Open Quantum Random Walks: Ergodicity, Hitting Times, Gambler’s Ruin and Potential Theory

Abstract

In this work we study certain aspects of open quantum random walks (OQRWs), a class of quantum channels described by Attal et al. (J Stat Phys 147: 832–852, 2012). As a first objective we consider processes which are nonhomogeneous in time, i.e., at each time step, a possibly distinct evolution kernel. Inspired by a spectral technique described by Saloff-Coste and Zúñiga (Stoch Proc Appl 117: 961–979, 2007), we define a notion of ergodicity for finite nonhomogeneous quantum Markov chains and describe a criterion for ergodicity of such objects in terms of singular values. As a second objective, and based on a quantum trajectory approach, we study a notion of hitting time for OQRWs and we see that many constructions are variations of well-known classical probability results, with the density matrix degree of freedom on each site giving rise to systems which are seen to be nonclassical. In this way we are able to examine open quantum versions of the gambler’s ruin, birth-and-death chain and a basic theorem on potential theory.

Appendix: Proofs

Proof of Proposition 4.2

We have for $B=(B_{ij})$ QTM, for all j,

$$\begin{aligned}&\sum _{i=1}^n \left\| B_{j}^{i*}B_{j}^i-\frac{I}{n}\right\| _2^2=\sum _{i=1}^n \left\langle B_{j}^{i*}B_{j}^i,B_{j}^{i*}B_{j}^i\right\rangle _2-2\sum _{i=1}^n\left\langle B_{j}^{i*}B_{j}^i,\frac{I}{n}\right\rangle _2+n\left\langle \frac{I}{n},\frac{I}{n}\right\rangle _2 \nonumber \\&\quad =\sum _{i=1}^ntr\big (\big (B_{j}^{i*}B_{j}^i\big )^2\big )-2\left\langle I,\frac{I}{n}\right\rangle _2+\frac{1}{n}\langle I,I\rangle _2=\sum _{i=1}^ntr\big (\big (B_{j}^{i*}B_{j}^i\big )^2\big )-\frac{1}{n}\langle I,I\rangle _2 \nonumber \\&\quad =\sum _{i=1}^n tr((B_{j}^{i*}B_{j}^i)^2) -\frac{k}{n} \end{aligned}$$

(10.1)

With (10.1) in mind, we perform another calculation. Consider an orthonormal basis of eigenstates for $\Phi ^*\Phi$ given by $\mathcal {F}=\{\eta _i\}_{i=1}^{N_\Phi }$, such that

$$\begin{aligned} \eta _1=\frac{1}{\sqrt{kn}}\left( I\otimes |1\rangle \langle 1|+\cdots +I\otimes |n\rangle \langle n|\right) , \;\;\;I=I_k\in M_k(\mathbb {C}), \end{aligned}$$

(10.2)

where $I=I_k$ is the order k identity matrix. Note that $\Vert \eta _1\Vert =1$. Recall that by Remark 2.1, whatever is the initial state $\rho$ on $\mathcal {H}\otimes \mathcal {K}$, the density $\Phi (\rho )$ is of the form $\sum _i\rho _i\otimes |i\rangle \langle i|$. This imposes a restriction on the kind of eigenstates present in $\mathcal {F}$. Define

$$\begin{aligned} \rho _1\,:\,=\begin{bmatrix} I&0&\dots&0\end{bmatrix}^T=\sum _{i=1}^{N_\Phi } d_i\eta _i,\, \, \, d_i\in \mathbb {C}, \end{aligned}$$

(10.3)

with $N_\Phi$ as in Remark 4.1. We have

$$\begin{aligned} \langle \Phi ^*\Phi \rho _1,\rho _1\rangle _2=\langle \Phi ^*\Phi \sum _{i} d_i\eta _i,\sum _{j} d_j\eta _j \rangle =\sum _{i,j} d_i\overline{d_j}\langle \Phi ^*\Phi \eta _i,\eta _j\rangle =\sum _{i=1}^{N_\Phi } |d_i|^2\sigma _i^2 \end{aligned}$$

