Clean-input isometry error bounds output trace distance

Put $A=V-W$, and let $(r_j)$ be an orthonormal basis of $R$. By [F1], diagonalizing $A^{\ast }A$ in a right singular basis shows that $(A\otimes I_R{)}^{\ast }(A\otimes I_R)=A^{\ast }A\otimes I_R$ has the same largest eigenvalue as $A^{\ast }A$: its eigenvectors are the products of right singular vectors of $A$ with the $r_j$, and its eigenvalues are the corresponding squared singular values. Therefore $\mathrm{\parallel }{V}_R-W_R{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=\mathrm{\parallel }A\otimes I_R{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \delta$. For every unit vector $\psi \in H\otimes R$, the vectors $a=V_R\psi$ and $b=W_R\psi$ have norm one and satisfy $\mathrm{\parallel }a-b\mathrm{\parallel }\le \delta$.

For $Z=V_R\rho V_R^{\ast }$, positivity follows from $y^{\ast }Zy=(V_R^{\ast }y{)}^{\ast }\rho (V_R^{\ast }y)\ge 0$ for every $y$; the same holds with $W_R$. For a vector $z\in K\otimes R$, the entry-sum formula gives $z^{\ast }({\mathrm{Tr}}_E\otimes I_R)Zz={\sum }_e(I_K\otimes \mathrm{\mid }e⟩\otimes I_{R}z{)}^{\ast }Z(I_K\otimes \mathrm{\mid }e⟩\otimes I_{R}z)\ge 0$, and likewise for $W_R$; hence both reduced outputs are positive. That formula also gives $\mathrm{tr}(({\mathrm{Tr}}_E\otimes I_R)Z)=\mathrm{tr}Z$. Since $V_R^{\ast }{V}_R=W_R^{\ast }{W}_R=I$, cyclicity of finite matrix trace gives $\mathrm{tr}Z=\mathrm{tr}\rho =1$ for both outputs. Thus ${\rho }_V,{\rho }_W$ are density matrices.

In an orthonormal basis of $\mathrm{span}\{a,b\}$, write $b=za+s{e}_2$, where $z=a^{\ast }b$, $s=\sqrt{1-\mathrm{\mid }z{\mathrm{\mid }}^2}$, and $e_2\perp a$ is unit if $s>0$. The matrix of $a{a}^{\ast }-b{b}^{\ast }$ on this span is Hermitian, has trace zero and determinant $-s^2$, so its eigenvalues are $s,-s$; outside the span it is zero. Thus $\frac{1}{2}\mathrm{\parallel }a{a}^{\ast }-b{b}^{\ast }{\mathrm{\parallel }}_1=s$. Also $s^2=1-\mathrm{\mid }z{\mathrm{\mid }}^2\le 2-2\mathrm{Re}z=\mathrm{\parallel }a-b{\mathrm{\parallel }}^2$, because the difference of the right side and left side is $\mathrm{\mid }1-z{\mathrm{\mid }}^2$. This proves $D(a{a}^{\ast },b{b}^{\ast })\le \delta$ for every pure input, including $s=0$.

Let $X$ be Hermitian on $K\otimes E\otimes R$, and set $Y=({\mathrm{Tr}}_E\otimes I_R)X$. For a matrix $B$ on $K\otimes R$, define $\tilde{B}$ on $K\otimes E\otimes R$ by applying $B$ to the $K,R$ coordinates and leaving the $E$ coordinate fixed. The entry-sum definition of partial trace gives $\mathrm{tr}(B^{\ast }Y)=\mathrm{tr}({\tilde{B}}^{\ast }X)$. Reordering the product basis from $K,E,R$ to $K,R,E$ conjugates $\tilde{B}$ to $B\otimes I_E$, so $\mathrm{\parallel }\tilde{B}{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$ by the same singular-basis argument as in step 1.1. For any square matrix $X=U\mathrm{\Sigma }{V}^{\ast }$, [F1] gives its singular value decomposition, and $\mathrm{\mid }\mathrm{tr}(C^{\ast }X)\mathrm{\mid }=\mathrm{\mid }\mathrm{tr}(V^{\ast }{C}^{\ast }U\mathrm{\Sigma })\mathrm{\mid }\le \mathrm{\parallel }C{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}{\sum }_j{\mathrm{\Sigma }}_{jj}=\mathrm{\parallel }C{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\mathrm{\parallel }X{\mathrm{\parallel }}_1,$ since every diagonal entry of a contraction has modulus at most one. Taking the supremum over $\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 1$, and using the equality case $B=U_Y{V}_Y^{\ast }$ in a singular value decomposition of $Y$, proves $\mathrm{\parallel }Y{\mathrm{\parallel }}_1\le \mathrm{\parallel }X{\mathrm{\parallel }}_1$. This dual formula also gives the triangle inequality and homogeneity for $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_1$ by applying it to a sum and scalar multiple.

For a pure input $\rho =\psi {\psi }^{\ast }$, step 2.1 bounds the trace distance before discarding $E$, and step 2.2 shows the partial trace cannot increase the trace norm of the Hermitian difference. Hence the reduced output distance is at most $\delta$. For general positive trace-one $\rho$, [F2] gives a spectral decomposition $\rho ={\sum }_j{p}_j{\psi }_j{\psi }_j^{\ast }$ with $p_j\ge 0$ and ${\sum }_j{p}_j=1$. The maps and partial trace are linear; the triangle inequality and homogeneity from step 2.2 therefore bound the reduced output distance by ${\sum }_j{p}_j\delta =\delta$.