Proving Completeness of SHO's Coherent States

[WisheDeom]

Homework Statement

I must prove that the set of coherent states \left\{ \left| \lambda \right\rangle \right\} of the quantum simple harmonic oscillator (SHO) is a complete set, i.e. it forms a basis for the Hilbert space of the SHO.

Homework Equations

The coherent states are defined as eigenkets of the creation operator with eigenvalue \lambda; in terms of the energy eigenkets they can be written

\left| \lambda \right\rangle = \exp \left( -\frac{|\lambda|^2}{2} \right) \sum_n \frac{\lambda^n}{\sqrt{n!}} \left| n \right\rangle

Completeness means the sum (infinite series in this case)

\sum_{\left\{ \left| \lambda \right\rangle \right\}} \left| \lambda \right\rangle \left\langle \lambda \right|

converges and is non-zero. Sites have told me the sum should converge to \pi, but I don't know how to compute that.

The Attempt at a Solution

I'm not even quite sure how to start. The eigenvalues are complex numbers, so I know the sum (integration) must be over the complex plane, but how should I do this? I tried parametrizing \lambda = x + iy, and then separately by \lambda = r e^{i \theta}, but both got very messy quickly, and I'm not sure what to do. Am I on the right track at all?

 

[dextercioby]

Can you show the series involved in the ket-bra converges ? What criteria do you know for an infinite series to converge ?

 

[WisheDeom]

Can you show the series involved in the ket-bra converges ? What criteria do you know for an infinite series to converge ?

I know there are a number of tests for series of numbers, but I'm not sure how to translate this to operators. My class didn't do any rigorous operator calculus; we sort of played it by ear. But in this case I'm not even sure how to start really.

If I assume it does converge, the sum should be

\int_{-\infty}^{\infty} d^2 \lambda e^{|\lambda|^2} \sum_m \sum_n \frac{\lambda*^m \lambda^n}{\sqrt{m! n!}} \left| m \right\rangle \left\langle n \right|

but I don't know how to evaluate this.

 

[dextercioby]

But you know that

|m\rangle\langle n| = \delta_{mn}