# expected value of eigenstate (time dependance)

Suppose you have a quantum system in which the initial wave function is a pure eigenstate of some observable to be measured, then when you measure at some time to, you will measure the eigenvalue of that eigenstate. Then suppose you set up a new system with the same characteristics as the first one, but now you measure at some other time t1. Will you obtain the same eigenvalue as in the first case or could it change?. It is my understanding that the expected values of eigenstates are their corresponding eigenvalues and therefore if the eigenstate is not stationary the value of the measurement could change, but I am not sure

The time-evolution of a system is determined by the Hamiltonian, and “stationary” refers to eigenstate of the Hamiltonian. Thus it is perfectly possible for the eigenstate $\vert\psi_A\rangle$ of an observable $\hat A$ not to be also an eigenstate of the Hamiltonian. As a result, $\vert\psi_A\rangle$ is not stationary and the expectation value of $\hat A$ will change in time.
The simplest example is possibly the operator $\sigma_x=\left(\begin{array}{cc} 0&1\\ 1&0\end{array}\right)$. The eigenstates are clearly $$\vert \pm \rangle = \frac{1}{\sqrt{2}}\left(\begin{array}{c} 1 \\ \pm 1\end{array}\right)$$ but the eigenstates of $H=\hbar\omega \sigma_z=\hbar\omega \left(\begin{array}{cc}1&0\\ 0 &-1\end{array}\right)$ are $\left(\begin{array}{c} 1 \\0\end{array}\right)$ and $\left(\begin{array}{c}0 \\ 1\end{array}\right)$ so a system prepared in one of the eigenstates of $\sigma_x$ evolves as $$\vert \pm (t)\rangle =\frac{1}{\sqrt{2}}\left(\begin{array}{c} e^{i\omega t} \\ \pm e^{-i\omega t}\end{array}\right)\, . \tag{1}$$ For some arbitrary $t$, the states of (1) are no longer eigenstates of $\sigma_x$.