Connection between Hubbard-Stratonovich and (generalized) coherent states

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.

Hubbard-Stratonovich transformation and mean-field approximation

Hubbard-Stratonovich transformation in the operator form

• Hi, have your got a better answer yet? By accident I'm looking at Gaussian states theses days and got back to your question again. It seems I did not understand your previous comments correctly -- sorry about that. In previous comments you said: "the vacuum of the quasiparticles/pairs - does not fix the corresponding bosonic operator". But since quasiparticles and original particles are related by an unitary transformation either in fermion or boson case, shouldn't the vacuum be unique in any basis? – Kite.Y Jun 8 '18 at 14:44
• I still agree with this statement! What I ment by "not fixing the corresponding bosonic operator" referred to pair-coherent states (see the paper I link above). There are many unitary operators for a pair of bosons that could be interpreted as a "pair" of coherent bosons. Why exactly this one (which corresponds to a two-mode squeezed state and has special and interesting entanglement properties)? – plan Jun 11 '18 at 10:39
• The energetic argument for pair condensation is probably the best explanation! But there are more complicated cases, like quantum spin liquids. Why is a two-mode squeezed state a good ansatz for pair-condensation? Is maybe another way of rephrasing what I'm still thinking about. I will try to answer this when I have the time! – plan Jun 11 '18 at 10:55

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.

• The Read-Sachdev paper was useful, thank you! Apparently the "displacement operator" for boson pair condensation is $\exp(-\sum_k f_k b^\dagger_k b^\dagger_{-k})$ but the condition you state for the fermionic ground state - that it is the vacuum of the quasiparticles/pairs - does not fix the corresponding bosonic operator I believe. There are many different bosonic operators $U$ such that $[U,a_k a_{-k}]=0$, for example a product state of two coherent states including modes $k, -k$ etc... – plan Dec 18 '17 at 14:01
• Not very familiar with displacement operator, I guess more used in quantum optics. The relation between Read-Sachdev's boson $|GS\rangle$ and the above fermion $|GS\rangle$ is simple, though: they are both the vacuum of a new set of quasiparticles obtained from Bogliubov transformation, and can be expressed in original particles as some coherent state describing condensation. In fermion case, if you, very roughly, define a pair-operator $C_{k} = c_{k\uparrow}c_{-k\downarrow}$, then $e^{g_kC_k}$ just represents a coherent state. – Kite.Y Dec 18 '17 at 15:25
• @plan Umm... Just ignore it if that's not what you are looking for.... – Kite.Y Dec 18 '17 at 15:38
• This clears some things up for me about condensation of Schwinger bosons, but because of my quantum optics background I am still curious as to why a simple exponentiation of the pair-operator is the right one! In quantum optics, there are many other ways to define pairs of entangled coherent states. – plan Dec 18 '17 at 18:46
• @plan I have condensed matter background. To my understanding, for both Schwinger boson and BCS mean field cases, the appearance of pair-operator, physically, is because the pairing mechanism can lower energy which can be seen from the Hamiltonian --- but I guess you already know this. Therefore for me, I would like to explain it from an energetic perspective -- the coherent state directly written in the pair-operator has the lowest energy -- other than an entangled perspective. – Kite.Y Dec 18 '17 at 20:18