A Partial Derandomization of PhaseLift Using Spherical Designs

Abstract

The problem of retrieving phase information from amplitude measurements alone has appeared in many scientific disciplines over the last century. PhaseLift is a recently introduced algorithm for phase recovery that is computationally tractable, numerically stable, and comes with rigorous performance guarantees. PhaseLift is optimal in the sense that the number of amplitude measurements required for phase reconstruction scales linearly with the dimension of the signal. However, it specifically demands Gaussian random measurement vectors—a limitation that restricts practical utility and obscures the specific properties of measurement ensembles that enable phase retrieval. Here we present a partial derandomization of PhaseLift that only requires sampling from certain polynomial size vector configurations, called $t$ -designs. Such configurations have been studied in algebraic combinatorics, coding theory, and quantum information. We prove reconstruction guarantees for a number of measurements that depends on the degree $t$ of the design. If the degree is allowed to grow logarithmically with the dimension, the bounds become tight up to polylog-factors. Beyond the specific case of PhaseLift, this work highlights the utility of spherical designs for the derandomization of data recovery schemes.

Appendix

Here we briefly state an elementary proof of Lemma 6. In the main text we proved this result using wiring diagrams. The purpose of this is to underline the relative simplicity of wiring diagram calculations. Indeed, the elementary proof below is considerably more cumbersome than its pictorial counterpart.

1.1 Elementary Proof of Lemma 6

Let us choose an arbitrary orthonormal basis $b_1, \ldots ,b_d$ of $V^d$. In the induced basis $\left\{ b_i \otimes b_j \right\} _{i,j=1}^d$ of $V^d \otimes V^d$ the transpositions then correspond to

$$\begin{aligned} \underline{1} = \mathbb {1}\otimes \mathbb {1}= \sum _{i=1}^d b_i b_i^* \otimes \sum _{j=1}^d b_j b_j^* \quad \text {and} \quad \sigma _{(1,2)} = \sum _{i,j=1}^d b_i b_j^* \otimes b_j b_i^*. \end{aligned}$$

This choice of basis furthermore allows us to write down ${{\mathrm{tr}}}_2 (A)$ for $A \in M^d \otimes M^d$ explicitly:

$$\begin{aligned} {{\mathrm{tr}}}_2 (A) = \sum _{i=1}^d \left( \mathbb {1}\otimes b_i^* \right) A \left( \mathbb {1}\otimes b_i \right) . \end{aligned}$$

Consequently we get for $A,B \in H^d$ arbitrary

$$\begin{aligned} {{\mathrm{tr}}}_2 \left( P_{{{\mathrm{Sym}}}^2} A \otimes B \right) = \frac{1}{2} {{\mathrm{tr}}}_2 \left( A \otimes B \right) + \frac{1}{2} {{\mathrm{tr}}}_2 \left( \sigma _{(1,2)} A \otimes B \right) . \end{aligned}$$

The latter term can be evaluated explicitly:

$$\begin{aligned} {{\mathrm{tr}}}_2 \left( \sigma _{(1,2)} A \otimes B \right)&= \sum _{k=1}^d \left( \mathbb {1}\otimes b_k^* \right) \sum _{i,j=1}^d b_i b_j^* \otimes b_j b_i^* A \otimes B \left( \mathbb {1}\otimes b_k \right) \\&= \sum _{i,j,k=1}^d b_i b_j^* A b_k^* b_j b_i^* B b_k = \sum _{i,j=1}^d \langle b_i, B b_j \rangle b_i b_j^* A \\&= \left( \sum _{i=1}^d b_i b_i ^* \right) B \left( \sum _{j=1}^d b_j b_j^* \right) A = \mathbb {1}B \mathbb {1}A = B A, \end{aligned}$$

and the desired result follows. Here we have used the basis representation of the identity, namely $\mathbb {1}= \sum _{i=1}^d b_i b_i^*$.