Connection between Hubbard-Stratonovich and (generalized) coherent states

Question

A simple-minded mean-field approximation for the Bose-Hubbard model consists in writing operators as $\hat{a}_i = \alpha_i + \hat{\delta \alpha}_i, \alpha_i \in \mathbb{C}$ and only include terms up to second order in $\hat{\delta \alpha}$. Using coherent states/displacement operators, this may be written as

$H(\left\{\hat{a}_i\right\}) = D(\left\{-\alpha_i\right\}) H((\left\{\alpha_i + \hat{\delta a}_i\right\})) D(\left\{-\alpha_i\right\})^\dagger = D(\left\{-\alpha_i\right\}) H^{(2)}((\left\{\alpha_i + \hat{\delta a}_i\right\})) D(\left\{-\alpha_i\right\})^\dagger$

where $H^{(2)}$ is quadratic, $D(\alpha) = \exp(\alpha \hat{a}-\alpha^* \hat{a}^\dagger)$ and $D(\left\{-\alpha_i\right\}) = \bigotimes_i D(\alpha_i)$. In this approximation, the ground state of $H^{(2)}$ will be "displaced" to the mean-field minimum and so we can have $\alpha_i = \langle \hat{a}_i\rangle \neq 0$. In a Bose-Hubbard model, this would be in the superfluid phase.

With a Hubbard-Stratonovich transformation, which is used for example in BCS theory, one also gets a quadratic Hamiltonian and $\langle \hat{c}_k \hat{c}_{-k}\rangle \neq 0$. Is there a similar "displacement" or a similar generalized coherent state in this case, which displaces entangled fermion pairs? I have looked at pair coherent states (see section 2 of https://arxiv.org/pdf/quant-ph/0607162.pdf) as a candidate. Please note that I'm aware of the BCS wavefunction - I want to understand it's (and other HS-decoupled solutions) relation to coherent states/displacement operators, regardless of fermionic/bosonic/etc statistics.

See also:

Hubbard-Stratonovich transformation and mean-field approximation

Hubbard-Stratonovich transformation in the operator form

Answer 1

Not completely sure if you are looking for this, but let me write down what I know.

From a Bogliubov transformation as bellow we diagonalize the Hamiltonian: \begin{align} H &= \left( \begin{array}{cc} c_{k,\uparrow}^{\dagger} & c_{-k,\downarrow} \end{array} \right) M_{k} \left( \begin{array}{c} c_{k,\uparrow}\\ c_{-k,\downarrow}^{\dagger} \end{array} \right) \nonumber \\ &= \left( \begin{array}{cc} \gamma_{k,\uparrow}^{\dagger} & \gamma_{-k,\downarrow} \end{array} \right) \Omega_{k}^{\dagger}M_{k}\Omega_{k} \left( \begin{array}{c} \gamma_{k,\uparrow}\\ \gamma_{-k,\downarrow}^{\dagger} \end{array} \right) \\ &= \left( \begin{array}{cc} \gamma_{k,\uparrow}^{\dagger} & \gamma_{-k,\downarrow} \end{array} \right) \Lambda_k \left( \begin{array}{c} \gamma_{k,\uparrow}\\ \gamma_{-k,\downarrow}^{\dagger} \end{array} \right) \end{align} where the transformation keeping fermion commutation relation is: \begin{align} \Omega_{k}&= \left( \begin{array}{cc} u_k & -v_k \\ v_k^* & u_k^* \end{array} \right) \end{align} for convenience we could define: \begin{align} u_k & = \frac{e^{-i\theta_k}}{\sqrt{1+ |g_k|^2}} \\ v_k & = \frac{g_ke^{i\theta_k}}{\sqrt{1+ |g_k|^2}} \end{align}

Then the ground state wavefunction could be written as: \begin{align} |GS\rangle \propto e^{\sum_{k}g_kc_{k,\uparrow}^{\dagger}c_{-k,\downarrow}^{\dagger}}|0\rangle \end{align} up to a normalization constant, where $|0\rangle$ is the vacuum of original fermion (electrons) $c_{k,\sigma}$.

The reason that this is the ground state (mean-field GS, more precisely), can be seen from the calculation for arbitrary $\gamma_{k, \sigma}$ that: \begin{align} \gamma_{k,\sigma}|GS\rangle = 0 \end{align} which suggests the $|GS\rangle$ is the vacuum of quasiparticles $\gamma_{k,\sigma}$.

This is the way to construct fermion pairs' condensation wavefunctions. Also, there is a similar way for boson pairs' condensation, which can be found in Eq.(3.8) of Phys. Rev. B 42, 4568. The structure difference, mathematically, just comes from the different commutation relations which affects the diagonalization process.