Calculating grand canonical partition function in QFT In thermal field theory, if one has a conserved current, it's convenient to work with the grand-canonical partition function
$$Z = \exp \left( H - \mu Q \right) ,$$
where $H$ is the hamiltonian operator, $Q$ the conserved charge operator, and $\mu$ the corresponding chemical potential. Is there a "canonical" way to calculate $Z$? I haven't been able to find any explicit calculations.
 A: As an example the following article by A.B. Balantekin describes in section III the structure of thermal grand canonical partition functions (i.e., with chemical potentials) for systems with Lie group symmetries.
The sum over the group characters originates from the presence of the chemical potentials, which are the analytical continuations of the coordinates of the maximal torus of the group. 
The Balantekin article describes the "sum over states" formulas of the partition functions in which the summation over the group representations is explicit (in contrast to the Feynman's "sum over paths" in which the summation over the representations is implicit)
I'll describe here a simple example of an explicit calculation of the grand canonical partition function over the two-sphere.
Let $-H$ be the scalar Laplacian on the two sphere. The state Hilbert space is spanned the spherical representations (with multiplicity 1), 
corresponding to integer spins only. The symmetry group $SU(2)$ is generated by the usual set of generators $[J_i, J_j] = \epsilon_{ijk} J_k$ 
The grand canonical partition function is given by:
$Z = \textrm{Tr}(\exp(-\beta H + \mu J_3)) = \sum_{j=0}^{\infty}\chi_j(i \mu)\exp(-\beta j(j+1))$
$= \sum_{j=0}^{\infty}\frac{\sinh((2j+1) \mu)}{\sinh(\mu)}\exp(-\beta j(j+1))$
$= \frac{e^{\frac{\beta}{4}}}{2\sinh(\mu) }\theta_1(\mu, \beta)$
The actual computation (of the $N=4$ SYM) referred to in the question needs much more work and it relies on several assumptions and approximations.
The introduction section of the following article by Yamada and Yaffe and references therein may be helpful.
