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.

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    $\begingroup$ Would you mind clarifying what you mean by "discerning discrete values for an observable?" and by the adjective "obtainable" as relating to average values? $\endgroup$ – joshphysics Feb 5 '14 at 4:40
  • $\begingroup$ 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. $\endgroup$ – A4Treok Feb 5 '14 at 4:45
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    $\begingroup$ 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 : $$ <\hat A> = \frac{\langle \psi \vert \hat A \vert \psi \rangle}{\langle \psi \vert \psi \rangle} $$ $\endgroup$ – user35952 Feb 5 '14 at 4:45
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    $\begingroup$ 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. $\endgroup$ – user35952 Feb 5 '14 at 4:50
  • $\begingroup$ 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 $O$? $\endgroup$ – 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 $a_k$.

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  • $\begingroup$ Hm, my book uses a different notation than yours, but I'm pretty sure I understand it all. $\endgroup$ – A4Treok Feb 5 '14 at 5:25
  • $\begingroup$ 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 $. $\endgroup$ – user35952 Feb 5 '14 at 5:28
  • $\begingroup$ 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. $\endgroup$ – Valter Moretti Feb 5 '14 at 7:26
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    $\begingroup$ @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. $\endgroup$ – user35952 Mar 24 '14 at 2:52

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