Valid to use <1/r3> to get spin-orbit correction to H? (perturbation)

  • #1
Happiness
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TL;DR Summary
<1/r^3> uses the standard wavefunctions ψ_nlm of hydrogen, which are not good states to use in perturbation theory because the Hamiltonian (under spin-orbit interaction) no longer commutes with L. So shouldn't we solve for the simultaneous eigenstates of L^2, S^2, J^2 and J_z first? And then use those to find <1/r^3>?
Below is the derivation of E1so, the first-order correction to the Hamiltonian due to spin-orbit coupling of the election in hydrogen atom. My question is whether it's valid to use [6.64] (see below). ##<\frac{1}{r^3}>## I believe is ##<\psi_{nlm}|\frac{1}{r^3}|\psi_{nlm}>##, but ##\psi_{nlm}## is NOT a good state to use in perturbation theory, because ##\psi_{nlm}## is an eigenstate of ##L_{z}## but H'so does not commute with ##L## (as mentioned in the paragraph above [6.62]-[6.63]).

Screenshot 2024-06-24 at 1.46.49 AM.png

Screenshot 2024-06-24 at 1.46.24 AM.png

For elaboration, the phrase "good state" relates to the following theorem:

Screenshot 2024-06-24 at 1.53.20 AM.png

Screenshot 2024-06-24 at 1.53.34 AM.png

Ordinary first-order perturbation theory means using [6.9] below, with ##\psi^{0}_{n}## replaced with a good state.

Screenshot 2024-06-24 at 1.57.15 AM.png
 
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  • #2
Happiness said:
Below is the derivation
Where is this from? Please give a reference.
 
  • #3
PeterDonis said:
Where is this from? Please give a reference.
Introduction to Quantum Mechanics, 2nd edition, by David J. Griffiths
 
  • #4
Indeed, the ##\psi_{nlm}## are not "good" functions, because of the ##m## index, and the "good" functions will be linear combinations of ##\psi_{nlm}## with different ##m##. However, the value of ##\braket{1/r^3}## is independent of ##m##, therefore you you know that the linear combination will have the given value, whichever ##m## states are combined.
 
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  • #5
DrClaude said:
Indeed, the ##\psi_{nlm}## are not "good" functions, because of the ##m## index, and the "good" functions will be linear combinations of ##\psi_{nlm}## with different ##m##. However, the value of ##\braket{1/r^3}## is independent of ##m##, therefore you you know that the linear combination will have the given value, whichever ##m## states are combined.

But how do you know that the "cross terms" are zero?

Suppose a good state ##\psi^{0}=\alpha\psi_{a}+\beta\psi_{b}## , where ##\psi_{a}## and ##\psi_{b}## are some ##\psi_{nlm}## .

##\braket{\frac{1}{r^3}}=\braket{\alpha\psi_{a}+\beta\psi_{b}|\frac{1}{r^3}|\alpha\psi_{a}+\beta\psi_{b}}##

##=\alpha^2\braket{\psi_{a}|\frac{1}{r^3}|\psi_{a}}+\beta^2\braket{\psi_{b}|\frac{1}{r^3}|\psi_{b}}+\alpha^*\beta\braket{\psi_{a}|\frac{1}{r^3}|\psi_{b}}+\alpha\beta^*\braket{\psi_{b}|\frac{1}{r^3}|\psi_{a}}##

How do you know the cross terms ##\braket{\psi_{a}|\frac{1}{r^3}|\psi_{b}}## and ##\braket{\psi_{b}|\frac{1}{r^3}|\psi_{a}}## are zero?
 
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  • #6
Happiness said:
How do you know the cross terms ##\braket{\psi_{a}|\frac{1}{r^3}|\psi_{b}}## and ##\braket{\psi_{b}|\frac{1}{r^3}|\psi_{a}}## are zero?
Ok, I've tried calculating them. They are indeed zero.
 
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