Physics Asked by user148792 on January 12, 2021

Suppose there’s a strong external magnetic field applied on a Hydrogen-like atom. The Hamiltonian due to spin-orbit coupling will have much less effect compared to the other Hamiltonians. I would like to use perturbation theory to calculate the energy shifts.

In an article (https://farside.ph.utexas.edu/teaching/qm/Quantum/node72.html), it mentioned:

The eigenstates of $ H_0+H_{B}$ are not eigenstates of $ H_{LS}$ . However, $ H_{LS}$ only couples non-degenerate eigenstates of $ H_0 + H_B$ . Hence, there is no danger of singular terms arising in the perturbation expansion to second order.

Why is this the case? I don’t understand why we don’t have to use degenerate perturbation theory and diagonalise the matrix.

I don't understand why we don't have to use degenerate perturbation theory and diagonalise the matrix.

See Wikipedia's "Approximate Hamiltonians":

"Perturbation theory is an important tool for describing real quantum systems, as it turns out to be very difficult to find exact solutions to the Schrödinger equation for Hamiltonians of even moderate complexity. The Hamiltonians to which we know exact solutions, such as the hydrogen atom, the quantum harmonic oscillator and the particle in a box, are too idealized to adequately describe most systems. Using perturbation theory, we can use the known solutions of these simple Hamiltonians to generate solutions for a range of more complicated systems.

...

The expressions produced by perturbation theory are not exact, but they can lead to accurate results as long as the expansion parameter, say α, is very small. Typically, the results are expressed in terms of finite power series in α that seem to converge to the exact values when summed to higher order. After a certain order n ~ 1/α however, the results become increasingly worse since the series are usually divergent (being asymptotic series). There exist ways to convert them into convergent series, which can be evaluated for large-expansion parameters, most efficiently by the variational method.

...

Continue reading "Strong perturbation theory" below on the same page.

The link you quoted says:

Note, finally, that the eigenstates of $ H_0 + H_{LS}$ are not eigenstates of $ H_B$ . However, $ H_B$ only couples eigenstates with different values of $ j$ . It follows that

such eigenstates are non-degenerate(because the eigenvalues of $ H_{LS}$ are different for states with different values of $ j$ --see Section 7.7). Hence, there is no danger of singular terms arising in the perturbation expansion to second order.Let us apply this theory to the sodium atom.

...

Note, finally, that the eigenstates of $ H_0+H_{B}$ are not eigenstates of $H_{LS}$ . However,

$H_{LS}$ only couples non-degenerate eigenstates of $H_0 + H_B$. Hence, there is no danger of singular terms arising in the perturbation expansion to second order.Let us apply the previous result to a sodium atom. In the presence of an intense magnetic field, the $ 3p$ states are split into five groups with ($ m_l,,m_s$ ) quantum numbers $ (1,1/2)$ , $ (0,1/2)$ , $ (1,-1/2)$ or $ (-1, 1/2)$ , $ (0,-1/2)$ , and $ (-1,-1/2)$ , respectively, in order of decreasing energy. The spin-orbit term increases the energy of the highest energy state (because $ langle r^{,-1},dV/drrangle>0$ ), does not affect the next highest energy state, decreases, but does not split, the energy of the doublet, does not affect the next lowest energy state, and increases the energy of the lowest energy state. (See Exercise 18.) The net result is that the five groups of states are no longer equally spaced in energy.

...

Same explanation, but for the Stark effect: A question about first order perturbation in the Stark effect.

For hydrogen-like atoms see the Dirac equation and beyond the Dirac equation later on that webpage.

Note: Your link calculates $B_{PB}backsimeq25$ Tesla while the hyperphysics.phy-astr.gsu.edu webpage calculates $m_BgB = (5.79 x 10^{-5} eV/T)2B = 0.0021 eV$ and $B = 18$ Tesla for sodium.

The website hyperphysics.phy-astr.gsu.edu also discusses the Paschen-Back effect for sodium, offering this additional information:

"Sodium was used as the basis of the model for convenience, but the fields required to create Paschen-Back conditions for sodium are unrealistically high. Lithium, on the other hand, has a spin-orbit splitting of only 0.00004 eV compared to 0.0021 eV for sodium. Such small energy values are sometimes expressed in "wavenumbers", or 1/l in cm$^{-1}$. In these units the lithium separation is about 0.3 cm$^{-1}$ and the sodium separation is about 17 cm$^{-1}$. The Paschen-Back conditions are met in some lithium spectra observed on the Sun, so this effect does have astronomical significance.".

And this useful graphic:

In the weak field case the vector model at left implies that the coupling of the orbital angular momentum L to the spin angular momentum S is stronger than their coupling to the external field. In this case where spin-orbit coupling is dominant, they can be visualized as combining to form a total angular momentum J which then precesses about the magnetic field direction.

In the strong-field case, S and L couple more strongly to the external magnetic field than to each other, and can be visualized as independently precessing about the external field direction.

Answered by Rob on January 12, 2021

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