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Let ${A, B}$ be two Hermitian ${n \times n}$ matrices. When ${A}$ and ${B}$ commute, we have the identity

$\displaystyle e^{A+B} = e^A e^B.$

When ${A}$ and ${B}$ do not commute, the situation is more complicated; we have the Baker-Campbell-Hausdorff formula

$\displaystyle e^{A+B} = e^A e^B e^{-\frac{1}{2}[A,B]} \ldots$

where the infinite product here is explicit but very messy. On the other hand, taking determinants we still have the identity

$\displaystyle \hbox{det}(e^{A+B}) = \hbox{det}(e^A e^B).$

Recently I learned (from Emmanuel Candes, who in turn learned it from David Gross) that there is another very nice relationship between ${e^{A+B}}$ and ${e^A e^B}$, namely the Golden-Thompson inequality

$\displaystyle \hbox{tr}(e^{A+B}) \leq \hbox{tr}(e^A e^B). \ \ \ \ \ (1)$

The remarkable thing about this inequality is that no commutativity hypotheses whatsoever on the matrices ${A, B}$ are required. Note that the right-hand side can be rearranged using the cyclic property of trace as ${\hbox{tr}( e^{B/2} e^A e^{B/2} )}$; the expression inside the trace is positive definite so the right-hand side is positive. (On the other hand, there is no reason why expressions such as ${\hbox{tr}(e^A e^B e^C)}$ need to be positive or even real, so the obvious extension of the Golden-Thompson inequality to three or more Hermitian matrices fails.) I am told that this inequality is quite useful in statistical mechanics, although I do not know the details of this.

To get a sense of how delicate the Golden-Thompson inequality is, let us expand both sides to fourth order in ${A, B}$. The left-hand side expands as

$\displaystyle \hbox{tr} 1 + \hbox{tr} (A+B) + \frac{1}{2} \hbox{tr} (A^2 + AB + BA + B^2) + \frac{1}{6} \hbox{tr} (A+B)^3$

$\displaystyle + \frac{1}{24} \hbox{tr} (A+B)^4 + \ldots$

while the right-hand side expands as

$\displaystyle \hbox{tr} 1 + \hbox{tr} (A+B) + \frac{1}{2} \hbox{tr} (A^2 + 2AB + B^2)$

$\displaystyle + \frac{1}{6} \hbox{tr} (A^3 + 3A^2 B + 3 A B^2+B^3) +$

$\displaystyle \frac{1}{24} \hbox{tr} (A^4 + 4 A^3 B + 6 A^2 B^2 + 4 A B^3 +B^4) + \ldots$

Using the cyclic property of trace ${\hbox{tr}(AB) = \hbox{tr}(BA)}$, one can verify that all terms up to third order agree. Turning to the fourth order terms, one sees after expanding out ${(A+B)^4}$ and using the cyclic property of trace as much as possible, we see that the fourth order terms almost agree, but the left-hand side contains a term ${\frac{1}{12} \hbox{tr}(ABAB)}$ whose counterpart on the right-hand side is ${\frac{1}{12} \hbox{tr}(ABBA)}$. The difference between the two can be factorised (again using the cyclic property of trace) as ${-\frac{1}{24} \hbox{tr} [A,B]^2}$. Since ${[A,B] := AB-BA}$ is skew-Hermitian, ${-[A,B]^2}$ is positive definite, and so we have proven the Golden-Thompson inequality to fourth order. (One could also have used the Cauchy-Schwarz inequality for the Frobenius norm to establish this; see below.)

Intuitively, the Golden-Thompson inequality is asserting that interactions between a pair ${A, B}$ of non-commuting Hermitian matrices are strongest when cross-interactions are kept to a minimum, so that all the ${A}$ factors lie on one side of a product and all the ${B}$ factors lie on the other. Indeed, this theme will be running through the proof of this inequality, to which we now turn.