How Does Spin-Orbit Coupling Affect Quantum Number Validity?

[kelly0303]

Hello! Assume I have a system containing intrinsic spin and orbital angular momentum and before coupling the two, $|J,J_z,S,S_z,L,L_z>$ is a good basis (i.e. all these quantum numbers are good), with $J=L+S$ and $J_z = S_z + L_z$. If I add a term of the form $S\cdot L$, $L_z$ and $S_z$ won't be good quantum numbers, as $S\cdot L$ doesn't commute with them. However I also have: $S\cdot L |J,J_z,S,S_z,L,L_z>$ $= \frac{(S+L)^2-S^2-L^2}{2} |J,J_z,S,S_z,L,L_z> = \frac{J^2-S^2-L^2}{2}|J,J_z,S,S_z,L,L_z> = \frac{J(J+1)-S(S+1)-L(L+1)}{2}|J,J_z,S,S_z,L,L_z>$. So it looks like $|J,J_z,S,S_z,L,L_z>$ is still an eigenstate of $S\cdot L$, which would mean that original Hamiltonian remains diagonal even after adding the $S\cdot L$ term. What am I missing?

 

[PeterDonis]

What am I missing?

What does the eigenvalue $[ J(J+1) - S(S+1) - L(L+1) ]/ 2$ work out to?

 

[Twigg]

Notice that $L_z$ and $S_z$ are absent from your eigenvalue, it's only $S^2$ and $L^2$ that appear. As you said, $L \cdot S$ doesn't commute with $L_z$ or $S_z$. Does $L \cdot S$ commute with $S^2$ and/or $L^2$?

 

[kelly0303]

Notice that $L_z$ and $S_z$ are absent from your eigenvalue, it's only $S^2$ and $L^2$ that appear. As you said, $L \cdot S$ doesn't commute with $L_z$ or $S_z$. Does $L \cdot S$ commute with $S^2$ and/or $L^2$?

But given that these old eigenstates are not good anymore, shouldn't the new hamiltonian, containing $L \cdot S$ have off-diagonal terms in the original basis? So I would expect $<1|L\cdot S|2>\neq 0$. But based on the calculations above $<1|L\cdot S|2> = <1|\frac{J(J+1)-S(S+1)-L(L+1)}{2}|2> = \frac{J(J+1)-S(S+1)-L(L+1)}{2}<1||2> =0$

 

[kelly0303]

What does the eigenvalue $[ J(J+1) - S(S+1) - L(L+1) ]/ 2$ work out to?

Doesn't that depends on the actual values of the state? For S=L=1/2 and J = 0 I would get -3/2, for S=L=1/2 and J=1 I would get 1/2.

 

[Twigg]

In the absence of any magnetic field to lift the degeneracy, $|J,L,L_z,S,S_z\rangle$ may look the same as $|J,J_z,L,S\rangle$. It's when you apply a magnetic field that you'll be able to see that $L_z$ and $S_z$ aren't good quantum numbers. Otherwise, as your calculation nicely shows, the energies only depend on $J^2$, $L^2$, and $S^2$. Does that make sense or is it more confusing now?

Edit: the above is assuming that the only quantity you're concerned with are the energies. You could tell the difference between $|J,L,L_z,S,S_z\rangle$ and $|J,J_z,L,S\rangle$ in the absence of magnetic fields by projecting the eigenstates onto the basis states $|J,L,L_z,S,S_z\rangle$. If $|J,L,L_z,S,S_z\rangle$ were good states, you'd expect to see 1, but because they're not good states you'll instead measure the Clebsch-Gordon coefficients $\langle J,J_z,L,S| J,L,L_z,S,S_z\rangle$.

 

[kelly0303]

In the absence of any magnetic field to lift the degeneracy, $|J,L,L_z,S,S_z\rangle$ may look the same as $|J,J_z,L,S\rangle$. It's when you apply a magnetic field that you'll be able to see that $L_z$ and $S_z$ aren't good quantum numbers. Otherwise, as your calculation nicely shows, the energies only depend on $J^2$, $L^2$, and $S^2$. Does that make sense or is it more confusing now?

