Irreducible representations

In summary, the conversation discusses finding irreducible representations of a group using linear combinations of permutations of a tensor. It is shown that any linear combination of permutations of a tensor transforms into similar permutations, regardless of the group. The example of adding three spin 1/2 particles in the group SU(2) is used to demonstrate this concept. The completely symmetric and mixed-symmetry combinations are explored, and it is determined that the remaining irreducible representations can be obtained by antisymmetrizing the first two indices and then symmetrizing. The conversation also briefly touches on the rotation group and the confusion surrounding the intervals of the angles.
  • #1
RedX
970
3
I don't understand how to find the irreducible representations of a group.

Under transformation U: [tex](T')^{ijk}=U^{il}U^{jm}U^{kn}T^{lmn} [/tex]

But suppose [tex](M')^{ijk}=(T')^{ijk}+\pi (T')^{jik} [/tex]

Then [tex](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}[/tex]

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

[tex](M)^{ijk}=(T)^{ijk}+\pi (T)^{jik} [/tex]

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:

[tex](M)^{ijk}=(T)^{ijk}+(T)^{jik}+... (\operatorname{all \: permutations}) [/tex]

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:

[tex]\begin{equation*}\begin{split}
|+++> \\
|++->+|+-+>+|-++> \\
|--+>+|-+->+|+--> \\
|--->
\end{split}\end{equation*} [/tex]

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

[tex]M^{ijk}=T^{ijk}+T^{jik}-T^{kji}-T^{jki} [/tex]

and indeed there are only 2 elements:

[tex]\begin{equation*}\begin{split}

|++->+|++->-|-++>-|+-+>=2|++->-|-++>-|+-+> \\
|--+>+|--+>-|+-->-|-+->=2|--+>-|+-->-|-+->
\end{split}\end{equation*}
[/tex]

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):

[tex]T^{ijk}+T^{jik}[/tex]

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

[tex](T^{ijk}+T^{jik})-(T^{kji}+T^{jki}) [/tex]

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.
 
Last edited:
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  • #2
Okay, after trial and error, you have to antisymmetrize the first two indices, and then symmetrize.

So first:

[tex]T^{ijk}-T^{jik} [/tex]

then symmetrize (say i and k):

[tex](T^{ijk}-T^{jik})+(T^{kji}-T^{jki}) [/tex]

This produces the 2 states:

[tex]\begin{equation*}\begin{split}
|-++> - |+-+> + |++-> - |++->=|-++> - |+-+> \\
|+--> - |-+-> + |--+> - |--+>=|+--> - |-+->
\end{split}\end{equation*}[/tex]

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

I guess it kinda makes sense.
 

1. What are irreducible representations?

Irreducible representations are mathematical tools used to describe the behavior and properties of a physical system. They represent the different ways in which a system can transform under specific operations, such as rotations or reflections.

2. What is the importance of irreducible representations?

Irreducible representations are important because they provide a way to simplify complex systems and understand their underlying symmetries. They also allow us to make predictions about the behavior of a system under different conditions.

3. How are irreducible representations used in chemistry?

In chemistry, irreducible representations are used to classify and analyze molecular vibrations. By determining the symmetry of a molecule and its corresponding irreducible representations, we can predict the types of vibrations that it can undergo.

4. Can an irreducible representation be reducible?

No, by definition, an irreducible representation cannot be reduced into smaller representations. It represents a fundamental and indivisible aspect of a system's symmetry and cannot be broken down further.

5. How are irreducible representations related to group theory?

Irreducible representations are closely related to group theory, as they are used to classify and analyze the symmetry properties of groups. In group theory, a group is a set of elements that can be combined together using a defined operation, and irreducible representations are used to describe the different ways in which a group can transform.

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