(10.4)

Also,

$$\begin{aligned} \Phi (\rho _1)=\begin{bmatrix} B_{1}^1IB_{1}^{1*}&B_{1}^2IB_{1}^{2*}&\cdots&B_{1}^nIB_{1}^{n*}\end{bmatrix}^T, \end{aligned}$$

(10.5)

so

$$\begin{aligned} \langle \Phi (\rho _1),\Phi (\rho _1)\rangle _2= & {} \left\langle \begin{bmatrix} B_{1}^1IB_{1}^{1*} \\ B_{1}^2IB_{1}^{2*} \\ \vdots \\ B_{1}^nIB_{1}^{n*}\end{bmatrix},\begin{bmatrix} B_{1}^1IB_{1}^{1*} \\ B_{1}^2IB_{1}^{2*} \\ \vdots \\ B_{1}^nIB_{1}^{n*}\end{bmatrix}\right\rangle _2=tr((B_{1}^{1*}B_{1}^1)^2)\nonumber \\&+tr((B_{1}^{2*}B_{1}^2)^2)+\cdots +tr((B_{1}^{n*}B_{1}^n)^2). \end{aligned}$$

(10.6)

Therefore on one hand we have

$$\begin{aligned} \sum _{i=1}^n tr((B_{1}^{i*}B_{1}^i)^2)=\sum _{i=1}^{N_\Phi } |d_i|^2\sigma _i^2, \end{aligned}$$

(10.7)

and on the other we obtained, by (10.1),

$$\begin{aligned} \sum _{i=1}^n \left\| B_{1}^{i*}B_{1}^i-\frac{I}{n}\right\| _2^2=\sum _{i=1}^n tr((B_{1}^{i*}B_{1}^i)^2) -\frac{k}{n} \end{aligned}$$

(10.8)

Finally, note that $d_1=\langle \rho _1,\eta _1\rangle =\frac{1}{\sqrt{kn}}tr(I)=\frac{k}{\sqrt{kn}}$ and so $|d_1|^2=\frac{k}{n}$. We can repeat an analogous reasoning where we define $\rho _2$, $\rho _3,\dots$ in a similar way as $\rho _1$ in (10.3). $\square$

Proof of Theorem 4.7

Assume that $\sigma =\max _{1,\dots , q}\sigma _2(Q_j)<1$. Let $\{B_i\}_{i=1}^\infty$ be a sequence of OQRWs such that

$$\begin{aligned} N_l=\#\{i\in \{1,\dots ,l\}:B_i\in \mathcal {Q}\} \end{aligned}$$

(10.9)

tends to infinity with l. By (4.9) we have that for every j,

$$\begin{aligned} \left( \sum _{i=1}^n \left\| \mathcal {B}_{0,l}(i,j)\mathcal {B}_{0,l}(i,j)^*-\frac{I}{n}\right\| _2^2\right) ^{1/2}\le \sigma ^{N_l}C(j,n) \end{aligned}$$

(10.10)

which tends to zero as $l\rightarrow \infty$. Conversely, assume that the pair $(\mathcal {Q},\rho _\pi )$ is ergodic. Then equation (4.6) holds for any sequence $(B_i)_1^\infty$ of QTMs with invariant measure $\rho _\pi$ such that $B_i\in \mathcal {Q}$ for infinitely many i’s. That is, the columns of the iterated product are becoming equal to I / n. By contradiction, assume that one of the $Q_i$, say $Q_1$ satisfies $\sigma _2(Q_1)=1$ and consider the following sequence of QTMs: $B_{2i+1}=Q_1$, $B_{2i}=Q_1^*$, $i=1,2,\dots$. Now we consider $Q_1Q_1^*$, for which $\sigma _2(Q_1)=1$ is an eigenvalue with algebraic and geometric multiplicity at least 2, i.e., $\mu _\Phi (1)=\gamma _\Phi (1)\ge 2$ (see Prop. 4.5).