I see your point about using a magnetic field to see the change. But for some reason (and it might be my lack of a proper understanding), I thought that if you start with a diagonal hamiltonian and add a term the messes up some quantum numbers, than the eigenstates that depend on these quantum numbers will also be affected explicitly i.e. the diagonal will have some off diagonal terms in the matrix form, even without using some external field. For example in the hydrogen atom, I thought that once you add the spin-orbit coupling the hamiltonian is not diagonal anymore and this is why we use perturbation theory to calculate the effect of the spin orbit coupling (at least this is how it is done in Griffiths). If the Hamiltonian is still diagonal why do we need to use perturbation theory, when we can take the change in energy levels directly from reading the diagonal of the new hamiltonian.

 

[PeterDonis]

Doesn't that depends on the actual values of the state?

Yes, it does, sorry, I realized after re-reading my question that that can't be where the problem is.

Here's a better suggestion: you are saying that $S \cdot L = \left( S + L \right)^2 - S^2 - L^2$. (With a factor of $1/2$, but that doesn't matter for the point I am making in this post.) But if this were correct, then since $\left( S + L \right)^2 = J^2$, and $J^2$, $L^2$, and $S^2$ all commute with $L_z$ and $S_z$, $S \cdot L$ would also have to commute with $L_z$ and $S_z$. But, as you note in your OP, it doesn't. So there has to be a problem with the equality $S \cdot L = \left( S + L \right)^2 - S^2 - L^2$.

I think the problem is that this equality assumes that $S$ and $L$ commute. (Can you see why? Hint: expand out $\left( S + L \right)^2$ explicitly.) But they don't. $S^2$ and $L^2$ commute, but that does not mean that $S$ and $L$ commute.

 

[Twigg]

Your understanding is correct. It's just that the degeneracy of the m-levels hides the fact that $L_z$ and $S_z$ aren't good quantum numbers anymore. Because sublevels with identical L,S, and J have the same energy, any combination of those sublevels has the same energy. $|J,L,L_z,S,S_z\rangle$ is just another combination of the $|J,J_z,L,S\rangle$'s.

As far as why Griffiths does this, can you share what section of Griffiths this is in so I can follow along? Been a while and I don't remember what's where.

 

[Twigg]

Found it! It's section 6.3.2. It's really subtle and honestly I think the choice to put this in the perturbation theory chapter is kind of arbitrary. The reason is that the spin-orbit Hamiltonian for hydrogen goes as $\frac{1}{r^3} \vec{L} \cdot \vec{S}$. It's the $\frac{1}{r^3}$ term that really has to be treated as perturbation. The angular part of this Hamiltonian has an exact solution, as you proved. But the $\langle\frac{1}{r^3} \rangle$ is evaluated with the unperturbed solutions of the hydrogen radial equation (eqn 6.64 in my second edition copy).

 

[PeterDonis]

I think the problem is that this equality assumes that $S$ and $L$ commute. (Can you see why? Hint: expand out $\left( S + L \right)^2$ explicitly.) But they don't. $S^2$ and $L^2$ commute, but that does not mean that $S$ and $L$ commute.

I just looked this up in Ballentine and this "solution" won't work either; $S$ and $L$ do commute.

 

[kelly0303]

Found it! It's section 6.3.2. It's really subtle and honestly I think the choice to put this in the perturbation theory chapter is kind of arbitrary. The reason is that the spin-orbit Hamiltonian for hydrogen goes as $\frac{1}{r^3} \vec{L} \cdot \vec{S}$. It's the $\frac{1}{r^3}$ term that really has to be treated as perturbation. The angular part of this Hamiltonian has an exact solution, as you proved. But the $\langle\frac{1}{r^3} \rangle$ is evaluated with the unperturbed solutions of the hydrogen radial equation (eqn 6.64 in my second edition copy).

Thanks a lot! So in general, if I add a term to my Hamiltonian that would make some quantum numbers not good anymore, even if the Hamiltonian is still diagonal, I should change the wavefunctions such that the eigenstates reflect the new quantum numbers, so that I don't have problems along the way (e.g. if I turn on some external field).

