Proving the inequality $|\mathrm{tr}(AU)|\le \mathrm{tr}|A|$ in Uhlmann's theorem

Question

In Nielsen and Chuang, in the Fidelity section, (Lemma 9.5, page 410 in the 2002 edition), they prove the following.

$$ \mathrm{tr}(AU) = |\mathrm{tr}(|A|VU)| = |\mathrm{tr}(|A|^{1/2}|A|^{1/2}VU)| $$ $$ |\mathrm{tr}(AU)| \leq \sqrt{\mathrm{tr}|A| \mathrm{tr}(U^{\dagger}V^{\dagger}|A|VU)} = \mathrm{tr}|A|$$

First, is $|A|$ the positive matrix in the polar decomposition of $A$? And second, apparently, the second equation comes from the first due to Cauchy Schwartz inequality of Hilbert-Schmidt.

How does the first expression lead to the second?

Answer 1

I'll give a couple of methods to do this:

1. (Using matrix inequalities) The idea is to use CS inequality in the form $$\newcommand{\tr}{\operatorname{Tr}}\lvert \sum_{ij}A_{ij}^* B_{ij}\rvert\le\sqrt{\sum_{ij} \left\lvert A_{ij}\right\rvert^2}\sqrt{\sum_{ij}\left\lvert B_{ij}\right\rvert^2},$$ which in matrix formalism reads $$\lvert\tr(A^\dagger B)\rvert\le\sqrt{\tr(A^\dagger A})\sqrt{\tr(B^\dagger B)}.$$ Therefore, $$\lvert\tr(AU)\rvert=\lvert\tr(\lvert A\rvert VU)\rvert =\lvert\tr(\lvert A\rvert^{1/2} \underbrace{\lvert A\rvert^{1/2} VU}_{D})\rvert \le\sqrt{\tr(\lvert A\rvert^{1/2}\lvert A\rvert^{1/2})}\sqrt{\tr(D^\dagger D)}, $$ and putting in the explicit expression for $D$ you get the result.

2. (Via SVD) Another approach is to leverage the SVD of the operators. Notice that, if the SVD of $A$ reads $$A = \sum_i a_i |a_i^L\rangle\!\langle a_i^R|,$$ then, for any unitary $U$, $AU$ is an operator which has the same singular values and left singular vectors as $A$. Similarly, $A$ and $UA$ have the same singular values and right singular vectors. The question to maximise $\lvert {\rm Tr}(AU)\rvert$ thus amounts to maximising $$\lvert{\rm Tr}(AU)\rvert = \left\lvert \sum_i a_i \langle u_i,a_i^L\rangle\right\rvert$$ over the sets of orthonormal bases $\{\lvert u_i\rangle\}_i$. The connection with the above is via $|u_i\rangle=U^\dagger \lvert a_i^R\rangle$. From the above expression, it is clear that Cauchy-Schwarz implies $$\lvert{\rm Tr}(AU)\rvert \le \sum_i a_i = {\rm Tr}|A|,$$ and that the maximum is achieved when $|u_i\rangle=|a_i^L\rangle$, that is, $U=\sum_i |a_i^R\rangle\!\langle a_i^L\rvert$.

3. (Via SVD, generalised result) Let us try now for a more general result. Relax the constraint of $U$ being unitary, and let it by an arbitrary (squared) matrix. The only difference compared to the previous case is that we can now have non-unit singular values, call these $u_i$. The trace we are interested in now reads $$\lvert{\rm Tr}(AU)\rvert = \left\lvert\sum_i a_i u_i \langle u_i,a_i^L\rangle\right\rvert.$$ We can again use Cauchy-Schwarz to get an upper bound, but this time the singular values $u_i$ are in the way. That is not that big of a problem though: observe that $\|U\|_{\rm op}=\max_i u_i$, and thus we can write $$\lvert{\rm Tr}(AU)\rvert \le \|U\|_{\rm op}{\rm Tr}|A|.$$ We can also, if we want, characterise the $U$ that achieves the maximum: this will be the (or a) unitary operator connecting right and left singular vectors of $A$, as in the previous case.