# 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?

• Would you mind clarifying what you mean by "discerning discrete values for an observable?" and by the adjective "obtainable" as relating to average values? – joshphysics Feb 5 '14 at 4:40
• Sure, by discrete I really just mean finding an exact value for the observable. A solution, I guess. For the obtainable, I also mean a solution. Any solution, as long as it's possible to determine with the integral. – A4Treok Feb 5 '14 at 4:45
• It depends on what you mean by discrete values !! However you are correct about the fact that average value of an observable is always obtainable using the formula : $$<\hat A> = \frac{\langle \psi \vert \hat A \vert \psi \rangle}{\langle \psi \vert \psi \rangle}$$ – user35952 Feb 5 '14 at 4:45
• Well from the above definition, if the wavefunction is eigenfunction of $\hat A$ it is quite obvious to see that value of measurement is the coressponding eigenvalue. i.e. $$\hat A \vert \psi \rangle = a \vert \psi \rangle$$, then $<\hat A> = a$ provided the wavefunction is normalised. – user35952 Feb 5 '14 at 4:50
• Hm, I guess I didn't explain that properly. I mean, if you have an eigenfunction $\psi$, and you had an eigenvalue operator $O$, if you you were to solve for the observable, would you only obtain a definite, fixed value for eigenfunction $\psi$, or is it possible to obtain a definite, fixed value for wave functions without a corresponding eigenvalue $O$? – A4Treok Feb 5 '14 at 4:53

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 $a_k$.

• Hm, my book uses a different notation than yours, but I'm pretty sure I understand it all. – A4Treok Feb 5 '14 at 5:25
• The dirac notation is a handy tool in these things. $\vert a \rangle$ is an abstract vector. The complex conjugate of the vector is defined as $\langle a \vert \equiv \vert a \rangle^\dagger$. – user35952 Feb 5 '14 at 5:28
• What you wrote is true for finite dimensional Hilbert spaces, otherwise it is generally false. The operator has to be self-adjoint and not only Hermitian. Moreover if the spectrum includes a continuous part, there are no proper eigenvectors. All that should be at least mentioned, because important operators (position and momentum) one immediately face studying QM show up these features. – Valter Moretti Feb 5 '14 at 7:26
• @V.Moretti : I am sorry to respond to the comment so late !! A4Treok : Yes, Moretti is right. In the case of infinite dimensional and rigged Hilbert spaces, the procedure is a little different !! This is not true in general. – user35952 Mar 24 '14 at 2:52