Szegedy Walk Unitaries for Quantum Maps

Abstract

Szegedy developed a generic method for quantizing classical algorithms based on random walks (Szegedy, in: 45th annual IEEE Symposium on Foundations of Computer Science, pp 32–41, 2004. https://doi.org/10.1109/FOCS.2004.53). A major contribution of his work was the construction of a walk unitary for any reversible random walk. Such unitary posses two crucial properties: its eigenvector with eigenphase 0 is a quantum sample of the limiting distribution of the random walk and its eigenphase gap is quadratically larger than the spectral gap of the random walk. It was an open question if it is possible to generalize Szegedy’s quantization method for stochastic maps to quantum maps. We answer this in the affirmative by presenting an explicit construction of a Szegedy walk unitary for detailed balanced Lindbladians—generators of quantum Markov semigroups—and detailed balanced quantum channels. We prove that our Szegedy walk unitary has a purification of the fixed point of the Lindbladian as eigenvector with eigenphase 0 and that its eigenphase gap is quadratically larger than the spectral gap of the Lindbladian. To construct the walk unitary we leverage a canonical form for detailed balanced Lindbladians showing that they are structurally related to Davies generators. We also explain how the quantization method for Lindbladians can be applied to quantum channels. We give an efficient quantum algorithm for quantizing Davies generators that describe many important dynamics of open quantum systems, for instance, the relaxation of a quantum system coupled to a bath. Our algorithm extends known techniques for simulating dynamics of quantum systems on a quantum computer.

Appendices

Appendix A: Quantum–Quantum Metropolis Algorithm and Its Limitations

We mentioned in the introduction that the quantum-quantum Metropolis algorithm (QQMA) as presented in [23] only works when the spectrum of the Hamiltonian is non-degenerate. The goal of this appendix is to discuss the reasons for this limitation.

One of the problems of QQMA in the degenerate case is the implementation of the projector $\Lambda _1$ defined below equation (6) in [23]. Before explaining this issue in detail, let us state some facts to put everything better into context.

It is stated that QQMA prepares a coherent encoding of the thermal state

$$\begin{aligned} \vert \alpha _0\rangle = \sum _{i=0}^{N-1} \sqrt{e^{-\beta E_i}/Z_\beta } \, \vert i \rangle \quad \text{ with } \quad \vert i \rangle \equiv \vert \varphi _i \rangle \otimes \vert {\tilde{\varphi }}_i \rangle , \end{aligned}$$

(A1)

where $\vert \varphi _i\rangle$ denote the eigenstates of the Hamiltonian with corresponding eigenvalues $E_i$. The tilde in $\vert {\tilde{\varphi }}_i\rangle$ denotes complex conjugation of the entries of the vectors $\vert \varphi _i\rangle$.

When the spectrum of the Hamiltonian is degenerate (and $\beta >0)$, it may seem at first that there could exist nonequivalent coherent encodings that depend on the particular chosen eigenbasis $\vert \varphi _i\rangle$. More precisely, let U be any unitary that commutes with the Hamiltonian H and set $\vert \psi _i\rangle = U\vert \varphi _i\rangle$ for $i=0,\ldots ,N-1$. Define the corresponding “alternative” coherent encoding

$$\begin{aligned} \vert \beta _0\rangle = \sum _{i=0}^{N-1} \sqrt{e^{-\beta E_i}/Z_\beta } \, \vert \psi _i \rangle \otimes \vert {\tilde{\psi }}_i \rangle . \end{aligned}$$

(A2)

Clearly, we have $(U\otimes {\tilde{U}}) \vert \alpha _0\rangle = \vert \beta _0\rangle$. However, it already holds that $\vert \alpha _0\rangle = \vert \beta _0\rangle$ so that the coherent encoding does not depend on the chosen eigenbasis.

