Invariant measure for quantum trajectories

Abstract

We study a class of Markov chains that model the evolution of a quantum system subject to repeated measurements. Each Markov chain in this class is defined by a measure on the space of matrices, and is then given by a random product of correlated matrices taken from the support of the defining measure. We give natural conditions on this support that imply that the Markov chain admits a unique invariant probability measure. We moreover prove the geometric convergence towards this invariant measure in the Wasserstein metric. Standard techniques from the theory of products of random matrices cannot be applied under our assumptions, and new techniques are developed, such as maximum likelihood-type estimations.

Appendices

Appendix A: Equivalence of (Pur) and contractivity

We assume ${\text {supp}}\mu \subset \mathrm {GL}_k({{\mathbb {C}}})$. Recall that $T_\mu$ is the smallest closed sub-semigroup of $\mathrm {GL}_k({{\mathbb {C}}})$ that contains ${\text {supp}}\mu$. It is said to be contracting if there exists a sequence $(a_n)_{n\in {\mathbb {N}}}\subset T_\mu$ such that $\lim _{n\rightarrow \infty } a_n/\Vert a_n\Vert$ exists and is a rank one matrix.

Proposition A.1

Assume ${\text {supp}}\mu \subset \mathrm {GL}_k({\mathbb {C}})$ and $T_\mu$ is strongly irreducible. Then $\mu$ verifies (Pur) if and only if $T_\mu$ is contracting.

Proof

By Proposition 2.2 the implication (Pur)$\Rightarrow$ contractivity follows by taking for $(a_n)$ a convergent subsequence of $(W_n(\omega ))$ for $\omega \in {\text {supp}}{{\mathbb {P}}}^{\mathrm {ch}}$.

We prove the opposite implication by contradiction. Following [12, Lemma 3], under the assumptions of the proposition, $T_\mu$ is contracting if and only if, for any two $\hat{x}, \hat{y}\in {{\mathrm P}({{\mathbb {C}}}^k)}$ there exists a sequence of matrices $(a_n)\subset T_\mu$ such that

$$\begin{aligned} \lim _{n\rightarrow \infty }d\left( a_n\cdot \hat{x},a_n\cdot \hat{y}\right) =0. \end{aligned}$$

Now, assume that contractivity holds but (Pur) does not. Namely, that $T_\mu$ is contracting but there exists an orthogonal projector $\pi$ of rank $\ge 2$, such that for any $a\in T_\mu$,

$$\begin{aligned} \pi a^*a\pi \propto \pi . \end{aligned}$$

Let x, y in the range of $\pi$ be orthonormal vectors. Then $\langle ax,ay\rangle =\langle x,y\rangle =0$, and $\Vert ax\Vert , \Vert ay\Vert$ are nonzero, so that $d(a\cdot \hat{x},a\cdot \hat{y})=1$. As this is true for any a in $T_{\mu }$, contractivity cannot hold. This contradiction yields the proposition. $\square$

Appendix B: Set of invariant measures under assumption (Pur)

A quantum channel is a map $\phi$ on $\mathrm {M}_k({{\mathbb {C}}})$ of a form

$$\begin{aligned} \phi (\rho ) = \int _{\mathrm {M}_k(\mathcal {C})} v \rho v^* \mathrm{d}\mu (v), \end{aligned}$$

where $\mu$ is a measure satisfying the normalization condition (1). The decomposition of quantum channels to irreducible components was derived in [2, 5, 22]. The space $\mathbb {C}^k$ is decomposed into orthogonal subspaces, one subspace is transient and in all other subspaces the map has a canonical tensor product structure. We recall these results.

There exists a decomposition

$$\begin{aligned} \mathbb {C}^k \simeq \mathbb {C}^{n_1} \oplus \dots \oplus \mathbb {C}^{n_d} \oplus \mathbb {C}^{D}, \quad k = n_1 + \dots + n_d + D \end{aligned}$$

with the following properties. We denote by $v^{(j)}$ the restriction of v to $\mathbb {C}^{n_j}$.

• (e1) All invariant states are supported in the subspace $L = \mathbb {C}^{n_1} \oplus \dots \oplus \mathbb {C}^{n_d} \oplus 0$,

• (e2) The restriction of v to this subspace is block diagonal,

  $$\begin{aligned} v|_L = v^{(1)} \oplus \cdots \oplus v^{(d)}\oplus 0 \quad \mu {\text {-}}\mathrm {a.e.}\end{aligned}$$

  (38)

