Average value of observable

1. Mar 2, 2014

LagrangeEuler

Why we define that average value of some observable $\hat{A}$ in state $\psi$ is
$(\psi,\hat{A}\psi)$
Why this isnot perhaps
$26(\psi,\hat{A}\psi)$?

2. Mar 2, 2014

Staff: Mentor

What if the state ψ was an eigenstate of the observable A?

3. Mar 2, 2014

LagrangeEuler

In that case $\hat{A}\psi=a\psi$,
so
$(\psi,\hat{A}\psi)=a$.

4. Mar 2, 2014

Staff: Mentor

Right, where $a$ is the eigenvalue of the observable. Not $26a$!

5. Mar 2, 2014

LagrangeEuler

Tnx.

6. Mar 16, 2014

LagrangeEuler

But still after I think. Ok. When the system is in some eigen- state $\varphi_n$, we measure energy $E_n$. So
$(\varphi_n,\hat{H}\varphi_n)=E_n$
but if system is in some other state, perhaps, $\phi(x)$ which is not eigenstate of observable $\hat{H}$ how I can be sure that average value of energy is
$(\phi(x),\hat{H}\phi(x))$?

7. Mar 16, 2014

kith

The expectation value can be defined as <H>=∑npnEn. If you use Born's rule for the pn you get <ψ|H|ψ>.

8. Mar 16, 2014

Staff: Mentor

You can express the general state $\phi(x)$ in terms of the eigenfunctions: $\phi(x)$ = $a_1\varphi_1$ + $a_2\varphi_2$ ...
where $a_n^*a_n$ represents the probability of measuring $E_n$. Thus the average value will be $(\phi(x),\hat{H}\phi(x))$.

(This is equivalent to what kith just said about using the Born rule.)