I am reading at the moment the paper http://arxiv.org/abs/1401.5203 and try to reproduce the results. One result is the proximity correction $S_{\Sigma}$ to the system

$$ e^{-S_{\Sigma}} =\frac{\int\prod_{j\sigma\omega}D\Psi_{j\sigma}\left(\omega\right)D\Psi_{j\sigma}^{\dagger}\left(\omega\right)e^{-S_{T} - S_{S}}}{\int\prod_{j\sigma\omega}D\Psi_{j\sigma}\left(\omega\right)D\Psi_{j\sigma}^{\dagger}\left(\omega\right)e^{-S_{S}}}\text{,} $$

where $S_{S}$ is the action for the superconductor and $S_{T} = \sum_{\omega}\sum_{\sigma}\left(-\gamma\Psi_{j=1,\sigma}^{\dagger}\psi_{D,\sigma} -\gamma\psi_{D,\sigma}^{\dagger}\Psi_{j=1,\sigma}\right)$ is the tunneling Hamiltonian between superconductor and the System.

After a Bogoliubov transformation the tunneling terms becomes

$$ S_{T} = \sum_{w}\frac{2}{\pi}\int_{0}^{\pi}dq-\gamma\sin\left(q\right)\left[\left(u_{q}\phi^{\dagger}_{+}\left(q,\omega\right) - v_{q}\phi_{-}\left(q,-\omega\right)\right)\psi_{D\uparrow}\left(\omega\right) + \left(v_{q}\phi_{+}\left(q,\omega\right) + u_{q}\phi_{-}^{\dagger}\left(q,-\omega\right)\right)\psi_{D\downarrow}\left(-\omega\right) + \text{ h.c.}\right] $$


$$ \prod_{j\sigma\omega}D\Psi_{j\sigma}\left(\omega\right)D\Psi_{j\sigma}^{\dagger}\left(q,\omega\right)\to\prod_{q\tau\omega}D\phi_{\tau}\left(q,\omega\right)D\phi_{\tau}^{\dagger}\left(\omega\right) $$

the final result for $S_{\Sigma}$ should be

$$ S_{\Sigma} = \sum_{\omega}\gamma^{2}\frac{2}{\pi}\int_{0}^{\pi}dq\sin^{2}\left(q\right)\left[\left(\frac{v_{q}^{2}}{i\omega + E_{q}} - \frac{u_{q}^{2}}{-i\omega + E_{q}}\right)\sum_{\sigma =\uparrow,\downarrow}\psi_{D\sigma}^{\dagger}\psi_{D\sigma} + \frac{2u_{q}v_{q}E_{q}}{\omega^{2} + E_{q}^{2}}\left(\psi_{D\downarrow}\left(-\omega\right)\psi_{D\uparrow}\left(\omega\right) + \text{ h.c.}\right)\right] $$

However, I can not reproduce this final result. I watch in the Books by Altland & Simons and Negele & Orland and find the Gaussian integral

$$ \int d\eta^{\dagger}d\eta e^{-\eta^{\dagger}H\eta + \psi^{\dagger}\eta + \psi\eta^{\dagger}} = det\left(H\right)e^{\psi^{\dagger}H^{-1}\psi} $$

which sould be solved my problem. But I do not known how I get from this integral the result which I want.

Edit: My problem is that if I write for $S_{T}$ each term and try to integrate out the quasiparticle operators I do not obtain the same result (without the factors in front):

$$ S_{T} \sim \sum_{w}\left[\left(u_{q}\phi^{\dagger}_{+}\left(q,\omega\right) - v_{q}\phi_{-}\left(q,-\omega\right)\right)\psi_{D\uparrow}\left(\omega\right) + \left(v_{q}\phi_{+}\left(q,\omega\right) + u_{q}\phi_{-}^{\dagger}\left(q,-\omega\right)\right)\psi_{D\downarrow}\left(-\omega\right) + \text{ h.c.}\right] = u_{q}\left(\phi^{\dagger}_{+}\left(q,\omega\right)\psi_{D\uparrow}\left(\omega\right) + \psi^{\dagger}_{D\uparrow}\left(\omega\right)\phi_{+}\left(q,\omega\right) + \phi^{\dagger}_{-}\left(q,-\omega\right)\psi_{D\downarrow}\left(-\omega\right) + \psi^{\dagger}_{D\downarrow}\left(-\omega\right)\phi_{-}\left(q,-\omega\right)\right) + v_{q}\left(\phi_{+}\left(q,\omega\right)\psi_{D\downarrow}\left(-\omega\right) + \psi^{\dagger}_{D\downarrow}\left(-\omega\right)\phi^{\dagger}_{+}\left(q,\omega\right) - \phi_{-}\left(q,-\omega\right)\psi_{D\uparrow}\left(\omega\right) - \psi^{\dagger}_{D\uparrow}\left(\omega\right)\phi^{\dagger}_{-}\left(q,-\omega\right)\right)\text{.} $$

