Suggestions to better formulate my question are also highly appreciated.
To better explain my question let me start with a very familiar case:
Suppose we have a two-band insulator described by
$$H = \sum_{\boldsymbol k, n} \varepsilon^n(\boldsymbol k)\ d^\dagger_n(\boldsymbol k)d_n (\boldsymbol k), \qquad n \in\{0,1\}, \qquad \varepsilon^1(\boldsymbol k) > \varepsilon^0(\boldsymbol k).$$
Here $d^\dagger_n(\boldsymbol k) $ creates an electron in the state $|\phi_n(\boldsymbol k)\rangle$.
The ground state of such an insulator $|\Psi\rangle = \prod_k d^\dagger_0(\boldsymbol k) |0\rangle$.
Suppose I have another insulator described by $H' = \sum_{\boldsymbol k, n} \varepsilon'^{n}(\boldsymbol k)\ d'^{\dagger}_{n}(\boldsymbol k)d'_n (\boldsymbol k)$, where $d'^\dagger_n(\boldsymbol k) $ creates an electron in the state $|\phi'_n(\boldsymbol k)\rangle$. The ground state of this primed Hamiltonian is $|\Psi'\rangle = \prod_k d'^\dagger_0(\boldsymbol k) |0\rangle$. The overlap between $|\Psi\rangle$ and $\Psi'\rangle$ is readily given in terms of the single particle wavefunctions as
$$\langle \Psi' | \Psi\rangle = \det \left[ \langle \phi'_0(\boldsymbol k') | \phi_0(\boldsymbol k)\rangle \right ].$$
This determinant is very efficient to compute, since there is no need to construct the many-body wave functions $|\Psi\rangle$ and $|\Psi' \rangle$.
My question is, suppose we have a similar question for superconductors. If one superconductor is described by
$$H_{BdG} = \sum_{\boldsymbol k, n} \varepsilon^n(\boldsymbol k)\ \gamma^\dagger_n(\boldsymbol k)\gamma_n (\boldsymbol k), \qquad n \in\{0,1\}, \qquad \qquad \varepsilon^1(\boldsymbol k) ,\ \varepsilon^0(\boldsymbol k) > 0, $$
where $\gamma^\dagger_n(\boldsymbol k)$ are the Bogoliubov quasiparticles. The superconducting ground state can be written as $|\Psi_{BdG}\rangle = \prod_k \gamma_1(\boldsymbol k)\gamma_0(\boldsymbol k) |0\rangle$. Considering another superconductor with $H'_{BdG} = \sum_{\boldsymbol k, n} \varepsilon'^n(\boldsymbol k)\ \gamma'^\dagger_n(\boldsymbol k)\gamma'_n (\boldsymbol k)$, and a ground state $|\Psi'_{BdG}\rangle = \prod_k\gamma'_1(\boldsymbol k) \gamma'_0(\boldsymbol k) |0\rangle$, if I ask what is $\langle \Psi'_{BdG} | \Psi_{BdG}\rangle$, is there an efficient way to compute this akin to the Slater determinant for the case of insulators? Given the quadratic nature of the BdG Hamiltonian, it seems that there should be an easy way of doing this, correct?
Best Answer
The analogue of the Slater determinant for BCS is a Pfaffian. I have a longish account here.
The usual wavefunction of an $n$-particle Fock-space state $|{\Psi}\rangle$ is $$ \Psi({\bf r}_1,{\bf r}_2,\ldots,{\bf r}_n)= \langle{0}| {\hat \psi({\bf r}_n)\ldots \hat \psi ({\bf r}_2)\hat\psi ({\bf r}_1)}|{\Psi}\rangle , $$ where $$ \hat\psi({\bf r})= \sum_i a_i \psi_i({\bf r}) $$ is the field operator and $\langle{0}|$ is the no-particle state.
In the BCS case the unnormalized $2n$-particle-projected wavefunction is
$$ \Psi(i_1,i_2,\ldots,i_{2n})=_a\langle{0}| a_{i_{2n}}\ldots a_{i_1} \exp\left\{\frac 12 a^\dagger_ia^\dagger_jS_{ij}\right\}|{0}\rangle _a\\ ={\rm Pf}(S_\sigma), \quad \sigma=\{i_1,i_2,\ldots,i_{2n}\}.\nonumber $$ Summing over all possible $2n$-element subsets $\{i_1,i_2,\ldots,i_{2n}\}$ of $\{1,2,\ldots, N\}$ and then summing over the particle number $2n$ gives $$ \langle {S}|{S}\rangle =\sum_{2n=0}^\infty \sum_{\{i_1,\ldots,i_{2n}\}} |\Psi(i_1,i_2,\ldots,i_{2n})|^2\nonumber\\ = \sum_{\sigma} |{\rm Pf}(S_\sigma)|^2,\nonumber $$
The normalization factor ${\mathcal N}$ is found from the fermionic version of Major MacMahon's Master Theorem: This asserts that the inner product of states of the form $$ |{S}\rangle = \exp\left\{\frac 12 a^\dagger_ia^\dagger _jS_{ij}\right\}|0\rangle _a $$ is given by $$ \langle S_1| S_2\rangle = {\rm det}^{1/2}(I+S^\dagger_1 S_2){=} {\rm Pf}(1+S_1^\dagger S_2)={\rm Pf}(1-S_1^* S_2). $$ Here, for skew $S$, $T$, the ``double Pfaffian'' is
$$ {\rm Pf}(1-ST)\equiv \sum_\sigma {\rm Pf}(S_\sigma){\rm Pf}(T_\sigma) $$ where $\sigma$ runs though all distinct even-order subsets of $\{1,2,\ldots, N\}$, including the empty set. The skew-symmetric sub-matrix $S_\sigma$ is assembled from the rows and columns of $S$ indexed by $\sigma$, and ${\rm Pf}(S_\sigma)$ is the usual Pfaffian