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Given the following state $$|\Psi^{\phi}\rangle=\prod_{\mathbf{k}}(u_{k}+v_{k}e^{i\phi}c_{k1}^{+}c_{-k-1}^{+})|\phi_{0}\rangle,$$ where $|\phi_{0}>$ is the vacuum, $u_{k}, v_{k}\in\mathbb{R}$, and $c^{+}$ (and $c$) are the usual second quantization creation and annihilation operators, with $1\equiv$ up spin, and $-1\equiv$ down spin. This is for fermions. I'm trying to find the quantity $$\langle\Psi^{\phi}|\Psi^{\phi'}\rangle$$ to study the phase overlap this involves. This affords $$\langle\phi_{0}|\prod_{k}(u_{k}+v_{k}e^{-i\phi}c_{-k-1}c_{k1})\prod_{l}(u_{l}+v_{l}e^{i\phi'}c_ {l1}^{+}c_{-l-1}^{+})|\phi_{0}\rangle,$$ and since only $k=l$ terms will contribute, this yields $$\langle\phi_{0}|\prod_{k}[u_{k}^{2}+v_{k}^{2}e^{i(\phi'-\phi)}c_{-k-1}c_{k1}c_{k1}^{+}c_{-l-1}^{+}+u_{k}v_{k}e^{i\phi'}c_{k1}^{+}c_{-k-1}^{+}+u_{k}v_{k}e^{-i\phi}c_{-k-1}c_{k1}]|\phi_{0}\rangle,$$ which reduces almost trivially for the phase overlap to $$\prod_{k}[u_{k}^{2}+v_{k}^{2}e^{i(\phi'-\phi)}].$$ I'm not very comfortable with this result. Is there anywhere I'm going wrong? Is this correct?

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When one has a general Bogoluibov transformation $$ b_\alpha = a_iu^*_{i\alpha}+ a^\dagger_i v^*_{i\alpha}\\ b^\dagger_\alpha = a^\dagger_iu_{i\alpha}+ a_i v_{i\alpha} $$ and want to find an expression for the vacuum $|0\rangle_b$ annihilated by the $b_i$ in terms of the vacuum $|0\rangle_a$ annihilated by the $a_i$ we see that this is equivent to $$ (a_i+a^\dagger_k v^*_{k\alpha}(u^*)^{-1}_{\alpha i})|{0}\rangle_b=0 $$ for all the $i$ labels (your $k,\pm 1$'s). The BdG equation that gives the $u$ and $v$'s makes the matrix $$ S_{ij}= v^*_{i\alpha}(u^*)^{-1}_{\alpha j} $$ automatically skew symmetric. We then have
$$ \exp\left\{\frac 12 a^\dagger_i a^\dagger_j S_{ij}\right\} a_k \exp\left\{-\frac 12 a^\dagger_i a^\dagger_jS_{ij}\right\} =a_k+a^\dagger_iS_{ik}. $$ so $$ |{0}\rangle_b ={\mathcal N} \exp\left\{\frac 12 a^\dagger_ia^\dagger_jS_{ij}\right\} |{0}\rangle_a $$ for some normalization constant $\mathcal N$. The new vacuum is therefore populated by pairs related to $S_{ij}$.

We now to have a formula for the general computation of ${\mathcal N}$. If $$ |S\rangle = \exp\left\{\frac 12 S_{ij}a_i^\dagger a_j^\dagger\right\}|0\rangle, $$ similarly $|T\rangle$. Then $$ \langle S|T\rangle =\sqrt{{\rm det}(1+S^\dagger T)}. $$

I see that it's going to take me longer than I have at the moment to TeX up the specific $S$ for your problem because the fact that your matrices are diagonal in $|k|$ and I need some thought to get the skew symmetry with respect to the spin indices correct. I'll try to do it tomorrow.

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  • $\begingroup$ I do not exacly see how this solves my problem :/ $\endgroup$ – KernelPanic May 16 at 19:58
  • $\begingroup$ Just plug your soution in my (now typo corrected) formula. I think that you will see that your expression is correct. Did you have some reason for thinking that it was not correct? $\endgroup$ – mike stone May 16 at 21:49
  • $\begingroup$ Not really. I just didn’t know if my procedure was completely fine. What is supposed to be $S_{ij}$ in my case? $\endgroup$ – KernelPanic May 16 at 21:52
  • $\begingroup$ I'll edid my answer to specify the $S_{ij}$. $\endgroup$ – mike stone May 16 at 22:06
  • $\begingroup$ yes, please. That will be greatly appreciated. $\endgroup$ – KernelPanic May 16 at 22:12

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