Now let $\Psi _r=(Q_1Q_1^*)^r$. It is clear that each $\Psi _r$ is a quantum channel with real spectrum in [0, 1], by the remarks preceding the statement of this theorem. By standard arguments such as the one seen in Novotný et al. [34, 35] (via Jordan blocks), the asymptotic behavior of $\Psi _r$ is determined by the peripheral spectrum, as contributions of eigenspaces associated to eigenvalues with norm less than 1 tend to disappear. Since 1 is the only eigenvalue in the unit circle associated to $\Psi _r$, for all r, it is clear that the limit of $\Psi _r$ as $r\rightarrow \infty$ exists. The QTM $B=\lim _{r\rightarrow \infty }\Psi _r=\lim _{r\rightarrow \infty }(Q_1Q_1^*)^r$ is such that there must be two matrices B(i, j) and B(l, m) which are distinct. Indeed, let $\rho _0$ be an eigenstate of B associated to eigenvalue 1. Suppose that $\rho _0$ is not the maximally mixed column. This assumption is possible since we have that the geometric multiplicity of 1 for $Q_1Q_1^*$ is at least 2. In particular, by writing $\rho _0=\sum _i \eta _i\otimes |i\rangle \langle i|$ we may assume there are k, l such that $\eta _k\ne \eta _l$. Now the fact that $B(\rho _0)=\rho _0$, corresponds to the system of equations

$$\begin{aligned} B(1,1)\eta _1 B(1,1)^{*}+\cdots +B(1,n)\eta _n B(1,n)^{*}=\eta _1 \end{aligned}$$

(10.11)

$$\begin{aligned} B(2,1)\eta _1 B(2,1)^{*}+\cdots +B(2,n)\eta _n B(2,n)^{*}=\eta _2 \end{aligned}$$

(10.12)

$$\begin{aligned} \vdots \end{aligned}$$

$$\begin{aligned} B(n,1)\eta _1 B(n,1)^{*}+\cdots +B(n,n)\eta _n B(n,n)^{*}=\eta _n \end{aligned}$$

(10.13)

If, on the contrary, all B(i, j) are equal, then by considering the k-th and l-th equation we conclude that $\eta _k=\eta _l$, which is absurd. Therefore the QTM $B=\lim _{r\rightarrow \infty }\Psi _r$ is such that there must be two matrices B(i, j) and B(l, m) which are distinct.

Moreover, there must be a row in B with two different entries, that is, two matrices $B(i,r)\ne B(i,s)$ for some i, r, s. In fact, suppose $i_1$ and $i_2$ are rows where we found two distinct elements of B, say, $B(i_1,j_1)$ and $B(i_2,j_2)$. If all entries of row $i_1$ are equal then these must be equal to the maximally mixed column. The same conclusion holds for row $i_2$. But then we would have $B(i_1,j_1)=B(i_2,j_2)$, which is absurd. We conclude there must be a row in B with two different entries. Hence we are able to obtain x, y, z such that

$$\begin{aligned} \lim _{r\rightarrow \infty } {\mathcal {B}}_{0,2r}(x,y)- {\mathcal {B}}_{0,2r}(x,z)\ne 0, \end{aligned}$$

(10.14)

as required. $\square$

Proof of Theorem 8.3

• (a) Gambler’s ruin. Let $A=\{0\}$ so that $h_i^A(\rho _i)$ is the ruin probability starting from i. Using Theorem 6.3 we have the system of equations

  $$\begin{aligned} \left\{ \begin{array}{ll} h_0^A(\rho _i)=1 &{} \\ h_i^A(\rho _i)=p(\rho _i)h_{i+1}^A\left( \frac{B_i^{i+1}\rho _i B_i^{i+1*}}{tr(B_i^{i+1}\rho _i B_i^{i+1*})}\right) +q(\rho _i)h_{i-1}^A\left( \frac{B_i^{i-1}\rho _i B_i^{i-1*}}{tr(B_i^{i-1}\rho _i B_i^{i-1*})}\right) &{} i=1,2,\dots \end{array} \right. \end{aligned}$$

