Explicit Expression of $S_2(\zeta)$ on a general Fock State?

Question

I would like to know the explicit expression of the two-mode squeezing operator $\hat{S}_2(\zeta)$ acting on a general Fock state $|p,q\rangle$, without including any additional operators in the expression.

$\hat{S}_2(\zeta)$ is the two-mode squeezing operator defined as: \begin{equation} \hat{S}_2(\zeta) = \exp\left( \zeta^* \hat{a} \hat{b} - \zeta \hat{a}^\dagger \hat{b}^\dagger\right),\\ \zeta = re^{i\theta} \end{equation}

To provide some context, I have found two specific examples where the operator $\hat{S}_2(\zeta)$ is applied to the vacuum state $|0,0\rangle$ and the number state $|p,0\rangle$:

Example (1). Apply $\hat{S}_2(\zeta) $ to vacuum state $|0,0\rangle$: \begin{align} |\Psi_{0,0}\rangle= \hat{S}_2(\zeta) |0,0\rangle = \sum_{k=0}^{\infty} \frac{(-e^{i\theta} \tanh r)^k}{\cosh r} |k, k\rangle \end{align}

Example (2). Apply $\hat{S}_2(\zeta) $ to the state $|p,0\rangle$: \begin{equation} |\Psi_{p,0}\rangle = \hat{S}_2(\zeta) |p, 0\rangle = \frac{1}{{(\cosh r)}^{p+1}}\sum_{k=0}^\infty \binom{p+k}{p}^{\frac{1}{2}} (-e^{i\theta} \tanh r)^k |p+k, k\rangle \end{equation}

Question:

What is the expression of $|\Psi\rangle$ in Fock basis when apply $\hat{S}_2(\zeta)$ to a general number state $|p, q\rangle$? Specifically, given \begin{equation} |\Psi_{p,q}\rangle=\hat{S}_2(\zeta)|p,q\rangle = \sum_{m=0}^\infty \sum_{n=0}^\infty T(m,n)|m,n\rangle, \end{equation} what are the coefficients $T(m,n)$? Are there any references that provide a detailed derivation or computation of this state $|\Psi\rangle$?

Any guidance, suggestions, or references on this topic would be greatly appreciated. Thank you very much in advance!

Answer 1

The operators $\hat a\hat b\sim K_-$ and $\hat a^\dagger \hat b^\dagger\sim K_+$ are ladder operators of the $su(1,1)$ algebra so one possibility is to use the $SU(1,1)$ disentanglement formula $$ D(\zeta)=e^{\zeta K_+-\zeta^*K_-}=e^{\eta K_+}e^{-\beta K_0}e^{\gamma K_-} \tag{1} $$ where $$ \zeta= -\frac12 \tau e^{-i\varphi}\, ,\quad \eta=-\tanh\frac\tau 2 e^{-i\varphi}\, ,\quad \beta=-2\ln\cosh\frac\tau 2\, ,\quad \gamma=-\eta^*\, . $$ You can find this in the textbook of

Peremolov, Generalized coherent states,

or more generally by searching for $SU(1,1)$ disentanglement formula. Another reference would be

A group-theoretic approach to the disentanglement of generalized squeezing operators, F.A. Raffaa, M. Rasettib, V. Penna, Physics Letters A 438 (2022) 128106

Using (1) you can work out what you need, although it's probably messy.

An alternative starts by assuming $p>q$ (the case $p<q$ follows a similar reasoning) and write $$ \vert p,q\rangle \sim (\hat a\hat b)^q \vert p-q,0\rangle\, . \tag{2} $$ You can then compute $$ e^{\zeta K_+-\zeta^*K_-}\vert p,q\rangle \sim e^{\zeta K_+-\zeta^*K_-} (\hat a\hat b)^q e^{-\zeta K_++\zeta^*K_-} \, e^{\zeta K_+-\zeta^*K_-}\vert p-q,0\rangle\\ \sim\left( e^{\zeta K_+-\zeta^*K_-} \hat a\hat b e^{-\zeta K_++\zeta^*K_-} \right)^q\,\times \, e^{\zeta K_+-\zeta^*K_-}\vert p-q,0\rangle\, . $$ Since $$ \left( e^{\zeta K_+-\zeta^*K_-} \hat a\hat b e^{-\zeta K_++\zeta^*K_-} \right)\sim \left( e^{\zeta K_+-\zeta^*K_-} K_- e^{-\zeta K_++\zeta^*K_-} \right)\, , $$ you get a closed form in terms of $\zeta$ and hyperbolic functions, which you can then expand (remembering that $K_+$ and $K_-$ do not commute) and apply to $e^{\zeta K_+-\zeta^*K_-}\vert p-q,0\rangle$, which you already know.

Note that I didn't bother to compute factors of $\frac12$ and various square roots of factorials as in Eq.(2); you'll have to figure that out by yourself.

Answer 2

I think that in case of general $p$, $q$ it may be impossible to obtain simple expressions and relations for $T(m,n,p,q)$. The operator $S_2(\zeta)$ is the exponential of the quadratic boson operator, taking into account the symmetry between $a$ and $b$ it can be written in the following form $$ S_2(\zeta) = A \exp(\alpha a^\dagger b^\dagger)\exp(\beta(a^\dagger a + b^\dagger b))\exp(\gamma ab)\tag{1} $$ Such equalities are related to the normal ordering of operators, they are discussed in some books, now I only remember F. Berezin's book on the method of secondary quantization. From (1) it follows $$ S_2(\zeta)|0,0\rangle = \sum_{k=0}^\infty A\alpha^k |k,k\rangle, \quad S_2(\zeta)|1,0\rangle = \sum_{k=0}^\infty Ae^\beta\alpha^k\sqrt{k+1}\ |k+1,k\rangle. $$ These formulas are consistent with what is written in the question when $$ A = \frac1{\cosh(r)}, \quad \alpha = -e^{i\theta}\tanh(r),\quad e^\beta = \frac1{\cosh(r)}. $$ Considering $\langle 0,0|S_2(\zeta)$ allows you to find $\gamma=e^{-i\theta}\tanh(r)$. Now, using (1), we can definitely obtain the following formula $$ S_2(\zeta)|p,q\rangle = \sum_{l = -\mbox{min}(p,q)}^\infty T(p+l,q+l,p,q)|p+l,q+l\rangle $$ with the coefficients $T$ in explicit form. I think, however, that the expression for the $T$ coefficients will contain a sum for which it is not obvious that it has a simple expression in terms of the binomial coefficients.