Hitting time expressions for quantum channels: beyond the irreducible case and applications to unitary walks

Abstract

In this work we make use of generalized inverses associated with quantum channels acting on finite-dimensional Hilbert spaces, so that one may calculate the mean hitting time for a particle to reach a chosen goal subspace. The questions studied in this work are motivated by recent results on quantum dynamics on graphs, most particularly quantum Markov chains. We focus on describing how generalized inverses and hitting times can be obtained, with the main novelties of this work with respect to previous ones being that (a) we are able to weaken the notion of irreducibility, so that reducible examples can be considered as well, and (b) one may consider arbitrary arrival subspaces for general positive, trace-preserving maps. Natural examples of reducible maps are given by unitary quantum walks. We also take the opportunity to explain how a more specific inverse, namely the group inverse, appears in our context, in connection with matrix algebraic constructions which may be of independent interest.

Appendix

1.1 Proof of Lemma 4.8

The proof is inspired by [19], Thm. 2.2], which is valid for stochastic matrices. For the setting of positive maps, we proceed as follows: first, use Theorem 4.6 in order to write

$$\begin{aligned} \lceil {T}\rceil = X\begin{bmatrix} I &{} \\ &{} B \end{bmatrix}X^{-1}, \end{aligned}$$

where I is an identity matrix of some dimension and B is a matrix with no eigenvalue equal to 1. Then, write

$$\begin{aligned} \frac{I+B+B^2+\cdots +B^{n-1}}{n}=\frac{(I-B^n)(I-B)^{-1}}{n}. \end{aligned}$$

From this, and with the factorizations

$$\begin{aligned} A=X\begin{bmatrix} O &{} \\ {} &{} I-B\end{bmatrix}X^{-1},\;\;\;A^\#=X\begin{bmatrix} O &{} \\ {} &{} (I-B)^{-1}\end{bmatrix}X^{-1}, \end{aligned}$$

we obtain that

$$\begin{aligned} \frac{I+T+\cdots +T^{n-1}}{n}=\frac{(I-T^n)A^\#}{n}+I-AA^\#. \end{aligned}$$

By Russo-Dye’s theorem [4], the positivity and trace preservation of T implies that $\Vert T^n\Vert =1$ for all n. Therefore,

$$\begin{aligned} \lim _{n\rightarrow \infty } \frac{(I-T^n)A^\#}{n}=0, \end{aligned}$$

and hence

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{I+T+\cdots +T^{n-1}}{n}=I-AA^\#. \end{aligned}$$

Now, note that $(I-T)(I-A^\#A)=A(I-A^\#A)=A-AA^\#A=A-A=0$, which implies that $(I-A^\#A)\rho$ is invariant for T for any choice of matrix $\rho$. Finally, if $\rho$ is a density matrix, so is $(I-A^\#A)\rho$. In fact, consider the sequence $T_n=(I+T+\cdots +T^{n-1})/n$, $n=1,2,\dots$, which approximates $(I-A^\#A)$ uniformly. We have that $T_n\rho$ is a density matrix for all n and, as the set of density matrices in $M_n$ is compact, the result follows.

1.2 Proof of Theorem 8.1

First, we recall a technical result described by Hunter [14]:

Theorem 11.1

[14] A necessary and sufficient condition for the equation $FXB=C$ to have a solution is that $FF^-CB^-B=C$, where $F^-$ and $B^-$ are any g-inverses for F and B, respectively. In this case, the general solution is given by one of the two equivalent forms: either

$$\begin{aligned} X=F^-CB^-+H-F^-DHBB^-, \end{aligned}$$

where H is an arbitrary matrix, or

$$\begin{aligned} X=F^-CB^-+(I-F^-F)U+V(I-B^-B), \end{aligned}$$

where U and V are arbitrary matrices.

Now we recall a basic lemma, regarding a conditioning on the first step reasoning. The probabilistic reasoning given in terms of the operator L has been discussed in [16] in the context of OQWs, but the same proof holds for the case of QMCs, also see [18] for the case of positive maps.

Lemma 11.2

(Adapted from [16], Lemma 2]). Let $\Phi$ be a QMC, let K be its hitting time operator and $D=\text {diag}(K_{11},\dots ,K_{nn})$, and define $L:=K-(K-D)\Phi$. Then for $\rho _j$ a density concentrated at site j, for all i, it holds that ${\text {Tr}}\big (L_{ij}(\rho _j)\big )={\text {Tr}}(\rho _j)=1$. As a consequence, ${\text {Tr}}\big (L_{ij}\rho )=Tr(\rho )$ for every $\rho \in M_n$.

Together with the results stated above, the following proof is inspired by ideas seen in [17, 19].

