**Theorem.** If and are positive semi-definite matrices then

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

**Proof.** First, note that

Next, note that the commutator is skew-Hermitian, that is, . In other words, for some Hermitian matrix . This means that all eigenvalues of are purely imaginary. When we square the matrix, they become negative, i.e., is negative semi-definite. From this we get

Using the cyclic property of the trace, this simplifies to

and we are done.

Here is an interesting open question.

**Christmas Conjecture.** Let and be density matrices with fixed spectrum: the eigenvalues of are while the eigenvalues of are . Then

*where the maximum is over all with the given spectrum.*

If the last two terms would not be there, then showing this would be quite straightforward. Indeed, the left-hand side becomes simply

where the maximum is over doubly-stochastic matrices . By the Birkhoff–von Neumann theorem (see Theorem 13.1 in John Watrous’ lecture notes), the maximal value is achieved when is a permutation matrix. In fact, by the rearrangement theorem, 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, , 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 .

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