sigma )=((1),(23))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_1\rangle \langle i_2 | i'_2\rangle \langle i_3 | i'_3\rangle \langle j_1 | j'_1\rangle \langle j_2 | j'_3\rangle \langle j_3 | j'_2\rangle \end{aligned}$$

  (4.10)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( DAX\right) {{\mathrm{Tr}}}_{}\left( BC\right) \langle i_1 | i'_1\rangle . \end{aligned}$$

  (4.11)

• (5). If $(\pi ,\sigma )=((1),(123))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_1\rangle \langle i_2 | i'_2\rangle \langle i_3 | i'_3\rangle \langle j_1 | j'_2\rangle \langle j_2 | j'_3\rangle \langle j_3 | j'_1\rangle \end{aligned}$$

  (4.12)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) {{\mathrm{Tr}}}_{}\left( BC\right) \langle i_1 | i'_1\rangle . \end{aligned}$$

  (4.13)

• (6). If $(\pi ,\sigma )=((1),(132))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_1\rangle \langle i_2 | i'_2\rangle \langle i_3 | i'_3\rangle \langle j_1 | j'_3\rangle \langle j_2 | j'_1\rangle \langle j_3 | j'_2\rangle \end{aligned}$$

  (4.14)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) {{\mathrm{Tr}}}_{}\left( BC\right) \langle i_1 | i'_1\rangle . \end{aligned}$$

  (4.15)

• (7). If $(\pi ,\sigma )=((12),(1))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_2\rangle \langle i_2 | i'_1\rangle \langle i_3 | i'_3\rangle \langle j_1 | j'_1\rangle \langle j_2 | j'_2\rangle \langle j_3 | j'_3\rangle \end{aligned}$$

  (4.16)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) \left\langle i_1 \left| BC \right| i'_1 \right\rangle . \end{aligned}$$

  (4.17)

• (8). If $(\pi ,\sigma )=((12),(12))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_2\rangle \langle i_2 | i'_1\rangle \langle i_3 | i'_3\rangle \langle j_1 | j'_2\rangle \langle j_2 | j'_1\rangle \langle j_3 | j'_3\rangle \end{aligned}$$

  (4.18)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) \left\langle i_1 \left| BC \right| i'_1 \right\rangle . \end{aligned}$$

  (4.19)

• (9). If $(\pi ,\sigma )=((12),(13))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_2\rangle \langle i_2 | i'_1\rangle \langle i_3 | i'_3\rangle \langle j_1 | j'_3\rangle \langle j_2 | j'_2\rangle \langle j_3 | j'_1\rangle \end{aligned}$$

  (4.20)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( ADX\right) \left\langle i_1 \left| BC \right| i'_1 \right\rangle . \end{aligned}$$

  (4.21)

• (10). If $(\pi ,\sigma )=((12),(23))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_2\rangle \langle i_2 | i'_1\rangle \langle i_3 | i'_3\rangle \langle j_1 | j'_1\rangle \langle j_2 | j'_3\rangle \langle j_3 | j'_2\rangle \end{aligned}$$

  (4.22)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( DAX\right) \left\langle i_1 \left| BC \right| i'_1 \right\rangle . \end{aligned}$$

  (4.23)

• (11). If $(\pi ,\sigma )=((12),(123))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_2\rangle \langle i_2 | i'_1\rangle \langle i_3 | i'_3\rangle \langle j_1 | j'_2\rangle \langle j_2 | j'_3\rangle \langle j_3 | j'_1\rangle \end{aligned}$$

  (4.24)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) \left\langle i_1 \left| BC \right| i'_1 \right\rangle . \end{aligned}$$

  (4.25)

• (12). If $(\pi ,\sigma )=((12),(132))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_2\rangle \langle i_2 | i'_1\rangle \langle i_3 | i'_3\rangle \langle j_1 | j'_3\rangle \langle j_2 | j'_1\rangle \langle j_3 | j'_2\rangle \end{aligned}$$

  (4.26)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) \left\langle i_1 \left| BC \right| i'_1 \right\rangle . \end{aligned}$$

  (4.27)

• (13). If $(\pi ,\sigma )=((13),(1))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_3\rangle \langle i_2 | i'_2\rangle \langle i_3 | i'_1\rangle \langle j_1 | j'_1\rangle \langle j_2 | j'_2\rangle \langle j_3 | j'_3\rangle \end{aligned}$$

  (4.28)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) \left\langle i_1 \left| CB \right| i'_1 \right\rangle . \end{aligned}$$

  (4.29)

• (14). If $(\pi ,\sigma )=((13),(12))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_3\rangle \langle i_2 | i'_2\rangle \langle i_3 | i'_1\rangle \langle j_1 | j'_2\rangle \langle j_2 | j'_1\rangle \langle j_3 | j'_3\rangle \end{aligned}$$

  (4.30)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) \left\langle i_1 \left| CB \right| i'_1 \right\rangle . \end{aligned}$$

  (4.31)

• (15). If $(\pi ,\sigma )=((13),(13))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_3\rangle \langle i_2 | i'_2\rangle \langle i_3 | i'_1\rangle \langle j_1 | j'_3\rangle \langle j_2 | j'_2\rangle \langle j_3 | j'_1\rangle \end{aligned}$$

  (4.32)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( ADX\right) \left\langle i_1 \left| CB \right| i'_1 \right\rangle . \end{aligned}$$

  (4.33)

• (16). If $(\pi ,\sigma )=((13),(23))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_3\rangle \langle i_2 | i'_2\rangle \langle i_3 | i'_1\rangle \langle j_1 | j'_1\rangle \langle j_2 | j'_3\rangle \langle j_3 | j'_2\rangle \end{aligned}$$

