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:

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1 Answer 1


$ \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 $a^\dagger a$.

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    $\begingroup$ 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.) $\endgroup$ Commented Jul 30, 2019 at 18:02

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