Finite Gibbs circuit marginal and trace-distance error

Multiply $\mathrm{\mid }{\tilde{\mathrm{\Psi }}}_{\beta }⟩$ by a phase so $s=⟨{\mathrm{\Psi }}_{\beta },{\tilde{\mathrm{\Psi }}}_{\beta }⟩$ is real and $0\le s\le 1$. If $s=1$, their rank-one projections agree. Otherwise write ${\tilde{\mathrm{\Psi }}}_{\beta }=s{\mathrm{\Psi }}_{\beta }+\sqrt{1-s^2}w$, with $w\perp {\mathrm{\Psi }}_{\beta }$ and $\mathrm{\parallel }w\mathrm{\parallel }=1$. On the span of ${\mathrm{\Psi }}_{\beta },w$, the difference of their rank-one projections has matrix $\left(\begin{array}{cc}1-s^2& -s\sqrt{1-s^2}\\ -s\sqrt{1-s^2}& -(1-s^2)\end{array}\right)$, whose eigenvalues are $\pm \sqrt{1-s^2}$; it vanishes on the orthogonal complement. By [F4] and [F5], the pure-state trace distance is $\sqrt{1-s^2}$. Since $\mathrm{\parallel }{\mathrm{\Psi }}_{\beta }-{\tilde{\mathrm{\Psi }}}_{\beta }{\mathrm{\parallel }}^2=2(1-s)$ and $1-s^2\le 2(1-s)$, this distance is at most the vector distance.

For finite matrices $C,D$, $\mathrm{Tr}(CD)={\sum }_{i,j}{C}_{ij}{D}_{ji}=\mathrm{Tr}(DC)$ by interchanging finite indices. Let $X$ be any finite Hermitian trace-zero matrix and write its positive and negative parts $X=X_{+}-X_{-}$. Their traces agree and equal $\mathrm{\parallel }X{\mathrm{\parallel }}_1\mathrm{/}2$. For every effect $0\le E\le I$, this cyclicity and square roots from [F8] give $0\le \mathrm{Tr}(E{X}_{\pm })=\mathrm{Tr}(X_{\pm }^{1\mathrm{/}2}E{X}_{\pm }^{1\mathrm{/}2})\le \mathrm{Tr}{X}_{\pm }$, hence $\mathrm{\mid }\mathrm{Tr}(EX)\mathrm{\mid }\le \mathrm{\parallel }X{\mathrm{\parallel }}_1\mathrm{/}2$. Put $Y={\mathrm{Tr}}_{R}X$. By setting $A=I$ in [F6], $\mathrm{Tr}Y=\mathrm{Tr}X=0$. Its positive spectral projection $E_Y$ is an effect and satisfies $\mathrm{Tr}(E_{Y}Y)=\mathrm{\parallel }Y{\mathrm{\parallel }}_1\mathrm{/}2$ by [F5]. Again by [F6], $\mathrm{Tr}(E_{Y}Y)=\mathrm{Tr}((E_Y\otimes I_R)X)$; diagonalizing $E_Y$ shows $E_Y\otimes I_R$ is an effect. Applying the preceding bound to this effect yields $\mathrm{\parallel }Y{\mathrm{\parallel }}_1\le \mathrm{\parallel }X{\mathrm{\parallel }}_1$.

Apply step 1.1 to the normalized output on the successful branch [F2] and ideal vector from [F3]. Their pure-state trace distance is at most ${\epsilon }_{\mathrm{o}\mathrm{u}\mathrm{t}}$. Set $X=\mathrm{\mid }{\tilde{\mathrm{\Psi }}}_{\beta }⟩⟨{\tilde{\mathrm{\Psi }}}_{\beta }\mathrm{\mid }-\mathrm{\mid }{\mathrm{\Psi }}_{\beta }⟩⟨{\mathrm{\Psi }}_{\beta }\mathrm{\mid }$, which is Hermitian with trace zero. Step 1.2 contracts its trace norm under ${\mathrm{Tr}}_R$. By [F1], the ideal system marginal is ${\rho }_{\beta }$; [F7] ensures both marginals are density operators, so [F4] applies and gives the claimed bound. The conclusion remains conditional on the successful branch and shift event.