Group theory - triangle rotation matrix problem? probably simple?

[jeebs]

I'm attempting to do some problems in a group theory exercise for the first time and am falling flat on my face. Here's the problem:

"the molecule 'triangulum' consists of 3 identical atoms arranged in an equilateral triangle. Using a basis which consists of a single localised orbital on each atom, x_a,x_b, and x_c, a Hamiltonian for the molecule can be written as H = \left(\begin{array}{ccc}e&d&d\\d&e&d\\d&d&e\end{array}\right)

Consider the operator R_2_\pi_/_3 which rotates the molecule through an angle 2\pi/3. We thus have:
R_2_\pi_/_3x_a = x_b
R_2_\pi_/_3x_b = x_c
R_2_\pi_/_3x_c = x_a
Use these results to obtain a 3x3 matrix representation of R_2_\pi_/_3."

So, I'm fairly lost here. I suspect this is probably straightforward but I've never seen this done before.

The first thing that bothers me is, this is an equilateral triangle molecule we are dealing with. That is a flat object that can be described in 2D, so why do we want a 3D representation of a rotation matrix?

Also I'm struggling to see what I'm supposed to do with the information I've been given. My immediate reaction was just to write down, say, Hx_a = Ex_a where, say, x_a = \left(\begin{array}{c}a_1\\a_2\\a_3\end{array}\right) so that Hx_a = \left(\begin{array}{c}ea_1 + da_2 + da_3\\da_1 + e_a2 + da_3 \\da_1 + da_2 + ea_3\end{array}\right) but I really haven't got a clue what I'm being asked to do here.

 

[grey_earl]

You don't need the Hamiltonian at all here, besides the fact that it reassures you the three atoms are equal.

The rotation matrix is 3x3 because you have three atoms here (and are working in the basis of their coordinates), it has nothing to do with 3D space at all.

Just make an ansatz
R_{2/3 \mathpi} = \begin{pmatrix} R_{aa} & R_{ab} & R_{ac} \\ R_{ba} & R_{bb} & R_{bc} \\ R_{ca} & R_{cb} & R_{cc} \end{pmatrix}
and then calculate R_{2/3 \mathpi} x_a etc. I think this should be all ...

 

[jeebs]

The rotation matrix is 3x3 because you have three atoms here (and are working in the basis of their coordinates), it has nothing to do with 3D space at all..

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I'm not sure what you mean by this. What is meant by the "basis of their coordinates"?
Surely coordinates have everything to do with a real 3D space?

 

[grey_earl]

The Matrices are 3x3 because your basis has 3 elements, which are x_a, x_b, x_c. Just forget at all about physical space, your atoms could live on a 2D surface or in 26-dimensional string theory space.
Instead of "Rotation Matrix", say "Magic Operator" and the abstract group-theoretical problem doesn't change a bit. You are given the action of the operator on the basis elements and need to find a matrix, such that multiplication of a vector by this matrix is equivalent to the operation of the operator on the element represented by the vector (for example, x_a =(1,0,0)).

 

[jeebs]

The Matrices are 3x3 because your basis has 3 elements, which are x_a, x_b, x_c. Just forget at all about physical space, your atoms could live on a 2D surface or in 26-dimensional string theory space.
Instead of "Rotation Matrix", say "Magic Operator" and the abstract group-theoretical problem doesn't change a bit. You are given the action of the operator on the basis elements and need to find a matrix, such that multiplication of a vector by this matrix is equivalent to the operation of the operator on the element represented by the vector (for example, x_a =(1,0,0)).

hmm, so what, I have no idea what these x_a,b,c are so I start with some general x_a = \left(\begin{array}{c}x_a_1\\x_a_2\\x_a_3\end{array}\right) (and the same thing for x_b,c) and multiply them out with that general R matrix...
then just mess around with all the equations I get until I have expressions for the individual R matrix elements?

or am I supposed to have something specific for my x_a,b,c?
what made you say x_a =(1,0,0)?

infact, if these atoms are identical, and we are ignoring their difference in position, then why isn't x_a = x_b = x_c ?

 

[grey_earl]

No, you don't have x_a = (x_a1, x_a2, x_a3). The three elements x_a, x_b, x_c form your basis, so for a general x you have x = (v_1,v_2,v_3) = v_1 x_a + v_2 x_b + v_3 x_c.

The atoms are identical, but we don't ignore their difference in position. x_a is describing a fixed position in space (namely, that of atom a), but it is completely irrelevant what this position is.

Think of the following: you have a group with three elements (a,b,c) and an operator R which is given by R a = b, R b = c, R c = a.
Then, obtaining a 3x3 matrix representation of R is: choose a vector representation for your basis elements. I choose a = (1,0,0), b = (0,1,0) and c = (0,0,1), but you could take other values (although I assume your instructor implicitely wants you to use this representation). Then calculate the matrix corresponding to R, since you know what R a, etc. must give.

Group theory is nice because it abstracts all details away: it doesn't matter that a corresponds to x_a, the position of atom a, or that R is a rotation of the molecule by 2/3 Pi, you just need to know that R a = b, etc. And that makes everything very simple, so simple, in fact, that it confuses you quite a lot :)

Still confused?