Perturbation Theory and Zeeman Splitting

[aliens123]

Suppose we have a hamiltonian $$H_0$$ and we know the eigenvectors/values:

$$H_0 |E_i \rangle = E_i|E_i \rangle $$

We then add to it another perturbing Hamiltonian:
$$H’$$ which commutes with $$H_0.$$ According to nondegenerate first order perturbation theory:
$$\langle H \rangle \approx \langle H_0 \rangle + \langle H’ \rangle.$$

Question: with the extra assumption that $$H’$$ commutes with $$H_0$$ is this an exact result?

If the answer is yes, then why is it that we need to consider different types of Zeeman splitting (weak, intermediate, strong)? Taking $$B_{ext}$$ to lie in the z direction gives:
$$H’_Z = \mu_B B_{ext}( L_Z + 2S_Z).$$
This commutes with the original Hamiltonian for the hydrogen atom, so we should be able to solve this new Hamiltonian exactly, and then add in fine-structure as a perturbation. But there would be no need to treat the Zeeman effect as a “perturbation.”

 

[anuttarasammyak]

H′Z=μBBext(LZ+2SZ).HZ′=μBBext(LZ+2SZ).​

H’_Z = \mu_B B_{ext}( L_Z + 2S_Z). This commutes with the original Hamiltonian for the hydrogen atom, so we should be able to solve this new Hamiltonian exactly, and then add in fine-structure as a perturbation. But there would be no need to treat the Zeeman effect as a “perturbation.”

As an operator $\mu$ is
$$i\frac{e\hbar}{2m}\mathbf{r}\times\nabla$$
that does not seem to commute with $H_0$.

 

[DrClaude]

Concerning perturbation theory and commuting Hamiltonians, the answer is that yes, the results obtained using perturbation theory can be exact, with the caveat that not all eigenstates of an operator are necessarily eigenstates of another operator with which it commutes. If eigenvalues are degenerate, then it is possible that one has to take linear combinations of the degenerate eigenstates to form eigenstates for both operators.

Now for the Zeeman effect, the base Hamiltonian should include the spin-orbit coupling, so the field Hamiltonian $H_Z'$ will not commute with the base Hamiltonian. Also, the result is not exact because the perturbation $H_Z'$ is only the first order term in the coupling between the electron and the field.

 

[PeterDonis]

As an operator $\mu$ is

I think you mean the operator $L$; $\mu$ is not an operator, it's a constant.

 

[aliens123]

As an operator $\mu$ is
$$i\frac{e\hbar}{2m}\mathbf{r}\times\nabla$$
that does not seem to commute with $H_0$.

The operator that we have though is (taking our magnetic field to be in the z direction)
$$-\vec{\mu} \cdot \vec{B}_{ext} = \mu_B B_{ext}(L_Z + 2S_Z) $$
This commutes with our original Hamiltonian.

 

[aliens123]

Concerning perturbation theory and commuting Hamiltonians, the answer is that yes, the results obtained using perturbation theory can be exact, with the caveat that not all eigenstates of an operator are necessarily eigenstates of another operator with which it commutes. If eigenvalues are degenerate, then it is possible that one has to take linear combinations of the degenerate eigenstates to form eigenstates for both operators.

Now for the Zeeman effect, the base Hamiltonian should include the spin-orbit coupling, so the field Hamiltonian $H_Z'$ will not commute with the base Hamiltonian. Also, the result is not exact because the perturbation $H_Z'$ is only the first order term in the coupling between the electron and the field.

Why not just call the base Hamiltonian
$$H=H_H + H_Z$$
Where these are “hydrogen” and “zeeman” respectively, solve this exactly, and then add in the fine structure spin-orbit coupling as a perturbation?

 

[DrClaude]

Why not just call the base Hamiltonian
$$H=H_H + H_Z$$
Where these are “hydrogen” and “zeeman” respectively, solve this exactly, and then add in the fine structure spin-orbit coupling as a perturbation?

It is a question of the relative strength of the couplings. The situation you describe is what happens in strong fields, where you get the Paschen-Bach effect.

 

[aliens123]

It is a question of the relative strength of the couplings. The situation you describe is what happens in strong fields, where you get the Paschen-Bach effect.

But my question is: if this can be done exactly why bother worrying about the relative strength of the couplings. Why not solve the part we can exactly, then add in the other field as the perturbation?

 

[anuttarasammyak]

But my question is: if this can be done exactly why bother worrying about the relative strength of the couplings. Why not solve the part we can exactly, then add in the other field as the perturbation?

Let me restate your question to know whether I understand it properly.

Say H1 and H2 are Hamiltonians of the state. {H1,H2}=0, so they share common eigenstates.
It seems there are two ways to handle,
1) Take one of them as main Hamiltonian and let the other one work as perturbation Hamiltonian to the main one.
2) Add them to make one Hamiltonian H=H1+H2 and let it work as usual.
Which way or the both work ?