  (10.15)

  For instance if $i=2$, using expression (5.9) and recalling that $B_i^{i+1}=R$, $B_i^{i-1}=L$,

  $$\begin{aligned} h_2^A(\rho )= & {} tr(B_2^3\rho B_2^{3*})\sum _{D\in \pi (3;A)} \frac{tr(DB_2^3\rho B_2^{3*}D^*)}{tr(B_2^3\rho B_2^{3*})}\nonumber \\&+\,tr(B_2^1\rho B_2^{1*})\sum _{C\in \pi (1;A)} \frac{tr(CB_2^1\rho B_2^{1*}C^*)}{tr(B_2^1\rho B_2^{1*})} \end{aligned}$$

  (10.16)

  We will also write $q(\rho )=tr(L\rho L^*)$ and $p(\rho )=tr(R\rho R^*)$. Let $\Phi _L(X)=LXL^*$ and $\Phi _R(X)=RXR^*$. Then if $LR=RL$ we have $\Phi _R\Phi _L=\Phi _L\Phi _R$ so let $\{\eta _i\}$ be a basis for $M_2(\mathbb {C})$ consisting of eigenstates for $\Phi _L$ and $\Phi _R$ and write $\Phi _L(\eta _i)=\lambda _i\eta _i$, $\Phi _R(\eta _i)=\mu _i\eta _i$. Then we have

  $$\begin{aligned} h_2^A(\eta _i)= & {} tr(R\eta _i R^*)\sum _{D\in \pi (3;A)} \frac{tr(DB_2^3\eta _i B_2^{3*}D^*)}{tr(B_2^3\eta _i B_2^{3*})}\\&+\,tr(L\eta _i L^*)\sum _{C\in \pi (1;A)} \frac{tr(CB_2^1\eta _i B_2^{1*}C^*)}{tr(B_2^1\eta _i B_2^{1*})} \end{aligned}$$
  $$\begin{aligned} tr(R\eta _i R^*)\sum _{D\in \pi (3;A)} \frac{tr(D\mu _i\eta _i D^*)}{\mu _i tr(\eta _i)}+tr(L\eta _i L^*)\sum _{C\in \pi (1;A)} \frac{tr(C\lambda _i\eta _i C^*)}{\lambda _i tr(\eta _i)} \end{aligned}$$
  $$\begin{aligned} =p(\eta _i)h_{3}^A(\eta _i)+q(\eta _i)h_{1}^A(\eta _i) \end{aligned}$$

  (10.17)

  In a simpler notation, let $\eta$ be such that $\Phi _L(\eta )=\lambda \eta$ and $\Phi _R(\eta )=\mu \eta$, $\lambda ,\eta \in \mathbb {C}$. Then for such choice the above equation becomes

  $$\begin{aligned} h_i^A(\eta )=\mu h_{i+1}^A(\eta )+\lambda h_{i-1}^A(\eta ). \end{aligned}$$

  (10.18)

  If $\mu \ne \lambda$ the recurrence relation has, for fixed $\eta$, a general solution

  $$\begin{aligned} h_i^A(\eta )=B+C\left( \frac{\lambda }{\mu }\right) ^i,\,\,\,\lambda =\lambda (\eta ),\, \mu =\mu (\eta ) \end{aligned}$$

  (10.19)

