Probability of getting a particular energy eigenvalue time dependance

Question

A system has a time independent Hamiltonian with an eigenbasis $|E_i\rangle$. Show that the probability of getting $E_k$ when you measure for the energy is time independent.

So if I let a system $|\psi\rangle=\sum a_i|E_i\rangle$ the probability of getting $E_k$ is:

$$ P=\langle E_ k|\psi\rangle\langle\psi|E_k\rangle=|\langle E_ k|\psi\rangle|^2 $$

and I will use the relations: $$ \frac{d}{dt}|\phi\rangle=-i/\hbar H |\phi\rangle $$ and its dual $$ \frac{d}{dt}\langle\phi|=i/\hbar\langle\phi|H $$

If we differentiate the probability using the product rule:

$$ \frac{dP}{dt}= \langle \frac{d}{dt}E_ k|\psi\rangle\langle\psi|E_k\rangle +\langle E_ k|\frac{d}{dt}\psi\rangle\langle\psi|E_k\rangle +\langle E_ k|\psi\rangle\langle\frac{d}{dt}\psi|E_k\rangle +\langle E_ k|\psi\rangle\langle\psi|\frac{d}{dt}E_k\rangle $$

$$ =i/\hbar( \langle E_ k|H|\psi\rangle\langle\psi|E_k\rangle- \langle E_ k|H|\psi\rangle\langle\psi|E_k\rangle+ \langle E_ k|\psi\rangle\langle\psi|H|E_k\rangle- \langle E_ k|\psi\rangle\langle\psi|H|E_k\rangle ) $$ $$ =0 $$

The second equality uses the relationships given above.

This proof, however, doesn't use the fact that the Hamiltonian is time independent and so I think I must have missed something.

Answer 1

You have used time-independence implicitly at the very beginning when you assume that there are energies "$E_k$" that themselves to not depend on time, i.e. the eigenvalues of the time-independent Hamiltonian. For a time-dependent Hamiltonian you'd have time-dependent eigenvalues $E_k(t)$.

However, I think you're not showing what

Show that the probability of getting $E_k$ when you measure for the energy is time-independent.

intends you to show: In your calculation, both $\lvert E_k\rangle$ and $\lvert \psi\rangle$ evolve in time. But that $\langle \psi \vert \phi \rangle$ is time-independent when we evolve both states in time just follow from time-evolution being unitary ($U^\dagger U =0$, hence $\langle \psi(t)\vert \phi(t)\rangle = \langle \psi(0)\vert U^\dagger(t) U(t)\vert \phi(0)\rangle = \langle \psi(0) \vert \phi(0)\rangle$), it holds for all states, always. So what should worry you is not that you haven't used time-independence, but that you haven't used that the $E_k$ are eigenvalues of the Hamiltonian.

What you actually have to show is that $\lvert \langle E_k \vert \psi(t) \rangle\rvert^2 = \lvert\langle E_k \vert U(t) \vert \psi(0)\rangle\rvert^2$ is time-independent, and for that you will have to use that $\lvert E_k\rangle$ is an eigenstate of the (time-independent) Hamiltonian.