The central problem of QQMA in the degenerate case is the implementation of the projector $\Lambda _1$ defined below equation (6) in [23]. The operator $\Lambda _1$ is defined there as

$$\begin{aligned} \Lambda _1 = \sum _{i=0}^{N-1} \vert i{\rangle }{\langle } i \vert \otimes \vert 0{\rangle }{\langle } 0 \vert = \sum _{i=0}^{N-1} \vert \varphi _i{\rangle }{\langle }\varphi _i \vert \otimes \vert {\tilde{\varphi }}_i{\rangle }{\langle }{\tilde{\varphi }}_i \vert \otimes \vert 0{\rangle }{\langle }0 \vert . \end{aligned}$$

(A3)

It is obvious that only energy measurements (or measurements of energy differences) are performed in QQMA. It is essential to observe that it is impossible to realize this projector with energy measurements alone if the spectrum is degenerate. The issue is that the above projector does depend on the chosen eigenbasis $\vert \varphi _i{\rangle }$. More precisely, let

$$\begin{aligned} \Lambda _1' = \sum _{i=0}^{N-1} \vert \psi _i{\rangle }{\langle }\psi _i\vert \otimes \vert {\tilde{\psi }}_i{\rangle }{\langle }{\tilde{\psi }}_i\vert \otimes \vert 0{\rangle }{\langle }0 \vert . \end{aligned}$$

(A4)

The problem is now that these two projectors are not equal, that is $\Lambda _1 \ne \Lambda _1'$. (A simple example that demonstrates inequality of the projectors $\Lambda _1$ and $\Lambda '_1$ is $\vert \phi _0{\rangle } = \vert 0{\rangle }, \vert \phi _1{\rangle } = \vert 1{\rangle }$ and $\vert \psi _0{\rangle } = \vert +{\rangle }, \vert \psi _1{\rangle } = \vert -{\rangle }$.) This is in contrast to the situation for the states $\vert \alpha _0{\rangle }$ and $\vert \beta _0{\rangle }$. Clearly, if one can only compare energies, then one cannot distinguish between different bases of a degenerate Hamiltonian. Thus, the authors implicitly assume that the Hamiltonian has non-degenerate spectrum in [23].

While projectors of the form $\Lambda _1$ are unphysical, it is possible to implement a projector (assuming perfect energy estimation) of the form

$$\begin{aligned} \sum _E P_E \otimes {\tilde{P}}_E, \end{aligned}$$

(A5)

where E ranges over the eigenvalues of the Hamiltonian and $P_E$ over the projectors onto the corresponding eigensubspaces.

Observe that in the classical setting there is a way out. Although some energies of the classical Hamiltonian are equal, there is always a unique preferred basis—namely, the computational basis. But such preferred eigenbasis does not exist in the quantum case for degenerate Hamiltonians unless some additional assumptions are made.

Appendix B: Gap Amplification

The appendix is an adaptation of [33, 17.2 How to quantize a Markov chain].

Definition 6

(Quantization). Let $Q\in \mathbbm {C}^{M\times M}$ be a Hermitian matrix, $\vert \varphi _1{\rangle },\ldots , \vert \varphi _M{\rangle }$ an orthonormal basis of $\mathbbm {C}^M$ consisting of eigenvectors of Q with eigenvalues $\lambda _1=1\gneqq \lambda _2 = 1 - \Delta \ge \lambda _3 \ge \ldots \ge \lambda _M \ge -1$. We refer to $\Delta$ as the classical gap.

Assume that $T\in \mathbbm {C}^{N\times M}$ is an isometry from $\mathbbm {C}^M$ to $\mathbbm {C}^N$ and $S\in \mathbbm {C}^{N\times N}$ is a reflection acting on $\mathbbm {C}^N$ such that

$$\begin{aligned} T^\dagger S T = Q. \end{aligned}$$

(B6)

Then, we define the quantization of Q to be

$$\begin{aligned} U=S(2\Pi - I)\in \mathbbm {C}^{N\times N}, \end{aligned}$$

(B7)

where $\Pi =TT^\dagger$ is an orthogonal projector in $\mathbbm {C}^{N\times N}$. Let $\mathcal {A}$ denote the image of T (or equivalently the image of $\Pi$). We refer to the value $\theta =\arccos (1-\Delta )$ as the phase gap of the quantization U of Q.

