Phase recovery, MaxCut and complex semidefinite programming

Abstract

Phase retrieval seeks to recover a signal $x \in {\mathbb {C}}^p$ from the amplitude $|A x|$ of linear measurements $Ax \in {\mathbb {C}}^n$. We cast the phase retrieval problem as a non-convex quadratic program over a complex phase vector and formulate a tractable relaxation (called PhaseCut) similar to the classical MaxCut semidefinite program. We solve this problem using a provably convergent block coordinate descent algorithm whose structure is similar to that of the original greedy algorithm in Gerchberg and Saxton (Optik 35:237–246, 1972), where each iteration is a matrix vector product. Numerical results show the performance of this approach over three different phase retrieval problems, in comparison with greedy phase retrieval algorithms and matrix completion formulations.

Appendix: Technical lemmas

We now prove two technical lemmas used in the proof of Theorem 1.

Lemma 3

Under the assumptions and notations of Theorem 1, we have

$$\begin{aligned} \Vert V_{PC}^{/\!\!/}-(Ax_0)(Ax_0)^*\Vert _2> 2C\Vert Ax_0\Vert _ 2\Vert b_\mathrm{n,PC}\Vert _2 \end{aligned}$$

Proof

We first give an upper bound of $\Vert V_{PC}-V_{PC}^{/\!\!/}\Vert _2$. We use the Cauchy–Schwarz inequality: for every positive matrix $X$ and all $x,y, |x^*Xy|\le \sqrt{x^*Xx}\sqrt{y^*Xy}$. Let $\{f_i\}$ be an hermitian base of $\mathrm{range}(A)$ diagonalizing $V_{PC}^{/\!\!/}$ and $\{g_i\}$ an hermitian base of $\mathrm{range}(A)^\perp$ diagonalizing $V_{PC}^\perp$. As $\{f_i\}\cap \{g_i\}$ is an hermitian base of ${\mathbb {C}}^n$, we have

$$\begin{aligned} \Vert V_{PC}-V_{PC}^{/\!\!/}\Vert _2^2&=\underset{i,i'}{\sum }|f_i^*(V_{PC}-V_{PC}^ {/\!\!/})f_{i'}|^2+\underset{i,j}{\sum }|f_i^*(V_{PC}-V_{PC}^{/\!\!/})g_j|^2\nonumber \\&\,\,\,\,\,+\underset{i,j}{\sum }|g_j^*(V_{PC}-V_{PC}^{/\!\!/})f_i|^2+ \underset{j,j'}{\sum }|g_j^*(V_{PC}-V_{PC}^{/\!\!/})g_{j'}|^2\nonumber \\&=2\underset{i,j}{\sum }|f_i^*(V_{PC})g_j|^2+\underset{i}{\sum }|g_i^*(V_{PC}^\perp ) g_i|^2\nonumber \\&\le 2\underset{i,j}{\sum }|f_i^*(V_{PC})f_i\Vert g_j^*(V_{PC})g_j|+\left( \underset{i}{ \sum } g_i^*(V_{PC}^\perp )g_i\right) ^2\nonumber \\&=2\mathbf{Tr}V_{PC}^{/\!\!/}\mathbf{Tr}V_{PC}^\perp +(\mathbf{Tr}V_{PC}^\perp )^2\nonumber \\&\le \left( \sqrt{2}\sqrt{\mathbf{Tr}V_{PC}^{/\!\!/}}\sqrt{\mathbf{Tr}V_{PC}^\perp }+ \mathbf{Tr}V_{PC}^\perp \right) ^2 \end{aligned}$$

(21)

Let us now bound $\mathbf{Tr}V_{PC}^\perp$. We first note that $\mathbf{Tr}V_{PC}^\perp =\mathbf{Tr}((\mathbf{I}-AA^\dag )V_{PC}(\mathbf{I}-AA^\dag ))=\mathbf{Tr}(V_{PC}(\mathbf{I}-AA^\dag ))=d_1(V_{PC},\mathcal {F})$ (according to Lemma 2). Let $u\in {\mathbb {C}}^n$ be such that, for all $i$, $|u_i|=1$ and $(Ax_0)_i=u_i|Ax_0|_i$. We set $b=|Ax_0|+b_\mathrm{n,PC}$ and $V=(b\times u)(b\times u)^*$. As $V\in \mathbf{H}_n^+\cap \mathcal {H}_b$ and $V_{PC}$ minimizes (13),

$$\begin{aligned} \mathbf{Tr}V_{PC}^\perp&= d_1(V_{PC},\mathcal {F})\le d_1(V,\mathcal {F})=d_1((Ax_0+b_\mathrm{n,PC}u)(Ax_0+b_\mathrm{n,PC}u)^*,\mathcal {F})\nonumber \\&= d_1((b_\mathrm{n,PC}u)(b_\mathrm{n,PC}u)^*,\mathcal {F})\nonumber \\&\le \Vert (b_\mathrm{n,PC}u)(b_\mathrm{n,PC}u)^*\Vert _1=\mathbf{Tr}(b_\mathrm{n,PC}u)(b_\mathrm{n,PC}u)^*=\Vert b_\mathrm{n,PC}\Vert _2^2 \end{aligned}$$