  Now, if $\mu <\lambda$ then the restriction $0\le h_i\le 1$ forces $C=0$ so $h_i=1$ for all i. That is, the ruin of the quantum gambler is certain. On the other hand, if $\mu >\lambda$ then we get the family of solutions

  $$\begin{aligned} h_{i}^{A}({\eta })=\left( \frac{\lambda }{\mu }\right) ^{i} +A\left( 1-\left( \frac{\lambda }{\mu }\right) ^{i}\right) \end{aligned}$$

  (10.20)

  For a nonnegative solution we must have $B\ge 0$ and so the minimal solution is

  $$\begin{aligned} h_{i}^{A}(\eta )=\left( \frac{\lambda }{\mu }\right) ^{i} \end{aligned}$$

  (10.21)

  Finally if $\mu =\lambda$, the recurrence relation has a general solution $h_i^{A}=B+Ci$ and again the restriction $0\le h_i^A(\eta )\le 1$ forces $C=0$, so $h_i^{A}(\eta )=1$ for all i.

• (b) Birth-and-death chain. We write $p_i(\rho )=tr(R_i{\rho } R_i^{*})$, $q_i(\rho )=tr(L_i\rho L_{i}^{*})$. We still assume commutativity of the translation operators (moves to the left or right). Let $\Phi _{L_{i}}(\eta _{j})=\lambda _{i;j}\eta _j$, $\Phi _{R_{i}}(\eta _{j})=\mu _{i;j}\eta _j$. Then for any eigenstate $\eta _j$ we have

  $$\begin{aligned} h_i^{A}(\eta _j)=p_i(\eta _j)h_{i+1}^A\left( \frac{R_i\eta _j R_i^{*}}{tr(R_i\eta _j R_i^{*})}\right) +q_i(\eta _j)h_{i-1}^A\left( \frac{L_i\eta _j L_i^{*}}{tr(L_i\eta _j L_i^*)}\right) , \end{aligned}$$

  (10.22)

  thus implying, due to the eigenvalue conditions $p_i(\eta _j)=\mu _{ij}$, $q_i(\eta _j)=\lambda _{ij}$ and using the same calculation used in the gambler’s ruin,

  $$\begin{aligned} h_i^A(\eta _j)=\mu _{i;j}h_{i+1}^A(\eta _j)+\lambda _{i;j} h_{i-1}^A(\eta _j). \end{aligned}$$

  (10.23)

  Let $u_i=h_{i-1}^A-h_i^A$, then

  $$\begin{aligned} \mu _{i;j}p_iu_{i+1}=\mu _{i;j}p_i(h_i-h_{i+1})=\mu _{i;j}(p_ih_i-p_ih_{i+1})=\mu _{i;j}(p_ih_i+q_ih_{i-1}-h_i), \end{aligned}$$

  (10.24)

  the last equality due to (10.22), by isolating $p_i$ (we omit the arguments for simplicity). Then

  $$\begin{aligned}&\mu _{i;j}p_iu_{i+1}=\mu _{i;j}(h_i(p_i-1)+q_ih_{i-1})=\mu _{i;j}(-h_iq_i+q_ih_{i-1})\nonumber \\&\quad =\mu _{i;j}q_i(h_{i-1}-h_i)=\mu _{i;j}q_iu_i \end{aligned}$$

  (10.25)

  Hence $\mu _{i;j}p_iu_{i+1}=\mu _{i;j}q_iu_i$ and so $p_iu_{i+1}=q_iu_i$. Therefore,

  $$\begin{aligned} u_{i+1}(\eta _j)=\left( \frac{q_i(\eta _j)}{p_i(\eta _j)}\right) u_i(\eta _j)=\left( \frac{q_i(\eta _j)q_{i-1}(\eta _j)\cdots q_1(\eta _j)}{p_i(\eta _j)p_{i-1}(\eta _j)\cdots p_1(\eta _j)}\right) u_1(\eta _j)=\gamma _i(\eta _j)u_1(\eta _j) \end{aligned}$$

