Primitivity for random quantum channels

Abstract

In this paper, we consider the empirical spectrum distribution of the output of an n-fold composition of random quantum channels. As a corollary, we show that the random quantum channel is generically primitive. Our method is the graphical Weingarten calculus introduced by Collins and Nechita.

Appendix A: Some simple examples

Let $\Phi$ be a random quantum channel given by Eq. (1), let us consider $\mathbb {E} (\textrm{Tr} [\Phi (X)]).$ It is clear that

$$\begin{aligned} \mathbb {E} (\textrm{Tr} [\Phi (X)]) = \textrm{Tr} [X]. \end{aligned}$$

Note that we assume that Y is a rank-one projection. Now we use the diagram language to calculate the expectation. The diagram for $\textrm{Tr} [\Phi (X)]$ is given in Fig. 6. Since there is only one pair of $(U, \overline{U})$ boxes, the removal of the diagram is simple, see Fig. 7. Now let us count the contribution of the loops as follows:

• (1) “$\Box U$”-loop, which contributes d;

• (2) “$\bigcirc U$”-loop, which contributes D;

• (3) “$\bullet U$”-loop, which contributes $\textrm{Tr}[X]$;

• (4) The Weingarten function ${\text {Wg}}(dD,\mathbbm {1}_1)$.

Putting the above ingredients together, we obtain the correct value. Note that we have used the fact that ${\text {Wg}}(dD,\mathbbm {1}_1)=1/dD$, where $\mathbbm {1}_1$ is the identity of $\mathcal S_1$ [20].

Fig. 6

Diagram for $\textrm{Tr} [\Phi (X)]$

Fig. 7

Removal of the diagram $\textrm{Tr}[\Phi (X)]$ related to $(\alpha , \beta )=(\mathbbm {1}_1,\mathbbm {1}_1)$

Fig. 8

Diagram for $\textrm{Tr}[\Phi (X)^2]$

Now we move to another simple example, $\mathbb {E} (\textrm{Tr}[\Phi (X)^2])$. We refer to Fig. 8 for the diagram for $\textrm{Tr}[\Phi (X)^2].$ Since there are two U-boxes and two $\overline{U}$-boxes in the diagram, then the removal depends on the elements of $\mathcal S_2.$ Firstly, we numerate the U-boxes from left to right and the same for $\overline{U}$-boxes in the diagram. Hence there are four kinds of combinatorics for $(\alpha , \beta ),$ i.e., $(\alpha , \beta ) = (\mathbbm {1}_2, \mathbbm {1}_2)$, $(\alpha , \beta ) = (\mathbbm {1}_2, (1,2))$, $(\alpha , \beta ) = ((1,2), \mathbbm {1}_2)$, and $(\alpha , \beta ) = ((1,2), (1,2))$. Here $\mathbbm {1}_2$ is the identity of $\mathcal S_2.$ We only give two terms related to $(\alpha , \beta ) = (\mathbbm {1}_2, \mathbbm {1}_2)$ and $(\alpha , \beta ) = (\mathbbm {1}_2, (1,2))$, see Fig. 9.

Fig. 9

Removal of the diagram $\textrm{Tr}[\Phi (X)^2]$ related to $(\alpha , \beta ) = (\mathbbm {1}_2,\mathbbm {1}_2)$ and $(\alpha , \beta ) = (\mathbbm {1}_2, (1,2))$

Let us consider the removal that relates to $(\alpha , \beta ) = (\mathbbm {1}_2, \mathbbm {1}_2)$, the contributions of the loops is as follows:

• (1) “$\Box U$”-loop, which contributes $d^2$;

• (2) “$\bigcirc U$”-loop, which contributes D;

• (3) “$\bullet U$”-loop, which contributes $\textrm{Tr}[X]^2$;

• (4) The Weingarten function ${\text {Wg}}(dD, \mathbbm {1}_2)$.

For the removal that relates to $(\alpha , \beta ) = (\mathbbm {1}_2, (1,2))$, the contribution of the loops is as follows:

• (1’) “$\Box U$”-loop, which contributes $d^2$;

• (2’) “$\bigcirc U$”-loop, which contributes D;

• (3’) “$\bullet U$”-loop, which contributes $\textrm{Tr}[X^2]$;

• (4’) The Weingarten function ${\text {Wg}}(dD, (1,2))$.

Similarly, we can deal with the removal for $(\alpha , \beta ) = ((1,2), \mathbbm {1}_2)$, which contributes $dD^2 \textrm{Tr} [X]^2 {\text {Wg}}(dD, (1,2)).$ And for $(\alpha , \beta ) = ((1,2), (1,2)),$ the contribution is $dD^2 \textrm{Tr} [X^2] {\text {Wg}}(dD, \mathbbm {1}_2).$ Thus, by summing all terms we obtain

$$\begin{aligned} \begin{aligned} \mathbb {E} (\mathrm{Tr[X]})^2&= d^2 D \textrm{Tr}[X]^2 {\text {Wg}}(dD, \mathbbm {1}_2) + d^2 D \textrm{Tr}[X^2]{\text {Wg}}(dD, (1,2)) \\&\quad + dD^2 \textrm{Tr} [X]^2 {\text {Wg}}(dD, (1,2))+ dD^2 \textrm{Tr} [X^2] {\text {Wg}}(dD, \mathbbm {1}_2)\\&= \frac{1}{(dD)^2-1} [D(d^2-1) \textrm{Tr}[X]^2 + d(D^2-1) \textrm{Tr}[X^2]], \end{aligned} \end{aligned}$$

where we have used the fact that ${\text {Wg}}(dD, \mathbbm {1}_2) = 1/[(dD)^2-1]$ and ${\text {Wg}}(dD, (1,2)) = -1/[dD((dD)^2-1)]$ [20].