What Is the Variance in Quantum Operator N?

[Piano man]

Homework Statement

N=a^+a

a=\frac{ip+mwx}{\sqrt{2m\hbar w}} \quad a^+=\frac{-ip+mwx}{\sqrt{2m\hbar w}}

|z>=e^{\frac{-|z|^2}{2}}\sum^{\infty}_{n=0}\frac{z^n}{\sqrt{n!}}|n>

where <n|n>=1

Show that the variance (uncertainty) in N, \Delta N is |z|

i.e. calculate (\Delta N)^2=<z|N^2|z>-<z|N|z>^2



2. The attempt at a solution

Taking <z|=e^{\frac{-|z|^2}{2}}\sum^{\infty}_{n=0}\frac{\bar{z}^n}{\sqrt{n!}}<n|

I subbed in appropriately and I got

e^{-|z|^2}\sum^{\infty}_{n=0}\frac{(\bar{z}z)^n}{n!}y-e^{-2|z|^2}\sum^{\infty}_{n=0}\frac{(\bar{z}z)^{2n}}{(n!)^2}y

where y=\frac{m^4w^4x^4+2p^2m^2w^2x^2+p^4}{4m^2\hbar^2 w^2}

This leads to e^{-|z|^2+z\bar{z}}y-e^{-2|z|^2+2z\bar{z}}y I believe.

From here, how do I show that this equals |z|^2?

 

[gabbagabbahey]

I subbed in appropriately and I got

e^{-|z|^2}\sum^{\infty}_{n=0}\frac{(\bar{z}z)^n}{n!}y-e^{-2|z|^2}\sum^{\infty}_{n=0}\frac{(\bar{z}z)^{2n}}{(n!)^2}y

where y=\frac{m^4w^4x^4+2p^2m^2w^2x^2+p^4}{4m^2\hbar^2 w^2}

No, a, a^{\dagger}, p and x are all operators. You can't treat them like scalars and pull them out of the inner product.

Use the fact that a|n\rangle= \sqrt{n}|n-1\rangle (for n\geq 1 ) and a^{\dagger}|n\rangle= \sqrt{n+1}|n+1\rangle to Calculate N|n\rangle and N^2|n\rangle and then use those results to calculate N|z\rangle and N^2|z\rangle and the variance.