Real phase retrieval and phaseless ambiguity — Examples and Counterexamples

Write $s=\sqrt{3}$. By [F1] and [F2], ${\omega }^2=(-1+is)\mathrm{/}2$, ${\omega }^3=1$, and $\mathrm{\mid }\omega {\mathrm{\mid }}^2=1$, hence $\mathrm{\mid }\omega \mathrm{\mid }=1$. Reindexing the finite sum with $j=n-1$ gives $\widehat{{\tau }_{1}f}(k)={\sum }_{j}f(j){\omega }^{k(j+1)}={\omega }^k\widehat{f}(k)$. Multiplicativity of the modulus [F1] gives $\mathrm{\mid }{\omega }^k\mathrm{\mid }=1$, so every Fourier magnitude is unchanged.

Since $\mathrm{\mid }\omega \mathrm{\mid }=1$, [F1] gives $\bar{\omega }={\omega }^{-1}$. Reindexing with $j=-n$ gives $\widehat{Rf}(k)={\sum }_{j}f(j){\omega }^{-kj}$. Because every $f(j)$ is real, this equals $\overline{{\sum }_{j}f(j){\omega }^{kj}}=\overline{\widehat{f}(k)}$ by the conjugation laws in [F1], so reflection preserves every Fourier magnitude.

The support of $f$ and $-f$ is $\{0,1\}$, while the supports of ${\tau }_{1}f$ and $Rf$ are $\{1,2\}$ and $\{0,2\}$ respectively. Thus neither transformed signal equals $f$ or $-f$. Combined with the invariances in steps 1.1 and 2.1, both transformed signals are distinct from the original up to global sign but have identical Fourier magnitude data.

Let $E_1=\{(t,0):t\in \mathbb{R}\}$ and $E_2=\{(0,t):t\in \mathbb{R}\}$; coordinatewise addition and scalar multiplication show these are subspaces. Since $a_1\in E_1$, minimality of the span gives $\mathrm{span}\{a_1\}\subseteq E_1$, while every $(t,0)=t{a}_1$ belongs to the span because it is a subspace containing $a_1$. Thus $\mathrm{span}({\mathrm{\Phi }}_{\{1\}})=E_1$. The same two inclusions for $a_2$ give $\mathrm{span}({\mathrm{\Phi }}_{\{2\}})=E_2$. Both are proper subspaces of ${\mathbb{R}}^2$, so [F2] says the complement property fails.

The vectors $(1,1)$ and $(-1,1)$ are neither equal nor negatives: equality fails in the first coordinate, while equality with the negative fails in the second coordinate. Thus they define distinct signal classes by [F3].

Their measurements are $((1,1)\cdot a_1,(1,1)\cdot a_2)=(1,1)$ and $((-1,1)\cdot a_1,(-1,1)\cdot a_2)=(-1,1)$. Squaring coordinatewise gives $(1,1)$ in both cases, so [F1] gives equal intensity data on the distinct classes from step 1.2. Together with the failed partition in step 1.1, this exhibits the required ambiguity when the complement property fails.

Since $m=2$, the canonical sign set is exactly $(+,+),(+,-)$ in the stated order. Both budgets permit one solve, so only $(+,+)$ is processed and the algorithm reports budget-exhausted.

For the processed target $y_{++}=(1,2)$, the squared signed residual is $(x-1{)}^2+(-x-2{)}^2=2(x+1\mathrm{/}2{)}^2+9\mathrm{/}2$, minimized uniquely at $x=-1\mathrm{/}2$ by [F2]. Its amplitudes are $(1\mathrm{/}2,1\mathrm{/}2)$, so $J(-1\mathrm{/}2)=(1\mathrm{/}2-1{)}^2+(1\mathrm{/}2-2{)}^2=5\mathrm{/}2$.

For the unprocessed target $y_{+-}=(1,-2)$, the squared signed residual is $(x-1{)}^2+(-x+2{)}^2=2(x-3\mathrm{/}2{)}^2+1\mathrm{/}2$, minimized uniquely at $x=3\mathrm{/}2$ by [F2]. Its amplitudes are $(3\mathrm{/}2,3\mathrm{/}2)$, so $J(3\mathrm{/}2)=(3\mathrm{/}2-1{)}^2+(3\mathrm{/}2-2{)}^2=1\mathrm{/}2<5\mathrm{/}2$. Thus the budget-limited incumbent is not the best candidate over all patterns.

For every $(u,v)\in {\mathbb{R}}^2$, $(u,v)=u{a}_1+v{a}_2=(u-v)a_1+v{a}_3=u{a}_3+(v-u)a_2,$ so each pair spans ${\mathbb{R}}^2$ by [F2]. For independence, a vanishing combination of $a_1,a_2$ has coordinates $(\alpha ,\beta )=(0,0)$; a vanishing combination $\alpha a_1+\beta a_3$ has coordinates $(\alpha +\beta ,\beta )=(0,0)$, forcing $\beta =\alpha =0$; and a vanishing combination $\alpha a_2+\beta a_3$ has coordinates $(\beta ,\alpha +\beta )=(0,0)$, again forcing $\alpha =\beta =0$. Thus every pair is linearly independent, so [F1] gives full spark.

Here $d=2$ and $m=3=2d-1$. Applying [F3] to step 1.1 proves phase retrieval on the stated global-sign quotient; the intensity coordinates are exactly the three squared inner products displayed in the example.

For arbitrary $(u,v)\in {\mathbb{C}}^2$, the three identities $(u,v)=u{a}_1+v{a}_2=(u-v)a_1+v{a}_3=u{a}_3+(v-u)a_2$ show that each of the pairs $\{a_1,a_2\}$, $\{a_1,a_3\}$, and $\{a_2,a_3\}$ spans ${\mathbb{C}}^2$: the span is a subspace containing the pair, so it contains the corresponding displayed combination for every $(u,v)$. Any partition of three labels has at least two labels on one side, and a set containing a spanning pair also spans; hence the complex complement property holds.

The measurements of $x=(1,i)$ are $(1,i,1+i)$ and those of $z=(1,-i)$ are $(1,-i,1-i)$. By [F3] and [F4], their squared moduli are $(1,1,2)$ in both cases.

If $z=\lambda x$ with $\mathrm{\mid }\lambda \mathrm{\mid }=1$, the first coordinate forces $\lambda =1$, while the second would then force $-i=i$, which is false in the field [F3]. Thus $x\approx ̸z$, and step 1.2 proves noninjectivity despite step 1.1. By [F1] and [F2], this does not contradict either real-field result: the ambient field, phase relation, and measurements differ.