For $j\in \{1,\ldots ,M\}$, define the states

$$\begin{aligned} \vert \chi _j{\rangle }=T \vert \varphi _j{\rangle } \in \mathbbm {C}^N. \end{aligned}$$

(B8)

Let ${\mathcal {B}}$ be the subspace ${\mathcal {A}}+S{\mathcal {A}}$, where $S{\mathcal {A}}=\{ S \vert \chi {\rangle }: \vert \chi {\rangle }\in {\mathcal {A}}\}$, and ${\mathcal {B}}^\perp$ the orthogonal complement of ${\mathcal {B}}$.

Lemma 9

The subspace spanned $\vert \chi _1{\rangle },\ldots , \vert \chi _M{\rangle }$ coincides with the subspace $\mathcal {A}$.

Proof

We have $\sum _{j\in [M]} \vert \chi _j{\rangle }{\langle }\chi _j \vert = \sum _{j\in [M]} T \vert \varphi _j{\rangle }{\langle }\varphi _j \vert T^\dagger = T T^\dagger = \Pi$. $\square$

Theorem 5

(Spectrum of quantization). The subspace ${\mathcal {B}}$ and its orthogonal complement ${\mathcal {B}}^\perp$ are invariant under U. The spectrum of U restricted to ${\mathcal {B}}$ is as follows:

• 1. For $j=1$, the one-dimensional subspace $\mathcal {V}_1$ spanned by $\vert \chi _1{\rangle }$ is invariant under U and the eigenvector of U in $\mathcal {V}_1$ is

  $$\begin{aligned} \vert \psi _1{\rangle } = \vert \chi _1{\rangle } \in {\mathcal {B}} \end{aligned}$$

  (B9)

  with eigenvalue 1.

• 2. For $j\ge 2$, the two-dimensional subspace $\mathcal {V}_j$ spanned by $\vert \chi _j{\rangle }$ and $S \vert \chi _j{\rangle }$ is invariant under U and the two orthogonal eigenvectors of U in $\mathcal {V}_j$ are

  $$\begin{aligned} \vert \psi ^{\pm }_j{\rangle } = \vert \chi _j{\rangle } - \mu ^{\pm }_j S \vert \chi _j{\rangle } \in {\mathcal {B}} \end{aligned}$$

  (B10)

  with corresponding eigenvalues $\mu ^{\pm }_j$, where

  $$\begin{aligned} \mu ^{\pm }_j = \lambda _j \pm i \sqrt{1-\lambda _j^2} = e^{\pm i \, \textrm{arccos}(\lambda _j)}. \end{aligned}$$

  (B11)

Proof

Let j be arbitrary. We have

$$\begin{aligned} U \vert \chi _j{\rangle }&= S (2\Pi - I) \vert \chi _j{\rangle } \end{aligned}$$

(B12)

$$\begin{aligned}&= S (T T^\dagger - I) T \vert \varphi _j{\rangle } \end{aligned}$$

(B13)

$$\begin{aligned}&= 2 S T \vert \varphi _j{\rangle } - ST \vert \varphi _j{\rangle } \end{aligned}$$

(B14)

$$\begin{aligned}&= S \vert \chi _j{\rangle } \end{aligned}$$

(B15)

and

$$\begin{aligned} U S \vert \chi _j{\rangle }&= S(2\Pi - I) S T \vert \varphi _j{\rangle } \end{aligned}$$

(B16)

$$\begin{aligned}&= S(2 T T^\dagger - I) ST \vert \varphi _j{\rangle } \end{aligned}$$

(B17)

$$\begin{aligned}&= (2S T T^\dagger S T - T) \vert \varphi _j{\rangle } \end{aligned}$$

(B18)

$$\begin{aligned}&= (2S T Q - T) \vert \varphi _j{\rangle } \end{aligned}$$

(B19)

$$\begin{aligned}&= (2\lambda _j S T - T) \vert \varphi _j{\rangle } \end{aligned}$$

(B20)

$$\begin{aligned}&= (2\lambda _j S - I) \vert \chi _j{\rangle }. \end{aligned}$$

(B21)

We see that the subspace spanned by $\vert \chi _j{\rangle }$ and $S\vert \chi _j{\rangle }$ is invariant under U for each j. Thus, we can find eigenvectors of U within this subspace.