  (10.26)

  where

  $$\begin{aligned} \gamma _i(\eta _j):=\frac{q_i(\eta _j)q_{i-1}(\eta _j)\cdots q_1(\eta _j)}{p_i(\eta _j)p_{i-1}(\eta _j)\cdots p_1(\eta _j)}=\frac{\lambda _{i;j}\lambda _{i-1;j}\cdots \lambda _{1;j}}{\mu _{i;j}\mu _{i-1;j}\cdots \mu _{1;j}}, \end{aligned}$$

  (10.27)

  the last equality due to the eigenvalue conditions. Then note that $u_1+\cdots +u_i=h_0^A-h_i^A$ and recalling that $h_0^A=1$ we can write

  $$\begin{aligned} h_i^A=h_0^A-u_1-u_2-\cdots - u_i=1-\gamma _0u_1-\gamma _1u_1-\cdots -\gamma _{i-1}u_1 \end{aligned}$$
  $$\begin{aligned} =1-B(\gamma _0+\cdots +\gamma _{i-1}),\,\,\,A=u_1,\,\,\,\gamma _0=1 \end{aligned}$$

  (10.28)

  Here B is yet to be determined. If $\sum _{i=0}^\infty \gamma _i(\eta _j)=\infty$ the restriction $0\le h_i^A\le 1$ forces $B=0$ and $h_i^A=1$ for all i. If $\sum _{i=0}^\infty \gamma _i(\eta _j)< \infty$ then we may take $B>0$ such that

  $$\begin{aligned} 1-B(\gamma _0+\cdots +\gamma _{i-1})\ge 0,\,\,\,\forall i \end{aligned}$$

  (10.29)

  So the minimal nonnegative solution occurs when $B=B(j)=(\sum _{k=0}^\infty \gamma _k(\eta _j))^{-1}$ and then for all j,

  $$\begin{aligned} h_i^A(\eta _j)=\frac{\sum _{k=i}^\infty \gamma _k(\eta _j)}{\sum _{k=0}^\infty \gamma _k(\eta _j)},\,\,\,\forall i=1,2,\dots \end{aligned}$$

  (10.30)

$\square$

Proof of Theorem 8.4

The proof is inspired by a classical description [33]. Item a) was proved before the statement of this theorem. b) Consider the expected cost up to time n:

$$\begin{aligned}&\phi _i(\rho ;n):=\sum _{X_0=i,X_1,X_2,\dots \in (-a,a), X_r\in \{-a,a\},r\le n}\left[ c(X_0)+c(X_1)+c(X_2)+\cdots +c(X_{r-1})\right. \nonumber \\&\qquad \qquad \qquad \left. +f(X_r)\right] P_i(X_1,\dots ,X_r;\rho ),\,\,\,n\ge 1 \end{aligned}$$

(10.31)

and we define $\phi _i(\rho ;0)=0$. Then it should be clear that $\phi _i(\rho ;n)\uparrow \phi _i(\rho )$ as $n\rightarrow \infty$. Similarly,

$$\begin{aligned} \phi _i(\rho ;n+1)=c(i)+\sum _{j\in I}p_{ij}\phi _j\left( \frac{B_i^j\rho B_i^{j*}}{tr(B_i^j\rho B_i^{j*})};n\right) ,\,\,\,n\ge 0 \end{aligned}$$

(10.32)

Now suppose $\psi \ge 0$ satisfies

$$\begin{aligned} \psi _i(\rho )\ge c(i)+\sum _j p_{ij}(\rho )\psi _j\left( \frac{B_i^j\rho B_i^{j*}}{tr(B_i^j\rho B_i^{j*})}\right) , \end{aligned}$$