  (4.34)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( DAX\right) \left\langle i_1 \left| CB \right| i'_1 \right\rangle . \end{aligned}$$

  (4.35)

• (17). If $(\pi ,\sigma )=((13),(123))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_3\rangle \langle i_2 | i'_2\rangle \langle i_3 | i'_1\rangle \langle j_1 | j'_2\rangle \langle j_2 | j'_3\rangle \langle j_3 | j'_1\rangle \end{aligned}$$

  (4.36)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) \left\langle i_1 \left| CB \right| i'_1 \right\rangle . \end{aligned}$$

  (4.37)

• (18). If $(\pi ,\sigma )=((13),(132))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_3\rangle \langle i_2 | i'_2\rangle \langle i_3 | i'_1\rangle \langle j_1 | j'_3\rangle \langle j_2 | j'_1\rangle \langle j_3 | j'_2\rangle \end{aligned}$$

  (4.38)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) \left\langle i_1 \left| CB \right| i'_1 \right\rangle . \end{aligned}$$

  (4.39)

• (19). If $(\pi ,\sigma )=((23),(1))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_1\rangle \langle i_2 | i'_3\rangle \langle i_3 | i'_2\rangle \langle j_1 | j'_1\rangle \langle j_2 | j'_2\rangle \langle j_3 | j'_3\rangle \end{aligned}$$

  (4.40)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) \langle i_1 | i'_{1}\rangle . \end{aligned}$$

  (4.41)

• (20). If $(\pi ,\sigma )=((23),(12))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_1\rangle \langle i_2 | i'_3\rangle \langle i_3 | i'_2\rangle \langle j_1 | j'_2\rangle \langle j_2 | j'_1\rangle \langle j_3 | j'_3\rangle \end{aligned}$$

  (4.42)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) \langle i_1 | i'_{1}\rangle . \end{aligned}$$

  (4.43)

• (21). If $(\pi ,\sigma )=((23),(13))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_1\rangle \langle i_2 | i'_3\rangle \langle i_3 | i'_2\rangle \langle j_1 | j'_3\rangle \langle j_2 | j'_2\rangle \langle j_3 | j'_1\rangle \end{aligned}$$

  (4.44)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( ADX\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) \langle i_1 | i'_1\rangle . \end{aligned}$$

  (4.45)

• (22). If $(\pi ,\sigma )=((23),(23))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_1\rangle \langle i_2 | i'_3\rangle \langle i_3 | i'_2\rangle \langle j_1 | j'_1\rangle \langle j_2 | j'_3\rangle \langle j_3 | j'_2\rangle \end{aligned}$$

  (4.46)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( DAX\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) \langle i_1 | i'_1\rangle . \end{aligned}$$

  (4.47)

• (23). If $(\pi ,\sigma )=((23),(123))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_1\rangle \langle i_2 | i'_3\rangle \langle i_3 | i'_2\rangle \langle j_1 | j'_2\rangle \langle j_2 | j'_3\rangle \langle j_3 | j'_1\rangle \end{aligned}$$

  (4.48)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) \langle i_1 | i'_1\rangle . \end{aligned}$$

  (4.49)

• (24). If $(\pi ,\sigma )=((23),(132))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_1\rangle \langle i_2 | i'_3\rangle \langle i_3 | i'_2\rangle \langle j_1 | j'_3\rangle \langle j_2 | j'_1\rangle \langle j_3 | j'_2\rangle \end{aligned}$$

  (4.50)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) \langle i_1 | i'_1\rangle . \end{aligned}$$

  (4.51)

• (25). If $(\pi ,\sigma )=((123),(1))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_2\rangle \langle i_2 | i'_3\rangle \langle i_3 | i'_1\rangle \langle j_1 | j'_1\rangle \langle j_2 | j'_2\rangle \langle j_3 | j'_3\rangle \end{aligned}$$

  (4.52)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( C\right) \left\langle i_1 \left| B \right| i'_1 \right\rangle . \end{aligned}$$

  (4.53)

• (26). If $(\pi ,\sigma )=((123),(12))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_2\rangle \langle i_2 | i'_3\rangle \langle i_3 | i'_1\rangle \langle j_1 | j'_2\rangle \langle j_2 | j'_1\rangle \langle j_3 | j'_3\rangle \end{aligned}$$

  (4.54)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( C\right) \left\langle i_1 \left| B \right| i'_1 \right\rangle . \end{aligned}$$

  (4.55)

• (27). If $(\pi ,\sigma )=((123),(13))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_2\rangle \langle i_2 | i'_3\rangle \langle i_3 | i'_1\rangle \langle j_1 | j'_3\rangle \langle j_2 | j'_2\rangle \langle j_3 | j'_1\rangle \end{aligned}$$

  (4.56)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( ADX\right) {{\mathrm{Tr}}}_{}\left( C\right) \left\langle i_1 \left| B \right| i'_1 \right\rangle . \end{aligned}$$

  (4.57)

• (28). If $(\pi ,\sigma )=((123),(23))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_2\rangle \langle i_2 | i'_3\rangle \langle i_3 | i'_1\rangle \langle j_1 | j'_1\rangle \langle j_2 | j'_3\rangle \langle j_3 | j'_2\rangle \end{aligned}$$

  (4.58)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( DAX\right) {{\mathrm{Tr}}}_{}\left( C\right) \left\langle i_1 \left| B \right| i'_1 \right\rangle . \end{aligned}$$

  (4.59)

• (29). If $(\pi ,\sigma )=((123),(123))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_2\rangle \langle i_2 | i'_3\rangle \langle i_3 | i'_1\rangle \langle j_1 | j'_2\rangle \langle j_2 | j'_3\rangle \langle j_3 | j'_1\rangle \end{aligned}$$

