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In the paper arXiv:1208.3469 (equation (85)) it is stated that the coherent state for two bosons with corresponding annihilation operators $a,b$ can be written as:

$$|\Psi_\lambda \rangle = \sqrt{1-|\lambda|^2} e^{-\lambda a^\dagger b^\dagger}|0\rangle,$$

where $|0\rangle$ is the vacuum.

I tried to check that this state is indeed normalised and tried to derive equation (86) in the paper but failed. My attempt is the following:

\begin{align} \langle \Psi_\lambda |\Psi_\lambda \rangle & = (1-|\lambda|^2)\langle 0|e^{-\lambda^\star ab} e^{-\lambda a^\dagger b^\dagger}|0\rangle \\ & = (1-|\lambda|)^2 \sum_{m,n=0}^\infty \langle 0|a^n (a^\dagger)^m b^m (b^\dagger)^n |0\rangle (-\lambda)^n (-\lambda^\star)^m. \end{align}

Then we see that only cases with $m=n$ survive and that each term simply gives $(n!)^2$ since e.g. $\langle 0|a^n (a^\dagger)^n|0\rangle = n!$.

What is wrong in my reasoning here?

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You forgot a factor of $1/n!$ in your exponential expansions, and then forgot to use the geometric series. The correct set of steps are: $$\langle \Psi_\lambda |\Psi_\lambda \rangle = (1-|\lambda|^2)\langle 0|e^{-\lambda^\star ab} e^{-\lambda a^\dagger b^\dagger}|0\rangle = (1-|\lambda|)^2 \sum_{m,n=0}^\infty \frac{\langle 0|a^n (a^\dagger)^m b^m (b^\dagger)^n |0\rangle}{n! m!} (-\lambda)^n (-\lambda^\star)^m.$$ And then as you point out, the expectation values are $\frac{\langle 0|a^n (a^\dagger)^m b^m (b^\dagger)^n |0\rangle} = \delta_{mn} n! m!$, and from there you get $$ \cdots = (1-|\lambda|)^2 \sum_{m,n=0}^\infty |\lambda|^{2n} = 1. $$ where the geometric series says that $\sum_{n=0}^\infty r^n = \frac{1}{1-r}$ for $r$ in the unit ball.

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