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 $a_k$.
No comments:
Post a Comment