  (4.60)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) {{\mathrm{Tr}}}_{}\left( C\right) \left\langle i_1 \left| B \right| i'_1 \right\rangle . \end{aligned}$$

  (4.61)

• (30). If $(\pi ,\sigma )=((123),(132))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_2\rangle \langle i_2 | i'_3\rangle \langle i_3 | i'_1\rangle \langle j_1 | j'_3\rangle \langle j_2 | j'_1\rangle \langle j_3 | j'_2\rangle \end{aligned}$$

  (4.62)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) {{\mathrm{Tr}}}_{}\left( C\right) \left\langle i_1 \left| B \right| i'_1 \right\rangle . \end{aligned}$$

  (4.63)

• (31). If $(\pi ,\sigma )=((132),(1))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_3\rangle \langle i_2 | i'_1\rangle \langle i_3 | i'_2\rangle \langle j_1 | j'_1\rangle \langle j_2 | j'_2\rangle \langle j_3 | j'_3\rangle \end{aligned}$$

  (4.64)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( B\right) \left\langle i_1 \left| C \right| i'_1 \right\rangle . \end{aligned}$$

  (4.65)

• (32). If $(\pi ,\sigma )=((132),(12))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_3\rangle \langle i_2 | i'_1\rangle \langle i_3 | i'_2\rangle \langle j_1 | j'_2\rangle \langle j_2 | j'_1\rangle \langle j_3 | j'_3\rangle \end{aligned}$$

  (4.66)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( B\right) \left\langle i_1 \left| C \right| i'_1 \right\rangle . \end{aligned}$$

  (4.67)

• (33). If $(\pi ,\sigma )=((132),(13))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_3\rangle \langle i_2 | i'_1\rangle \langle i_3 | i'_2\rangle \langle j_1 | j'_3\rangle \langle j_2 | j'_2\rangle \langle j_3 | j'_1\rangle \end{aligned}$$

  (4.68)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( ADX\right) {{\mathrm{Tr}}}_{}\left( B\right) \left\langle i_1 \left| C \right| i'_1 \right\rangle . \end{aligned}$$

  (4.69)

• (34). If $(\pi ,\sigma )=((132),(23))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_3\rangle \langle i_2 | i'_1\rangle \langle i_3 | i'_2\rangle \langle j_1 | j'_1\rangle \langle j_2 | j'_3\rangle \langle j_3 | j'_2\rangle \end{aligned}$$

  (4.70)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( DAX\right) {{\mathrm{Tr}}}_{}\left( B\right) \left\langle i_1 \left| C \right| i'_1 \right\rangle . \end{aligned}$$

  (4.71)

• (35). If $(\pi ,\sigma )=((132),(123))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_3\rangle \langle i_2 | i'_1\rangle \langle i_3 | i'_2\rangle \langle j_1 | j'_2\rangle \langle j_2 | j'_3\rangle \langle j_3 | j'_1\rangle \end{aligned}$$

  (4.72)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) {{\mathrm{Tr}}}_{}\left( B\right) \left\langle i_1 \left| C \right| i'_1 \right\rangle . \end{aligned}$$

  (4.73)

• (36). If $(\pi ,\sigma )=((132),(132))$, then

  $$\begin{aligned}&\sum _{i_2,i_3,j_1,j_2,j_3,i'_2,i'_3,j'_1,j'_2,j'_3} A_{j_1,j'_2}B_{i'_2,i_3}X_{j_3,j'_3}C_{i'_3,i_2}D_{j_2,j'_1} \nonumber \\&\quad \langle i_1 | i'_3\rangle \langle i_2 | i'_1\rangle \langle i_3 | i'_2\rangle \langle j_1 | j'_3\rangle \langle j_2 | j'_1\rangle \langle j_3 | j'_2\rangle \end{aligned}$$

  (4.74)
  $$\begin{aligned}&\quad = {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) {{\mathrm{Tr}}}_{}\left( B\right) \left\langle i_1 \left| C \right| i'_1 \right\rangle . \end{aligned}$$

  (4.75)

  Combing the 36 cases together gives the desired conclusion:

  $$\begin{aligned}&\int UAU^\dagger BUXU^\dagger CUDU^\dagger \mathrm {d}\mu (U)= \mu _1\cdot \mathbb {1}_d+\mu _2\cdot BC+\mu _3\cdot CB \nonumber \\&\quad +\,\mu _4\cdot B + \mu _5\cdot C, \end{aligned}$$

  (4.76)

  where the coefficients $\mu _j(j=1,\ldots ,5)$ are given below:

  $$\begin{aligned} \mu _1:= & {} \mathrm {Wg}(1,1,1){{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( BC\right) \nonumber \\&+\,\mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( BC\right) \nonumber \\&+\,\mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( ADX\right) {{\mathrm{Tr}}}_{}\left( BC\right) \nonumber \\&+\,\mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( DAX\right) {{\mathrm{Tr}}}_{}\left( BC\right) \nonumber \\&+\,\mathrm {Wg}(3){{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) {{\mathrm{Tr}}}_{}\left( BC\right) \nonumber \\&+\,\mathrm {Wg}(3){{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) {{\mathrm{Tr}}}_{}\left( BC\right) \nonumber \\&+\,\mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) \nonumber \\&+\,\mathrm {Wg}(3){{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) \nonumber \\&+\,\mathrm {Wg}(3){{\mathrm{Tr}}}_{}\left( ADX\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) \nonumber \\&+\,\mathrm {Wg}(1,1,1){{\mathrm{Tr}}}_{}\left( DAX\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) \nonumber \\&+\,\mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) \nonumber \\&+\,\mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) , \end{aligned}$$

