Quantum variance (uncertainty)

[Piano man]

Homework Statement

[tex]N=a^+a[/tex]

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

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

where [tex]<n|n>=1[/tex]

Show that the variance (uncertainty) in N, [tex]\Delta N[/tex] is [tex]|z|[/tex]

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



2. The attempt at a solution

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

I subbed in appropriately and I got

[tex]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[/tex]

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

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

From here, how do I show that this equals [tex]|z|^2[/tex]?

 

[gabbagabbahey]

I subbed in appropriately and I got

[tex]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[/tex]

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

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

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