A stupidly basic question about expectation value

[HAYAO]

TL;DR

You can find expectation value for a system even if the operator is not operated on an eigenfunction. How do you prove this without using specific examples?

I'm teaching Quantum mechanics to freshmen in my college, but I'm stuck on a very basic concept that I always took for granted. I'm a chemist major, not a physicist, but I thought it's something I should clearly understand. I am embarrassed to ask this question, but I'd rather be embarrassed now than forever.

The "expectation value" formula still holds for some operators that is not necessarily the eigenfunction of the operator.
For example, applying the position operator on wavefunction derived from solving the energies for "particle in an infinitely deep well" system, doesn't yield a scalar times the wavefunction. But you can still solve the expectation value, yielding L/2.

Is there any way I can prove that wavefunction doesn't necessarily have to be an eigenfunction of the operator to measure expectation value? Like I can do it by presenting an example above, but I imagine students would ask "but why". People may ask intuitive question like "if an operator on a wavefunction doesn't even yield a value (measured value), then how can you find an expectation value, which is an average of the value yielded for the same measurement infinitely done?"

Sometimes, I've seen lecture notes on "deriving" the expectation value formula (not really derivation because it's a definition, but I supposed they made that lecture note because students would ask why the formula works in finding the average value). They rely on the fact that if the wavefunction in question is an eigenfunction of an operator, then the operator is self-adjoint, making the transition from $\int_{-∞}^{∞}\hat{A}Ψ^{*}Ψdx = \int_{-∞}^{∞}Ψ^{*}\hat{A}Ψdx$ possible. Indeed position operator and momentum operators are Hermitian (thus self-adjoint). Nonetheless, this doesn't really answer the question.

I am certain that I am making some sort of major miscomprehension, but I would like an explanation.

 

[mad mathematician]

No.
Observables are assumed to be Hermitian operators and thus their eigenvalues will be real.


These lectures are in QC, but the beginning lecture should clarify what is an observable.
https://en.wikipedia.org/wiki/Observable
As for your opening question, simply by the definition of a scalar product in the specific function space we are working on. Which is basically rigged Hilbert space $L^2$.

 

[PeroK]

TL;DR Summary: You can find expectation value for a system even if the operator is not operated on an eigenfunction. How do you prove this without using specific examples?

The point is that the eigenfunctions of a Hermitian operator, $\hat O$, generally form a complete, orthonormal set. That means that any state can be expressed as a linear combination of the eigenfunctions ($\psi_n$):
$$\psi = \sum_{n = 1}^{\infty} c_n\psi_n$$Then the expectation value of the operator is given by:
$$E(\hat O, \psi) = \langle \psi|\hat O|\psi \rangle =\sum_{n = 1}^{\infty}|c_n|^2\lambda_n$$Where $\lambda_n$ is the eigenvalue associated with the nth eigenfunction.

 

[DrClaude]

Then the expectation value of the operator is given by:
$$E(\hat O, \psi) = \langle \psi|\hat O|\psi \rangle =\sum_{n = 1}^{\infty}|c_n|^2\lambda_n$$Where $\lambda_n$ is the eigenvalue associated with the nth eigenfunction.

To add to that: note that the final result is "sum over outcomes of probability times result", which is exactly the statistical definition of the expectation value of a random variable, see https://en.wikipedia.org/wiki/Expected_value#Random_variables_with_finitely_many_outcomes

 

[PeterDonis]

Is there any way I can prove that wavefunction doesn't necessarily have to be an eigenfunction of the operator to measure expectation value?

This seems like an odd question, because the only time you need to compute an expectation value is when the wave function is not an eigenfunction of the operator. If the wave function is an eigenfunction, you don't need to compute anything: you just use the eigenvalue.

 

[HAYAO]

I'm sorry if my question wasn't clear. I already know all of what you guys mentioned. That was not my question. My question doesn't ask why Hermitian operator can give expectation value. I was asking how non-Hermitian operator can give expectation value.

I was asking a possible intuitive question on how is it possible to compute an "expectation value" when wavefunction is not an eigenfunction of the operator. I mean if an operator on a wavefunction don't give you an eigenvalue, then you can't "observe" it, but you can still get an expectation value. That sounds very odd classically intuitively speaking.

Thus, I wanted to show my students that the "expectation value" can still be computed even if the operator is not applied to an eigenfunction of that operator. I can present a specific example like i said above above about position operator on particle in a box wavefunction, but I wanted to show a general derivation. My question is how can I do that?

 

[Nugatory]

I was asking how non-Hermitian operator can give expectation value

I was asking … how is it possible to compute an "expectation value" when wavefunction is not an eigenfunction of the operator.

Those are different questions.
The first doesn’t make a lot of sense because we’re talking about the expectation value for measurement results, and if the operator isn’t Hermitian it doesn’t represent any measurement.
To answer the second question (and your original post) you can take the Born rule as a postulate. That gives you the probability of getting each possible measurement result, and the expectation value follows from that.

 

[Demystifier]

I mean if an operator on a wavefunction don't give you an eigenvalue, then you can't "observe" it, but you can still get an expectation value.

This is not true. You can observe an observable represented by operator $A$ even if there is no number $a$ such that $A\psi(x)=a\psi(x)$. However, due to quantum uncertainty, you cannot predict (compute) what that result of observation will be. What you can compute is the probability $p(a)$ that the result of observation will be $a$. And if you know the probability of each $a$, then you can compute the average value of the observable $A$. The formula for computing the average value is
$$\langle A\rangle = \int dx \psi^*(x)A\psi(x)$$
The average value can also be experimentally obtained by performing many measurements of $A$ and taking the average over all measurements. For some historical reasons the average value is also called "expectation" value, but this name is confusing because the "expectation" value is not what you expect to obtain in a single measurement. Instead, this is what you expect to obtain as an average value, when you perform many measurements.

Or to make the long story short, if an operator on a wavefunction does not give you an eigenvalue, you can still measure it, but the result of measurement is random, so you don't expect any specific result of measurement. Nevertheless, if you repeat the measurement many times, you can predict very well what will be the average value, and this predicted average value is called expectation value.

 

[HAYAO]

Or to make the long story short, if an operator on a wavefunction does not give you an eigenvalue, you can still measure it, but the result of measurement is random, so you don't expect any specific result of measurement. Nevertheless, if you repeat the measurement many times, you can predict very well what will be the average value, and this predicted average value is called expectation value.

This answers my question. Thank you.