We also have $\mathbf{Tr}V_{PC}^{/\!\!/}=\mathbf{Tr}V_{PC}-\mathbf{Tr}V_{PC}^\perp$. This equality comes from the fact that, if $\{f_i\}$ is an hermitian base of $\mathrm{range}(A)$ and $\{g_i\}$ an hermitian base of $\mathrm{range}(A)^\perp$, then

$$\begin{aligned} \mathbf{Tr}V_{PC}&= \underset{i}{\sum }f_iV_{PC}f_i^*+\underset{i}{\sum }g_iV_{PC}g_i^*= \underset{i}{\sum }f_iV_{PC}^{/\!\!/}f_i^*+\underset{i}{\sum }g_iV_{PC}^\perp g_i^*\\&= \mathbf{Tr}V_{PC}^{/\!\!/}+\mathbf{Tr}V_{PC}^\perp \end{aligned}$$

As $V_{PC}^\perp \succeq 0$, $\mathbf{Tr}V_{PC}^{/\!\!/}\le \mathbf{Tr}V_{PC}=\Vert |Ax_0|+b_\mathrm{n,PC}\Vert _2^2$ and, by combining this with relations (21) and (22), we get

$$\begin{aligned} \Vert V_{PC}-V_{PC}^{/\!\!/}\Vert _2&\le \sqrt{2}\Vert |Ax_0|+b_\mathrm{n,PC}\Vert _2\Vert b_\mathrm{n,PC}\Vert _2+\Vert b_\mathrm{n,PC}\Vert _2^2\\&\le \sqrt{2}\Vert Ax_0\Vert _2\Vert b_\mathrm{n,PC}\Vert _2+(1+\sqrt{2})\Vert b_\mathrm{n,PC}\Vert _2^2 \end{aligned}$$

And, by reminding that we assumed $\Vert b_\mathrm{n,PC}\Vert _2\le \Vert Ax_0\Vert _2$,

$$\begin{aligned} \Vert V_{PC}^{/\!\!/}-(Ax_0)(Ax_0)^*\Vert _2&\ge \Vert V_{PC}-(Ax_0)(Ax_0)^*\Vert _2- \Vert V_{PC}^{/\!\!/}-V_{PC}\Vert _2\\&> D\Vert Ax_0\Vert _2\Vert b_\mathrm{n,PC}\Vert _2-\sqrt{2}\Vert Ax_0\Vert _2\Vert b_\mathrm{n,PC} \Vert _2\\&\quad -(1+\sqrt{2})\Vert b_\mathrm{n,PC}\Vert _2^2\\&\ge (D-2\sqrt{2}-1)\Vert Ax_0\Vert _2\Vert b_\mathrm{n,PC}\Vert _2=2C\Vert Ax_0\Vert _2\Vert b_\mathrm{n,PC}\Vert _2 \end{aligned}$$

which concludes the proof.$\square$

Lemma 4

Under the assumptions and notations of Theorem 1, we have $\Vert b_\mathrm{n,PL}\Vert _2\le 2\Vert b_\mathrm{n,PC}\Vert$.

Proof

Let $e_i$ be the $i$-th vector of ${\mathbb {C}}^n$’s canonical base. We set $e_i=f_i+g_i$ where $f_i\in \mathrm{range}(A)$ and $g_i\in \mathrm{range}(A)^\perp$.

$$\begin{aligned} V_{PC\,ii}&=e_i^*V_{PC}e_i\\&=f_i^*V_{PC}^{/\!\!/}f_i+2\mathrm{Re }(f_i^*V_{PC}g_i)+g_i^*V_{PC}^\perp g_i\\&=V_{PC\,ii}^{/\!\!/}+2\mathrm{Re }(f_i^*V_{PC}g_i)+V_{PC\,ii}^\perp \end{aligned}$$

Because $|f_i^*V_{PC}g_i|\le \sqrt{f_i^*V_{PC}f_i}\sqrt{g_i^*V_{PC}g_i}=\sqrt{V_{PC\,ii}^{/\!\!/}}\sqrt{V_{PC\,ii}^\perp }$,

$$\begin{aligned} \left( \sqrt{V_{PC\,ii}^{/\!\!/}}-\sqrt{V_{PC\,ii}^\perp }\right) ^2\le V_{PC\,ii}\le \left( \sqrt{V_{PC\,ii}^{/\!\!/}}+\sqrt{V_{PC\,ii}^\perp }\right) ^2\\ \quad \Rightarrow \sqrt{V_{PC\,ii}^{/\!\!/}}-\sqrt{V_{PC\,ii}^\perp }\le \sqrt{V_{PC\,ii}}\le \sqrt{V_{PC\,ii}^{/\!\!/}}+\sqrt{V_{PC\,ii}^\perp } \end{aligned}$$

So

$$\begin{aligned} |b_{\mathrm{n,PL,}i}|&=|\sqrt{V_{PC\,ii}^{/\!\!/}}-|Ax_0|_i|\\&\le |\sqrt{V_{PC\,ii}^{/\!\!/}}-\sqrt{V_{PC\,ii}}|+|\sqrt{V_{PC\,ii}}-|Ax_0|_i|\\&\le \sqrt{V_{PC\,ii}^\perp }+b_{\mathrm{n,PC,}i} \end{aligned}$$

and, by (22),

$$\begin{aligned} \Vert b_\mathrm{n,PL}\Vert _2&\le \left\| \left\{ \sqrt{V_{PC\,ii}^\perp }\right\} _i\right\| _2+\Vert b_\mathrm{n,PC}\Vert _2\\&=\sqrt{\mathbf{Tr}V_{PC}^\perp }+\Vert b_\mathrm{n,PC}\Vert _2\le 2\Vert b_\mathrm{n,PC}\Vert _2 \end{aligned}$$

which concludes the proof.$\square$