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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?

Qmechanic
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Dr. user44690
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1 Answers1

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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.

LPZ
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  • I think the point of the Bogoliubov transformation is that it mixes creation and annihilation operators. – lcv Nov 30 '23 at 16:50
  • Yes exactly, that’s what I’m addressing, unless my answer isn’t clear enough – LPZ Nov 30 '23 at 18:24
  • Frankly I am not understanding. In your answer the vacuum of the $a$ and the vacuum of the $\tilde{a}$ are the same, which is not the case for a BT. – lcv Nov 30 '23 at 18:30
  • Not quite, the vacuum are related by the unitary transformation: $|\tilde 0\rangle =U|0\rangle$ – LPZ Nov 30 '23 at 18:45
  • Your first equation is definitely not a Bogoliubov transformation. It's also hard to make sense of it in general. One can assume that $a$ is a vector of annihilation operators $a=(a_{x_1},a_{x_2},\ldots )$, what is then $U$? – lcv Nov 30 '23 at 21:45
  • Yes it is, I’ve added some details for clarity. The construction of $U$ is sketched in my linked answer and there are more resources in the thread. The general existence follows from Stone-von Neumann – LPZ Nov 30 '23 at 22:14
  • @LPZ But there are non-unitary implementable BT, no (say, infinitely many creation/annihilation operators)? Do you know how/if this generalizes in this context? – Tobias Fünke Dec 03 '23 at 15:43
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    When there are infinitely many, this is the realm of QFT. Mathematically, things are harder to define, but if I recall correctly Stone-von Neumann does not hold anymore. However the usual BT we encounter in physics like for BCS can be implemented because the BT is block diagonal with finite dimensional blocks (corresponding to the modes). – LPZ Dec 03 '23 at 18:09