# Spin orbit and double group representations

1. Feb 2, 2014

### vabite

I am reading a text about the splitting of the energy levels in crystals caused by the spin orbit interaction. In particular, the argument is treated from the point of view of the group theory.

The text starts saying that a representation (TxD) for the double group can be obtained from the direct product of an irreducible representation T of the single group and an irreducible representation of the full rotational group D. Then, it concludes that:
_If the spin orbit term is neglected in the Hamiltonian, then (TxD) is irreducible, and every eigenvalue that was p-degenerate without considering the spin becomes 2p-degenerate.
_If the spin orbit term is included in the Hamiltonian, then (TxD) can be in general reducible, and the number and dimensions of the irreducible representations contained in this reducible representation determines the splitting of the energy levels due to the spin-orbit interaction.

Now the question is: given that the double group of the Hamiltonian operator and its representation (TxD) are the same with and without considering the spin orbit term, how can it be that (TxD) is irreducible neglecting the spin orbit term, and can become reducible by considering it?

2. Feb 2, 2014

### OhYoungLions

Can you give details on what book you've found this?

3. Feb 3, 2014

### OhYoungLions

The reason why I want to see the text is because it sounds completely messed up!

Usually when you discuss double groups in the context of spin-orbit coupling, the operators in your group are whatever is in the single group (T), plus time reversal (R) - not the full rotation group (D). The Hamiltonian associated with spin-orbit coupling (say $\mathbf{S}\cdot\mathbf{L}$) commutes with time-reversal, but not generally with an arbitrary rotation, so spin-orbit coupling will usually lower the symmetry of the total Hamiltonian.

In other words, with spin-orbit coupling, the Hamiltonian will not transform as any representation of (TxD) at all, let along a reducible or irreducible one. It will only transform as a representation of (TxR), which in this context can be considered a subgroup of (TxD).

In addition, the statement that every eigenvalue that is p-degenerate becomes 2p-degenerate only seems to apply exactly to the case where the total spin is S = 1/2. But, for this case, the 2p-degeneracy is retained even in the presence of spin-orbit coupling, because it is protected by time reversal R. (This is called Kramer's theorem.)

So, I am hoping to read the text myself.

Last edited: Feb 3, 2014
4. Feb 3, 2014

### vabite

Of course I can give you the details of the book. The book is: "Optical Orientation", Chapter 7, Section 2.2.
The first part of this chapter is well explained in "Group theory in physics", by Cornwell. Unfortunately the latter book does not discuss the topic of double groups.

If I did write about a 'reducible group' it is my fault. In that case I was speaking about a representation of that group.

5. Feb 3, 2014

### OhYoungLions

I apologize! I've reread you post, and found the text you were referring to, and now understand the notation and the context.

First I will explain my first post (so I am not too confusing), then I will answer your question as best I can:

In the copy of "Optical Orientation" on Google Books, on page 311 they introduce a symmetry element $\bar{E}$ whose associated operator acts only on the spin-space, and gives $\bar{E}|\uparrow\rangle = |\downarrow\rangle$ for example. This symmetry element is usually referred to as time reversal, $R = \bar{E}$. The symmetry elements of the double group are all of the elements in the space group {G} and all those elements combined with time reversal {GR}, so this is why there are twice the number.

For the special case of only one spin, S=1/2, and the (spin) rotation group corresponds to SU(2). In this case, the element associated with SU(2) is just R, or $\bar{E}$. Notice, this is not true for states of higher total angular momentum. For example, two spins may form a triplet state, in which the ms=+1 and ms=-1 states are related to eachother by time-reversal, but not to the ms=0 state. In the absence of spin-orbit coupling, spin rotational symmetry (which is not SU(2), since S = 1) holds that these are all degenerate, but spin-orbit coupling only respects time-reversal so you may split the ms=0 and ms=+/-1 states. For this reason, I made an objection in my first post. This is relevant because if you couple spin and orbital angular momentum, you end up with states for which the full rotation group is not equal to R.

I will first remark that the text does not provide a particularly transparent discussion, so I had to read it many times before I figured out what it was saying. The point is that in the absence of spin-orbit coupling, the Hamiltonian has an additional "symmetry" that the orbital and spin parts are completely decoupled which leads to accidental degeneracies that are not necessarily suggested by the double group representation. The text has a very bad way of saying this due to poor notation.

I've attached a character table for Td, which is the example they refer to in the text. You have three degenerate (p-bands?) which alone transform as $\Gamma_{15}$ at k=0, provided one does not consider the spin. When you include the spin, there are now 6 states at k=0. If there is no spin-orbit coupling, the text suggests that one should think about the symmetry group of the Hamiltonian as being Td x SU(2), in which case your six states transform as $\Gamma_{15}$ (3-fold degenerate) x D1/2 (two-fold degenerate). The symmetry elements of this "fictitious" group are all the G plus R, but we make a rule that R acts only on the spin part, and G acts only on the orbital part. Since $\Gamma_{15}$ and D1/2 are irreducible in their respective parts of the total group, the texts calls this irreducible.

When spin-orbit coupling is included, these rules don't make any sense, because the spin and orbital parts are no longer separable. In this case, the symmetry of the Hamiltonian is lowered, and one must consider the general double group, which we might call Td* to distinguish it. I've attached a character table for this group. In this group, as described above, one has G, and all GR. The six states have combined characters (in order) $\Gamma_{15} \times \Gamma_{6}$: 6 -6 0 0 0 -√2 √2 0. In this case, it is reducible into $\Gamma_7+\Gamma_8$.

So, in a sense, the reason why there is a change is because the symmetry group of the Hamiltonian is NOT the same with and without spin-orbit coupling.

(If you know anything about atomic spectra, the former is similar to Russel-Saunders coupling, and the latter is similar to j-j coupling.)

Anyway, as a final help, you can see in the character table that I attached a list of local functions and operators that transform according to that representation. I happen to be typing these out last week, so I have a few of them handy if it will help you.

$|3/2,1/2\rangle = \sqrt{\frac{2}{3}} |p_0,\uparrow\rangle + \sqrt{\frac{1}{3}}|p_+,\downarrow\rangle \\ |1/2,1/2\rangle = \sqrt{\frac{1}{3}} |p_0,\uparrow\rangle - \sqrt{\frac{2}{3}}|p_+,\downarrow\rangle$

(In each case, the first number corresponds to the total angular momentum J = L+S, and the second to mj; p0 = pz, and p+ and p- are appropriate combinations of px and py)

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6. Feb 4, 2014

### vabite

In fact, the text in some parts is not so transparent. Unfortunately, it is the only text that treats so in depth the optical orientation, that is in my domain of interest.