 

[Twigg]

Yep! That's a good procedure. However, it is true that you can lift the degeneracy in ways that make $J_z$ not a good quantum number. Maybe some maniac decides to put your atom in a 5 tesla B-field, and then the Zeeman interaction ends up overpowering the spin-orbit interaction and $L_z$ and $S_z$ wind up being good quantum numbers again (Griffith's Section 6.4.2)! Or maybe some smart-alec turns on an E-field, and that lifts the degeneracy such that $J_z$, $L_z$, or $S_z$ are all bad (Griffith's Problem 6.36 and others). That's kind of the key takeaway from degenerate perturbation theory. The good quantum numbers aren't set in stone until you lift the degeneracy.

 

[vanhees71]

Yes, it does, sorry, I realized after re-reading my question that that can't be where the problem is.

Here's a better suggestion: you are saying that $S \cdot L = \left( S + L \right)^2 - S^2 - L^2$. (With a factor of $1/2$, but that doesn't matter for the point I am making in this post.) But if this were correct, then since $\left( S + L \right)^2 = J^2$, and $J^2$, $L^2$, and $S^2$ all commute with $L_z$ and $S_z$, $S \cdot L$ would also have to commute with $L_z$ and $S_z$. But, as you note in your OP, it doesn't. So there has to be a problem with the equality $S \cdot L = \left( S + L \right)^2 - S^2 - L^2$.

I think the problem is that this equality assumes that $S$ and $L$ commute. (Can you see why? Hint: expand out $\left( S + L \right)^2$ explicitly.) But they don't. $S^2$ and $L^2$ commute, but that does not mean that $S$ and $L$ commute.

The spin-orbit term is $\propto \hat{\vec{S}} \cdot \hat{\vec{L}}$, and that clearly doesn't commute with $\hat{L}_z$ and $\hat{S}_z$ but with $\hat{J}_z$! That's leading to the fine structure of atomic spectral lines:

https://en.wikipedia.org/wiki/Fine_structure#Spin–orbit_coupling

Including the spin-orbit term in the Hamiltonian means that only $\hat{\vec{J}}^2$, $\hat{J}_z$, $\hat{\vec{L}}^2$ and $\hat{\vec{S}}^2$ commute with the Hamiltonian, i.e., you can build a system of common (orthonormal) eigenstates for these observables. That's why you call the corresponding spin quantum numbers, $j$, $j_z$, $\ell$, and $s$ "good".

Neglecting the spin-orbit term leads to additional "good quantum numbers" $m_z$ and $s_z$, which implies that your energy eigenvalues have higher degeneracies than with the spin-orbit term, i.e., including the spin-orbit term (partially) lifts this degeneracy (particularly for the hydrogen atom, which has a large O(4) symmetry, and the energy eigenvalues only depend on the principal quantum number $n \in \{1,2,\ldots\}$, such that each energy eigenvalue is degenerate to degree $2n^2$. The spin-orbit term is part of the fine structure, i.e., the corresponding splitting of spectral lines lifting the degneracy. Other effects are other relativistic corrections, like the corretion to the kinetic energy of order $p^4$.

 

[PeterDonis]

Including the spin-orbit term in the Hamiltonian means that only $\hat{\vec{J}}^2$, $\hat{J}_z$, $\hat{\vec{L}}^2$ and $\hat{\vec{S}}^2$ commute with the Hamiltonian

Actually, since, as the OP shows, $S \cdot L$ is a linear combination of $J^2$, $L^2$, and $S^2$, $S \cdot L$ also commutes with the Hamiltonian, so you can use eigenvalues of $S \cdot L$ as quantum numbers instead of eigenvalues of one of the other three operators.

that original Hamiltonian remains diagonal even after adding the $S\cdot L$ term.

In view of what is stated just above, this is because there is more than one basis in which the Hamiltonian is diagonal, whether the $S \cdot L$ term is included or not. You have four operators, $J^2$, $L^2$, $S^2$, and $S \cdot L$, which are not linearly independent; anyone can be expressed as a linear combination of the other three. So you have four possible choices of basis, corresponding to which one of the four operators you leave out (the other three being your chosen basis and providing the eigenvalues that become your quantum numbers to label states).