I got confused by the concept of the average value of an observable.
I know that when we measure a physical quantity $A$ in a specific state described by $\psi_1$, we only get the eigenvalue $a$ of the operator $\hat{A}$.
So theoretically the value of the physical quantity $A$ that we are measuring is equal to the eigenvalue of $\hat{A}\psi_1=a\psi_1$.
Then there is a formula to find the average value $\langle A\rangle$. And this is also an expectation value.
Does it mean that normally $\langle A\rangle=a$?
Then I get confused by the world ''average'' in ''average value $\langle A\rangle$''. Is this the average value of the physical quantity $A$ in all its the states?
And finally, are we assuming here that my system has only one system described by $\psi$? For example there is only one particle. Is it possible that I have three systems (three particles) described by different eigenfunction $\psi$ and eigenvalue $a$?
I am confusing everything.
Best Answer
You are confusing a lot all right : )
Start with getting this equation $ \hat{A}|\psi_1 \rangle =a |\psi_1\rangle$ nailed down first.
You apply a Hermitian operator, $\hat{A}$ to a ket $|\psi_1 \rangle$ and it gives you an appropriate eigenvalue $a $.
So the $ \hat{X}$ operator gives you the position observable $x$, likewise applying $ \hat{P}$ gives you the real observable momentum $p $, associated with $|\psi_1 \rangle$.
I am assuming you know what using a Hermitian operator implies.
You should read this website article later: Average Value for a fuller description but a summary for now is:
No, $<A>$ is not equal to $a$, in general.
$<A>$ is a prediction of the average value of repeated measurements, whereas $ \hat{A}|\psi_1 \rangle =a |\psi_1\rangle$ is the value of 1 (just 1) single measurement.
For example, say you have a wave function that you know describes the ground state of a particle $m $.
Now you want to find the average momentum $$<P> = \langle \psi_1 | \hat P|\psi_1\rangle$$
For our purposes, expectation value and average value are the same idea.
So when you work out $<P> = \langle \psi_1 | \hat P|\psi_1\rangle$ in the 1D ground state simple box potential, you are going to find that it's equal to 0.
That's because the average momentum of the particle in these conditions is zero. The particle has the same momentum in two opposite directions, so it's average value is zero.
No.... It's the average value in one state, where you sandwich the A between, for example the functions for the ground state.
If you then want to find the average value for the first excited state, you sandwich the A between the functions that describe the first excited state.
You then keep doing this for any energy state that that you know the functions for.
Image Source: Energy Levels
And finally, are we assuming here that my system has only one system described by $\psi$? For example there is only one particle.
Yes, there is just 1 particle. No offence, but I would absolutely, definitely, completely stick to learning about one particle state for the moment.
No. I would not worry at this time about 3 particle systems, that is normally further down the course, that is weeks or months way if you are just starting off.