  (4.77)
  $$\begin{aligned} \mu _2:= & {} \mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) \nonumber \\&+\,\mathrm {Wg}(1,1,1){{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) +\mathrm {Wg}(3){{\mathrm{Tr}}}_{}\left( ADX\right) \nonumber \\&+\,\mathrm {Wg}(3){{\mathrm{Tr}}}_{}\left( DAX\right) +\mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) \nonumber \\&+\,\mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) , \end{aligned}$$

  (4.78)
  $$\begin{aligned} \mu _3:= & {} \mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) \nonumber \\&+\,\mathrm {Wg}(3){{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) +\mathrm {Wg}(1,1,1){{\mathrm{Tr}}}_{}\left( ADX\right) \nonumber \\&+\,\mathrm {Wg}(3){{\mathrm{Tr}}}_{}\left( DAX\right) +\mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) \nonumber \\&+\,\mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) , \end{aligned}$$

  (4.79)
  $$\begin{aligned} \mu _4:= & {} \mathrm {Wg}(3){{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( C\right) \nonumber \\&+\,\mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( C\right) \nonumber \\&+\,\mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( ADX\right) {{\mathrm{Tr}}}_{}\left( C\right) \nonumber \\&+\,\mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( DAX\right) {{\mathrm{Tr}}}_{}\left( C\right) \nonumber \\&+\,\mathrm {Wg}(1,1,1){{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) {{\mathrm{Tr}}}_{}\left( C\right) \nonumber \\&+\,\mathrm {Wg}(3){{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) {{\mathrm{Tr}}}_{}\left( C\right) , \end{aligned}$$

  (4.80)
  $$\begin{aligned} \mu _5:= & {} \mathrm {Wg}(3){{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( B\right) \nonumber \\&+\,\mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( B\right) \nonumber \\&+\,\mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( ADX\right) {{\mathrm{Tr}}}_{}\left( B\right) \nonumber \\&+\,\mathrm {Wg}(2,1){{\mathrm{Tr}}}_{}\left( DAX\right) {{\mathrm{Tr}}}_{}\left( B\right) \nonumber \\&+\,\mathrm {Wg}(3){{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) {{\mathrm{Tr}}}_{}\left( B\right) \nonumber \\&+\,\mathrm {Wg}(1,1,1){{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) {{\mathrm{Tr}}}_{}\left( B\right) . \end{aligned}$$

  (4.81)

We are done. $\square$

Remark 4.1

$\mathrm {Wg}:=\frac{1}{(k!)^2}\sum _{\lambda \vdash k}\frac{\dim (\mathbf {P}_\lambda )^2}{\dim (\mathbf {Q}_\lambda )}\chi _\lambda$ is called the Weingarten function [4]. In particular, for $\lambda \vdash 3$, we have:

$$\begin{aligned} \mathrm {Wg}(1,1,1)= & {} \frac{d^2-2}{d(d^2-1)(d^2-4)}=\left( d-\frac{2}{d}\right) \cdot \frac{1}{N_d}, \end{aligned}$$

(4.82)

$$\begin{aligned} \mathrm {Wg}(2,1)= & {} -\frac{1}{(d^2-1)(d^2-4)}=(-1)\cdot \frac{1}{N_d}, \end{aligned}$$

(4.83)

$$\begin{aligned} \mathrm {Wg}(3)= & {} \frac{2}{d(d^2-1)(d^2-4)} = \frac{2}{d}\cdot \frac{1}{N_d}, \end{aligned}$$

(4.84)

where $N_d=(d^2-1)(d^2-4)$. With these coefficients, we then have

$$\begin{aligned} N_d\mu _1:= & {} \left( d-\frac{2}{d}\right) {{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( BC\right) \nonumber \\&+\,\left( d-\frac{2}{d}\right) {{\mathrm{Tr}}}_{}\left( DAX\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) \nonumber \\&+\,\frac{2}{d}{{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) {{\mathrm{Tr}}}_{}\left( BC\right) \nonumber \\&+\,\frac{2}{d}{{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) {{\mathrm{Tr}}}_{}\left( BC\right) +\frac{2}{d}{{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) \nonumber \\&+\,\frac{2}{d}{{\mathrm{Tr}}}_{}\left( ADX\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) \nonumber \\&-{{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( BC\right) \nonumber \\&-{{\mathrm{Tr}}}_{}\left( ADX\right) {{\mathrm{Tr}}}_{}\left( BC\right) -{{\mathrm{Tr}}}_{}\left( DAX\right) {{\mathrm{Tr}}}_{}\left( BC\right) \nonumber \\&-{{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) -{{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) \nonumber \\&-{{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) {{\mathrm{Tr}}}_{}\left( B\right) {{\mathrm{Tr}}}_{}\left( C\right) , \end{aligned}$$

(4.85)

$$\begin{aligned} N_d\mu _2:= & {} \left( d-\frac{2}{d}\right) {{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) \nonumber \\&+\,\frac{2}{d}{{\mathrm{Tr}}}_{}\left( ADX\right) +\frac{2}{d}{{\mathrm{Tr}}}_{}\left( DAX\right) \nonumber \\&-{{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) -{{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) -{{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) , \end{aligned}$$

(4.86)

$$\begin{aligned} N_d\mu _3:= & {} \frac{2}{d}{{\mathrm{Tr}}}_{}\left( DAX\right) + \frac{2}{d}{{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) \nonumber \\&+\,\left( d-\frac{2}{d}\right) {{\mathrm{Tr}}}_{}\left( ADX\right) \nonumber \\&-{{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) -{{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) -{{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) , \end{aligned}$$

