# Exponential of ladder operators acting on a Fock state

I'm trying to evaluate

$$(\hat{a}+\hat{a}^\dagger)^k|n\rangle$$

Where $$\hat{a}$$ and $$\hat{a}^\dagger$$ are ladder operators and $$|n\rangle$$ the $$n$$th Fock state.

For this, I separated the problem in three parts: $$k, $$k=n$$ and $$k>n$$. In this part I'm a bit confused: Is it possible to evaluate such operations? Some tips on how to start would be really great.

• Do you need to know this state for any $n$ and $k$?
– Gec
Sep 17, 2020 at 18:57
• Yes, if it helps, I just need the projection of them on <0I, but I was curious if it's possible to evaluate analytically Sep 17, 2020 at 20:59
• I think I can write a simple formula for $\langle 0|(\hat{a} + \hat{a}^\dagger )^k |n\rangle$. But the analytical formula for the general matrix element $\langle m|(\hat{a} + \hat{a}^\dagger )^k |n\rangle$ is too cumbersome.
– Gec
Sep 18, 2020 at 6:18
• If you could share your progress with that or anything I would be very grateful to see. Sep 18, 2020 at 15:03

Let's use some math tricks to rewrite $$(\hat{a}+\hat{a}^\dagger)^k$$ in a normally ordered form. First, recall equalities $$\frac1{2\pi}\int\limits_{-\pi}^\pi e^{i(k'-k)x} dx = \delta_{k\ k'},\quad \exp(A) = \sum_{n=0}^\infty \frac1{n!} A^n.$$ Then we have $$(\hat{a}+\hat{a}^\dagger)^k = k! \sum_{n=0}^\infty \delta_{k\ n} \frac1{n!} (\hat{a}+\hat{a}^\dagger)^n = \frac{k!}{2\pi}\int\limits_{-\pi}^\pi e^{-ikx} \sum_{n=0}^\infty \frac1{n!} e^{inx} (\hat{a}+\hat{a}^\dagger)^n\ dx =$$ $$= \frac{k!}{2\pi}\int\limits_{-\pi}^\pi e^{-ikx}\exp\left(e^{ix} (\hat{a}+\hat{a}^\dagger)\right)\ dx.$$ Due to the commutation relation $$[\hat{a},\hat{a}^\dagger] = 1$$ we have $$\exp\left(e^{ix} (\hat{a}+\hat{a}^\dagger)\right) = \exp(e^{ix}\hat{a}^\dagger) \exp(e^{ix}\hat{a}) \exp(e^{2ix}/2)$$ Hence we further obtain $$(\hat{a}+\hat{a}^\dagger)^k = \frac{k!}{2\pi}\int\limits_{-\pi}^\pi e^{-ikx} \sum_{l,l',m=0}^\infty \frac1{l!l'!m!2^m}^\infty (\hat{a}^\dagger)^l (\hat{a})^{l'} e^{i(l+l'+2m)x}\ dx =$$ $$=\sum_{l,l',m=0}^\infty \delta_{k\ l+l'+2m} \frac{k!}{l!\ l'!\ m!\ 2^m} (\hat{a}^\dagger)^l (\hat{a})^{l'}\quad (*)$$ In the last expression, operators $$\hat{a}^\dagger$$ and $$\hat{a}$$ are normally ordered. Now it is easy to find $$\langle 0|(\hat{a}+\hat{a}^\dagger)^k|n\rangle$$. Due to $$\langle 0|(\hat{a}^\dagger)^l(\hat{a})^{l'}|n\rangle = \sqrt{n!}\ \delta_{n\ l'}$$ we get from $$(*)$$: if $$m = (k-n)/2$$ is non-negative integer, then $$\langle 0|(\hat{a}+\hat{a}^\dagger)^k|n\rangle = \frac{k!\sqrt{n!}}{n!\ m!\ 2^m},$$ else $$\langle 0|(\hat{a}+\hat{a}^\dagger)^k|n\rangle = 0$$ I think it is possible to use (*) to find any matrix element $$\langle n'|(\hat{a}+\hat{a}^\dagger)^k|n\rangle$$.