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I am reviewing an intro quantum chemistry book and I came across the line "If $\hat{A}$ corresponds to a physical property of the system, whether or not $\psi$ is an eigenfunction of $\hat{A}$, the average value of that property is given by the expectation value of the operator." Doesn't $\psi$ have to be an eigenfunction of the operator in order to obtain an expectation value?

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    $\begingroup$ No. The wave function will in general be a linear combination of some or all of the eigenfunctions. It does not have to be a specific eigenfunction. If it is a linear combination of eigenfunctions, than the expectation value will be a linear combination of the expectation values of the eigenfunctions. $\endgroup$ – Jon Custer Mar 9 '16 at 23:07
  • $\begingroup$ If the wavefunction is a combination of only some of the eigenfunctions, than the rest of the functions that make up the wavefunction are not eigenfunctions? What are these other functions? $\endgroup$ – Stuff Mar 9 '16 at 23:22
  • $\begingroup$ No, you have it backwards. The eigenfunctions are the orthogonal basis for the solution space. Any wave function in that solution space will be some linear combination of the eigenfunctions. That is why they are called 'eigenfunctions'. $\endgroup$ – Jon Custer Mar 9 '16 at 23:26
  • $\begingroup$ Okay so if I understand correctly the wavefunction is a linear combination of basis functions (which are eigenfunctions) and this linear combination can be in a sense "incomplete" only consisting of a fraction of the basis functions, thus leading the wavefunction to not be an eigenfunction of an operator? $\endgroup$ – Stuff Mar 9 '16 at 23:38
  • $\begingroup$ Well, life gets really complicated, because eigenfunctions are not unique. Given one set of functions that are a basis for a space, you can generate an infinite number of different, orthogonal basis sets through appropriate combinations of the first set. But, as an example, in chemistry one will often work with the eigenfunctions for atomic electron orbitals. Taken all together, they are a complete basis set. You can express any electron state around the hydrogen atom with some linear combination of them. But, that does not imply that some state has to have a contribution from the 5s orbital. $\endgroup$ – Jon Custer Mar 9 '16 at 23:48
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No, assuming that wavefunction is normalized, the expectation value $o$ of the operaror $\mathcal O $ is obtained as $\langle \Psi | \mathcal O | \Psi \rangle $.

EDIT

In response at your comment: I am not asking how to calculate the expectation value, I am asking why the wavefunction does not have to be an eigenfunction of the operator in order to calculate an expectation value

The fact that the expectation value is obtained with $\langle \Psi | \mathcal O | \Psi \rangle $ implies that there is no need that the wavefunction be an eigenfunction of the hermitian operator $\mathcal O$.

If it is an eigenfunction, then it is possible to get a simple expresion with:

$$ \langle \Psi | \mathcal O | \Psi \rangle = \langle \Psi | \mathcal O \Psi \rangle = \langle \Psi | \mathcal o \Psi \rangle = o \langle \Psi | \mathcal \Psi \rangle = o$$

If not the case, then you can not make the substiution $\mathcal O \Psi = o \Psi$. So, the application of the map $\mathcal O$ to $\Psi$ will be some function $\Omega$. Then, the expectation value will be $\langle \Psi | \mathcal \Omega \rangle$ that is an scalar that you can compute as $$ \int d\mu \Psi^*\Omega$$

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  • $\begingroup$ I am not asking how to calculate the expectation value, I am asking why the wavefunction does not have to be an eigenfunction of the operator in order to calculate an expectation value $\endgroup$ – Stuff Mar 10 '16 at 0:34
  • $\begingroup$ @Stuff Your welcome $\endgroup$ – user1420303 Mar 10 '16 at 1:34

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