(4.87)

$$\begin{aligned} N_d\mu _4:= & {} \left( d-\frac{2}{d}\right) {{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) {{\mathrm{Tr}}}_{}\left( C\right) +\frac{2}{d}{{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( C\right) \nonumber \\&+\,\frac{2}{d}{{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) {{\mathrm{Tr}}}_{}\left( C\right) \nonumber \\&- {{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( C\right) - {{\mathrm{Tr}}}_{}\left( ADX\right) {{\mathrm{Tr}}}_{}\left( C\right) \nonumber \\&- {{\mathrm{Tr}}}_{}\left( DAX\right) {{\mathrm{Tr}}}_{}\left( C\right) , \end{aligned}$$

(4.88)

$$\begin{aligned} N_d\mu _5:= & {} \frac{2}{d}{{\mathrm{Tr}}}_{}\left( AD\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( B\right) +\frac{2}{d}{{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( DX\right) {{\mathrm{Tr}}}_{}\left( B\right) \nonumber \\&+\,\left( d-\frac{2}{d}\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( AX\right) {{\mathrm{Tr}}}_{}\left( B\right) \nonumber \\&- {{\mathrm{Tr}}}_{}\left( A\right) {{\mathrm{Tr}}}_{}\left( D\right) {{\mathrm{Tr}}}_{}\left( X\right) {{\mathrm{Tr}}}_{}\left( B\right) \nonumber \\&- {{\mathrm{Tr}}}_{}\left( ADX\right) {{\mathrm{Tr}}}_{}\left( B\right) - {{\mathrm{Tr}}}_{}\left( DAX\right) {{\mathrm{Tr}}}_{}\left( B\right) . \end{aligned}$$

(4.89)

Appendix 2: The proof of Theorem 2.2

By Lemma 2.1,

$$\begin{aligned}&\int U^\dagger M_{ij}UTU^\dagger M_{jl}UTU^\dagger M_{li}U\mathrm {d}\mu (U) \\&\quad = f(i,j,l)\cdot \mathbb {1}_d + g(i,j,l)\cdot T + h(i,j,l)\cdot T^2, \end{aligned}$$

where

$$\begin{aligned} f(i,j,l)= & {} \mathrm {Wg}(1,1,1)d^2_A\delta _{jl}{{\mathrm{Tr}}}_{}\left( T^2\right) \\&+\,\mathrm {Wg}(2,1)d^3_A\delta _{ij}\delta _{jl}\delta _{li}{{\mathrm{Tr}}}_{}\left( T^2\right) \\&+\,\mathrm {Wg}(2,1)d_A\delta _{ij}\delta _{jl}\delta _{li}{{\mathrm{Tr}}}_{}\left( T^2\right) \\&+\,\mathrm {Wg}(2,1)d_A{{\mathrm{Tr}}}_{}\left( T^2\right) +\mathrm {Wg}(3)d^2_A\delta _{ij}{{\mathrm{Tr}}}_{}\left( T^2\right) \\&+\,\mathrm {Wg}(3)d^2_A\delta _{il}{{\mathrm{Tr}}}_{}\left( T^2\right) \\&+\,\mathrm {Wg}(2,1)d^2_A\delta _{jl}{{\mathrm{Tr}}}_{}\left( T\right) ^2+\mathrm {Wg}(3)d^3_A\delta _{ij}\delta _{jl}\delta _{li}{{\mathrm{Tr}}}_{}\left( T\right) ^2 \\&+\,\mathrm {Wg}(3)d_A\delta _{ij}\delta _{jl}\delta _{li}{{\mathrm{Tr}}}_{}\left( T\right) ^2\\&+\,\mathrm {Wg}(1,1,1)d_A{{\mathrm{Tr}}}_{}\left( T\right) ^2+\mathrm {Wg}(2,1)d^2_A\delta _{ij}{{\mathrm{Tr}}}_{}\left( T\right) ^2 \\&+\,\mathrm {Wg}(2,1)d^2_A\delta _{il}{{\mathrm{Tr}}}_{}\left( T\right) ^2,\\ g(i,j,l)= & {} \mathrm {Wg}(3)d^2_A\delta _{jl}{{\mathrm{Tr}}}_{}\left( T\right) \\&+\,\mathrm {Wg}(2,1)d^3_A\delta _{ij}\delta _{jl}\delta _{li}{{\mathrm{Tr}}}_{}\left( T\right) \\&+\,\mathrm {Wg}(2,1)d_A\delta _{ij}\delta _{jl}\delta _{li}{{\mathrm{Tr}}}_{}\left( T\right) \\&+\,\mathrm {Wg}(2,1)d_A{{\mathrm{Tr}}}_{}\left( T\right) +\mathrm {Wg}(1,1,1)d^2_A\delta _{ij}{{\mathrm{Tr}}}_{}\left( T\right) \\&+\,\mathrm {Wg}(3)d^2_A\delta _{il}{{\mathrm{Tr}}}_{}\left( T\right) \\&+\,\mathrm {Wg}(3)d^2_A\delta _{jl}{{\mathrm{Tr}}}_{}\left( T\right) \\&+\,\mathrm {Wg}(2,1)d^3_A\delta _{ij}\delta _{jl}\delta _{li}{{\mathrm{Tr}}}_{}\left( T\right) \\&+\,\mathrm {Wg}(2,1)d_A\delta _{ij}\delta _{jl}\delta _{li}{{\mathrm{Tr}}}_{}\left( T\right) \\&+\,\mathrm {Wg}(2,1)d_A{{\mathrm{Tr}}}_{}\left( T\right) +\mathrm {Wg}(3)d^2_A\delta _{ij}{{\mathrm{Tr}}}_{}\left( T\right) \\&+\,\mathrm {Wg}(1,1,1)d^2_A\delta _{il}{{\mathrm{Tr}}}_{}\left( T\right) ,\\ h(i,j,l)= & {} \mathrm {Wg}(2,1)d^2_A\delta _{jl} \\&+\,\mathrm {Wg}(1,1,1)d^3_A\delta _{ij}\delta _{jl}\delta _{li} \\&+\,\mathrm {Wg}(3)d_A\delta _{ij}\delta _{jl}\delta _{li}\\&+\,\mathrm {Wg}(3)d_A+\mathrm {Wg}(2,1)d^2_A\delta _{ij} \\&+\,\mathrm {Wg}(2,1)d^2_A\delta _{il}, \\&+\,\mathrm {Wg}(2,1)d^2_A\delta _{jl} \\&+\,\mathrm {Wg}(3)d^3_A\delta _{ij}\delta _{jl}\delta _{li} \\&+\,\mathrm {Wg}(1,1,1)d_A\delta _{ij}\delta _{jl}\delta _{li}\\&+\,\mathrm {Wg}(3)d_A+\mathrm {Wg}(2,1)d^2_A\delta _{ij} \\&+\,\mathrm {Wg}(2,1)d^2_A\delta _{il}. \end{aligned}$$

