Time dependent quantum state probability calculation

[kdlsw]

For part a I have (H₀-ω[itex]\hbar[/itex]m)|nlm>, which I think the (H₀-ω[itex]\hbar[/itex]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/[itex]\hbar[/itex]}. Do I simply add the e^{-iE_nt/[itex]\hbar[/itex]} term after each ψ state? And then what? Please help me with this, thanks

 

[DrClaude]

For part a I have (H₀-ω[itex]\hbar[/itex]m)|nlm>, which I think the (H₀-ω[itex]\hbar[/itex]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/[itex]\hbar[/itex]}. Do I simply add the e^{-iE_nt/[itex]\hbar[/itex]} 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.