Is the expectation value always an eigenvalue?

Question

Must the expectation value of an observable always be equal to an eigenvalue of the corresponding operator? I already know that 0 is not an eigenvalue, but are there any other examples?

Answer 1

I would actually expect this to be rare, and only generically true when the state of the system corresponds to an eigenstate. This simply because, for a state $\psi = \sum a_{n}\lvert\phi_{n}\rangle$ with eigenvalues $V_{n}$, you would have $\langle V\rangle = \sum V_{n}\lvert a_{n}\rvert^{2}$, which is not constrained to be equal to one of the $V_{n}$. It's easy to check this for a two state system with the two values of $V_{n}$ different.

Answer 2

A specific quantum mechanical example to show the contrary is spin-$\frac{1}{2}$ systems. If you are in an eigenstate of the $S_{z}$ operator, the expectation value of $S_{x}$ is $0$, but it has eigenvalues $\frac{1}{2} \hbar$ and $-\frac{1}{2} \hbar$.

Answer 3

No. Expectation valve is the mean value of the observable for a given state. The state your system assumes is a superposition (linear combination) of the eigenstates. Say, for example: you have a infinite potential well of width a, the then the energy/momentum eigenstates are: $\Phi_n=\sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a}).e^{-iE_n t/\hbar}$. You can create any other state of the system(at $t=0$) using the linear combination of these eigenfunctions (Hint: Fourier half sine series). Say, your system is in state $\Psi$, then:

\begin{equation} \Psi=\sum_{n=1}^{m}c_n \Phi_n \quad ...(a) \end{equation} $\quad \quad$ (read about Fourier's trick to find $c_n$)

$\quad$ where $c_n$ is some complex number such that $|c_n|^2$ gives the probability of the system being in the state $\Phi_n$ when the measurement is made. When you make a measurement on $\Psi$ (say, energy for example), You will get $one$ of the values $E_n$(energy eigenvalues) corresponding to the eigenstate $\Phi_n$ in $(a)$, and the probability of getting $E_n$ is $|c_n|^2$. Clearly all $|c_n|^2$s in $(a)$ should add up to give $1$. Clearly the average value of $E$(the expectation value): \begin{equation}\langle E \rangle=\sum_{n=1}^{m}|c_n|^2 E_n \quad ...(b) \end{equation} The inherent randomness of Quantum Mechanics is that you can know the values$(E_1,E_2,E_3...E_m)$ the system will give when the measurement is made, you can know the probability of getting $E_n$ when the measurement is made, but you cannot say with absolute certainty that you will get a particular $E_n$ before making the measurement.

Finally to answer your question, looking at $(b)$ you can tell that the expectation value is not always equal to the eigenvalue(it may be, perhaps by pure chance the summation can still add up to give one of the $E_n$), But if $\Psi=\Phi_n$ itself(compare this to $(b)$ and see how $|c|^2 =1$) then no matter how many measurements you make you'll always get $E_n$, so $\langle E \rangle=E_n$.

Answer 4

For the eigenstate the expectation value was the eigenvalue.

For an operator with a continuous spectrum, the range of the expectation value could align with the range of the eigenvalue.

However, there's also case where the expectation value not contained in the range of the eigenvalue.

Consider an operator $H$ with the integer eigenvalue $i$ of $|i\rangle$. Setting $a=\frac{1}{\pi}$ and $r=(1-\frac{1}{\pi})$, where $|\psi\rangle =\sum_{i=0}^\infty \sqrt{a r^i} |i\rangle$. The states was normalized, but the expectation value $\langle \psi|H|\psi\rangle=\frac{1}{1-r}=\pi$, which was an irrational number.

This was an supplemental answer to pat's and Jerry Schirmer's.