# When does the identity $\|\mathrm{Tr}_B UXU^\dagger\|_1=\|\mathrm{Tr}_B X\|_1$, with $U$ unitary, hold?

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.

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

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.

• 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 $X=\sigma_z$.) Commented Nov 24, 2019 at 17:05
• @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.
– glS
Commented Nov 24, 2019 at 17:45
• Ok. BTW, a hermitian counterexample would probably be stronger. Commented Nov 24, 2019 at 19:45
• @NorbertSchuch I agree, thanks for the suggestion. I added a counterexample on the lines of your proposal
– glS
Commented Nov 24, 2019 at 20:10