• (e3) For each $j=1, \dots ,d$ there is a decomposition $\mathbb {C}^{n_j} = \mathbb {C}^{k_j} \otimes \mathbb {C}^{m_j}, \, n_j = k_j m_j$, a unitary matrix $U_j$ on $\mathbb {C}^{n_j}$ and a matrix $\tilde{v}^{(j)}$ on $\mathbb {C}^{k_j}$ such that

  $$\begin{aligned} v^{(j)} = U_j \left( \tilde{v}^{(j)} \otimes \mathrm{Id}_{{{\mathbb {C}}}^{m_j}}\right) U_j^* \quad \mu -a.s. \end{aligned}$$

  (39)

• (e4) There exists a full rank positive matrix $\rho _j$ on $\mathbb {C}^{k_j}$ such that

  $$\begin{aligned} 0 \oplus \cdots \oplus U_j \left( \rho _j \otimes \mathrm{Id}_{{{\mathbb {C}}}^{m_j}}\right) U_j^* \oplus \cdots \oplus 0 \end{aligned}$$

  (40)

  is a fixed point of $\phi$.

It follows from (e3) and (e4) that the set of fixed points for $\phi$ is

$$\begin{aligned} U_1\big (\rho _1\otimes M_{m_1}({{\mathbb {C}}})\big )U_1^*\oplus \cdots \oplus U_d\big (\rho _d\otimes M_{m_d}({{\mathbb {C}}})\big )U_d^*\oplus 0_{M_D({{\mathbb {C}}})}. \end{aligned}$$

The decomposition simplifies under the purification assumption.

Proposition B.1

Assume (Pur) holds. Then there exists a set $\{\rho _j\}_{j=1}^d$ of positive definite matrices and an integer D such that the set of $\phi$ fixed points is

$$\begin{aligned} {{\mathbb {C}}}\rho _1\oplus \cdots \oplus {{\mathbb {C}}}\rho _d\oplus 0_{M_D({{\mathbb {C}}})}. \end{aligned}$$

Proof

The statement follows from the discussion preceding the proposition if we show that (Pur) implies $m_1 = \dots = m_d =1$. Assume that one of the $m_j$, e.g. $m_1$, is greater than 1. Let x be a norm one vector in $\mathbb {C}^{k_1}$. Then $\pi = U_1\pi _{\hat{x}} \otimes \mathrm{Id}_{\mathbb {C}^{m_1}} U_1^*\oplus 0 \oplus \dots \oplus 0$ is a projection with rank bigger than 1, and by Eq. (39) we have, in the notation of (38) and (39),

$$\begin{aligned} \pi v_1^*\ldots v_n^* v_n\ldots v_1 \pi = \left\| \tilde{v}^{(1)}_n\ldots \tilde{v}^{(1)}_1 x\right\| ^2 \pi \end{aligned}$$

for $\mu ^{\otimes n}$-almost all $v_1,\ldots ,v_n$. This contradicts (Pur). $\square$

It is clear from Eq. (38) that to each extremal fixed point $0 \oplus \dots \oplus \rho _j \oplus \dots \oplus 0$ corresponds a unique invariant measure $\nu _j$ supported on its range $F_j$. The converse is the subject of the next proposition.

Proposition B.2

Assume (Pur) holds. Then any $\Pi$-invariant probability measure is a convex combination of the measures $\nu _j$, $j=1,\ldots ,d$.

Proof

Let $\nu$ be a $\Pi$-invariant probability measure. Let f be a continuous function. From Lemma 2.3,

$$\begin{aligned} {{\mathbb {E}}}_\nu (f)=\lim _{n\rightarrow \infty }{{\mathbb {E}}}_\nu \big (f\left( U_n\cdot \hat{z}\right) \big ). \end{aligned}$$

Proposition 2.1 implies

$$\begin{aligned} {{\mathbb {E}}}_\nu (f)=\lim _{n\rightarrow \infty }{{\mathbb {E}}}^{\rho _\nu }\big (f\left( U_n\cdot \hat{z}\right) \big ) \end{aligned}$$

with $\rho _\nu \in {\mathcal {D}}_k$ a fixed point of $\phi$. By Proposition B.1, (Pur) implies that there exist non negative numbers $t_1,\ldots ,t_d$ summing up to one such that $\rho _\nu =t_1\rho _1\oplus \cdots \oplus t_d\rho _d\oplus 0_{M_D({{\mathbb {C}}})}$. From the definition of ${{\mathbb {P}}}^{\rho _\nu }$,

$$\begin{aligned} {{\mathbb {P}}}^{\rho _\nu }=t_1{{\mathbb {P}}}^{\rho _1}+\cdots +t_d{{\mathbb {P}}}^{\rho _d} \end{aligned}$$

where we used the abuse of notation $\rho _j\equiv 0\oplus \cdots \oplus \rho _j\oplus \cdots \oplus 0$. Using Proposition 2.1, it follows that

$$\begin{aligned} {{\mathbb {E}}}_\nu (f)=\lim _{n\rightarrow \infty } t_1{{\mathbb {E}}}_{\nu _1}\big (f\left( U_n\cdot \hat{z}\right) \big )+\cdots +t_d{{\mathbb {E}}}_{\nu _d}\big (f\left( U_n\cdot \hat{z}\right) \big ). \end{aligned}$$