NB As a spin off this topic,
Say H1 and H2 are Hamiltonians of the state. {H1,H2}$\neq$0. It seems there are two ways to handle,
1) Take one of them as main Hamiltonian and let the other one work as perturbation Hamiltonian to the main one.
2) Add them to make one Hamiltonian H=H1+H2 and try to solve Shrodinger equation exactly.
Is it assured that 1) and 2) coincide?
Example : The sate of electron in hydrogen atom. H1=kinetic energy, H2=potential Coulomb energy.
We know the exact solution by Shrodinger for this case.

 

[PeterDonis]

our original Hamiltonian

What do you think the "original" Hamiltonian is?

 

[PeterDonis]

The original Hamiltonian

I would actually like @aliens123 to answer the question, not someone else.

However, if his answer is the same as yours, the obvious follow-up question is: the Hamiltonian on that wiki page ignores spin-orbit coupling. But as @DrClaude has already pointed out in this thread, you can't ignore spin-orbit coupling for this case.

 

[anuttarasammyak]

Addition to my post #9

Forgetting about spin for simplicity (ref. https://en.wikipedia.org/wiki/Hydrogen_atom), eigenvectors |n,l,m> satisfy
$$H_H|n,l,m>=-\frac{const.}{n^2}|n.l.m>$$ and
$$H_Z|n.l.m>=\mu_B B m|n,l,m>$$
adding side by side
$$(H_H+H_Z)|n,l,m>=( -\frac{const.}{n^2}+\mu_B B m ) |n,l,m>$$
Obviously 2) works. We do not need the way 1).

Similarly I think even with spin considered if $H_H$ and $H_Z$ commute, 2) works and we do not need 1), perturbation. I should appreciate if you show us explicit formula of $H_H$ with spin so that we can see they commute.

 

[aliens123]

I will try to be more explicit.

Ignoring spin orbit coupling, we can write the Schrodinger equation for the Hydrogen atom:
$$\frac{1}{2mr^2} \left[ -\hbar^2 \frac{\partial}{\partial r} \left( r^2 \frac{\partial}{\partial r}\right) + L^2 \right]\Psi + V \Psi = E\Psi$$
I'll rewrite this:
$$H_0 \Psi = E\Psi $$

Now there are two corrections or "perturbations" to consider: the fine structure correction, and the Zeeman Effect. I will call these $$H_F$$ and $$H_Z$$. The fine structure correction includes a relativistic term and a spin-orbit coupling term. The Zeeman effect is due to a (uniform) magnetic field which we will assume is in the z direction.

If the magnetic field is zero, then we can treat $$H_F$$ as a perturbation and get the familiar results for the fine structure correction. I will call this "procedure A."

If the magnetic field is nonzero then the classical approach to the problem is to consider three different limiting cases: the Weak-Field Zeeman Effect, the Strong-Field Zeeman Effect, and the Intermediate-Field Zeeman Effect.

For the Weak-Field Zeeman effect we treat the Zeeman effect as a perturbation on top of "procedure A."
For the Strong-Field Zeeman effect we take as our unperturbed energies: $$E_{n m_l m_s} = -\frac{13.6}{n^2} + \mu_B B_{ext} (m_l + 2m_s)$$
and then add in the fine-structure correction:
$$E_{fs} = \langle n l m_l m_s | (H_Z | n l m_l m_s \rangle $$

My question is: it looks like the steps we take for the Strong-Field Zeeman effect are exact until we get to the part where we add in the fine-structure correction. So why not just do it this way for all three regimes?

 

[DrClaude]

My question is: it looks like the steps we take for the Strong-Field Zeeman effect are exact until we get to the part where we add in the fine-structure correction. So why not just do it this way for all three regimes?

Because the description you would get then should not be useful. The point of all this is not to calculate energies, which we could do much better using other approaches (numerical), but to be able to put instructive labels on levels and have a simple picture to explain the spectra of atoms.

When you add $H_Z$ first, you start labelling levels with $m_l$ in addition to $n$ and $l$. But spin-orbit coupling mixes up those states so much that it doesn't make sense to think of the new levels in terms of the $m_l$ levels that make them up.

You can't turn off spin-orbit coupling. The only spectra you can compare are that of an atom with spin-orbit coupling and no field and the same atom with spin-orbit coupling and a field.

 

[aliens123]

Because the description you would get then should not be useful. The point of all this is not to calculate energies, which we could do much better using other approaches (numerical), but to be able to put instructive labels on levels and have a simple picture to explain the spectra of atoms.

When you add $H_Z$ first, you start labelling levels with $m_l$ in addition to $n$ and $l$. But spin-orbit coupling mixes up those states so much that it doesn't make sense to think of the new levels in terms of the $m_l$ levels that make them up.

You can't turn off spin-orbit coupling. The only spectra you can compare are that of an atom with spin-orbit coupling and no field and the same atom with spin-orbit coupling and a field.

Ah, I see. Thank you!