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

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• fog37
In summary, you calculate the expectation value for an observable by taking the integral of the wavefunction corresponding to the state the system is in. The state you are interested in must be prepared into by the user. The system can be in any state when the expectation value is calculated.
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

fog37 said:
Does it matter what wavefunction ##\Psi(x)## we use to calculate this expectation value integral?
Yes
Can the system be in any state/wavefunction ##\Psi(x)## when we calculate the expectation value?
Yes. The result will be the expectation value for the operator A in that particular state.
different functions ##\Psi(x)## would produce different values for the integral.
generally, yes. There are exceptions ('degeneracy').
The wavefunction ##\Psi(x)## could be an eigenfunction of a certain operator, of the eigenfunction of other operator, etc
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:
bhobba
Ok, thanks BvU.

I am missing the important fact that

BvU said:
Yes. The result will be the expectation value for the operator A in that particular state.

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

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)## ?

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?

fog37 said:
But what state do we want to prepare the system in?

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

@fog37 :
BvU said:
Did you already encounter some examples of a ##\Psi(x,t)## or a ##\Psi(x)## ?
Do you understand the the wave function IS a (complex) probability density amplitude ?
fog37 said:
But what state do we want to prepare the system in?
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}>##.

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

fog37 said:
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?
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?"

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...

fog37 said:
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...
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.

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?

## 1. What is the expectation value of an operator in quantum mechanics?

The expectation value of an operator, also known as the average value, is a measure of the average outcome of a measurement for a physical observable in a quantum mechanical system. It represents the most probable result of measuring the observable, taking into account all possible outcomes and their probabilities.

## 2. How do you calculate the expectation value of an operator?

The expectation value of an operator is calculated by taking the inner product of the state vector and the operator, and then multiplying it by the complex conjugate of the state vector. This can be written mathematically as ⟨A⟩ = ⟨ψ|A|ψ⟩, where A is the operator and ψ is the state vector.

## 3. What is the significance of the expectation value in quantum mechanics?

The expectation value is a crucial concept in quantum mechanics as it allows us to make predictions about the outcome of measurements for physical observables. It provides a way to connect the mathematical formalism of quantum mechanics to observable quantities in the physical world.

## 4. Can the expectation value of an operator be negative?

Yes, the expectation value of an operator can be negative. This means that the average outcome of a measurement for that observable is negative, but it does not necessarily mean that the observable itself is negative. In quantum mechanics, observables can have both positive and negative eigenvalues, which can result in a negative expectation value.

## 5. How does the expectation value change with different states of a quantum system?

The expectation value of an operator can change with different states of a quantum system. This is because the state vector, ψ, is dependent on the quantum state of the system, and the expectation value is calculated using the state vector. As the state of the system changes, the state vector and thus the expectation value can also change.

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