Then Lemma 2.3 and the $\Pi$-invariance of each measure $\nu _j$ yield the proposition. $\square$

Appendix C: Products of special unitary matrices

Proposition C.1

Assume ${\text {supp}}\mu \subset \mathrm {SU}(k)$. Let G be the smallest closed subgroup of $\mathrm {SU}(k)$ such that ${\text {supp}}\mu \subset G$. For any $\hat{x}\in {{\mathrm P}({{\mathbb {C}}}^k)}$, let $[\hat{x}]_G$ be the orbit of $\hat{x}$ with respect to G and the action $G\times {{\mathrm P}({{\mathbb {C}}}^k)}\ni (v,\hat{x})\mapsto v\cdot \hat{x}$. Namely, $[\hat{x}]_G:=\{\hat{y}\in {{\mathrm P}({{\mathbb {C}}}^k)}\ |\ \exists v\in G \text{ s.t. } \hat{y}=v\cdot \hat{x}\}$. Then, for any $\hat{x}$, there exists a unique $\Pi$-invariant probability measure supported on $[\hat{x}]_G$, and this unique invariant measure is uniform in the sense that for any $v\in G$ it is invariant by the map $\hat{x}\mapsto v\cdot \hat{x}$.

Corollary C.2

With the assumption and definitions of the last proposition, if $G=\mathrm {SU}(k)$, $\Pi$ has a unique invariant probability measure and this probability is the uniform one on ${{\mathrm P}({{\mathbb {C}}}^k)}$.

Proof

The corollary being a trivial consequence of $G=\mathrm {SU}(k)\Rightarrow [\hat{x}]_G={{\mathrm P}({{\mathbb {C}}}^k)}\ \forall \hat{x}\in {{\mathrm P}({{\mathbb {C}}}^k)}$, we are left with proving the proposition.

Let $P_\mu$ be the Markov kernel on G defined by the left multiplication: $P_\mu f(v)=\int _G f(uv)d\mu (u)$. Since G is compact as a closed subset of $\mathrm {SU}(k)$, following [1, Proposition 4.8.1, Theorem 4.8.2], the unique $P_\mu$-invariant probability measure $\mu _G$ on G is the normalized Haar measure on G. Since G is compact, Prokhorov’s theorem implies that for any $u\in G$,

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{n}\sum _{k=1}^n\delta _uP_\mu ^k=\mu _G\quad \text{ weakly. } \end{aligned}$$

(41)

Let $\hat{x}\in {{\mathrm P}({{\mathbb {C}}}^k)}$. Since ${\text {supp}}\mu \subset G$, for any $\hat{y}\in [\hat{x}]_G$, $\Pi (\hat{y}, [\hat{x}]_G)=1$. Then, $[\hat{x}]_G$ being compact, there exists a $\Pi$-invariant measure $\nu$ supported on $[\hat{x}]_G$.

Let f be a continuous function on $[\hat{x}]_G$. Then,

$$\begin{aligned} \nu (f)=\frac{1}{n}\sum _{k=1}^n\nu \Pi ^k f=\frac{1}{n}\sum _{k=1}^n\int _{G^k\times [\hat{x}]_G} f\left( v_k \ldots v_1\cdot \hat{y}\right) \mathrm{d}\mu ^{\otimes k}(v_1,\ldots ,v_k)\mathrm{d}\nu (\hat{y}). \end{aligned}$$

For each $\hat{y}\in [\hat{x}]_G$ let $u_y\in G$ be such that $\hat{y}=u_y\cdot \hat{x}$. The map $v\mapsto vu_y\cdot \hat{x}$ being continuous, setting $u=u_y$, the weak convergence (41) and Lebesgue’s dominated convergence theorem imply,

$$\begin{aligned} \nu (f)=\int _{G}f\left( v\cdot \hat{x}\right) \mathrm{d}\mu _G(v). \end{aligned}$$

It follows that $\nu$ is the image measure of $\mu _G$ by the application $v\mapsto v\cdot \hat{x}$. The left multiplication invariance of the Haar measure $\mu _G$ yields the invariance of $\nu$ by the map $\hat{x}\mapsto v\cdot \hat{x}$ for any $v\in G$. $\square$

Example C.3

Let $\mu =\frac{1}{2}(\delta _{v_1}+\delta _{v_2})$ with,

$$\begin{aligned} v_1=\begin{pmatrix} e^i&{}\quad 0\\ 0&{}\quad e^{-i} \end{pmatrix}\quad \text{ and }\quad v_2=\begin{pmatrix} \cos 1&{}\quad i\sin 1\\ i\sin 1 &{}\quad \cos 1 \end{pmatrix}. \end{aligned}$$

Then $G=\mathrm {SU}(2)$ and the uniform measure on ${\mathrm P}({{\mathbb {C}}}^2)$ is the unique $\Pi$-invariant probability measure.

