Time dependent quantum state probability calculation

[kdlsw]

For part a I have (H₀-ω$\hbar$m)|nlm>, which I think the (H₀-ω$\hbar$m) part is the eigenvalue of the Hamiltonian, also is the energies?

And mainly, I am not sure how to approach part b, the time variable is not in any of the states. I saw this in our lecture notes: ψ(r,t)=∑C_nψ_n(r) e^{-iE_nt/$\hbar$}. Do I simply add the e^{-iE_nt/$\hbar$} term after each ψ state? And then what? Please help me with this, thanks

 

[DrClaude]

For part a I have (H₀-ω$\hbar$m)|nlm>, which I think the (H₀-ω$\hbar$m) part is the eigenvalue of the Hamiltonian, also is the energies?

$H_0$ is the Hamiltonian for the hydrogen atom in the absence of the external magnetic field. You should be able to find its eigenvalues.

And mainly, I am not sure how to approach part b, the time variable is not in any of the states. I saw this in our lecture notes: ψ(r,t)=∑C_nψ_n(r) e^{-iE_nt/$\hbar$}. Do I simply add the e^{-iE_nt/$\hbar$} term after each ψ state?

Basically, yes. The time evolution of a state $\psi$ (with a time-independent Hamiltonian $H$) is given by
$$
\psi(t) = e^{-i H t / \hbar} \psi(0)
$$
For $\phi_n$ and eigenstate of $H$ with eigenvalue $E_n$,
$$
e^{-i H t / \hbar} \phi_n = e^{-i E_n t / \hbar} \phi_n
$$
Therefore, if $\psi$ is a superposition of eigenstates $\phi_n$,
$$
\psi(0) = \sum_n c_n \phi_n
$$
then
$$
\psi(t) = \sum_n c_n e^{-i E_n t / \hbar} \phi_n
$$

And then what?

When you've done a correctly, you'll have $E_{nlm}$ for each state $| n l m \rangle$, and therefore can get $|\Psi(t)\rangle$ for any time $t$. You will then have to calculate projections like $\langle \psi_1 | \Psi(t) \rangle$, from which you can calculate the probabilities.