Proof of Theorem 8.1. We start with the definition $L:=K-(K-D)\Lambda$, where $D=K_d$, and rearrange it to obtain

$$\begin{aligned} K(I-\Lambda )=L-D\Lambda . \end{aligned}$$

(11.1)

Operator L is well-defined due to Assumption I (also see the remark at the end of Sect. 8). Now define $A:=I-\Lambda$ and we consider its group inverse $A^\#$, which exists by Proposition 4.7. Now, in Theorem 11.1, take $F=I$, $X=K$, $B=A$, and $C=L-D\Lambda$, so we have that Eq. (11.1) has a solution K if, and only if, $F^-FCB^-B=C$, which is equivalent to $(L-D\Lambda )A^\#A=L-D\Lambda$, that is,

$$\begin{aligned} (L-D\Lambda )(I-A^\#A)=0 \quad \Longleftrightarrow \quad KA(I-A^\#A)=0. \end{aligned}$$

But it is clear that this last equation is satisfied, due to the property $AA^\#A=A$. So, we have that the solution to (11.1) can be written as

$$\begin{aligned} K=(L-D\Lambda )A^\#+V(I-AA^\#), \end{aligned}$$

(11.2)

where V is an arbitrary matrix.

By Lemma 4.8, the operator $I-A^\#A$ projects onto the space of fixed points of $\Lambda$. Moreover, we note that for each i, vector $(I-A^\#A)|e_i\rangle$ is the i-th column of $I-A^\#A$, and that these are fixed points for $\Lambda$, due to the property $AA^\#A=A$. Also, by the fact that $\Lambda$ is a QMC acting on two vertices, we may write

$$\begin{aligned} I-A^\#A=\begin{bmatrix} X &{} X\\ Y &{} Y \end{bmatrix}, \end{aligned}$$

(11.3)

for certain $X, Y\in M_{k^2}$. The reason for this is simple: note that for any choice of $\rho _1$, $\rho _2$,

$$\begin{aligned} \begin{bmatrix} X &{} X\\ Y &{} Y \end{bmatrix}\begin{bmatrix} \rho _1 \\ \rho _2\end{bmatrix}=\begin{bmatrix} X(\rho _1+\rho _2) \\ Y(\rho _1+\rho _2)\end{bmatrix}, \end{aligned}$$

(11.4)

so that any choice of density will be projected onto the image of X and Y, with respect to vertices 1 and 2, respectively. It is then clear that the analogous behavior will apply to $V(I-AA^\#)$, for any matrix V, that is, these will be matrices of form (11.3) accordingly. Due to these facts, if we define

$$\begin{aligned} E:=\begin{bmatrix} I_{k^2} &{} I_{k^2} \\ I_{k^2} &{} I_{k^2} \end{bmatrix}, \end{aligned}$$

we will have $(I-AA^\#)_d E = (I-AA^\#)$ and $\big (V(I-AA^\#)\big )_d E = V(I-AA^\#)$.

Define $B:=\big (V(I-AA^\#)\big )_d$ and substitute $V(I-AA^\#)=BE$ in (11.2) in order to write

$$\begin{aligned} K=(L-D\Lambda )A^\#+BE. \end{aligned}$$

(11.5)

We take the diagonal and obtain

$$\begin{aligned} D=K_d =(LA^\#)_d -D(\Lambda A^\#)_d + B \quad \Longrightarrow \quad B=D+D(\Lambda A^\#)_d-(LA^\#)_d. \end{aligned}$$

Now we define $W:=L-D(I-AA^\#)$ and use it to eliminate L in the equation for B above, so we get:

$$\begin{aligned} B&=D+D(\Lambda A^\#)_d - (WA^\#)_d- \big (D(I-AA^\#)A^\#\big )_d \nonumber \\&=D+D(\Lambda A^\#)_d - (WA^\#)_d, \end{aligned}$$

(11.6)

where one term was canceled since $(I-AA^\#)A^\#=(I-A^\#A)A^\#=A^\#-A^\#AA^\#=0$. Substituting B from (11.6) in (11.5), and eliminating L using W, we have

$$\begin{aligned} K&=\Big (W+D(I-AA^\#)-D\Lambda \Big )A^\#+DE+D(\Lambda A^\#)_dE -(WA^\#)_dE \nonumber \\&=D\Big [-\Lambda A^\#+E+(\Lambda A^\#)_dE\Big ] +WA^\#-(WA^\#)_dE, \end{aligned}$$

(11.7)

where again a term has vanished since $(I-AA^\#)A^\#=0$. Now consider $H:=(I-AA^\#)_d$. We know $HE=I-AA^\#$, so

$$\begin{aligned} I-AA^\#=I-A^\#+\Lambda A^\# = HE, \end{aligned}$$

and hence, taking the diagonal we obtain

$$\begin{aligned}&I-(A^\#)_d+(\Lambda A^\#)_d=H \nonumber \\ \Longrightarrow&E-(A^\#)_dE+(\Lambda A^\#)_dE= HE = I-A^\#+\Lambda A^\# \nonumber \\ \Longrightarrow&-\Lambda A^\#+E+(\Lambda A^\#)_dE = I-A^\#+A^\#_dE. \end{aligned}$$

(11.8)

By replacing (11.8) into (11.7), we obtain

$$\begin{aligned} K=D\Big [I-A^\#+A^\#_dE\Big ]+WA^\#-(WA^\#)_dE. \end{aligned}$$

(11.9)