(10.33)

in D and $\psi _i(\rho )\ge f(i)$ in $\partial D$. Correspondingly, we call (10.33) the open quantum version of $\psi \ge c+P\psi$. Note that for all $\rho$, $\psi _i(\rho )\ge 0=\phi _i(\rho ;0)$ so

$$\begin{aligned} \psi _i(\rho )\ge & {} c(i)+\sum _j p_{ij}(\rho )\psi _j\left( \frac{B_i^j\rho B_i^{j*}}{tr(B_i^j\rho B_i^{j*})}\right) \ge c(i)+\sum _j p_{ij}(\rho )\phi _j\left( \frac{B_i^j\rho B_i^{j*}}{tr(B_i^j\rho B_i^{j*})};0\right) \nonumber \\= & {} \phi _i(\rho ;1) \end{aligned}$$

(10.34)

in D. By induction we conclude $\psi _i(\rho )\ge \phi _i(\rho ;n)$ and hence $\psi _i(\rho )\ge \phi _i(\rho )$ for all $\rho$, all i. For simplicity we write $\psi \ge \phi$.

c) Now we assume $P_i(T<\infty )=1$ for all i, i.e., we assume the probability of ever reaching $\partial D$ equals 1. Note that a choice of initial density $\rho$ is implicit here. We would like to show that the open quantum version of the problem $\phi =c+P\phi$ in D, $\phi =f$ in $\partial D$ has at most one bounded solution. Let $\psi _i(\rho )$ be another solution. For $i\in D$ we have, writing $p_{ij}=p_{ij}(\rho )=tr(B_i^j\rho B_i^{j*})$ for simplicity,

$$\begin{aligned} \psi _i(\rho )= & {} c(i)+\sum _{j\in I}p_{ij}\psi _j\left( \frac{B_i^j\rho B_i^{j*}}{tr(B_i^j\rho B_i^{j*})}\right) =c(i)+\sum _{j\in \partial D}p_{ij}f(j)\nonumber \\&+\sum _{j\in D}p_{ij}\psi _j\left( \frac{B_i^j\rho B_i^{j*}}{tr(B_i^j\rho B_i^{j*})}\right) \end{aligned}$$

(10.35)

By performing a repeated substitution we get

$$\begin{aligned}&\psi _i(\rho )= c(i)+\sum _{j\in \partial D}p_{ij}(\rho )f(j)+\sum _{j\in D}p_{ij}(\rho )\psi _j\left( \frac{B_i^j\rho B_i^{j*}}{tr(B_i^j\rho B_i^{j*})}\right) \\&\quad =c(i)+\sum _{j\in \partial D}p_{ij}(\rho )f(j)+\sum _{j\in D}p_{ij}(\rho )\left[ c(j)+\sum _{j_1\in \partial D}p_{jj_1}\left( \frac{B_i^j\rho B_i^{j*}}{tr(B_i^j\rho B_i^{j*})}\right) f(j_1)\right. \nonumber \\&\qquad \left. +\sum _{j_1\in D}p_{jj_1}\left( \frac{B_i^j\rho B_i^{j*}}{tr(B_i^j\rho B_i^{j*})}\right) \psi _{j_1}\left( \frac{B_j^{j_1}\frac{B_i^j\rho B_i^{j*}}{tr(B_i^j\rho B_i^{j*})} B_j^{j_1*}}{tr(B_j^{j_1}\frac{B_i^j\rho B_i^{j*}}{tr(B_i^j\rho B_i^{j*})} B_j^{j_1*})}\right) \right] \nonumber \\&\quad =c(i)+\sum _{j\in \partial D}p_{ij}(\rho )f(j)+\sum _{j\in D}p_{ij}(\rho )\left[ c(j)+\sum _{j_1\in \partial D}p_{jj_1}\left( \frac{B_i^j\rho B_i^{j*}}{tr(B_i^j\rho B_i^{j*})}\right) f(j_1)\right. \nonumber \\&\qquad \left. +\sum _{j_1\in D}p_{jj_1}\left( \frac{B_i^j\rho B_i^{j*}}{tr(B_i^j\rho B_i^{j*})}\right) \psi _{j_1}\left( \frac{B_j^{j_1}B_i^j\rho B_i^{j*} B_j^{j_1*}}{tr(B_j^{j_1}B_i^j\rho B_i^{j*} B_j^{j_1*})}\right) \right] \nonumber \\&\quad =c(i)+\sum _{j\in \partial D}p_{ij}(\rho )f(j)+\sum _{j\in D}p_{ij}(\rho )c(j)+\sum _{j\in D}\sum _{j_1\in \partial D}p_{ij}(\rho )p_{jj_1}\left( \frac{B_i^j\rho B_i^{j*}}{tr(B_i^j\rho B_i^{j*})}\right) f(j_1)\nonumber \\ \end{aligned}$$

