Pattern branch probabilities and zero-probability transcripts

For $0\le j\le q$, put $R_j:=V\setminus \{m_1,\dots ,m_j\}$ and let ${\mathrm{\Pi }}_j:{\mathcal{H}}_V\to {\mathcal{H}}_{m_1}\otimes \cdots \otimes {\mathcal{H}}_{m_j}\otimes {\mathcal{H}}_{R_j}$ be the wire permutation listing the first $j$ measured vertices in measurement order and the remainder in inherited order. For a prefix $\mathbf{c}=(c_1,\dots ,c_j)$, let $B_{\mathbf{c}}:={⨂}_{t=1}^j{}_{{\alpha }_t(c_{<t})}\text{\hspace{0.17em}⁣}⟨c_t{\mathrm{\mid }}_{m_t}$ be the tensor map of its linear measurement bras, with its scalar output ${\mathbb{C}}^{\otimes j}$ identified with $\mathbb{C}$, and define $K_{\mathbf{c}}:=(B_{\mathbf{c}}\otimes I_{R_j}){\mathrm{\Pi }}_j{U}_G{\iota }_I:{\mathcal{H}}_I\to {\mathcal{H}}_{R_j}$. For $j=0$, the empty tensor map and permutation are identities, so $K_{\mathrm{\varnothing }}=U_G{\iota }_I$. If $j<q$, let $Q_{j+1}:{\mathcal{H}}_{R_j}\to {\mathcal{H}}_{m_{j+1}}\otimes {\mathcal{H}}_{R_{j+1}}$ list the next measured vertex first, and put $L_{b\mathrm{\mid }\mathbf{c}}:=({}_{{\alpha }_{j+1}(\mathbf{c})}\text{\hspace{0.17em}⁣}⟨b{\mathrm{\mid }}_{m_{j+1}}\otimes I_{R_{j+1}})Q_{j+1}$. Under canonical associations, $B_{\mathbf{c}b}=B_{\mathbf{c}}\otimes {}_{{\alpha }_{j+1}(\mathbf{c})}\text{\hspace{0.17em}⁣}⟨b{\mathrm{\mid }}_{m_{j+1}}$ and ${\mathrm{\Pi }}_{j+1}=(I_{{\mathcal{H}}_{m_1}\otimes \cdots \otimes {\mathcal{H}}_{m_j}}\otimes Q_{j+1}){\mathrm{\Pi }}_j$. Tensor-map factorwise action and permutation composition therefore give $K_{\mathbf{c}b}=L_{b\mathrm{\mid }\mathbf{c}}{K}_{\mathbf{c}}$. At $j=q$, $R_q=O$, ${\mathrm{\Pi }}_q={\mathrm{\Pi }}_{M,O}$, and this recursive map is exactly the full branch map in the Definition. Set ${\sigma }_{\mathbf{c}}:=K_{\mathbf{c}}\rho K_{\mathbf{c}}^{†}$. At the root it is nonnegative since it is a congruence of $\rho$ by $U_G{\iota }_I$. At every prefix, nonnegativity of ${\sigma }_{\mathbf{c}}$ gives $⟨L_{b\mathrm{\mid }\mathbf{c}}{\sigma }_{\mathbf{c}}{L}_{b\mathrm{\mid }\mathbf{c}}^{†}v,v⟩=⟨{\sigma }_{\mathbf{c}}{L}_{b\mathrm{\mid }\mathbf{c}}^{†}v,L_{b\mathrm{\mid }\mathbf{c}}^{†}v⟩\ge 0$, so each child state is nonnegative. Orthonormality in [F2] gives ${\sum }_{b=0}^1{L}_{b\mathrm{\mid }\mathbf{c}}^{†}{L}_{b\mathrm{\mid }\mathbf{c}}=I_{R_j}$. By trace cyclicity [F4], the two child traces sum to $\mathrm{tr}({\sigma }_{\mathbf{c}})$.

The root state is $U_G{\iota }_I\rho {\iota }_I^{†}{U}_G^{†}$, where ${\iota }_I$ tensors the unit $\mathrm{\mid }+⟩$ ancillas and $U_G$ is the finite edge-gate product in [F1]. By [F8], ${\iota }_I^{†}{\iota }_I=I$ and $U_G^{†}{U}_G=I$; trace cyclicity [F4] therefore gives root trace $\mathrm{tr}(\rho )=1$. Repeatedly applying the trace identity of step 1.1 down the finite binary tree preserves the sum of leaf traces. By the terminal identity in step 1.1, the leaf state is the stated $K_{\mathbf{b}}\rho K_{\mathbf{b}}^{†}$, so ${\sum }_{\mathbf{b}}\mathrm{tr}(K_{\mathbf{b}}\rho K_{\mathbf{b}}^{†})=1$. When there are no measurements, the unique empty transcript has the same trace as this root state.

For a pure input $\rho =\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }$, the branch output is $\mathrm{\mid }{K}_{\mathbf{b}}\psi ⟩⟨K_{\mathbf{b}}\psi \mathrm{\mid }$ by [F6], so its trace is $\mathrm{\parallel }{K}_{\mathbf{b}}\mathrm{\mid }\psi ⟩{\mathrm{\parallel }}^2$ by the coordinate formula in [F6]. For a density input the branch probability is the branch-output trace in step 2.1.

Each branch output $K_{\mathbf{b}}\rho K_{\mathbf{b}}^{†}$ is nonnegative because $⟨K_{\mathbf{b}}\rho K_{\mathbf{b}}^{†}v,v⟩=⟨\rho K_{\mathbf{b}}^{†}v,K_{\mathbf{b}}^{†}v⟩\ge 0$ by [F3]. Its trace is therefore nonnegative by [F5]. If that trace is zero, the nonnegative eigenvalues in [F5] sum to zero by [F4], so each is zero and the output operator is zero. For a pure input, step 3.1 then gives $K_{\mathbf{b}}\mathrm{\mid }\psi ⟩=0$. In either case [F7] leaves no normalized conditional state; for positive probability the indicated division is defined.