# Coherent States of the Harmonic Oscillator

1. Jul 31, 2008

### tshafer

Alright, I'm sure I'm missing something extremely simple, but in Griffiths (and another text I'm reading) coherent states are mentioned as eigenfunctions of the annihilation operator.

I just don't understand:
a) how you can have an eigenfunction of the annihilation operator (other than |0>) if the whole point is it knocks you down a level from |n> to |n-1>

b) why the creation operator is described as not having ay eigenfunctions if you can have eigenfunctions of the annihilation operator

any help would be great, thanks!
tom

2. Jul 31, 2008

### Avodyne

An individual $|n\rangle$ state is obviously not an eigenstate (if $n$ is not zero), but a linear combination of them can be. Define a state $|\alpha\rangle$ as
$$|\alpha\rangle \equiv \sum_{n=0}^\infty{ \alpha^n\over\sqrt{n!}}|n\rangle$$
and act on it with the annihilation operator $a$; using $a|n\rangle=\sqrt{n}|n{-}1\rangle$, we get
$$a|\alpha\rangle = \sum_{n=0}^\infty{ \alpha^n\over\sqrt{n!}}\sqrt{n}|n{-}1\rangle = \sum_{n=1}^\infty{ \alpha^n\over\sqrt{(n{-}1)!}}|n{-}1\rangle.$$
Now replace $n$ with $n{+}1$, and we have
$$a|\alpha\rangle = \sum_{n=0}^\infty{ \alpha^{n+1}\over\sqrt{n!}}|n\rangle = \alpha\sum_{n=0}^\infty{ \alpha^{n}\over\sqrt{n!}}|n\rangle = \alpha|\alpha\rangle.$$
Well, the same trick doesn't work for the creation operator. Another way to do it is to work in the position basis, where $a$ becomes something like $x+d/dx$ (with various constants left out), and $a^\dagger$ becomes something like $x-d/dx$. The first has an eigenfunction $\exp[-(x-\alpha)^2/2]$ with eigenvalue $\alpha$, and the second has an eigenfunction $\exp[+(x-\alpha)^2/2]$. But this eigenfunction is not normalizable, so is not allowed.

3. Jul 31, 2008

### tshafer

Nice, that's a cute trick. I'll work through what you just said so I can get it for myself. Thanks!

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