# Representation theory question

1. Jul 15, 2012

### VantagePoint72

I'm self-studying representation theory for finite groups using "Group Theory and Quantum Mechanics" by Michael Tinkham. Most of it makes sense to me, but I'm having difficulty understanding what is meant by saying a function "belongs to a particular irreducible representation", or "has the symmetry of a particular irreducible representation". I assume this has something to do with a particular set of functions acting as a basis for an irreducible representation, but I still don't understand what is meant by "the symmetry of an irreducible representation". To make my question a bit more concrete, what do the last columns two columns in this character table (labeled "linear, rotations" and "quadratic") mean?

I have a couple other questions, but they're more specific to this book. I understand it's a fairly popular text, so if someone has a copy of it handy and wouldn't mind entertaining another question or two, I'll post them in a follow up comment.

Thanks!

2. Jul 16, 2012

### VantagePoint72

Alright, I'm just going to go ahead and post my follow up question and hope it catches someone's eye. I'm actually feeling a bit more comfortable with what I originally asked about, with the exception of one thing. Here's the theorem I'm trying to understand:

"Matrix elements of an operator H which is invariant under all operations of a group vanish between functions belonging to different irreducible representations or to different rows of the same representation."

So, that would mean the (possibly) non-vanishing matrix elements are between functions belonging to the same row of the same irreducible rep. But... aren't just those just the same function? Once you choose a basis, I thought that the index of a particular irreducible representation and a row index within that rep uniquely identify a single function. Isn't that the point of the rep index and row index being "good quantum numbers"? The only way I can think of it kind of making sense of it is if the two functions correspond to a different choice of basis. But that doesn't really make sense either: suppose I have a three dimensional irreducible representation and I choose basis functions {ψ123}. According to the theorem, <ψ1|H|ψ2> = 0. However, I could choose a "new" basis {$\phi$1,$\phi$2,$\phi$3} = {ψ213}. Now, going purely indices, ψ1 and $\phi$1 would be considered belonging to the same row of the same irreducible rep. This would seem to imply, again by the theorem, that <ψ1|H|$\phi$1> = <ψ1|H|ψ2> does not necessarily vanish. So, clearly my understanding of the notion "belong to a particular row of a particular irreducible representation" is not correct.

I hope I've explained the issue in a way clear to someone without access to the same text as me. However, for anyone who does, the theorem in question is in section 4.9 of Tinkham and the relevant material is developed in 3.8.

Can anyone help?

3. Jul 19, 2012

### The_Duck

Yes.

I think your theorem is Schur's lemma, one statement of which is:

"Any operator that commutes with all operations of a group must be proportional to the identity within any given irreducible representation."

That is, if H commutes with all group operators, and |ψ> belongs to some irreducible representation R of the group, then H|ψ> = C(R)|ψ>, where C(R) is a number (not an operator) that depends on the particular irreducible representation R (but does not depend on the row within the representation).

I suspect you're trying to make this more complicated than it is. The point is that any operator that commutes with the whole group must act essentially trivially within any irreducible representation.

A generalization of this is Wigner-Eckhart theorem, which in part says that if H is an operator that belongs to a particular irreducible representation R1, and |ψ> is a state that belongs to a particular irreducible representation R2, then H|ψ> belongs to the direct product representation R1 x R2. In your case R1 is the trivial representation, so R1 x R2 = R2, that is H|ψ> belongs to the same representation as |ψ>.

4. Jul 19, 2012

### VantagePoint72

Thank you for the response, but I'm afraid I didn't understand it very well. I mean... I mostly follow what you're saying (though I'm not sure I understand what it means for an operator to belong to a representation), I'm just not making the connection between it and my question :-S Could you elaborate? I'm sorry, I wish I just I could ask a more specific question, but I'm just not seeing it. Could you start by explaining what it means for a function to belong to a particular row of an irr. representation? I thought I understood that, but from your response it seems that maybe I don't...

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