# Formulating BCS theory in the functional integral with a real order parameter

Often in BCS theory, people take the order parameter $\Delta$ to be real. I tried to construct a BCS theory with a real order parameter from the start and ran into some trouble. I'd be interested to see if anyone can point out why what I'm doing is wrong or why its a misguided idea in the first place (I have heard that a real scalar field cannot couple to a vector potential, so maybe this is a hint that I shouldn't be doing it this way).

We start with a generic, attractive four Fermi interaction in the path integral \begin{align} \mathcal{Z} = \int \mathcal{D}[\bar\psi~ \psi] \exp \left[\int dx ~ \bar\psi_\sigma (G_0^{-1})\psi_\sigma + g \int dx~ \bar\psi_\uparrow\bar\psi_\downarrow \psi_\downarrow\psi_\uparrow \right]~~, \end{align} and then cook up a resolution of unity to insert into the partition function: \begin{align} 1 = \int \mathcal{D}[\Delta] \exp \left[\int dx~ {\Delta^2 \over 2g} \right]~~, \end{align} where $\Delta$ is a real scalar field. We then shift $\Delta$ as \begin{align} \Delta \rightarrow \Delta + g(\psi_\downarrow\psi_\uparrow + \bar\psi_\uparrow\bar\psi_\downarrow) \end{align} which is a transformation with a Jacobian of unity and so after doing some commutator algebra, we find \begin{align} {\Delta^2 \over 2g} \rightarrow& {\Delta^2 \over 2g} + \Delta(\psi_\downarrow\psi_\uparrow + \bar\psi_\uparrow\bar\psi_\downarrow) \\ &~~ + g~\bar\psi_\uparrow \bar\psi_\downarrow \psi_\downarrow \psi_\uparrow~~. \end{align} giving a new action \begin{align} S_{F + HS}[\Delta~ \bar\psi~ \psi] = \int dx ~\Big[\bar\psi_\sigma (G_0^{-1})\psi_\sigma + \Delta(\psi_\downarrow\psi_\uparrow + \bar\psi_\uparrow\bar\psi_\downarrow) \Big] \end{align} This can be recast with Nambu spinors $\bar\Psi = (\bar\psi_\uparrow,~ \psi_\downarrow)$ as \begin{align} S_{F + HS}[\Delta~ \bar\Psi~ \Psi] = \int dx~ \left[ \bar\Psi \mathcal{G}^{-1}\Psi +{\Delta^2 \over 2g} \right] \end{align} where \begin{align} \mathcal{G}^{-1} = \begin{pmatrix} [G_0^{(P)}]^{-1} & \Delta \\ \Delta & [G_0^{(h)}]^{-1} \end{pmatrix} \end{align} and $[G_0^{(h)}]^{-1}(k) = -G_0(-k)$.

The Nambu spinors are then integrated out, giving \begin{align} \mathcal{Z} = \int \mathcal{D}[\Delta] \exp \left[-\int dx {\Delta^2 \over 2g} + \mathrm{TrLog}\left[-\beta \mathcal{G}^{-1}\right] \right]~~. \end{align}

The saddle point condition is given by \begin{align} {\partial S_{Eff}[\Delta] \over \partial \Delta } = 0 = - {\Delta \over g} + \mathrm{Tr}\left[ \mathcal{G}{\partial \mathcal{G} \over \partial \Delta} \right] \end{align} which gives \begin{align} \Delta &= g \sum_k {2 \Delta \over G_0^{-1}(k)G_0^{-1}(-k) + \Delta^2}~~, \end{align} whereas the stanrdard BCS result is \begin{align} \Delta &= g \sum_k {\Delta \over G_0^{-1}(k)G_0^{-1}(-k) + |\Delta|^2}~~. \end{align} Is this difference in by a factor of two significant? Is what I've done here non-sense?

• The shift of $\Delta$ by the fermion bilinears feels suspicious; naively I don't think $\Delta$ is a real scalar field anymore if you do that because the fermion bilinears are complex. If that's the case, I feel like it would mess up your resolution of the identity by a factor of 2. – Aaron Aug 7 '18 at 17:12
• Well, that bilinear that I shifted by is equal to its conjugate, but even if not I’m not that matters. When doing a real Gaussian integral (not functional) you can shift by a complex number and that causes no problems. – Mason Aug 8 '18 at 13:56
• Though its true that the fermion bilinears equals the conjugate, while doing the path integral these are all treated as independent fields so this is no longer strictly true. Also, while its true that you can shift by a complex number, its not necessarily true that the integral remains the same because they were not in your original domain of integration. That is to say, shifting by the bilinears should not preserve the identity because the "limits" now involve the bilinears. – Aaron Aug 8 '18 at 15:20
• Carrying out the saddle point approximation is equivalent to approximating the $\Delta$ integral, so it seems reasonable to me that there are effects due to the integral if you're not careful regarding the integration properties. I agree, though, that it's a bit mysterious on physical grounds what exactly these subtleties are doing on physical grounds. – Aaron Aug 13 '18 at 18:16
• I deleted my answer since you were unhappy with it but the way you asked for it could have been more polite and was very rude. The reason I posted it as an answer was because at the time I was not able to post comments yet as I did not have enough points yet. You should be ashamed for being so rude... – Michael Aug 14 '18 at 19:04