2
$\begingroup$

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.

| cite | improve this question | | | | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ 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) $\endgroup$ – Medulla Oblongata May 7 '18 at 9:07
  • $\begingroup$ If so, can I use both these $\rho$ and not need a multiplicative factor of $1/2$? $\endgroup$ – Medulla Oblongata May 7 '18 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 '18 at 9:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.