Quantum tomography benchmarking

Abstract

Recent advances in quantum computers and simulators are steadily leading us toward full-scale quantum computing devices. Due to the fact that debugging is necessary to create any computing device, quantum tomography (QT) is a critical milestone on this path. In practice, the choice between different QT methods faces the lack of comparison methodology. Modern research provides a wide range of QT methods, which differ in their application areas, as well as experimental and computational complexity. Testing such methods is also being made under different conditions, and various efficiency measures are being applied. Moreover, many methods have complex programming implementations; thus, comparison becomes extremely difficult. In this study, we have developed a general methodology for comparing quantum-state tomography methods. The methodology is based on an estimate of the resources needed to achieve the required accuracy. We have developed a software library (in MATLAB and Python) that makes it easy to analyze any QT method implementation through a series of numerical experiments. The conditions for such a simulation are set by the number of tests corresponding to real physical experiments. As a validation of the proposed methodology and software, we analyzed and compared a set of QT methods. The analysis revealed some method-specific features and provided estimates of the relative efficiency of the methods.

Appendix: Equivalence between random unitary error and depolarization

Let us show the validity of (8) for the state $|0\rangle$. (Due to the unitary invariance of the distribution of random states $|g\rangle$, the proof for any state $|\varphi \rangle =U_\varphi |0\rangle$ with some unitary operator $U_\varphi$ is carried out in a similar way.)

Theorem 1

Let the action of a random unitary operator $W_{g,a}$ on the state $|0\rangle$ in a Hilbert space of dimension d be given by the expression

$$\begin{aligned} W_{g,a}|0\rangle = a|0\rangle + \sqrt{1-a^2}\frac{|g\rangle -g_0|0\rangle }{\sqrt{1-|g_0 |^2}}, \end{aligned}$$

where a is a fixed non-negative parameter and $g_0=\langle {0}|{g}\rangle$. If vectors $|g\rangle$ are uniformly distributed according to the Haar measure, then the following equality holds

$$\begin{aligned} \int {W_{g,a} |{0}\rangle \langle {0}| W_{g,a}^\dagger dg} = (1-p_a)|{0}\rangle \langle {0}| + p_aI_d/d, \end{aligned}$$

where $p_a = \frac{d}{d-1}(1-a^2)$.

Proof

The matrix representation of the vector $W_{g,a}|0\rangle$ in the computational basis has the form $\left( a \ \, \sqrt{1-a^2}\tilde{\mathbf {g}}^T\right) ^T$, where the column vector $\tilde{\mathbf {g}}$ specifies the amplitudes of a uniformly distributed random vector (according to the Haar measure) in the space of dimension $d-1$. The corresponding density matrix has the form

$$\begin{aligned} \begin{pmatrix} a^2 &{} \quad &{} a\sqrt{1-a^2}\tilde{\mathbf {g}}^\dagger \\ a\sqrt{1-a^2}\tilde{\mathbf {g}} &{} \quad &{} (1-a^2)\tilde{\mathbf {g}}\tilde{\mathbf {g}}^\dagger \end{pmatrix}. \end{aligned}$$

Since the amplitudes of the vector $\tilde{\mathbf {g}}$ are given by normalized complex random variables with standard normal distribution, the expected value of $\tilde{\mathbf {g}}$ is equal to the zero vector. Moreover, the averaging over $\tilde{\mathbf {g}}\tilde{\mathbf {g}}^\dagger$ is proportional to the identity matrix. Thus,

$$\begin{aligned} \int {W_{g,a} |{0}\rangle \langle {0}| W_{g,a}^\dagger dg} = \int {\begin{pmatrix} a^2 &{} \quad &{} a\sqrt{1-a^2}\tilde{\mathbf {g}}^\dagger \\ a\sqrt{1-a^2}\tilde{\mathbf {g}} &{} \quad &{} (1-a^2)\tilde{\mathbf {g}}\tilde{\mathbf {g}}^\dagger \end{pmatrix}d\tilde{g}} = \begin{pmatrix} a^2 &{} \quad &{} {\mathbf {0}}^T \\ {\mathbf {0}} &{} \quad &{} \frac{1-a^2}{d-1}{\mathbf {I}}_{d-1} \end{pmatrix}. \end{aligned}$$

$\square$

In our case, $a=1-\xi ^2/2$, where $\xi \sim \text {norm}(0, \sigma )$, in which $\sigma$ is a small parameter characterizing the error level ($\sigma$ should be small enough to make a positive almost certainly). Having calculated the expected value ${p = \langle p_a \rangle _\xi = \frac{d}{d-1}(\sigma ^2 - \frac{3}{4} \sigma ^4)}$, we obtain equality (8).