I'm studying an example of the Hubbard-Stratonovich transformation in Altland and Simons' Condensed Matter Field Theory (2nd ed.), pp. 246-247.

In it they say that...

one is frequently confronted with situations where more than one Hubbard-Stratonovich field is needed to capture the full physics of the problem. To appreciate this point, consider the Coulomb interaction in momentum space. $$S_{int}[\bar{\psi},\psi] = \tfrac{1}{2} \sum_{p_1,...,p_4} \bar{\psi}_{\sigma, p_1} \bar{\psi}_{\sigma', p_3} V(\textbf{p}_1-\textbf{p}_2) \psi_{\sigma', p_4} \psi_{\sigma, p_2} \delta_{p_1-p_2+p_3-p_4}.$$ In principle, we can decouple this interaction in any of the three channels...

discussed in the previous page. If one chooses to decouple in all three channels then the action becomes ...

$$ S_{int}[\bar{\psi},\psi] \simeq \tfrac{1}{2} \sum_{p,p',q} ( \bar{\psi}_{\sigma, p} \psi_{\sigma,p+q} V(\textbf{q}) \bar{\psi}_{\sigma', p'} \psi_{\sigma',p'-q} - \bar{\psi}_{\sigma, p} \psi_{\sigma',p+q} V(\textbf{p'}-\textbf{p}) \bar{\psi}_{\sigma', p'} \psi_{\sigma,p'} - \bar{\psi}_{\sigma, p} \bar{\psi}_{\sigma', -p+q} V(\textbf{p'}-\textbf{p}) \psi_{\sigma,p'} \psi_{\sigma',-p'+q} )$$

where the first term is decoupled via the

direct channel $\rho_{d,q} \sim \sum_{p} \bar{\psi}_{\sigma,p} \psi_{\sigma,p+q}$, second in the exchange channel $\rho_{x,\sigma\sigma',q} \sim \sum_{p} \bar{\psi}_{\sigma,p} \psi_{\sigma',p+q}$, and third in the Cooper channel $\rho_{c,\sigma\sigma',q} \sim \sum_{p} \bar{\psi}_{\sigma,p} \bar{\psi}_{\sigma',-p+q}$.

It's generally a good strategy to decouple in all available channels when one is in doubt, then let the mean-field analysis sort out the relevant fields.

My question is, if we choose to decouple the quartic term via 3 different channels (for example) is it necessary to multiply the resulting terms by a factor of $\tfrac{1}{3}$? This isn't discussed in the textbook and I'm confused by the liberal use of $\sim$ and $\simeq$ in the examples.


1 Answer 1


No. You should not add a factor of $1/3$. As you can see in page 244 of Altland and Simons, the HS transformation is done by multiplying by a unity expressed as a functional integral over an auxiliary field. In this case, they just choose to introduce 3 different fields - 1 for each term.

  • $\begingroup$ Great, thanks. Just to clarify, for some action $S= \bar{\psi}_\alpha(t) \psi_\beta(t) V \bar{\psi}_\gamma(t') \psi_\delta(t')$, decoupling via the direct channel is when $\rho=\bar{\psi}_\alpha(t) \psi_\beta(t)$, exchange channel: $\rho=\bar{\psi}_\alpha(t) \psi_\delta(t')$, and Cooper channel: $\rho=\bar{\psi}_\alpha(t)\bar{\psi}_\gamma(t')$. If I want to decouple via the exchange channel then I can let either $\rho=\bar{\psi}_\alpha(t) \psi_\delta(t')$ or $\rho=\bar{\psi}_\gamma(t') \psi_\beta(t)$. Is this correct? (tbc) $\endgroup$ May 7, 2018 at 9:07
  • $\begingroup$ If so, can I use both these $\rho$ and not need a multiplicative factor of $1/2$? $\endgroup$ May 7, 2018 at 9:07
  • $\begingroup$ What do you mean by using both? If you do an HS transformation using 1 of them, S is already decoupled so you can't do another HS with the other one. If you pick one of them and decouple with it, you do not need a 1/2. $\endgroup$
    – tsufli
    May 8, 2018 at 9:43

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