What Is the Variance of an Eigenvector in a Hermitian Operator?

[foxjwill]

Homework Statement

On page 5 of http://arcsecond.wordpress.com/2009...uality-and-heisenbergs-uncertainty-principle/ the author states (w/o proof) that if [tex]\psi[/tex] is an eigenvector (say with eigenvalue [tex]\lambda[/tex]) of an Hermitian operator A (I don't think the Hermitian-ness matters here), then its variance is 0; that is, [tex]\langle \psi| A^2\psi\rangle = \langle \psi| A\psi\rangle^2[/tex]. However, I've not been able to show this.

Homework Equations

The Attempt at a Solution

I keep getting

[tex]\langle \psi|A^2\psi\rangle = \lambda\langle \psi|A\psi\rangle = \lambda^2\langle\psi |\psi\rangle[/tex]​

and

[tex]\langle \psi|A\psi\rangle^2 = \left(\lambda\langle \psi|A\psi\rangle\right)^2 = \lambda^2\langle\psi |\psi\rangle^2.[/tex]​

Where am I going wrong?

 

[gabbagabbahey]

Eigenvectors are typically normalized, so [itex]\langle\psi\vert\psi\rangle=1[/itex]

 

[foxjwill]

Eigenvectors are typically normalized, so [itex]\langle\psi\vert\psi\rangle=1[/itex]

D'oh! Of course! Thanks!