# Exponential of ladder operators acting on vacuum state [closed]

How would I solve expressions of the following nature:

$$<0|e^{Vt(a+a^\dagger)}|0>$$ and $$<0|e^{\omega aa^\dagger t}|0>~?$$

My intuition is that I have to expand the exponent as a power series but I can't understand how to deal with the pre-factors.

## closed as off-topic by AccidentalFourierTransform, Cosmas Zachos, Jon Custer, heather, ZeroTheHeroJun 29 '18 at 20:26

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Note that $aa^{\dagger}|0>=a|1>=|0>$. With this, its easy to see that $<0|aa^{\dagger}|0> = 1$, and then $<0|(aa^{\dagger})^2|0>=1$ and so on. you can generalize this for any $N \geq 0$: $$<0|\left( \omega t a a^{\dagger} \right)^{N}|0> \ = \ (\omega t)^N <0|\left(a a^{\dagger} \right)^{N}|0> \ = \ (\omega t)^N$$
This means that: $$<0|e^{\omega t a a^{\dagger}}|0> \ = \ \sum_{N=0}^{\infty} \frac{<0|\left( \omega t a a^{\dagger} \right)^{N}|0>}{N!} \ = \ \sum_{N=0}^{\infty} \frac{(\omega t)^N}{N!} \ = \ e^{\omega t}$$
The process is the same for $<0|e^{Vt(a+a^{\dagger})}|0>$: Find an expression valid for any $N \geq 0$ for $<0|\left(Vt(a+a^{\dagger})\right)^N|0>$ and plug it into the power series.