A cute trace inequality

Theorem. If A and B are positive semi-definite matrices then

\mathrm{Tr} (AB)^2 \leq \mathrm{Tr} (A^2 B^2)

Note that both sides contain the same matrices the same number of times, just in a different order (ABAB versus AABB). Here is a cute proof due to Bellman.

Proof. First, note that

(AB-BA)^2 = (AB)^2 + (BA)^2 - A B^2 A - B A^2 B.

Next, note that the commutator AB-BA is skew-Hermitian, that is, (AB-BA)^\dagger = -(AB-BA). In other words, AB-BA = iH for some Hermitian matrix H. This means that all eigenvalues of AB-BA are purely imaginary. When we square the matrix, they become negative, i.e., (AB-BA)^2 is negative semi-definite. From this we get

0 \geq \mathrm{Tr} (AB-BA)^2    = \mathrm{Tr} \Bigl[ (AB)^2 + (BA)^2 - A B^2 A - B A^2 B \Bigr].

Using the cyclic property of the trace, this simplifies to

0 \geq 2 \mathrm{Tr}(AB)^2 - 2 \mathrm{Tr} (A^2 B^2)

and we are done.

Here is an interesting open question.

Christmas Conjecture. Let A and B be n \times n density matrices with fixed spectrum: the eigenvalues of A are \alpha_1 \geq \alpha_2 \geq \dots \geq \alpha_n while the eigenvalues of B are \beta_1 \geq \beta_2 \geq \dots \geq \beta_n. Then

\max_{A,B} \mathrm{Tr} \Bigl[ AB + A^2 B^2 - (AB)^2 \Bigr]  = \sum_{i=1}^n \alpha_i \beta_i.

where the maximum is over all A,B with the given spectrum.

If the last two terms A^2 B^2 - (AB)^2 would not be there, then showing this would be quite straightforward. Indeed, the left-hand side becomes simply

\max_P \sum_{i,j=1}^n \alpha_i P_{ij} \beta_j

where the maximum is over doubly-stochastic matrices P. By the Birkhoff–von Neumann theorem (see Theorem 13.1 in John Watrous’ lecture notes), the maximal value is achieved when P is a permutation matrix. In fact, by the rearrangement theorem, P must be the identity matrix, in which case the two lists of eigenvalues are sorted in the same order.

However, we are dealing with a slightly more complicated expression in the conjecture. As we just saw above, \mathrm{Tr} \bigl[ A^2 B^2 - (AB)^2 \bigr] \geq 0, so the last two terms can only increase the left-hand side and thus make the inequality harder to prove. Nevertheless, I conjecture that this stronger inequality still holds, and I can prove it for n = 2.

Update: I recently received an e-mail from Jianxin Chen containing a proof of the Christmas Conjecture!

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