Consider two states of the type $|\alpha,\xi \rangle = \hat{D}(\alpha) \hat{S}(\xi) |0\rangle$, where $D$ and $S$ are the displacement and squeeze operators, respectively, and $|0\rangle$ is a 1D harmonic oscillator vacuum state.

My question is: Is there a closed formula for $\langle \alpha, \xi | \beta, \eta \rangle$?

I know how to calculate this for two coherent states ($\xi = \eta = 0$), but since the commutator of $[a^2,a] \neq I$ (which comes from $S$) the same strategy I use in that case does not work (i.e., using the Zassenhaus formula).

I saw that there is a way to express the wave function of this state in position representation, so I could calculate this as $\int dx \langle \alpha, \xi | x \rangle \langle x \beta, \eta \rangle$, but this seems really unwieldly. Is there a simpler way analogous to the coherent case?

  • 2
    $\begingroup$ Please show your attempt, i.e. whatever you tried with the Zassenhaus formula. $\endgroup$
    – DanielSank
    Jan 9, 2019 at 19:42
  • $\begingroup$ Can inserting $\int_{z \in \mathbb{C}}^{}\frac{dz_{}^{*}dz_{}^{}}{2\pi i}| z\rangle\langle z|$ help? Atleast it will reduce the problem to overlap of coherent state and squeezed vaccum state and perhaps a gaussian integral. $\endgroup$
    – Sunyam
    Jan 9, 2019 at 20:19
  • $\begingroup$ @DanielSank I tried using it to move all the $a$'s in the exponentials to the right side (to act over the vacuum), but since the commutator I posted was not central I could not find a simple way to relate $e^A e^B$ to $e^B e^A$. Using the Zassenhaus formula didn't get me very far. $\endgroup$ Jan 9, 2019 at 21:07
  • 1
    $\begingroup$ @Sunyam Didn't try it, but maybe it helps. Thanks for the suggestion. $\endgroup$ Jan 9, 2019 at 21:13

1 Answer 1


For the pure squeezing case where $$ S(z) \equiv e^{{\textstyle \frac12}( z {a^\dagger}^2- z^*a^2)}, \quad z= e^{\theta} |z| $$ $$ = e^{ {\textstyle \frac12} e^{i\theta} \tanh |z| {a^\dagger}^2}e^{ - \ln \cosh |z| \left(a^\dagger a+\textstyle \frac12\right)} e^{- {\textstyle \frac12} e^{-i\theta} \tanh |z|{a}^2} $$ and defining $\alpha= e^{i\theta} \tanh |z|$ we can use the formula $$ \hat S(\alpha_2) \hat S(\alpha_1)= \hat S(\alpha_3) \exp\{i\chi(\alpha_1,\alpha_2) (a^\dagger a +{\textstyle \frac 12})\} $$ where $$ \alpha_3= \frac{\alpha_1+\alpha_2}{1+\alpha_1\alpha_2^*}, \quad \exp\{2i\chi\}= \frac{1+\alpha_1^*\alpha_2}{1+\alpha_1\alpha_2^* }. $$ to compute the overlap. I have not tried to add in the displacement operators, but it should not be too hard.

  • $\begingroup$ Thanks! I guess this solves my problem, since it is just a question of commuting $S$'s and $D$'s now in the right way. Do you have a reference for the formula you posted? $\endgroup$ Jan 9, 2019 at 21:03
  • $\begingroup$ They are bit complicated as the use the failthful but non-unitary 2d representation of the $\mathfrak{sp}(2)$ lie algebra. I think that I learned of them from Prelemov's book on generalized coherent states. $\endgroup$
    – mike stone
    Jan 9, 2019 at 21:05

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