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For an interacting quantum system, Hubbard-Stratonovich transformation and mean-field field approximation are methods often used to decouple interaction terms in the Hamiltonian. In the first method, auxiliary fields are introduced via an integral identity, and then approximated by their saddle-point values. In the second method, operators are directly replaced by their mean values, e.g. $c_i^\dagger c_jc_k^\dagger c_l \rightarrow \langle c_i^\dagger c_j\rangle c_k^\dagger c_l + c_i^\dagger c_j \langle c_k^\dagger c_l\rangle$. In both methods, order parameters can then be solved self-consistently to yield the decoupled Hamiltonian.

Are these two methods equivalent? If not, how are they related?

leongz
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2 Answers2

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They are indeed equivalent. It is quite illuminating and satisfying to see the proof of this. Consider a general action of the form $$ Z = \int \exp \left( -S_0[\varphi]+\frac{\lambda}{2} \;\mathcal O[\varphi] \mathcal O[\varphi] \right) \; \mathrm D[\varphi] $$ where $S_0$ is the action which we perturb around (we will not presume this to be a free theory).

1. Mean field theory. The essential assumption is that $$\mathcal O([\varphi]) = M + \underbrace{(\mathcal O([\varphi]) -M)}_{= \textrm{ small}}.$$ Squaring this gives us $\mathcal O[\varphi] \mathcal O[\varphi] \approx 2M \mathcal O[\varphi] - M^2$. Plugging this into the partition function gives us the mean field approximation: $$ \boxed{ Z_\textrm{mf}[M] = e^{- \lambda M^2 / 2} \int \exp \left( -S_0[\varphi]+\lambda M\mathcal O[\varphi] \right) \; \mathrm D[\varphi] }$$ with the self-consistency relation $$ M = \langle \mathcal O[\varphi] \rangle_\textrm{mf} = \frac{1}{\lambda} \partial_M \ln \left( e^{\lambda M^2/2} Z_\textrm{mf}[M] \right). $$ Note that the latter is equivalent to $\boxed{ \partial_M \ln Z_\textrm{mf}[M] = 0 }$ (which one can interpret as saying that the mean field extremizes the free energy).

2. Hubbard-Stratonovich transformation. The essential identity is $$ \exp\left( \frac{\lambda}{2} \; \mathcal O[\varphi] \mathcal O[\varphi] \right) = \int \exp \left( - \frac{\lambda \mathcal M^2}{2} + \lambda \mathcal M \mathcal O[\varphi] \right) \; \mathrm D[\mathcal M]. $$ Plugging this into the original partition function, we get $$ \boxed{ Z = \int \mathrm D[\mathcal M] \; e^{-\lambda \mathcal M^2/2} \int \mathrm D[\varphi] e^{-S_0[\varphi] + \lambda \mathcal M \mathcal O[\varphi]} = \int e^{-S_\textrm{HS}[\mathcal M]} \mathrm D [\mathcal M] }, $$ with the Hubbard-Stratonovich action $$ \boxed{ S_\textrm{HS}[\mathcal M] = \frac{\lambda \mathcal M^2 }{2} - \ln \left( \int e^{-S_0[\varphi] + \lambda \mathcal M \mathcal O[\varphi]} \; \mathrm D[\varphi] \right) }. $$

3. Connection between mean field and Hubbard-Stratonovich. From the above, we can directly see that $S_\textrm{HS}[\mathcal M ] = - \ln Z_\textrm{mf}[\mathcal M] $. Equivalently, we can write $$ \boxed{ Z = \int Z_\textrm{mf}[\mathcal M] \; \mathrm D[\mathcal M] } \; . $$ The saddle-point approximation is thus $$ \boxed{ Z \approx e^{-S_\textrm{HS}[\mathcal M_0]} = Z_\textrm{mf}[\mathcal M_0] \qquad \textrm{with } S'_\textrm{HS}[\mathcal M_0] = 0 },$$ but note that the Hubbard-Stratonovich action being extremal is exactly equivalent to the extremality of the mean field free energy, i.e., $\mathcal M_0 = M$. QED.

4. Beyond mean field theory. What do we gain from this? We can directly incorporate the quadratic corrections around this saddle-point approximation. I.e., a better approximation is $$ Z \approx \frac{Z_\textrm{mf}[M]}{\sqrt{S''_\textrm{HS}[M]}}. $$ In fact, this can be used to test how reliable/stable the mean field appoximation is.

Ruben Verresen
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They are equivalent. But one might say that that Hubbard-Stratanovich transformation are more systematic, as it might be easier to figure out how to go beyond mean-field. It might also be easier to combine different kinds of channels (for instance, you have selected a particle-hole channel in your example, whereas in the case of superconductivity, one would select a particle-particle channel $c^\dagger c^\dagger c c\to c^\dagger c^\dagger\langle c c \rangle+ \langle c^\dagger c^\dagger \rangle c c$). But one should keep in mind that HS transformation are arbitrary (you can combine an arbitrary number of them), and the different mean-field theories one gets from them give different results (even though if one could do the calculation exactly, the results would be the same).

Choosing the appropriate HS transformation is always an educated guess.

Adam
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