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.