Proof

Following Proposition C.1, it is sufficient to prove that any element of $\mathrm {SU}(2)$ is the limit of a sequence of products of $v_1$ and $v_2$.

Let $\sigma _1,\sigma _2,\sigma _3$ be the usual Pauli matrices:

$$\begin{aligned} \sigma _1:=\begin{pmatrix} 0&{}\quad 1\\ 1&{}\quad 0 \end{pmatrix},\quad \sigma _2:=\begin{pmatrix} 0&{}\quad -i\\ i&{}\quad 0 \end{pmatrix}\quad \text{ and }\quad \sigma _3=\begin{pmatrix} 1&{}\quad 0\\ 0&{}\quad -1 \end{pmatrix}. \end{aligned}$$

The Pauli matrices being generators of $\mathrm {SU}(2)$ in its fundamental representation, for any $u\in \mathrm {SU}(2)$, there exist three reals $\theta _1,\theta _2,\theta _3\in {{\mathbb {R}}}$ s.t.,

$$\begin{aligned} u=\exp (i(\theta _1\sigma _1+\theta _2\sigma _2+\theta _3\sigma _3)). \end{aligned}$$

Especially, $v_1=\exp (i\sigma _3)$ and $v_2=\exp (i\sigma _1)$. Since for any $j=1,2,3$, $\exp (i\theta _j\sigma _j)=\exp (i(\theta _j+2\pi )\sigma _j)$, taking limits of sequences of powers of $v_1$ or $v_2$, for any $\theta \in {{\mathbb {R}}}$, both

$$\begin{aligned} e^{i\theta \sigma _1}\quad \text{ and }\quad e^{i\theta \sigma _3} \end{aligned}$$

are elements of G. It remains to show that any $u\in \mathrm {SU}(2)$ is a product of elements equal to $\exp (i\theta \sigma _1)$ or $\exp (i\theta \sigma _3)$ with $\theta$ real.

Fix $(\theta _1,\theta _2,\theta _3)\in {{\mathbb {R}}}^3$. Then using spherical coordinates in ${{\mathbb {R}}}^3$, there exist $r\in {{\mathbb {R}}}_+$, $\theta \in [0,\pi ]$ and $\varphi \in [0,2\pi [$ such that $\theta _1=r\cos \theta$, $\theta _2=r\sin \theta \cos \varphi$ and $\theta _3=r\sin \theta \sin \varphi$. Then by direct computation,

$$\begin{aligned} e^{i(\theta _1\sigma _1+\theta _2\sigma _2+\theta _3\sigma _3)} =e^{-i\frac{\varphi }{2}\sigma _1}e^{i\frac{\theta }{2}\sigma _3}e^{ir\sigma _1} e^{-i\frac{\theta }{2}\sigma _3}e^{i\frac{\varphi }{2}\sigma _1}. \end{aligned}$$

It follows that as a product of elements of G, $e^{i(\theta _1\sigma _1+\theta _2\sigma _2+\theta _3\sigma _3)}\in G$, hence $G=\mathrm {SU}(2)$ and the example holds. $\square$

Example C.4

Let $\mu =\frac{1}{2}(\delta _{v_1}+\delta _{v_2})$ with,

$$\begin{aligned} v_1=\begin{pmatrix} i&{}\quad 0\\ 0&{}\quad -i \end{pmatrix}\quad \text{ and }\quad v_2=\begin{pmatrix} 0&{}\quad i\\ i &{}\quad 0 \end{pmatrix}. \end{aligned}$$

Then $G=\{\pm \mathrm{Id}_{{{\mathbb {C}}}^2}, \pm v_1, \pm v_2, \pm v_1v_2\}$. For $z\in {{\mathbb {C}}}$, let $e_z=(1,z)^\mathsf {T}$ and $e_\infty =(0,1)^\mathsf {T}$. With the conventions $\infty ^{-1}=0$, $0^{-1}=\infty$ and $-\infty =\infty$, for any $z\in {{\mathbb {C}}}\cup \{\infty \}$, $[\hat{e}_z]_G=\{\hat{e}_z, \hat{e}_{z^{-1}}, \hat{e}_{-z}, \hat{e}_{-z^{-1}}\}$ and the measure $\frac{1}{4}(\delta _{\hat{e}_z}+\delta _{\hat{e}_{-z}}+\delta _{\hat{e}_{z^{-1}}}+\delta _{\hat{e}_{-z^{-1}}})$ is a $\Pi$-invariant probability measure.

The proof of this example is obtained by an explicit computation.