Quantum Theory: Completeness of Coherent States of the Simple Harmonic Oscillator

1. Oct 9, 2012

WisheDeom

1. The problem statement, all variables and given/known data
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.

2. Relevant 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.

3. 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?

2. Oct 9, 2012

dextercioby

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

3. Oct 10, 2012

WisheDeom

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.

4. Oct 11, 2012

dextercioby

But you know that

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