First, we consider the case $j=1$. We have

$$\begin{aligned} 1 = {\langle }\varphi _1 \vert Q \vert \varphi _1{\rangle } = {\langle }\varphi _1 \vert T^\dagger S T \vert \varphi _1{\rangle } = {\langle }\chi _1 \vert S \vert \chi _1{\rangle }, \end{aligned}$$

(B22)

implying that $\vert \chi _1{\rangle }$ is an eigenvector of S with eigenvalue 1. Therefore, the subspace $\mathcal {V}_1$ is one-dimensional. Moreover, we have

$$\begin{aligned} U \vert \chi _1{\rangle } = S(2\Pi - I) \vert \chi _1{\rangle } = S \vert \chi _1{\rangle } = \vert \chi _1{\rangle }, \end{aligned}$$

(B23)

where we used $\Pi \vert \chi _1{\rangle }=T T^\dagger T \vert \varphi _1{\rangle }= T \vert \varphi _1{\rangle }= \vert \chi _1{\rangle }$, implying that $\vert \chi _1{\rangle }$ is an eigenvector of U with eigenvalue 1.

Second, we consider the case $j\ge 2$, that is, $\lambda _j$ is bounded away from 1 by at least the spectral gap $\Delta$. Let

$$\begin{aligned} \vert \psi ^{\pm }_j{\rangle } = \vert \chi _j{\rangle } - \mu ^{\pm }_j S \vert \chi _j{\rangle } \end{aligned}$$

(B24)

be an ansatz for the eigenvectors of U supported on $\mathcal {V}_j$. We have

$$\begin{aligned} U \vert \psi ^\pm _j{\rangle }&= S \vert \chi _j{\rangle } - \mu ^\pm _j (2\lambda _j S - I) \vert \chi _j{\rangle } \end{aligned}$$

(B25)

$$\begin{aligned}&= \mu ^\pm _j \vert \chi _j{\rangle } - (2\lambda _j \mu ^\pm _j - 1) S \vert \chi _j{\rangle }. \end{aligned}$$

(B26)

Therefore, $\vert \psi ^\pm _j{\rangle }$ is an eigenvector of U with eigenvalue $\mu ^\pm _j$ provided that

$$\begin{aligned} (\mu ^\pm _j)^2 - 2 \lambda _j\mu ^\pm _j + 1 = 0, \end{aligned}$$

(B27)

that is,

$$\begin{aligned} \mu ^\pm _j = \lambda _j \pm i \sqrt{1 - \lambda _j^2} = e^{\pm i \, \textrm{arccos}(\lambda _j)}. \end{aligned}$$

(B28)

This implies that the subspaces $V_j$ with $j \ge 2$ are two-dimensional.

The union of all the subspaces $\mathcal {V}_j$ is $\mathcal {B}$, implying that $\mathcal {B}$ and its orthogonal complement $\mathcal {B}^\perp$ are invariant under U. $\square$

Lemma 10

(Lower bound on quantum gap). Let $\Delta$ denote the classical gap of Q. The quantum gap of the corresponding unitary U is $\theta =\arccos (1-\Delta )$ and is quadratically larger than the classical gap because

$$\begin{aligned} \theta \ge \sqrt{2\Delta }. \end{aligned}$$

(B29)

Proof

Write the second largest eigenvalue as $\lambda _2 = 1 - \Delta = \cos (\theta )$ for a suitable $\theta$. We have

$$\begin{aligned} 1 - \Delta = \cos (\theta ) \ge 1 - \frac{\theta ^2}{2}, \end{aligned}$$

(B30)

which implies the bound $\theta \ge \sqrt{2\Delta }$. $\square$