I know that

$$ \int\prod_{j\sigma\omega}D\Psi_{j\sigma}\left(\omega\right)D\Psi_{j\sigma}^{\dagger}\left(\omega\right)e^{-S_{S}} \sim i\omega + E_{q} $$

However, with this integral I have trouble:

$$ \int\prod_{j\sigma\omega}D\Psi_{j\sigma}\left(\omega\right)D\Psi_{j\sigma}^{\dagger}\left(\omega\right)e^{-S_{T} - S_{S}} $$

The terms with the $u_{q}$ are the integral which I know

$$ \int d\eta^{\dagger}d\eta e^{-\eta^{\dagger}H\eta + \psi^{\dagger}\eta + \psi\eta^{\dagger}} = det\left(H\right)e^{\psi^{\dagger}H^{-1}\psi} $$

But how I can solve this with the $v_{q}$'s?

Edit2: I see my Substitution was wrong. Now I obtain

$$ \frac{1}{-i\omega + E_{q}}\left(\left(v_{q}\psi_{\downarrow}\left(-\omega\right) + u_{q}\psi_{\uparrow}^{\dagger}\left(\omega\right)\right)\left(v_{q}\psi_{\downarrow}^{\dagger}\left(-\omega\right) + u_{q}\psi_{\uparrow}\left(\omega\right)\right) + \left(u_{q}\psi_{\downarrow}^{\dagger}\left(-\omega\right) - v_{q}\psi_{\uparrow}^{\dagger}\left(\omega\right)\right)\left(u_{q}\psi_{\downarrow}\left(-\omega\right) - v_{q}\psi_{\uparrow}^{\dagger}\left(\omega\right)\right)\right) = \frac{1}{-i\omega + E_{q}}\left(\left(u_{q}^{2} - v_{q}^{2}\right)\left(\psi_{D\uparrow}^{\dagger}\left(\omega\right)\psi_{D\uparrow}\left(\omega\right) + \psi_{D\downarrow}^{\dagger}\left(-\omega\right)\psi_{D\downarrow}\left(-\omega\right)\right) +2u_{q}v_{q}\left(\psi_{D\uparrow}^{\dagger}\left(\omega\right)\psi_{D\downarrow}^{\dagger}\left(-\omega\right) + \psi_{D\downarrow}\left(-\omega\right)\psi_{D\uparrow}\left(\omega\right)\right) - 2v_{q}^{2}\right)\text{,} $$

which is very nice but not the same as in the paper. What I overlook at the moment?

  • $\begingroup$ Is $\psi_D$ is spinor? If not, what does $\Psi^\dagger \psi_D$ means? (I assume that $\Psi$ is a Nambu spinor?) I think that doing the integral directly with the Nambu spinor might be easier. In the quoted final result, we can easily recognize the $G$ and $F$ functions. $\endgroup$ – Adam Jun 10 '14 at 14:03
  • $\begingroup$ Ok, I just had a look at the reference, it seems now that your $\Psi$ is not a Nambu spinor (though people usually use this notation). Then, you should write everything as Nambu spinors (you'll have to make spinors with your $\psi_D$), and then do a Gaussian integral. Tell us what does not work in your calculation so we can help you. $\endgroup$ – Adam Jun 10 '14 at 14:08
  • 1
    $\begingroup$ You should do it as for any gaussian integral: Here, $\eta$ corresponds to $\phi_\pm$, and $H$ is diagonal (does not couple $\phi_\pm$, so you just need to read what are $\psi$ and $\psi^\dagger$. For $\phi_+$, it is $u_q \psi^\dagger_{D,+}(\omega)+v_q \psi_{D,-}(-\omega)$, etc. $\endgroup$ – Adam Jun 10 '14 at 19:52
  • $\begingroup$ Be careful that $\phi_\pm$ have energy $\pm E_q$ (I guess, since you don't give their action). And your second line of lhs of your edit has a problem (I this a $\dagger$ should not be there). PS: You should write a comment starting with @Adam if you want to be sure that I see your next edit. $\endgroup$ – Adam Jun 10 '14 at 22:10
  • $\begingroup$ Congrats. Now, you could write an answer about it and accept it, so that people could have an example of how to do this calculation. $\endgroup$ – Adam Jun 11 '14 at 13:00

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