# Creation operator on coherent state and issue with commutation relations

If our bosonic annihilation and creation operators are $$[a,a^\dagger] = 1$$, then for any complex number $$\varphi$$ we can define the (unnormalized) coherent state $$| \varphi \rangle \equiv e^{\varphi a^\dagger}|0\rangle = \sum_{n=0}^\infty \frac{\varphi^n}{\sqrt{n!}} |n\rangle.$$

Part of the beauty is that when acting on such coherent states, we can replace $$a \to \varphi$$ and $$a^\dagger \to \partial_\varphi$$. For example, to prove this: $$\boxed{ a^\dagger |\varphi\rangle } = \sum_{n=0}^\infty \frac{\varphi^{n} \sqrt{n+1} }{\sqrt{n!}} |n+1\rangle = \sum_{n=0}^\infty \frac{\varphi^{(n+1)-1} (n+1)}{\sqrt{(n+1)!}} |n+1\rangle = \boxed{\partial_\varphi |\varphi\rangle}.$$

My question is now the following: how is this consistent with the canonical commutation relations? The issue: $$\boxed{ [a,a^\dagger] |\varphi\rangle } = (aa^\dagger-a^\dagger a) |\varphi\rangle =(\varphi \partial_\varphi - \underbrace{\partial_\varphi \varphi}_{= 1 + \varphi \partial_\varphi}) |\varphi\rangle= \boxed{ - |\varphi\rangle }.$$ I.e., this would imply that $$[a,a^\dagger] =-1$$ instead of $$[a,a^\dagger] =1$$. Where have I gone astray?

Curiously, Altland and Simons claim all is good in their book, but they magically insert a minus sign halfway, which seems like a typo if it weren't for the fact that it gives the desired result! Here is a screenshot of p159 of the second edition:

$$\newcommand{\ket}[1]{\vert #1 \rangle}$$ The issue is that $$a^\dagger\ket{\phi} = \partial_\phi\ket\phi$$ no longer is a coherent state. Hence $$aa^{\dagger}\ket\phi \neq \phi\partial_\phi\ket\phi$$ but rather $$aa^{\dagger}\ket\phi = \partial_\phi \phi\ket\phi$$ which looks totally counter-intuitive, but can be easily verified using the Taylor expansion. It also coincides with the first term given in the formula from Altland.
I'll let you work out $$a^\dagger a$$.
• Makes a lot of sense, thanks! Also fun that the number operator then corresponds to the dilatation operator, i.e., $\hat n = a^\dagger a = \varphi \partial_\varphi$. (I still feel that the excerpt from Altland and Simons---although factually correct---is very misleading.) Commented Jul 30, 2019 at 18:02