# Exponential of a ladder operator justification

Considering the ladder operators of quantum mechanics, specifically the creation operator,can we take its exponential? I was looking at this post: Annihilation operator in the exponent? and in the answer it states that it is possible to have the exponential of such an operator. But the creation operator is not limited in the sense that we can apply it various times in a state.

$$e^{\hat{a}^{\dagger}} \left|0\right\rangle = \sum_{n=0}^{\infty} \frac{({\hat{a}^{\dagger}})^n}{n!} \left|0\right\rangle$$

Given this, the above series must converge in order for the exponential of the creation operator to be well-defined. How do we prove that it always converges?

• My bad, I wanted the creation operator not the annihilation one. Feb 23, 2021 at 21:08
• You are completely comfortable with coherent states, though, or are you having residual misgivings? Feb 23, 2021 at 21:30
• Im not completely comfortable with those states. I just dont know if I can apply the exponential of this creation operator to a state, given that the summation in the Taylor Series goes up to infinity. Maybe I run into trouble with the convergence of this series Feb 23, 2021 at 21:54
• Most convergence problems here are bypassable... As Napoleon said, "On s'engage, et puis on voit".... Feb 23, 2021 at 22:22
• How does one bypass them? Feb 24, 2021 at 0:50

## 2 Answers

You can evaluate the sum of that series explicitly. Noting that $$(a^\dagger)^n|0\rangle = \sqrt{n!}|n\rangle$$, we have

$$e^{ a^\dagger}|0\rangle= \sum_{n=0}^\infty \frac{(a^\dagger)^n}{n!} |0\rangle = \sum_{n=1}^\infty \frac{1}{\sqrt{n!}} |n\rangle$$

which is a state with norm $$\sum_n \frac{1}{n!} = e^1$$.

That being said, I get the impression that you're asking how to show that $$e^{a^\dagger}|\psi\rangle$$ converges for an arbitrary state $$|\psi\rangle$$. This you cannot do, because generically it doesn't converge. $$a^\dagger$$ is not a bounded operator, which means that there are some $$|\psi\rangle$$ in your Hilbert space$$^\ddagger$$ on which $$a^\dagger$$ cannot act; the same is true for $$e^{a^\dagger}$$.

More explicitly, note that $$|n\rangle = \frac{(a^\dagger)^n}{\sqrt{n!}} |0\rangle$$, so $$(a^\dagger)^m|n\rangle = \sqrt{\frac{(n+m)!}{n!}}|n+m\rangle$$. As a result,

$$e^{a^\dagger} |n\rangle = \sum_{m=0}^\infty \frac{(a^\dagger)^m}{m!} |n\rangle = \sum_{m=0}^\infty \frac{\sqrt{(n+m)!}}{m!}|n+m\rangle = \sum_{\ell=n}^\infty \frac{\sqrt{\ell !}}{(\ell-n)!} |\ell\rangle$$

which is a state with norm $$\sum_{\ell=n}^\infty \frac{\ell!}{[(\ell-n)!]^2} = n! _1F_1(n+1;1;1)$$ via WolframAlpha. Though well-defined for any particular $$|n\rangle$$, this increases rapidly - faster than $$n!$$. As a result, if $$|\psi\rangle = \sum_n c_n |n\rangle$$ such that $$\sum_n \bigg(n!_1 F_1(n+1;1;1)\bigg)^2|c_n|^2$$ does not converge, then $$|\psi\rangle$$ is outside the domain of $$e^{a^\dagger}$$.

$$^\ddagger$$For example, $$|\psi\rangle = \sum_{n=0}^\infty \frac{1}{n+1}|n\rangle$$. This state is square-normalizable (it has norm $$\frac{\pi}{\sqrt 6}$$) but if you naively act with $$a^\dagger$$, you obtain $$\sum_{n=1}^\infty \frac{1}{\sqrt{n}}|n\rangle$$ whose norm is $$\sum_{n=1}^\infty \frac{1}{n} \rightarrow \infty$$; as a result, $$a^\dagger |\psi\rangle$$ is not in the Hilbert space, so $$|\psi\rangle$$ is not in the domain of $$a^\dagger$$.

• For the state of norm $\frac{\pi}{\sqrt{6}}$, do we now have to impose a factor of $\frac{6}{ \pi^2}$ into the completness relation in order to obtain probabilities? Feb 24, 2021 at 18:52
• @MicrosoftBruh Feel free to normalize $|\psi\rangle$ for convenience - just let $|\psi\rangle = \frac{\sqrt{6}}{\pi} \sum_{n=0}^\infty \frac{1}{n+1}|n\rangle$. Also, I fixed a typo, but this doesn't affect the answer. Feb 24, 2021 at 19:32

Functions of operators are usually defined as expansion in series, so there is not much to prove here - it is a formal way to write the series on the right.