# I Expectation of an operator (observable) how to calculate it

1. Jan 31, 2017

### fog37

Hello Forum,

I understand that in order to calculate the average of a certain operator (observable), whatever that observable may be that we are interested in, we need to prepare many many many identical copies of the same state and apply the operator of interest to those identical state. By probability, we would then obtain several different values of that observable (these values are different eigenvalues of the operator). We would then take the total average of those many values to obtain the expectation value.

The expectation value of a particular observable is given by the integral:

$$<\hat A>=\int \Psi^*(x) \hat A \Psi(x) dx$$
where we sandwich the operator between the wavefunction $\Psi(x)$ and its complex conjugate wavefunction $\Psi^*(x)$.

Does it matter what wavefunction $\Psi(x)$ we use to calculate this expectation value integral? Can the system be in any state/wavefunction $\Psi(x)$ when we calculate the expectation value? I don't think so since different functions $\Psi(x)$ would produce different values for the integral.
The wavefunction $\Psi(x)$ could be an eigenfunction of a certain operator, of the eigenfunction of other operator, etc.

In essence, what wavefunction do we use inside the integral?

thanks for any clarification.

Fog37

2. Jan 31, 2017

### BvU

Yes
Yes. The result will be the expectation value for the operator A in that particular state.
generally, yes. There are exceptions ('degeneracy').
It could well be. In case of an eigenfunction $\Psi_a(x)$ of operator A with eigenvalue a the integral gives that eigenvalue a as a result.

Last edited: Jan 31, 2017
3. Jan 31, 2017

### fog37

Ok, thanks BvU.

I am missing the important fact that

And which state should we prepare the system in to later calculate the various expectation values for different observables (operators)?

4. Jan 31, 2017

### BvU

Not sure I understand what you are asking. In principle you can calculate the expectation value for any observable at any time you want with the expression in post #1.

Did you already encounter some examples of a $\Psi(x,t)$ or a $\Psi(x)$ ?

5. Jan 31, 2017

### fog37

Well, you mentioned that result of the expectation value of the operator depends on the state the system is in. So, to calculate the expectation value, we choose to prepare many identical copies of the system all in the same state and we apply on each the operator corresponding to the observable.

But what state do we want to prepare the system in?

6. Jan 31, 2017

### Staff: Mentor

Any state - they just have to all be the same.

Its central to the Ensemble interpretation of QM I hold to (actually its the ignorance ensemble but its just a minor change no need to go into here - start a new thread if interested).

As your QM studies advance you will want a copy of Ballentine as per my Signature. He explains that interpretation, which interestingly was Einstein's (yes - that's a misconception about Einstein promulgated by the popular press that Einstein did not believe in QM - he believed it incomplete - not incorrect and the Ensemble interpretation points to the exact issue - but again that's another thread).

Thanks
Bill

7. Feb 1, 2017

### BvU

@fog37 :
Do you understand the the wave function IS a (complex) probability density amplitude ?
The state you are interested in. Example: if you want to measure the energy of the ground state for hydrogen, you make sure you have hydrogen in the ground state and you measure. And if you have an idea for the wave function of that state, you calculate $<E_{n=1}>$.

8. Feb 1, 2017

### fog37

thanks BvU. Yes, I know that the wavefunction is a complex function and we can obtain the probability density from it.

You are saying that, based on the observable we are interested in, we can prepare the system in a certain state. How do we tell the system to be/to go in that particular state?
For example, if we were interested in learning about the energy of the system (its average value, etc.) why don't we directly prepare the system to be in one of the eigenstates of the energy operator? Can we really do that? Or are we only able to prepare the system in a superposition of eigenstates of the energy operator? If so, why?

I know that after the measurement the system will go into one of the eigenstates of the energy operator with a fixed energy value (eigenvalue associated to that eigenstate)...

Fog 37

9. Feb 2, 2017

### Staff: Mentor

Because then we wouldn't need to measure the system to determine its energy - we'd already know the answer because it was determined by the preparation procedure. Your question is the equivalent of "If we want to know the length of a piece of rope, why didn't we just cut it to a known length in the first place?"

10. Feb 2, 2017

### fog37

That is true :)

But I read somewhere else that is not possible to "choose" and prepare the state the system is in. I think there are experimental procedures to force the system in a particular state, correct? But if that is true, i.e. we can push the system is a particular state, we could also push into an eigenstate even if we would already know the measurement output at that point...

11. Feb 2, 2017

### Staff: Mentor

The way to "push" a system into an energy eigenstate is to measure its energy. In fact, that's how you'd go about preparing an ensemble of systems all in a particular energy eigenstate: you'd measure the energy of each one, and discard the ones that don't have the desired energy value.

But suppose you want to know what the energy of a particle is after it goes through some interesting interaction? If you're going to find out by measuring, you'll prepare a large number of particles by putting them through that interaction; and then you'll measure their energy. (Preparing them any other way and then measuring the energy is going to be answering a different question, namely what is the energy after a different interaction that you're not interested in). If you're going to find out by calculation, you'll use your knowledge of the interaction to calculate the wave function of a particle coming out of the interaction (which will, in general, not be an energy eigenstate) and then you'll plug it into the expectation value integral that you started this thread with.

If you did everything right, that expectation value will be close to the average of the measurements.

12. Feb 2, 2017

### fog37

Great. That was my missing piece.

In order:
a) We first prepare a large number of identical particles. How do we know, practically speaking, that the particles are identical?
b) We then put all those particles through the interaction;
c) We measure the energy of each particle. We measure the energy by applying the energy operator to each one of them. Is that how I would carry out a measurement practically speaking? Will the measuring device act on each particle and the interaction will theoretically be equivalent to the action of the energy operator?
d) We then compute the expectation value integral with the wave function that we have calculated from the knowledge of the interaction. Could you elaborate on that? What does it really mean to find the wavefunction from the knowledge of the interaction?