Note that the meaning of the notation $\mathrm {Wg}(*)$ can be found in Appendix. Thus for $f:=\sum ^{d_B}_{i,j,l=1}f(i,j,l)$, $g:=\sum ^{d_B}_{i,j,l=1}g(i,j,l)$, and $h:=\sum ^{d_B}_{i,j,l=1}h(i,j,l)$, we have

$$\begin{aligned} f= & {} \left( [\mathrm {Wg}(1,1,1)+2\mathrm {Wg}(3)]d^2+ \mathrm {Wg}(2,1)(dd^2_A+d+dd^2_B)\right) {{\mathrm{Tr}}}_{}\left( T^2\right) \\&+\left( 3\mathrm {Wg}(2,1)d^2+\mathrm {Wg}(3)(dd^2_A+d)+\mathrm {Wg}(1,1,1)dd^2_B\right) {{\mathrm{Tr}}}_{}\left( T\right) ^2,\\ g= & {} \left( [4\mathrm {Wg}(3) + 2\mathrm {Wg}(1,1,1)]d^2+ 2\mathrm {Wg}(2,1)(dd^2_A + d +dd^2_B)\right) {{\mathrm{Tr}}}_{}\left( T\right) ,\\ h= & {} 6\mathrm {Wg}(2,1)d^2 + [\mathrm {Wg}(1,1,1)+\mathrm {Wg}(3)]dd^2_A \\&+\,[\mathrm {Wg}(1,1,1)+\mathrm {Wg}(3)]d+2\mathrm {Wg}(3)dd^2_B. \end{aligned}$$

Hence, for $T=\mathbb {1}_A\otimes \mathbb {1}_B/d_B - \rho _{AB}$, ${{\mathrm{Tr}}}_{}\left( T\right) =d_A-1$ and ${{\mathrm{Tr}}}_{}\left( T^2\right) =\frac{d_A-2}{d_B}+{{\mathrm{Tr}}}_{}\left( \rho ^2_{AB}\right)$, then

$$\begin{aligned} f= & {} \frac{d\left( d^2-d^2_A-d^2_B+1\right) }{(d^2-1)(d^2-4)}{{\mathrm{Tr}}}_{}\left( T^2\right) \nonumber \\&+\,\frac{d^2\left( d^2_B-3\right) +2\left( d^2_A-d^2_B+1\right) }{(d^2-1)(d^2-4)}{{\mathrm{Tr}}}_{}\left( T\right) ^2,\end{aligned}$$

(4.90)

$$\begin{aligned}= & {} \frac{d_A\left( d^2_A-1\right) \left( d^2_B-1\right) }{(d^2-1)(d^2-4)}\left( d_A+d_B{{\mathrm{Tr}}}_{}\left( \rho ^2_{AB}\right) -2\right) \end{aligned}$$

(4.91)

$$\begin{aligned}&+\,\frac{\left( d^2-2d^2_A-2\right) \left( d^2_B-1\right) }{(d^2-1)(d^2-4)}(d_A-1)^2,\end{aligned}$$

(4.92)

$$\begin{aligned} g= & {} \frac{2d\left( d^2-d^2_A-d^2_B+1\right) }{(d^2-1)(d^2-4)}{{\mathrm{Tr}}}_{}\left( T\right) \nonumber \\= & {} \frac{2d\left( d_A-1\right) \left( d^2_A-1\right) \left( d^2_B-1\right) }{(d^2-1)(d^2-4)},\end{aligned}$$

(4.93)

$$\begin{aligned} h= & {} \frac{d^2\left( d^2_A-5\right) +4d^2_B}{(d^2-1)(d^2-4)} = \frac{\left( d^2_A-1\right) \left( d^2_A-4\right) d^2_B}{(d^2-1)(d^2-4)}. \end{aligned}$$

(4.94)

Therefore,

$$\begin{aligned} \int U^\dagger \left[ \Gamma (UTU^\dagger )\right] ^2U\mathrm {d}\mu (U) =f\cdot \mathbb {1}_d + g\cdot T + h\cdot T^2, \end{aligned}$$

(4.95)

implying that

$$\begin{aligned} a_2= & {} f + g\cdot {{\mathrm{Tr}}}_{}\left( \rho _{AB}T\right) + h\cdot {{\mathrm{Tr}}}_{}\left( \rho _{AB}T^2\right) \nonumber \\= & {} f + g\cdot \left( \frac{1}{d_B}-{{\mathrm{Tr}}}_{}\left( \rho ^2_{AB}\right) \right) + h\cdot \left( \frac{1}{d^2_B} - \frac{2}{d_B}{{\mathrm{Tr}}}_{}\left( \rho ^2_{AB}\right) +{{\mathrm{Tr}}}_{}\left( \rho ^3_{AB}\right) \right) \nonumber \\= & {} \left( f + g\frac{1}{d_B} + h\frac{1}{d^2_B}\right) - \left( g+h\frac{2}{d_B}\right) {{\mathrm{Tr}}}_{}\left( \rho ^2_{AB}\right) + h{{\mathrm{Tr}}}_{}\left( \rho ^3_{AB}\right) . \end{aligned}$$

