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I was reading the Chapter 14 of the textbook by Philip Phillips on Advanced Solid State Physics, when he introduced the mean-field treatment of the quantum rotor model, he used the method first used in this paper: Quantum fluctuations in two-dimensional superconductors.

Here my question is related to the Hubbard-Stratonovich transformation (H-S transformation) used here to decouple the "spin" interaction:

$$ T_{\tau}\exp \left( \int_0^{\beta} d\tau\ \sum_k J_k \mathbf{S}_k(\tau)\cdot\mathbf{S}_k^*(\tau) \right)$$

here the $\mathbf{S}_k(\tau)$ is the operator written in the interaction picture, then he introduced the auxiliary field $\psi_k$:

$$T_{\tau}\exp \left( \int_0^{\beta} d\tau\ \sum_k J_k \mathbf{S}_k(\tau)\cdot\mathbf{S}_k^*(\tau) \right)=T_{\tau}\int D \psi_k(\tau) \ e^{-\int d\tau \sum_k \psi_k^*(\tau)\psi_k(\tau)-2\int_0^{\beta} d \tau \sum_k \psi_k(\tau)\cdot\mathbf{S}_{-k}(\tau) } $$

because I learned H-S transformation in the context of path integral, where we write the partition function in terms of functional integral over the fields ("numbers"). So here I want to ask:

  1. Mathematically if it is well defined to do the H-S transformation on top of operators?
  2. It seems that the imaginary time ordering operator does not affect the auxiliary field $\psi_k(\tau)$ (can be seen from later on derivations in the book), what's the reason for that?
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  • $\begingroup$ Just a quick comment on this question, indeed that the H-S transformation does not really hold when operators are involved; however, notice that there is a time ordering operator $T_{\tau}$ in front! If we check the original paper by Hubbard: journals.aps.org/prl/abstract/10.1103/PhysRevLett.3.77 we will see that actually, the first version of H-S transformation is in the operator form (write the partition function in the interaction picture). And in this paper, he mentioned a seminal paper by R. Feynman: journals.aps.org/pr/abstract/10.1103/PhysRev.84.108 $\endgroup$ – Chuan Chen Jun 29 '17 at 5:41
  • $\begingroup$ In the paper by Feynman, he discussed the "fact" that we can generalize the identities which are true when everything is in the c-number form to the case of operators when we have the time ordering operator in front (of course, the operators are in the interaction or Heisenberg representation and carries a time index). My understanding about this "theorem" is that because the property of c-number we use to prove some equivalence between two expressions is that c-numbers $\textbf{commute}$; now generalize to the operator form: $\endgroup$ – Chuan Chen Jun 29 '17 at 5:50
  • $\begingroup$ when we say that to expressions are the same when operators are c-numbers means that we can write the two expressions into a form where they are only different by some permutation of the operators, but with the presence of time ordering operator in front of both expressions, they all need to be aligned in the same "time" order (if they are bosonic under time ordering, which means that there is no additional minus sign when we permute them)! Of course, if we have many operators with the same time index, we need to make sure that operators with same time index commute with each other! $\endgroup$ – Chuan Chen Jun 29 '17 at 5:55
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Well-defined mathematically depends on how much one wants to be rigorous. Here, I think a physicist would be satisfied with an argument like : working with eigenvectors of ${\bf S_k}$, one shows that the formal manipulation with operators is true when working with the eigenvalues, so it is fine to do that in terms of the operators. I don't know how much effort is needed to make that satisfying for a mathematician.

Concerning the second question : the auxiliary field are (complex) numbers, so they all commute. Therefore, there isn't any ambiguity in their ordering. By construction, in a path integral formalism, the fields are always "time-ordered".

Update : to give a bit more details on the HS transformation for operators, I will use a toy model. Let $\hat A$ be an operator, with eigenstates $|a\rangle$ associated to the eigenvalue $a$. Then $$ e^{\frac12\hat A^2}|a\rangle = e^{\frac12a^2}|a\rangle,\\ =\int dx e^{-\frac12x^2+a x}|a\rangle,\\ = \int dx e^{-\frac12x^2+\hat A x}|a\rangle.$$ Thus (assuming that the eigenstates of $\hat A$ form a basis and so on), we can say that at the operator level, $ e^{\frac12\hat A^2}=\int dx e^{-\frac12x^2+\hat A x}$ (up to a normalization constant).

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  • $\begingroup$ Can you be more specific about "working with eigenvectors of $\mathbf{S}_k$"? I don't really understand your explanation on the first question. thanks. $\endgroup$ – Chuan Chen May 3 '17 at 9:43
  • $\begingroup$ See my update. I think it should be clearer. $\endgroup$ – Adam May 3 '17 at 11:35
  • $\begingroup$ I see, thanks for your answer. Actually, motivated by your answer above, one can do a brute force calculation, which is to expand the $\exp(\hat{A}x)$ term with $\sum_n \frac{1}{n!}\hat{A}^n\ x^n$, then use the Wick theorem over $x$ to get the final expression. $\endgroup$ – Chuan Chen May 3 '17 at 13:02

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