Coherent States: Ʃ(|α|2)n = 1?

[Hazzattack]

Hi guys,

Just a quick question, is the following statement true (it seems to be implied in the article I'm looking at);

Ʃ(|α|²)ⁿ = 1

(The sum over n=0 to infinity)

Thanks to anyone who takes a look.

 

[fzero]

You don't post the reference, so I will refer to generalities. Typically the coherent state for the harmonic oscillator is written as

$$ | \alpha \rangle = c \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} | n\rangle.$$

The norm of this state is

$$\langle \alpha | \alpha \rangle = |c|^2 \sum_{n=0}^\infty \frac{|\alpha|^{2n}}{n!} = |c|^2 e^{|\alpha|^2}, $$

so we can normalize the state by choosing $c = \exp ( - |\alpha|^2/2 )$. If we had instead written

$$ | \tilde{\alpha} \rangle = \sum_{n=0}^\infty \tilde{\alpha}^n | n\rangle,$$

then the normalization condition is indeed

$$\sum_{n=0}^\infty |\tilde{\alpha}|^{2n} =1.$$

 

[Hazzattack]

Great! That's what i was hoping, but for some reason i was confusing myself. Will post the workings i was referencing later. Thanks for taking a look!

 

[naima]

We have to say that in the relation of hazzattack alpha is not the complex eigenvalue of the annihilation operator..