It remains to show that ${\text {Tr}}\Big ((WA^\#)_{12}\rho \Big )$ and ${\text {Tr}}\Big (\big [(WA^\#)_dE\big ]_{12}\rho \Big )$ are zero for any $\rho$. Note that by multiplying (11.1) on the right by any $\Lambda$-invariant vector $|\rho \rangle$, we obtain $L|\rho \rangle = D|\rho \rangle$, whence

$$\begin{aligned} {\text {Tr}}\big (K_{11}|\rho _1\rangle \big ) = {\text {Tr}}\big ((L|\rho \rangle )_1\big )=\sum _j {\text {Tr}}\big (L_{1j}|\rho _j\rangle \big ) = \sum _j {\text {Tr}}(\rho _j), \end{aligned}$$

(11.10)

where in the third equality we have used Lemma 11.2. Also recall that for any vector $|\rho \rangle$, the new vector $(I-AA^\#)|\rho \rangle$ will be a state which is invariant by $\Lambda$, with

$$\begin{aligned} \sum _j {\text {Tr}}\Big (\big ((I-AA^\#)|\rho \rangle \big )_j\Big )&= {\text {Tr}}\Big ((I-AA^\#)|\rho \rangle \Big ) \nonumber \\ {}&=\langle e_{I^2_{k^2}}|(I-AA^\#)|\rho \rangle =\langle e_{I^2_{k^2}}|\rho \rangle \nonumber \\&=\sum _j \langle e_{I_{k^2}}|\rho _j\rangle = \sum _j {\text {Tr}}(\rho _j). \end{aligned}$$

(11.11)

If $|\rho \rangle$ is a vector concentrated at site m, i.e., if it is of the form

$$\begin{aligned} |\rho \rangle = \begin{bmatrix} \rho \\ 0 \end{bmatrix} \text { or } \begin{bmatrix} 0 \\ \rho \end{bmatrix}, \end{aligned}$$

then Eq. (11.11) reduces to

$$\begin{aligned} \sum _j {\text {Tr}}\Big (\big ((I-AA^\#)|\rho \rangle \big )_j\Big ) = {\text {Tr}}\big (\rho \big ). \end{aligned}$$

(11.12)

Now we proceed to show that the terms involving W in (11.9) will have trace zero. First, we have that

$$\begin{aligned} {\text {Tr}}\Big ((WA^\#)_{1r}\rho \Big )&= {\text {Tr}}\big ((LA^\#)_{1r}\rho \big )-{\text {Tr}}\Big (\big (D(I-AA^\#)A^\#\big )_{1r}\rho \Big ) \nonumber \\&=\sum _m \left[ {\text {Tr}}\Big (L_{1m}A^\#_{mr}\rho \Big )-{\text {Tr}}\Big (K_{11}(I-AA^\#)_{1m}A^\#_{mr}\rho \Big )\right] \nonumber \\&=\sum _m \left[ {\text {Tr}}\Big (A^\#_{mr}\rho \Big )-{\text {Tr}}\Big (K_{11}(I-AA^\#)_{1m}A^\#_{mr}\rho \Big )\right] , \end{aligned}$$

(11.13)

where again Lemma 11.2 was applied for L, and index r can be either 1 or 2. Note that

$$\begin{aligned} (I-AA^\#)_{1m}A^\#_{mr}\rho = \big [(I-AA^\#)|\rho \rangle \big ]_1, \end{aligned}$$

where $|\rho \rangle$ is the vector with $A^\#_{mr}\rho$ concentrated at site m. So, for this choice of $|\rho \rangle$,

$$\begin{aligned} {\text {Tr}}\Big (K_{11}(I-AA^\#)_{1m}A^\#_{mr}\rho \Big )&={\text {Tr}}\Big (K_{11}\big [(I-AA^\#)|\rho \rangle \big ]_1\Big ) \nonumber \\&=\sum _j{\text {Tr}}\Big (\big [(I-AA^\#)|\rho \rangle \big ]_j\Big )\nonumber \\&={\text {Tr}}\Big (A^\#_{mr}\rho \Big ), \end{aligned}$$

(11.14)

where in the second equality we used Eq. (11.10), and in the last equality we used (11.12). Inserting (11.14) back into (11.13) we get

$$\begin{aligned} {\text {Tr}}\Big ((WA^\#)_{1r}\rho \Big ) = \sum _m \left[ {\text {Tr}}\Big (A^\#_{mr}\rho \Big )-{\text {Tr}}\Big (A^\#_{mr}\rho \Big )\right] =0. \end{aligned}$$

It is immediate from the above with $r=2$ that ${\text {Tr}}\Big ((WA^\#)_{12}\rho \Big )=0$. But also for $r=1$ it gives us

$$\begin{aligned} {\text {Tr}}\Big (\big ((WA^\#)_dE\big )_{12}\rho \Big )={\text {Tr}}\Big ((WA^\#)_{11}\rho \Big )=0. \end{aligned}$$

Therefore, when we calculate ${\text {Tr}}\big (K_{12}\rho \big )$ using (11.9), the terms involving W vanish and the result follows.