# Probabilities in non-stationary states

I'm confusing myself. Let's represent some state in the eigenbasis for Hydrogen:

$$|\psi\rangle = \sum_{n,l,m}|n,l,m\rangle\langle n,l,m|\psi\rangle.$$

Now denote the initial state by $$\psi(t=0)\equiv\psi_o$$, and hit this thing with time evolution:

$$U|\psi\rangle = \sum_{n,l,m}e^{-iE_nt/\hbar}|n,l,m\rangle\langle n,l,m|\psi_o\rangle.$$

I'm wanting to know what the probability is that I measure some specific $$(l^*,m^*)$$ at some later time $$t$$. Looking at this, we have

$$P(t,l=l^*,m=m^*)=\sum_n|\langle n,l^*,m^*|U|\psi\rangle|^2 \\ = \sum_n|\langle n,l^*,m^*|\psi_o\rangle|^2.$$

This has no time dependence, and I feel I'm missing something obvious. For example, say we prepare the state to initially be $$|\psi\rangle = a|1,0,0\rangle+b|2,1,1\rangle+c|3,1,1\rangle$$, where all constants are real. This would imply from the above, after normalization, that

$$P(l=1,m=1) = (b^2+c^2)/(a^2+b^2+c^2),$$

independent of time. What am I missing here? Obviously the probability density function has cross terms, so I do not see why this should physically be the case, thus sparking my question.

==================================================================== Closure:

As pointed out by user 'The Vee', my confusion stemmed from this observable being an integral of the eigenbasis representation. I had internally generalized the time dependence of observable expectations, when this is not the case if that observable is also being used as a quantum number in the state representation. The general time evolution of some observable $$\Omega$$ in this basis would be

$$\langle\Omega (t)\rangle = \langle \psi|U^{\dagger} \Omega U|\psi\rangle \\ = \sum_{n',l',m'}\sum_{n,l,m}e^{i(E_n'-E_n)t/\hbar}\langle n',l',m'|\Omega|n,l,m\rangle\langle n',l',m'|\psi_o\rangle^*\langle n,l,m|\psi_o\rangle.$$

If $$\Omega = L^2$$ or $$L_z$$, then orthogonality reduces this to

$$\langle L^2\rangle = \sum_{n,l,m}\hbar^2 l(l+1)|\langle n,l,m|\psi_o\rangle|^2 \\ \langle L_z\rangle = \sum_{n,l,m}\hbar m|\langle n,l,m|\psi_o\rangle|^2$$

No time dependence of the expectations, hence no time dependence of observation probability; all is well. If $$[H,\Omega]\neq 0$$, then all of those cross terms do not drop out, and we see the oscillation in the exponential depending on the energy difference of states. I've kept it in this basis to provide consistency with the above question, but we can see how this generalizes to whatever CSCO we use, as user 'gented' does in his answer by using a collective notation $$|a\rangle$$.

This is in general true whenever you calculate the projection onto an eigenstate (and not a combination thereof). Let $$\left\{|a\rangle\right\}_{a\in A}$$ be a set of eigentstates for the Hamiltionian $$\hat{H}$$, a state at time $$t$$ can be written as $$|\psi(t)\rangle = \sum_{a}\hat{U}(t)|a\rangle\langle a |\psi_0\rangle.$$ Its projection onto an eigenstate $$|a'\rangle$$ is $$\langle a'| \psi(t)\rangle = \langle a'| \Big(\sum_{a}\hat{U}(t)|a\rangle\langle a |\psi_0\rangle\Big)=\hat{U}(t)_{a' a'}\langle a'|\psi_0\rangle$$ whose norm does not depend on time as long as $$\hat{U}(t)$$ only picks up a phase factor when acting onto eigenstates. This is because once a state collapses into an eigenstate, it remains there indefinitely.

• Accepted because it nicely shows the generality of this. It should be noted that you are explicitly assuming $|a\rangle$ to be collective notation for an eigenstate of the hamiltonian. Thanks for the answer! – dsm Mar 18 '19 at 20:04
• Also, you dropped the $a$ from your bra in the summation. I tried to edit, but it won't let me. Perhaps you can modify it. It is clear from the top equation, but it could be confusing. – dsm Mar 18 '19 at 20:37
• Yes, I noticed and fixed it :) – gented Mar 18 '19 at 21:41

You just happened to consider an observable (or rather, a pair of observables) that is, in fact, an integral (integrals) of motion of the system. In other words, the probability of measuring any value of $$(l,m)$$ is in fact not expected to change during time evolution.

This is not true for other observables in general, but it does hold for any time-independent $$A$$ which commutes with the Hamiltonian. Since both $$L^2$$ and $$L_z$$ have this property, both $$l$$ and $$m$$ are integrals of motion and your result follows. (They also commute with each other which enables you to use both the measured values simultaneously.)

For a counterexample, you may consider the probability of measuring something that is not an eigenstate of the Hamiltonian, like $$|\varphi\rangle := (|1,0,0\rangle + |2,0,0\rangle)/\sqrt2$$. I won't try to come up with an observable of which this is an eigenvector – that would only obscure the idea and at the end of the day you only need the eigenvector anyway. If you want, examples of common observables that don't commute with the hydrogen Hamiltonian are any component of position or of momentum, but there the direct calculation is complicated by the fact that these do not have eigenvalues.

• Hi there, thanks for the answer, I appreciate it. That was exactly my confusion. I added a 'closure' to the bottom mathematically showing this. – dsm Mar 18 '19 at 20:33
• This is actually a good description of why at the end of the day integral of motion must commute with the Hamiltonian (because otherwise not being diagonal on the eigenstates, the probabilities may not be independent of time). – gented Mar 18 '19 at 21:57