Discrete Values for Observables vs Average Values (Quantum Mechanics)

When considering observables and their corresponding operators, would it be correct to believe that discerning discrete values for an observable is possible ONLY when $\psi$ is an eigenfunction of the operator? Alternatively, would it also be correct to believe that the average value of an observable is ALWAYS obtainable regardless if $\psi$ is an eigenfunction of the operator?

Thanks for your help.

Answer

I will try to answer from what I have understood so far.

Every Hermitian operator has a set of Linearly independent eigenvectors and hence we can use it construct a basis(provided it spans the space). Lets say say operator is $\hat A$ and their eigenvector set $\{\vert a_i \rangle\}$ with the eigenvalue equation,

$$\hat A \vert a_i \rangle = a_i \vert a_i \rangle\ $$ Now we have an arbitrary state $\vert \psi \rangle\ $and expand in the basis $\{\vert a_i \rangle\}$

$$\vert \psi \rangle\ = \sum_i c_i\vert a_i \rangle$$ $$\langle \psi \vert = \sum_i c_i^*\langle a_i \vert$$ Now normalising $\psi$ would require the condition $$\sum_i |c_i|^2 = 1$$ With that now we can what $c_i$ would mean physically, $|c_i|^2$ is probability of finding $\vert \psi \rangle\ $ in the eigen state $\vert a_i \rangle\ $. Hence the sum of probabilities is one(from the above equation).

When you do a measurement on $\vert \psi \rangle\ $of the observable $\hat A$, i.e.

$$\hat A \vert \psi \rangle\ = \hat A \sum_i c_i\vert a_i \rangle =\sum_i c_i \hat A \vert a_i \rangle = \sum_i c_ia_i\vert a_i \rangle$$ and $$\langle \psi \vert \hat A \vert \psi \rangle = \sum_i |c_i|^2a_i$$ with the interpretation that $|c_i|^2$ as the probability this would become the average value of the observable.

It is important to realise, when you do a single measurement the outcome is such that you will obtain a value $a_i$ with a probability $|c_i|^2$.

Remember your measurement will yield only a single value, this value is obtained by a number of measurements and averaging over them.

Now if the wavefunction is in an eigenstate of the observable, say $\vert \psi \rangle = \vert a_k \rangle$, if you do a measurement of the operator $\hat A$ you will always obtain the value $a_k$.