Partition function and coherent state path integral I have been working through the derivation of the partition function expressed as a path integral in terms of coherent states, following the many-body condensed-matter field theory books of Altland & Simons and Piers Coleman, and while I can follow the mathematical arguments, I am struggling with some of the concepts involved. I would very much appreciate some help with several points, which I have been unable to clarify to my satisfaction, despite referring to numerous other textbooks. I apologize in advance if these questions seem too basic.
Question: Starting from the very first step of the derivation, where the partition function is expressed as the trace of the operator $\exp[-\beta(\hat{H}-\mu\hat{N})]$, firstly in terms of a complete set of Fock space states and then coherent states, is it necessary for the basis to be chosen such that the Hamiltonian is diagonal in this basis? I know that the trace, and hence the partition function, is the sum of the diagonal matrix elements, but does the matrix need to be diagonal for this to be evaluated and make sense?
 A: While the trace is invariant under a transform to another basis, you need to take into account here that the coherent state basis is not an orthogonal basis and it is overcomplete. We can evaluate the trace of an operator $A$ by inserting identity operators in front of and after the operator and then using resolution of identity in terms of the coherent basis vectors. We then need to use that the resolution of identity is now given as:
$$I = \frac{1}{\pi}\int d^2\alpha \left|\alpha\right\rangle\left\langle\alpha\right|$$
So, there is an extra factor  of $\frac{1}{\pi}$ due to overcompleteness. We can thus write:
$$\operatorname{Tr}A = \sum_n\left\langle n\left|A\right|n\right\rangle =  \frac{1}{\pi^2}\sum_{n}\int d^2\alpha d^2\beta\left\langle n\right|\left.\alpha\right\rangle\left\langle\alpha\right|A\left|\beta\right\rangle\left\langle\beta\right|\left.n\right\rangle$$
If you now sum over the (complete, orthonormal) basis $\left|n\right\rangle$, you get using completeness in this basis:
$$\operatorname{Tr}A =\frac{1}{\pi^2}\int d^2\alpha d^2\beta\left\langle\beta\right|\left.\alpha\right\rangle\left\langle\alpha\right|A\left|\beta\right\rangle$$
You can then evaluate this by using the fact that the overlap between two coherent basis states is given by:
$$\left\langle\beta\right|\left.\alpha\right\rangle = \exp\left[-\frac{1}{2}\left(\left|\alpha\right|^2+\left|\beta\right|^2-2\beta^*\alpha\right)\right]$$
A: No, you don't need to work in the basis where the Hamiltonian is diagonal.  It's a fact of linear algebra that the sum of the diagonal elements of a matrix is the same no matter what basis you're in, so you can easily evaluate the trace in whichever basis is convenient.
