Squeezed Coherent State Overlap and Expectation Values

Question

I am unfamiliar with quantum optics and have recently been struggling to evaluate some expectation values involving squeezed states. Any help and guidance would be greatly appreciated!

The matrix elements I am considering boil down to the following form.

$$ \langle \Psi_1| \exp\left(\sum_{mk}\beta^*_{mk}b_m b_k\right) \text{($b$ ..$b^\dagger$..)} \exp\left(\sum_{ln}\beta_{l-d, n-d} b^\dagger_l b^\dagger_n\right)|\Psi_2\rangle $$

where $|\Psi_1\rangle$ and $| \Psi_2\rangle$ are coherent states defined as follows.

$$ |\Psi_1\rangle = \exp\left(\sum_j \xi_{j} b^\dagger_m - \text{h.c}\right)|0\rangle $$

$$ |\Psi_2\rangle = \exp\left(\sum_j \xi_{j-d} b^\dagger_m - \text{h.c}\right)|0\rangle $$

Here the indices $m,k,l,n, j$ are cyclic mod $N$ and $d$ refers to some constant shift of the indices. $\beta_{mk}$ can be assumed to be symmetric $(\beta_{mk} = \beta_{km})$.

Usually my strategy with coherent state expectation values has been to normal order the operator string sandwiched between the bra and ket and to act these directly onto the coherent states to the right or left. However, I have been unable to use this naive approach in evaluating the above matrix element.

Does the type of expression above have a closed form? If so, what would be the most convenient or clearest way of obtaining closed form expressions of the expectation value? I have been looking at some so called disentangling formulas and so forth but they remain pretty opaque to me at the moment.

Answer 1

First, let's rewrite the scalar product of two vectors as follows $$ \langle\Psi_1|\exp\left(\sum_{mk}\beta^*_{mk}b_mb_k\right) \exp\left(\sum_{ln}\beta_{l-d\ n-d}b_l^\dagger b_n^\dagger\right)|\Psi_2\rangle = $$ $$ = A \langle\tilde{\Psi}_1|\exp\left(\sum_{mk}\beta^*_{mk}b_mb_k\right) \exp\left(\sum_{ln}\beta_{l-d\ n-d}b_l^\dagger b_n^\dagger\right)|\tilde{\Psi}_2\rangle \tag{1} $$ where $$ A = \exp\left(-\sum_m\xi^*_m\xi_m\right),\quad |\tilde{\Psi}_1\rangle = \exp\left(\sum_m \xi_m b^\dagger_m \right)|0\rangle, \quad |\tilde{\Psi}_2\rangle = \exp\left(\sum_m \xi_{m-d} b^\dagger_m \right)|0\rangle. $$ According to the literature (Berezin, Itzykson and Drouffe and many others), the scalar product on the right-hand side of equality (1) can be written as a Gaussian integral $$ \langle\tilde{\Psi}_1|\exp\left(\sum_{mk}\beta^*_{mk}b_mb_k\right) \exp\left(\sum_{ln}\beta_{l-d\ n-d}b_l^\dagger b_n^\dagger\right)|\tilde{\Psi}_2\rangle = \int\!\ldots\!\int\prod_j d \overline{z}_j dz_j \tag{2} $$ $$ \exp\left(-\sum_m\overline{z}_mz_m+\sum_m\xi^*_m \overline{z}_m + \sum_{mk}\beta^*_{mk}\overline{z}_m \overline{z}_k +\sum_l\xi_{l-d}z_l+\sum_{l n}\beta_{l-d\ n-d}z_l z_n \right) $$ where $\overline{z}_j = x_j-iy_j$, $z_j = x_j + i y_j$, $x_j, y_j \in R$ and $$ \int d \overline{z}_j dz_j\equiv \iint\limits_{-\infty}^\infty\frac{dx_jdy_j}{\pi} $$ It is well known how the values ​​of Gaussian integrals are expressed through a matrix corresponding to a quadratic form in the exponential: $$ \int\!\ldots\!\int dx_1\ldots dx_n \exp\left(-\frac12\sum_{j,j'=1}^n M_{j,j'}x_jx_{j'} + \sum_{j=1}^n b_jx_j\right) = $$ $$ =\frac1{\sqrt{\det(M/2\pi)}} \exp\left(\frac12\sum_{j,j'=1}^n M^{-1}_{j,j'}b_j b_{j'} \right) $$ The matrix M corresponding to integral (2) has special properties. For example, if I remember correctly, in the case $\xi_m = 0$ the integral (2) is equal to $$ \exp\left(-\frac12\mbox{Tr}\ln\left(I - 4\beta^+\tilde{\beta}\right) \right) = \frac1{\sqrt{\det(I - 4\beta^+\tilde{\beta})}} $$ where $\beta^+$, $\tilde{\beta}$, $I$ are matrices with elements $\beta^+_{mn} = \beta^*_{nm}$, $\tilde{\beta}_{mn} = \beta_{m-d\ n-d}$, $I_{mn} = \delta_{mn}$.

If the matrix $\beta$ is cyclic, that is, $\beta_{mn} = f(m-n)$, then the formulas are significantly simplified. In the case of $N=2$, explicit expressions can also be written.

To consider the case with additional operator factors $(b_m\ldots b^\dagger_{m'}\ldots)$ between two exponentials, it suffices to differentiate the scalar product (2) with respect to $\xi^*_m,\ldots$ and $\xi_{m'-d},\ldots$.