(4.96)

We make further analysis of the term $a_n$, although we have already known the fact that $\lim _{n\rightarrow \infty }a_n=0$. Let $\varphi _\rho (X)={{\mathrm{Tr}}}_{}\left( \rho X\right)$. Apparently, $\varphi _\rho$ is a positive unital linear mapping (in fact, it is a positive unital linear functional from the set of $n\times n$ Hermitian matrices to $\mathbb {R}$). It is easily seen that $f(x)=x^n$ is a convex function from $\mathbb {R}$ to $\mathbb {R}$. By using [10, Theorem 4.15, pp 147], we see that

$$\begin{aligned} a_n= & {} \varphi _\rho \left( \int \left[ U^\dagger \Gamma (UTU^\dagger )U\right] ^n\mathrm {d}\mu (U)\right) = \varphi _\rho \left( \int f(U^\dagger \Gamma (UTU^\dagger )U)\mathrm {d}\mu (U)\right) \\= & {} \int \mathrm {d}\mu (U)(\varphi _\rho \circ f)(U^\dagger \Gamma (UTU^\dagger )U)\geqslant \int \mathrm {d}\mu (U) f[\varphi _\rho (U^\dagger \Gamma (UTU^\dagger )U)]\\\geqslant & {} \int \mathrm {d}\mu (U) \left[ {{\mathrm{Tr}}}_{}\left( \rho U^\dagger \Gamma (UTU^\dagger )U\right) \right] ^n\geqslant \left( {{\mathrm{Tr}}}_{}\left( \rho \int \mathrm {d}\mu (U)U^\dagger \Gamma (UTU^\dagger )U\right) \right) ^n. \end{aligned}$$

By (2.4), we have

$$\begin{aligned}&{{\mathrm{Tr}}}_{}\left( \rho \int \mathrm {d}\mu (U)U^\dagger \Gamma (UTU^\dagger )U\right) = \frac{d_A-1}{d^2-1} \\&\quad \left[ (1+dd_B) - (d+d_B){{\mathrm{Tr}}}_{}\left( \rho ^2_{AB}\right) \right] =a_1. \end{aligned}$$

Thus

$$\begin{aligned} a_n\geqslant & {} a^n_1. \end{aligned}$$

(4.97)

Then

$$\begin{aligned} a_1=\frac{(d_A-1)(d_B-1)}{d+1}+\frac{(d_A-1)(d+d_B)}{d^2-1}\mathrm {S}_L(\rho _{AB}), \end{aligned}$$

where $\mathrm {S}_L(\rho _{AB})\in \left[ 0,1-\frac{1}{d}\right]$. Clearly

$$\begin{aligned} \frac{(d_A-1)(d_B-1)}{d+1}\leqslant a_1\leqslant \frac{(d_A-1)\left( 1+\frac{1}{d_A}\right) }{d+1}<1. \end{aligned}$$

(4.98)

Now,

$$\begin{aligned} \int \mathrm {S}(\rho '_A)\mathrm {d}\mu (U)=\sum ^\infty _{n=1}\frac{a_n}{n}\geqslant \sum ^\infty _{n=1}\frac{a^n_1}{n} = -\ln (1-a_1). \end{aligned}$$

(4.99)

We see from the above lower bound, i.e., $-\ln (1-a_1)$, that when the purity of $\rho _{AB}$ decreases, $a_1$ increases. Hence such lower bound will be tighter.

Appendix 3: The proof of Proposition 3.2

Clearly, the first inequality is easily obtained

$$\begin{aligned} \int I(A:B)_{\rho '}\mathrm {d}\mu (U)\geqslant & {} \mathrm {S}\left( \rho _{AB}||\int \ln \left[ U^\dagger (\rho '_A\otimes \rho '_B)U\right] \mathrm {d}\mu (U)\right) \end{aligned}$$

(4.100)

$$\begin{aligned}= & {} \mathrm {S}\left( \rho _{AB}||c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right) . \end{aligned}$$

(4.101)

In order to show the second inequality, note that, for any two density matrices $\rho$ and $\sigma$,

$$\begin{aligned} \mathrm {F}(\rho ,\sigma )\geqslant {{\mathrm{Tr}}}_{}\left( \sqrt{\rho }\sqrt{\sigma }\right) \geqslant {{\mathrm{Tr}}}_{}\left( \rho \sigma \right) . \end{aligned}$$

(4.102)

Then, for $\rho '_{AB}=U\rho _{AB}U^\dagger$, we have

$$\begin{aligned} \mathrm {F}(\rho '_{AB},\rho '_A\otimes \rho '_B)\geqslant {{\mathrm{Tr}}}_{}\left( \rho '_{AB}\rho '_A\otimes \rho '_B\right) = {{\mathrm{Tr}}}_{}\left( \rho _{AB}U^\dagger \left( \rho '_A\otimes \rho '_B\right) U\right) ,\quad \quad \end{aligned}$$

(4.103)

implying that

$$\begin{aligned} \int \mathrm {F}(\rho '_{AB},\rho '_A\otimes \rho '_B)\mathrm {d}\mu (U)\geqslant & {} \int {{\mathrm{Tr}}}_{}\left( \rho _{AB}U^\dagger \left( \rho '_A\otimes \rho '_B\right) U\right) \mathrm {d}\mu (U) \nonumber \\= & {} c_0+c_1{{\mathrm{Tr}}}_{}\left( \rho ^2_{AB}\right) +c_2{{\mathrm{Tr}}}_{}\left( \rho ^3_{AB}\right) . \end{aligned}$$

