What is the average value of an observable in a quantum system?

[LagrangeEuler]

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

 

[Doc Al]

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

 

[LagrangeEuler]

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

 

[Doc Al]

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

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

 

[LagrangeEuler]

Tnx.

 

[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
$(\phi(x),\hat{H}\phi(x))$?

 

[kith]

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

 

[Doc Al]

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
$(\phi(x),\hat{H}\phi(x))$?

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.)