(10.36)

$$\begin{aligned}&\qquad +\sum _{j\in D}\sum _{j_1\in D}p_{ij}(\rho )p_{jj_1}\left( \frac{B_i^j\rho B_i^{j*}}{tr(B_i^j\rho B_i^{j*})}\right) \psi _{j_1}\left( \frac{B_j^{j_1}B_i^j\rho B_i^{j*} B_j^{j_1*}}{tr(B_j^{j_1}B_i^j\rho B_i^{j*} B_j^{j_1*})}\right) =\cdots \nonumber \\&\qquad \cdots =c(i)+\sum _{j\in \partial D} p_{ij}(\rho )f(j)+\sum _{j\in D} p_{ij}(\rho )c(j) \nonumber \\&\qquad +\cdots +\sum _{j_1\in D}\cdots \sum _{j_{n-1}\in D} p_{ij}p_{jj_1}\cdots p_{j_{n-2}j_{n-1}}c(j_{n-1})\nonumber \\&\qquad +\sum _{j_1\in D}\cdots \sum _{j_{n-1}\in D}\sum _{j_n\in \partial D} p_{ij}p_{jj_1}\cdots p_{j_{n-1}j_n}f(j_n) \nonumber \\ \end{aligned}$$

(10.37)

$$\begin{aligned}&\qquad +\sum _{j_1\in D}\cdots \sum _{j_n\in D} p_{ij_1}p_{jj_1}\cdots p_{j_{n-1}j_n}\psi _{j_n}\nonumber \end{aligned}$$

(10.38)

Note that in the last equality we have omitted the dependence of the $p_{ij}$ on the density matrices. A convenient cancellation of terms occur in each of them. For instance, in the last term, $p_{ij_1}\cdots p_{j_{n-1}j_n}$ simplifies to

$$\begin{aligned} p_{ij_1}\cdots p_{j_{n-1}j_n}= & {} tr(B_i^{j_1}\rho B_i^{j_1*})tr\left( \frac{B_{j_1}^{j_2}B_i^{j_1}\rho B_i^{j_1*}B_{j_1}^{j_2*}}{tr(B_i^{j_1}\rho B_i^{j_1*})}\right) \cdots \nonumber \\= & {} tr(B_{j_{n-1}}^{j_n}\cdots B_i^{j_1}\rho B_i^{j_1*}\cdots B_{j_{n-1}}^{j_n*}) \end{aligned}$$

(10.39)

Now suppose $P_i(T<\infty )=1$ for all i and that $|\psi _i|\le M$ then as $n\rightarrow \infty$,

$$\begin{aligned} \left| \sum _{j_1\in D}\cdots \sum _{j_n\in D} p_{ij_1}\cdots p_{j_{n-1}j_n}\psi _{j_n}\right| \le MP_i(T\ge n)\rightarrow 0, \end{aligned}$$

(10.40)

which means the last term in (10.39) vanishes when $n\rightarrow \infty$. Recalling definition (10.31) we obtain in the case of equality in Eqs. (10.36)–(10.39) that $\psi _i=\lim _{n\rightarrow \infty }\phi _i(\rho ;n)=\phi _i$. $\square$