(4.104)

By the concavity of fidelity, we have

$$\begin{aligned} \int \mathrm {F}\left( \rho '_{AB},\rho '_A\otimes \rho '_B\right) \mathrm {d}\mu (U)= & {} \int \mathrm {F}\left( \rho _{AB},U^\dagger \left( \rho '_A\otimes \rho '_B\right) U\right) \mathrm {d}\mu (U) \nonumber \\\leqslant & {} \mathrm {F}\left( \rho _{AB},c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right) .\quad \quad \quad \quad \end{aligned}$$

(4.105)

Therefore,

$$\begin{aligned} c_0+c_1{{\mathrm{Tr}}}_{}\left( \rho ^2_{AB}\right) +c_2{{\mathrm{Tr}}}_{}\left( \rho ^3_{AB}\right)\leqslant & {} \int \mathrm {F}\left( \rho '_{AB},\rho '_A\otimes \rho '_B\right) \mathrm {d}\mu (U) \nonumber \\\leqslant & {} \mathrm {F}\left( \rho _{AB},c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right) .\nonumber \\ \end{aligned}$$

(4.106)

This completes the proof.

Appendix 4: The proof of Theorem 3.4

Note that $\rho '_{AB}=U\rho _{AB}U^\dagger$. We see from (3.8) that

$$\begin{aligned} \int I(A:B)_{\rho '}\mathrm {d}\mu (U)\geqslant \mathrm {S}\left( \rho _{AB} ||c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right) . \end{aligned}$$

(4.107)

Since $I(A:B)_{\rho '} = \mathrm {S}(\rho '_A)+\mathrm {S}(\rho '_B)-\mathrm {S}(\rho _{AB})$, it follows that

$$\begin{aligned}&\int \left( \mathrm {S}(\rho '_A)+\mathrm {S}(\rho '_B)-\mathrm {S}(\rho _{AB})\right) \mathrm {d}\mu (U)\geqslant \mathrm {S}\nonumber \\&\quad \left( \rho _{AB} ||c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right) . \end{aligned}$$

(4.108)

That is,

$$\begin{aligned} \langle S_A+S_B\rangle \geqslant \mathrm {S}(\rho _{AB})+\mathrm {S}\left( \rho _{AB} ||c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right) . \end{aligned}$$

(4.109)

This confirms the first inequality. Besides, by Eq. (3.4), we get

$$\begin{aligned} \mathrm {S}\left( c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right)= & {} \mathrm {S}\left( \int U^\dagger (\rho '_A\otimes \rho '_B)U \mathrm {d}\mu (U)\right) \end{aligned}$$

(4.110)

$$\begin{aligned}\geqslant & {} \int \mathrm {S}\left( U^\dagger (\rho '_A\otimes \rho '_B)U\right) \mathrm {d}\mu (U) \nonumber \\= & {} \int \mathrm {S}(\rho '_A)\mathrm {d}\mu (U)+\int \mathrm {S}(\rho '_B)\mathrm {d}\mu (U).\qquad \qquad \end{aligned}$$

(4.111)

This confirms the second inequality. Therefore, we have

$$\begin{aligned}&\mathrm {S}(\rho _{AB})+\mathrm {S}\left( \rho _{AB} ||c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right) \\&\quad \leqslant \mathrm {S}\left( c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right) . \end{aligned}$$

This is equivalent to the following

$$\begin{aligned}&\mathrm {S}\left( \rho _{AB} ||c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right) \\&\quad \leqslant \mathrm {S}\left( c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right) - \mathrm {S}(\rho _{AB}). \end{aligned}$$

Next, we show that $\mathrm {S}\left( c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right) = \mathrm {S}(\rho _{AB})$ if and only if $\rho _{AB}$ is maximally mixed state. Clearly, if $\rho _{AB}$ is maximally mixed state, i.e., $\mathrm {S}(\rho _{AB})=\ln (d)$, since $\mathrm {S}\left( c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right) - \mathrm {S}(\rho _{AB})\geqslant 0$, then $\mathrm {S}\left( c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right) \geqslant \ln (d)$, apparently $\mathrm {S}\left( c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right) \leqslant \ln (d)$, thus $\mathrm {S}(\rho _{AB})=\mathrm {S}\left( c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right) =\ln (d)$, the maximum of von Neuman entropy. Reversely, if $\mathrm {S}\left( c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right) = \mathrm {S}(\rho _{AB})$, then by the obtained inequality, we have

$$\begin{aligned} \mathrm {S}\left( \rho _{AB} ||c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}\right) =0, \end{aligned}$$

which holds if and only if $\rho _{AB} = c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}$. This means that for any eigenvalue $\lambda (\geqslant 0)$ of $\rho _{AB}$ must satisfy that

$$\begin{aligned} c_2\lambda ^2 +(c_1-1)\lambda +c_0=0. \end{aligned}$$

Solving this equation, we get

$$\begin{aligned} \lambda = \frac{(1-c_1)-\sqrt{(1-c_1)^2-4c_0c_2}}{2c_2}=\frac{1}{d}. \end{aligned}$$

Note that we have dropped another root being larger than one. Thus $\rho _{AB}$ is maximally mixed state. In fact, we get that $\rho _{AB} = c_0\cdot \mathbb {1}_d + c_1\cdot \rho _{AB} + c_2\cdot \rho ^2_{AB}$ if and only if $\rho _{AB}$ is maximally mixed state.