Repeatedly measuring the same observable on the same system Suppose we have got some operator $A$ which does not commute with the Hamiltonian, for example $\hat{p}_{x}$ and $\hat{H}$, where $\hat{p}_{x}$ does not commute with the $\hat{H}$ and we make the first measurement and we get $p_{x}^{\prime}$ and immediately after the first measurement the system's momentum is in the eigenstate $|p_{x}^{\prime}>$ according to  a postulate of quantum mechanics. What if I continue to measure the same operator $\hat{p}_{x}$ again and again. Shouldn't the state remain in the same eigenstate if I keep repeating the measurement for an indefinite length of time as long as I keep measuring the same operator ie. $\hat{p}_{x}$. How can this occur because the operator doesn't commute with the Hamiltonian, so we should see the system evolve or change with time in the Schrodinger picture?
 A: Suppose the time between each measurement is $\Delta t$. Your first measurement puts the system in some eigenstate $|a\rangle$. After a time $\Delta t$, the state has time-evolved to the state $|a'\rangle=e^{-i\hat{H}\Delta t/\hbar}|a\rangle$. This has an easy expression in the basis of energy eigenstates $|E_n\rangle$:
$$|a'\rangle=e^{-i\hat{H}\Delta t/\hbar}|a\rangle=\sum_ne^{-iE_n\Delta t/\hbar}\langle E_n|a\rangle|E_n\rangle$$
Suppose you're interested in the probability that you measure your observable to be $b$, which is associated with the eignenstate $|b\rangle$. The probability of measuring your observable to be $b$ is then:
$$|\langle b|a'\rangle|^2=\bigg|\sum_ne^{-iE_n\Delta t/\hbar}\langle E_n|a\rangle\langle b|E_n\rangle\bigg|^2$$
Now suppose that you want to see what the probability is that you'll still measure the observable to be $a$ after a time $\Delta t$. Assigning $b=a$ in the above expression, we see that, in general, the probability is a sum of a bunch of interfering terms with different angular frequencies, which means that you might not measure the same value after a time $\Delta t$.
