# Superconductivity and phase overlap

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?

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.
• Not really. I just didn’t know if my procedure was completely fine. What is supposed to be $S_{ij}$ in my case? – KernelPanic May 16 at 21:52
• I'll edid my answer to specify the $S_{ij}$. – mike stone May 16 at 22:06