Bosonic coherent state normalization

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?

• Apr 5 at 16:54
• @EmilioPisanty Thank you for the remark. I will do it in the future! Apr 5 at 16:56

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.