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

Happiness
Messages
686
Reaction score
30
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
 
Physics news on Phys.org
Happiness said:
Below is the derivation
Where is this from? Please give a reference.
 
PeterDonis said:
Where is this from? Please give a reference.
Introduction to Quantum Mechanics, 2nd edition, by David J. Griffiths
 
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.
 
  • Like
Likes Happiness and pines-demon
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?
 
Last edited:
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.
 
  • Like
Likes pines-demon
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Is it possible, and fruitful, to use certain conceptual and technical tools from effective field theory (coarse-graining/integrating-out, power-counting, matching, RG) to think about the relationship between the fundamental (quantum) and the emergent (classical), both to account for the quasi-autonomy of the classical level and to quantify residual quantum corrections? By “emergent,” I mean the following: after integrating out fast/irrelevant quantum degrees of freedom (high-energy modes...
Back
Top