Are partition functions invariant under Bogoliubov transformations?

Question

Consider a Hamiltonian $H(a_i, a^{\dagger}_i)$ as a function of some ladder operators $a_i, a^{\dagger}_i$. Now, consider a partition function $H(a'_i, a'^{\dagger}_i)$ where $a', a'^{\dagger}$ are related to $a, a^\dagger$ by Bogoliubov transformation. Then is the partition function in each case the same?

Answer 1

Yes they are the same. It all boils down to how operators transform under conjugation. A general Bogoliubov transformation is a linear transformation between $a,a^\dagger$ to $\tilde a,\tilde a^\dagger$ (changed notation to avoid the prime cluttering the expressions): $$ \tilde a_i=A_{ij}a_j+B_{ij}a_j^\dagger\\ \tilde a_i^\dagger= B_{ij}^*a_j+A_{ij}^*a_j^\dagger\\ $$ Note that for compatibility of the CCR, you’ll need the additional constraints: $$ A A^\dagger -BB^\dagger =1\\ AB^T-BA^T =0 $$ You can implement these transformations as unitary conjugations. Check out this question thread for more on that. Namely, for any BT, there exists a unitary operator $U$ such that: $$ \tilde a = UaU^\dagger $$ from which you can deduce by conjugation: $$ \tilde a^\dagger = Ua^\dagger U^\dagger $$ If your original Hamiltonian is: $$ H = h(a,a^\dagger) $$ your new Hamiltonian is: $$ \tilde H = h(\tilde a,\tilde a^\dagger) $$ Typically, $h$ is polynomial, which is compatible with conjugacy, so: $$ \tilde H = UHU^\dagger $$ To compute your partition function, you notice that the exponential is also compatible with conjugation: $$ e^{-\beta \tilde H} = Ue^{-\beta H}U^\dagger $$ and finally, using that the trace is invariant by conjugacy, you get: $$ \text{Tr}[e^{-\beta \tilde H}] = \text{Tr}[e^{-\beta H}] $$ i.e. both partition functions are equal.

Hope this helps.