Irreducible representations

Main Question or Discussion Point

I don't understand how to find the irreducible representations of a group.

Under transformation U: $$(T')^{ijk}=U^{il}U^{jm}U^{kn}T^{lmn}$$

But suppose $$(M')^{ijk}=(T')^{ijk}+\pi (T')^{jik}$$

Then $$(M')^{ijk}=U^{il}U^{jm}U^{kn}T^{lmn} +\pi U^{jl}U^{im}U^{kn}T^{lmn}=U^{il}U^{jm}U^{kn}(T^{lmn}+\pi T^{mln}) =U^{il}U^{jm}U^{kn}M^{lmn}$$

In other words, any linear combination of permutations of a tensor, such as:

$$(M)^{ijk}=(T)^{ijk}+\pi (T)^{jik}$$

transform into similar permutations. This seems to be a property of tensors in general and not on any specific group such as SU(N), as long as the tensors have all upper indices (or all have lower indices).

So if you can find particular linear combinations whose total members is less than N^(tensor rank), then you will have reduced the tensor product into a smaller block.

Now apply this to the addition of three spin 1/2 particles, which will be the group SU(2). There are 2^3=8 total members (e.g., |+++>,|++->,|+-->, etc.)

If you choose the completely symmetric combination:

$$(M)^{ijk}=(T)^{ijk}+(T)^{jik}+... (\operatorname{all \: permutations})$$

then there are a total of 4 of them, so you have reduced the 8 total members into a sub-block consisting of the basis:

$$\begin{equation*}\begin{split} |+++> \\ |++->+|+-+>+|-++> \\ |--+>+|-+->+|+--> \\ |---> \end{split}\end{equation*}$$

The other choice of linear combination seems to be something called mixed symmetry:

$$M^{ijk}=T^{ijk}+T^{jik}-T^{kji}-T^{jki}$$

and indeed there are only 2 elements:

$$\begin{equation*}\begin{split} |++->+|++->-|-++>-|+-+>=2|++->-|-++>-|+-+> \\ |--+>+|--+>-|+-->-|-+->=2|--+>-|+-->-|-+-> \end{split}\end{equation*}$$

So you've broken up 8=4+2+ ????

My guess is that the ???? is 2, but in principle it can be 1+1.

What other linear combination am I missing besides the completely symmetric one, and the mixed-symmetry one?

I understand the mixed-symmetry corresponds to symmetrizing two indices (say i and j):

$$T^{ijk}+T^{jik}$$

and then antisymmetrizing this sum in two of its indices (say k and i):

$$(T^{ijk}+T^{jik})-(T^{kji}+T^{jki})$$

which is what I used above.

What else is there to get the remaining irreducible representations?

In principle any linear combination will transform only among itself, but how to find one that only has two elements?

Also, if angular momentum is SO(3), and spin is SU(2), how many parameters does the rotation group have, and what are the intervals of the angles? It seems one parameter has an interval [0,2pi), and the other is [0,4pi), so I'm a bit confused at treating SO(3) and SU(2) as one group and adding the generators to make one total generators. Does the group with generators J=L+S go from [0,2pi) or [0,4pi)?

My group theory is not any good so any help is appreciated.

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Okay, after trial and error, you have to antisymmetrize the first two indices, and then symmetrize.

So first:

$$T^{ijk}-T^{jik}$$

then symmetrize (say i and k):

$$(T^{ijk}-T^{jik})+(T^{kji}-T^{jki})$$

This produces the 2 states:

$$\begin{equation*}\begin{split} |-++> - |+-+> + |++-> - |++->=|-++> - |+-+> \\ |+--> - |-+-> + |--+> - |--+>=|+--> - |-+-> \end{split}\end{equation*}$$

that are orthogonal to the other 2 mixed symmetry states, so 8=4+2+2.

I guess it kinda makes sense.