The Golden-Thompson Inequality and the Upper bound of Quantum Partition Function

Let $A$ and $B$ be two Hermitian matrices. The famous Baker-Campell-Hausdorff formula says, $$\exp\big( A + B \big) = \exp \big( A \big) \exp\big( B \big) \exp\big( - [A, B]/2 \big) \exp\big( [A, [A, B]]/6 + [B, [A,B]]/3 \big)\cdots$$ In general, $\exp(A+B) \ne \exp (A) \exp(B)$. Applying the trace-determinant relation $\det \exp A = \exp \mathrm{tr} A$, however, we still have $$\det\exp(A+B) = \det \exp(A) \exp(B).$$

In a similar fashion, the Golden-Thompson theorem states, $$\mathrm{tr}\exp(A+B) \le \mathrm{tr} \exp(A) \exp(B),$$ the equality holds iff $[A, B] = 0$.

As an application, let's consider $-\beta^{-1}A = \hat p^2/(2m), -\beta^{-1}B = V(\hat q)$, and $[\hat q, \hat p] = i$. $-\beta^{-1}(A + B) = \hat p^2/(2m) + V(\hat q) = \hat H$ is the single-particle Hamiltonian. Apparently, $Z(\beta) = \mathrm{tr} \exp( -\beta \hat H )$ is the partition function of the one-body system. Now consider the right-hand side: $$\begin{split} & \mathrm{tr} \exp(-\beta \hat p^2/(2m)) \exp(-\beta V(\hat q) ) \\ & = \int dq \int \frac{dp}{2\pi} \langle q | \exp(-\beta \hat p^2/(2m)) | p \rangle \langle p | \exp(-\beta V(\hat q) ) | q \rangle \\ & = \int dx \int \frac{dp}{2\pi} \exp(-\beta p^2/(2m))  \exp(-\beta V(q) )  \langle q | p \rangle \langle p | q \rangle \\ & = \int dx \int \frac{dp}{2\pi} \exp(-\beta ( p^2/(2m) + V(q)) ) \\ & =  \int dx \int \frac{dp}{2\pi} \exp( -\beta H ) = Z_\text{cl}(\beta). \end{split}$$ The last line is the partition function of a classical single-particle system.

Therefore, the Golden-Thompson theorem says, the classical partition function is the upper bounds of the quantum partition function, $Z(\beta) \le Z_\text{cl}(\beta)$. Replace the single-particle operator and single-particle basis with the many-body ones, or even second quantized fields, it is straight-forward to show the general relation between quantum and classical partition functions.