Query regarding representations and Clebsch-Gordan series

  • #1
Hello all,

I'm stuck on understanding part of a discussion of representations and Clebsch-Gordan series in the book 'Groups, representations and Physics' by H F Jones. I'd be grateful to anyone who can help me out.

For starters, this discussion is in the SU(2) case. I don't know how to draw Young Tableau here, so I'll just go with the CG series. We have the result that [tex]3\otimes 2 = 4\oplus 2[/tex]. The 2 on the right-hand side is the part that I'll be concerned with here, and it corresponds to a Young diagram with three boxes, two in the first row, one in the second. This CG series corresponds to a product of spinors of the form [tex]\psi_a\phi_b\chi_c[/tex] (to save myself typing too much, I'll just write out the indices from now on i.e. [tex]bac=\psi_b\phi_a\chi_c[/tex]). The author writes that the explicit decomposition corresponding to this CG series is
[tex]
3 \{ab\}c = (\{ab\}c + \{bc\}a+\{ca\}b) + (\{ab\}c-\{cb\}a) + (\{ab\}c-\{ca\}b)
[/tex]
where I'm using [tex]\{ab\}[/tex] to represent the symmetric term [tex]ab+ba[/tex]. So the first term on the RHS is totally symmetric and corresponds to the 4 in the CG series above. Then there are two terms. And here's my problem; I think the Young diagram described above (first row has two boxes, second row has one) should be symmetric in a and b, antisymmetric in a and c. So why are there two extra (not fully symmetric) terms on the RHS in the equation above instead of just one? If you take them together as a single term then they're symmetric in a and b, but not antisymmetric in a and c. If you just consider the middle term its antisymmetric in a and c but not symmetric in a and b. Is there a typo in the book? Is my understanding faulty? Is it just too late at night and I've gorged myself on too much easter chocolate? Any help please? Thanks in advance
 

Answers and Replies

  • #2
Meir Achuz
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I think the Young diagram described above (first row has two boxes, second row has one) should be symmetric in a and b, antisymmetric in a and c.

The 2 dimension state is not antisymmetric in a and c.
It is a mixed symmetry state which is not purely symmetric or antisymmetric in a and c.
Interchanging a and c gives a linear combination of states.
 
Last edited:
  • #3
The 2 dimension state is not antisymmetric in a and c.
It is a mixed symmetry state which is not purely symmetric or antisymmetric in a and c.
Interchanging a and c gives a linear combination of states.

Okay, in that case how does one construct the correct algebraic expression (in terms of [tex]\psi_a[/tex]s, etc.)? The totally symmetric 4 dimension state in my problem corresponds to the diagram labelled [tex]S_3[/tex] in this figure;
t3007_young-diagramm.gif

and it's easy to write the corresponding expression - just sum over all the permutations of a, b, and c. The dimension 2 state corresponds to [tex]M_3[/tex]. Now I know that the first column (in the SU(2) case we're discussing) is totally antisymmetric, so it can technically be excluded from the diagram. But this diagram is symmetrized on the two boxes in the top row, and antisymmetrized on the two boxes in the first column.

So my question then is, how does one construct the corresponding algebraic expression, the equation that I included in my original question? How does the symmetization and antisymmetrization between the boxes in the diagram manifest in the algebraic expression? If you could show the steps explicitly please, I'd appreciate it, because I'd like to understand this well enough to be able to generalise to other diagrams, and to SU(3). Thank you very much in advance.
 
  • #4
Bill_K
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Mixed symmetry is a perpetual source of confusion. The only way I know to construct the states is by use of stepping operators. The highest level in the totally symmetric state (all spins up) is |↑↑↑>. If you step this down, you get |↑↑↓> + |↓↑↑> + |↑↓↑>.

For the mixed representation, you want a state which is orthogonal to this state and symmetric on the first two spins. The only such state is 2|↑↑↓> - |↓↑↑> - |↑↓↑>. To write this in terms of wavefunctions, look at the case a = b = ↑ and c = ↓. The state corresponds to 2(abc + bac) - (cab + cba) - (acb + bca) = {ab}c - {ca}b + {ab}c - {cb}a, which is what you have.
 
  • #5
Meir Achuz
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I too prefer the method described by BK.
You could do it your way by first antisymmetrizing the second and third state, and then symmetrizing this with respect to the first and second state.
 
  • #6
Thank you both for the replies, and I'm sorry I didn't say so sooner (Easter got in the way).

BK, I found your explanation helpful and easy to follow, if a little disappointing. Is there really no way (that you're aware of) to create the appropriate expression from the Young diagrams directly? The procedure you described would seem to require you to step through all the states in the totally symmetric case, then construct orthogonal vectors to these via the Gram-Schmidt process (or some equivalent). Is this the most direct way to go?

A couple of questions still puzzle me. Firstly, the descriptions I have read state that symmetrisation is performed before antisymmetrisation, meaning that the symmetrisation will, in general be destroyed and the antisymmetrisation persists. So wouldn't it be appropriate to look for a state which is antisymmetric on the first and third spins (which would exclude [tex]\left|\uparrow\downarrow\uparrow\right\rangle[/tex] for example), rather than one symmetric on the first and second spins? Secondly, the book I'm looking at states that the diagram [tex]M_3[/tex] corresponds to the state [tex]abc+bac-cba-bca[/tex], so I still have trouble reconciling this with the earlier expression I stated (and which you reproduced).

Thanks again in advance.
 
  • #7
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I am not sure I understand all the comments. However, I consider the original question to have a simple answer.

First it is important to note that Young diagrams are intended for U(n) not SU(n) groups. A U(2) irrep (irreducible representation) is labelled by two positive integers (a,b) and can be represented by a 2-row Young diagram with 'a' boxes in row 1 and 'b boxes in row 2. The corresponding SU(2) irrep is labelled by a single positive integer (a-b).

For the problem at hand the SU(2) tensor product 3 x 2 is given by the U(2) tensor product
(3,0) x (2,0) which has the decomposition into irreps
(3,0) x (2,0) = (5,0) + (4,1) + (3,2)
and on restriction to SU(2) gives
(3) x (2) = (5) + (3) + (1).

In familiar angular momentum coupling language, this corresponds to
J1 x J2 = J1+J2 + ... + |J1-J2|
which is a result that Jones quotes in his book.

This result certainly differs from that quoted in the question, i.e., 3 x 2 = 4 + 2. Is this really a result given by J.P. Jones?

Hope my comments help
David R
 

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