Two-path coherence Gram spectrum and Kraus weights

Set $r=\mathrm{\mid }\chi \mathrm{\mid }$ and choose a unit complex number $u$ with $\chi =ru$; when $r=0$ choose any such $u$. Direct multiplication gives $G(u,\pm 1{)}^T=(1\pm r)(u,\pm 1{)}^T$. The normalized vectors $(u,\pm 1{)}^T\mathrm{/}\sqrt{2}$ are orthonormal and span ${\mathbb{C}}^2$, so these are all eigenvalues. Since $0\le r\le 1$, both are nonnegative.

Put $A_{\pm }=\sqrt{p(1\pm r)\mathrm{/}2}(u{F}_0\pm F_1)$. Expanding the two terms gives $\mathrm{\Phi }(X)=A_{+}X{A}_{+}^{†}+A_{-}X{A}_{-}^{†}$: the diagonal coefficients are $p$, and the coefficient of $F_{0}X{F}_1^{†}$ is $pru=p\chi$. Also $\mathrm{Tr}(F_j^{†}{F}_k)={\delta }_{jk}$, hence $\mathrm{Tr}(A_{\pm }^{†}{A}_{\pm })=p(1\pm r)$ and $\mathrm{Tr}(A_{+}^{†}{A}_{-})=0$. For any $A:H\to K$, the diagonal entries of $A^{†}A$ are $\mathrm{\parallel }A{e}_j{\mathrm{\parallel }}^2$, so $\mathrm{\parallel }A{\mathrm{\parallel }}_{HS,E}^2={\sum }_{j=0}^1\mathrm{\parallel }A{e}_j{\mathrm{\parallel }}^2=\mathrm{Tr}(A^{†}A)$. Thus the displayed family is a Kraus representation in the sense of [F5], orthogonal in the sense of [F6], with the asserted squared Hilbert–Schmidt norms. Zero operators are allowed when a weight vanishes.

The path orthonormality gives $F_j^{†}{F}_k=0$ for $j\mathrm{\ne }k$ and $F_0^{†}{F}_0+F_1^{†}{F}_1=I_H$. Consequently $A_{+}^{†}{A}_{+}+A_{-}^{†}{A}_{-}=p{I}_H$. Put $V=F_0+F_1$, $B=\sqrt{1-p}V$ and $\mathrm{\Psi }(X)=BX{B}^{†}$; then $V^{†}V=I_H$ and $B^{†}B=(1-p)I_H$. For any $C:H\to K$, the finite block amplification of $X\mapsto CX{C}^{†}$ sends $Y\ge 0$ on $H\otimes {\mathbb{C}}^m$ to $(C\otimes I_m)Y(C\otimes I_m{)}^{†}\ge 0$: its quadratic form at $z$ is the nonnegative quadratic form of $Y$ at $(C\otimes I_m{)}^{†}z$. Thus [F9] makes $\mathrm{\Phi }$ and $\mathrm{\Psi }$ completely positive, including their zero cases. Finite coordinate summation gives $\mathrm{Tr}(CX{C}^{†})=\mathrm{Tr}(C^{†}CX)$, so $\mathrm{Tr}\mathrm{\Phi }(X)=p\mathrm{Tr}X$ and $\mathrm{Tr}\mathrm{\Psi }(X)=(1-p)\mathrm{Tr}X$ for every $X$. Hence $(\mathrm{\Phi },\mathrm{\Psi })$ is a two-outcome instrument by [F8], for all $0\le p\le 1$.

Expanding the Choi definition in [F3] on the matrix units gives $J(\mathrm{\Phi })={\sum }_{i,j=0}^1\mathrm{\mid }{e}_i⟩⟨e_j\mathrm{\mid }\otimes {\sum }_{s\in \{+,-\}}\mathrm{\mid }{A}_s{e}_i⟩⟨A_s{e}_j\mathrm{\mid }={\sum }_{s\in \{+,-\}}\mathrm{\mid }{A}_s⟩\text{\hspace{0.17em}⁣}⟩⟨\text{\hspace{0.17em}⁣}⟨A_s\mathrm{\mid }$. These vectors are orthogonal and have squared norms $p(1\pm r)$ by the same coordinate trace calculation, so the nonzero Choi spectrum has exactly the stated values, with any zero values omitted from the nonzero spectrum. This calculation applies directly to the branch map; it does not assume that the branch alone is trace-preserving.

For input $\mathrm{\mid }{e}_j⟩⟨e_j\mathrm{\mid }$, the branch output is $p\mathrm{\mid }{f}_j⟩⟨f_j\mathrm{\mid }$, whose trace is $p$. This input is nonnegative with trace one, so [F8] and step 3.1 identify $p$ as its probability for the $\mathrm{\Phi }$ outcome. If $p>0$, the weight sum is $2p$, and the smaller weight divided by this sum is $(1-r)\mathrm{/}2$. At $p=0$ this quotient is $0\mathrm{/}0$ and is undefined. The minus direction is the difference of the two paths after the phase adjustment $F_0\mapsto u{F}_0$; when $r=0$ and $p>0$, the two nonzero Choi eigenvalues coincide and any orthonormal basis of their two-dimensional eigenspace is valid. When $p=0$, step 3.2 gives $J(\mathrm{\Phi })=0$, whose zero eigenspace is all of $H\otimes K$. Rephasing conjugates the coefficient matrix by a diagonal unitary, leaving $r$ and the spectrum unchanged. Finally, [F4] supplies the displayed matrix properties for the stated phase-average specialization; the form of $\mathrm{\Phi }$ remains an additional hypothesis.