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In page 293 of Altland and Simons' "Condensed Matter Field Theory", just above equation (6.38), in the process of deriving the London equations from the BCS path integral, the authors say, "The integration over the field components $θ_q$ is now straightforward and leads to an effective action for A". I cannot get my mind around how the integral over the $\theta$ field configurations is computed.

Specifically, we have a path integral like $S=\int \int D[\textbf{A}] D[\theta] e^{-S[\theta,\textbf{A}]}$, where the action is

\begin{equation} S[\theta,\textbf{A}] = \frac{\beta}{2} \sum_{\textbf{q}} \left[\frac{n_s}{m}(\theta_qq^2\theta_{-q}-2i\theta_q\textbf{q}\cdot\textbf{A}_{-q}+\textbf{A}_{q}\cdot\textbf{A}_{-q}) + (\textbf{q}\times\textbf{A}_q)\cdot(\textbf{q}\times\textbf{A}_{-q})\right]\text{.}\tag{p.293} \end{equation}

And after the $\theta$ integration, $e^{-s[\textbf{A}]}\equiv\int D[\theta]e^{-S[\theta, \textbf{A}]}$, the effective action becomes

\begin{equation} S[\textbf{A}]=\frac{\beta}{2}\sum_{\textbf{q}}\left[\frac{n_s}{m}\left(\textbf{A}_{q}\cdot\textbf{A}_{-q}-\frac{(\textbf{q}\cdot\textbf{A}_q)(\textbf{q}\cdot\textbf{A}_{-q})}{q^2}\right)+(\textbf{q}\times\textbf{A}_{q})(\textbf{q}\times\textbf{A}_{-q})\right]\text{.}\tag{p.293} \end{equation}

This new action now only depends on the gauge field $\textbf{A}$ and it has gained a mass proportional to the superfluid density by absorbing the phase field.

But how was the integral calculated explicitly? I see that after completing the square in the exponential, the action can be put in a form that resembles a Gaussian in the $\theta_{q}$ components, like $e^{\text{const}\times\sum_q\theta_{q}q^2\theta_{-q}+\text{other terms}}$, but I still do not know how to approach that integral because I cannot calculate the determinant of the $q^2$ matrix. I see that if I had something with unit determinant I would get the correct expression, but I cannot see how this can be done.

Edit: This is the expression one obtains after completing the square in the $S[\theta, \textbf{A}]$ action:

\begin{equation}S[\theta,\textbf{A}] = \frac{\beta}{2}\sum_{\textbf{q}}\left[\left(\theta_q+i\frac{\textbf{q}}{q^2}\textbf{A}_q\right)\frac{q^2n_s}{m}\left(\theta_{-q}-i\frac{\textbf{q}}{q^2}\textbf{A}_{-q}\right) \\-\frac{n_s}{mq^2}(\textbf{q}\cdot\textbf{A}_q)(\textbf{q}\cdot\textbf{A}_{-q})+\frac{n_s}{m}(\textbf{A}_{q}\cdot\textbf{A}_{-q})+(\textbf{q}\times\textbf{A}_{q})(\textbf{q}\times\textbf{A}_{-q})\right]\text{.} \end{equation}

Notice that the expression that I am looking for is precisely this but without the first term. But, again, I do not see how it goes away after integrating.

Qmechanic
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laura_legesen
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    @laura_legesen , for me, it seems that you can organize the argumen in exp as $(\theta_q, \theta_{-q})^T\cdot M\cdot (\theta_q,\theta_{-q})$, where $M$ contains $q^2$ on anti-diagonal. Next, you can organize linear term as $(-iq\cdot A_{-q},+iq\cdot A_{q})\cdot(\theta_q,\theta_{-q})$. Finally, you can easily find inverse matrix $M^{-1}$ and perform formal Gaussian inegration with source term. In addition, for my view the presented formulas for action relate to each other up to constant, which contains determinant and (strictly speaking) can be divergent, but it is not interesting – Artem Alexandrov Mar 06 '23 at 19:32

1 Answers1

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  1. After shifting the integration variable $\theta_q\to \theta^{\prime}_q=\theta_q+i\frac{\textbf{q}}{q^2}\textbf{A}_q$, the integration contour becomes complex. However we can move the integration contour back to the real line due to Cauchy's integral theorem.

  2. As long as we are not interested in the dependence of the remaining parameter $\beta n_s/m$ in the resulting Gaussian functional integral, we may consider it as part of the normalization of the path integral.

  3. Concerning techniques to actually evaluate the functional integral, see e.g. this Phys.SE post.

Qmechanic
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