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Let $\mathcal{H}_{AB}$ be a bipartite, complex Euclidean space, and let $U\colon\mathcal{H}_{AB}\to\mathcal{H}_{AB}$ be a unitary operator. Define the trace norm as $$ \lVert X\rVert_1 = \text{Tr}(\sqrt{XX^\dagger}) $$ where $X$ is a linear operator on $\mathcal{H}_{AB}$, and $X^\dagger$ denote the complex conjugate of $X$. I know $\lVert UXU^\dagger\rVert_1 = \lVert X\rVert_1$, but does the following identity hold: $$ \lVert \text{Tr}_B UXU^\dagger\rVert_1 = \lVert \text{Tr}_B X\rVert_1 $$ Assuming finite dimensionality is fine for my application.

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  • $\begingroup$ Welcome to Physics SE! have you tried anything so far? $\endgroup$ Commented Nov 19, 2019 at 16:53
  • $\begingroup$ Yea, I realized it is true, when X is positive semi-definite, but not in general. $\endgroup$
    – user114158
    Commented Nov 19, 2019 at 22:12
  • $\begingroup$ then i'd recommend you post that as an answer :) $\endgroup$ Commented Nov 20, 2019 at 10:08
  • $\begingroup$ What is your question? (For all U? Does there exist U? etc. etc.) $\endgroup$ Commented Nov 24, 2019 at 17:09

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If $X$ is positive, then $\|X\|_1=\operatorname{Tr}(X)$. Moreover, if $X$ is positive, then also $\operatorname{Tr}_B(X)$ is. Therefore, if $X$ is positive, then $\|\mathrm{Tr}_B(X)\|_1=\operatorname{Tr}(X)$. Finally, if $X$ is positive and $U$ unitary, also $UXU^\dagger$ is positive, and therefore $$\|\mathrm{Tr}_B(UXU^\dagger)\|_1=\mathrm{Tr}(UXU^\dagger)=\operatorname{Tr}(X)=\|\mathrm{Tr}_B X\|_1.$$

The statement is false when $X$ is not positive. As a counter-example consider $$X=\begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$ Then, $$\operatorname{Tr}_B(X)=\begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix}$$ and thus $\|\mathrm{Tr}_B(X)\|_1=2$. However, taking $$U = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix},$$ we have $$ UXU^\dagger = \begin{pmatrix} 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, \qquad \mathrm{Tr}_B(UXU^\dagger)=\begin{pmatrix}1&1\\0&1\end{pmatrix}, $$ and thus clearly $$\|\mathrm{Tr}_B(UXU^\dagger)\|_1 \simeq 2.23 \neq 2=\|\mathrm{Tr}_B(X)\|_1. $$

Following Norbert's suggestion in the comments, a better example using normal matrices might be given by a matrix of the form $X=A\otimes I/d$ with $A$ Hermitian and $U$ swapping the two spaces, so that $UXU^\dagger = I/d\otimes A$. Then, $\operatorname{Tr}_2(X)=A$, $\|\mathrm{Tr}_2(X)\|_1=\|A\|_1$, whereas $\|\mathrm{Tr}_2(U XU^\dagger)\|_1=\operatorname{Tr}(A)$, and $\operatorname{Tr}(A)\neq\|A\|_1$ unless $A\ge0$.

This question on math.SE might also be of interest.

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  • $\begingroup$ How would even ||X||=tr(X) hold for normal matrices with non-positive eigenvalues? -- I mean, just take $\sigma_z\otimes I$ and let $U$ swap the two components: In one case, you get $4$, in the other case, you get $0$! (With "case" referring to the two sides of the equation in the original question. But your claim also fails for $X=\sigma_z$.) $\endgroup$ Commented Nov 24, 2019 at 17:05
  • $\begingroup$ @NorbertSchuch definitely. An even easier way to see it is that $\|A\|_1$ equals the sum of the singular values, so for for $A$ normal non-positive we have the sum of the absolute values of the eigenvalues on one side and the sum of the eigenvalues on the other. I'm not sure why I wrote "normal" rather than "positive" in the answer. $\endgroup$
    – glS
    Commented Nov 24, 2019 at 17:45
  • $\begingroup$ Ok. BTW, a hermitian counterexample would probably be stronger. $\endgroup$ Commented Nov 24, 2019 at 19:45
  • $\begingroup$ @NorbertSchuch I agree, thanks for the suggestion. I added a counterexample on the lines of your proposal $\endgroup$
    – glS
    Commented Nov 24, 2019 at 20:10

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