# Variance of an eigenvector

1. Aug 28, 2009

### foxjwill

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

2. Relevant equations

3. The attempt at a solution
I keep getting
$$\langle \psi|A^2\psi\rangle = \lambda\langle \psi|A\psi\rangle = \lambda^2\langle\psi |\psi\rangle$$​
and
$$\langle \psi|A\psi\rangle^2 = \left(\lambda\langle \psi|A\psi\rangle\right)^2 = \lambda^2\langle\psi |\psi\rangle^2.$$​
Where am I going wrong?

2. Aug 28, 2009

### gabbagabbahey

Eigenvectors are typically normalized, so $\langle\psi\vert\psi\rangle=1$

3. Aug 28, 2009

### foxjwill

D'oh! Of course! Thanks!