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  "Puzzling" Graph for irreducible Represntation

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14 views

From reading this very short note at p.12 and p.13, I see a strange convention to obtain the product of representations.

enter image description here

There are two derivations. Each derivation has four equalities/identities.

For both two derivations, I can follow all the equalities, except the 3rd equalities/identities between the left and right hand side.

Can some one focus on explaining what is going on very precisely?

To make sure this works, can one derive that

$$ 3/2 \otimes 1=? $$ based on this convention? Thanks.

This post imported from StackExchange Physics at 2020-10-29 20:03 (UTC), posted by SE-user annie marie heart
asked Sep 30, 2017 in Mathematics by annie marie heart (500 points) [ no revision ]

1 Answer

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In case of $SU(2)$, this convention is really not that useful, since there's already a generic formula for tensoring $SU(2)$ irreps: $$ j_1 \otimes j_2 = \left| j_1 - j_2 \right| \oplus \dots \oplus \left( j_1 + j_2 \right). $$

But I'll describe the generic algorithm anyway, since the same convention also applies to $SU(3)$ where it is actually very useful.

Note that I'm giving a general overview and thus all of what follows is mathematically imprecise.

  1. For every complex simple Lie algebra, there's a maximal Abelian subalgebra called the Cartan subalgebra. The dimensionality of the Cartan subalgebra is called the rank of the Lie algebra. The rank of $\mathfrak{su}_2$ is $1$, and the Cartan subalgebra is conventionally chosen as $\mathfrak{c} = \text{Span}(J_z)$ (a linear span of the $J_z$ generator of $\mathfrak{su}_2$. The rank of $\mathfrak{su}_3$ is 2, and the Cartan subalgebra is conventionally chosen as $\mathfrak{c} = \text{Span}(\lambda_3, \lambda_8)$ (with $\lambda_1 \dots \lambda_8$ the Gell-Mann matrices). Observe that $\lambda_3$ and $\lambda_8$ indeed commute.

  2. A specially interesting basis in the Lie algebra is given by the Cartan-Weyl basis, consisting of:

    • A basis in the Cartan subalgebra called $c_{\alpha}$. Let $n$ be the rank of the Lie algebra, then $\alpha$ runs from $1$ to $n$.
    • A basis consisting of the eigenvectors of the adjoint action of $c_{\alpha}$ on the Lie algebra. These are called roots and denoted $r_e$ with $\vec{e}$ a vector in the $n$-dimensional root space. By definition, $$ \left[ c_{\alpha}, r_e \right] = e_{\alpha} r_e, $$ where $e_{\alpha}$ are the components of the vector $\vec{e}$ in the root space. This formula just says that $r_e$ is an eigenvector of the adjoint action of $c_{\alpha}$ with eigenvalues $e_{\alpha}$.
  3. Clearly, the Lie algebra decomposes into Cartan generators and roots. Roots form the root diagram of the Lie algebra. Finite-dimensional irreps of the Lie algebra can be classified by weight diagrams, which are symmetric patterns in the root space consisting of points, with root vectors connecting the neighboring points. Note that root diagram is also a weight diagram, for the adjoint representation. There's two extremely important facts about weight diagrams:

    • Coordiates of the weight (a point in the pattern) represent the eigenvalues of the weight with respect to Cartan subalgebra generators (the same goes for roots and the adjoint action).
    • Vector addition of the root vector and the radius-vector of the weight represents the action of the Lie algebra element on the element of the representation.

For $SU(2)$, $n = 1$ and weight diagrams are 1-dimensional. The roots are $J_1 \pm i J_2$ and their root vectors are $-1$ and $1$ (remember that they are 1-dimensional). The number of weights in the diagram classifies irreps completely (it is naturally the dimension of the irrep and is related to spin by $\text{dim}(j) = 2j + 1$).

For $SU(3)$, $n = 2$ and weight diagrams are 2-dimensional. They form pretty triangular patterns described in the end of the presentation that you are interested in.

Now to the algorithm of tensoring the irreps.

You start with two irreps, represented by two weight diagrams. Now imagine a small weight diagram for irrep $B$ instead of the point for every weight of the weight diagram of $A$. You now expand the size of the weight diagram until the neighboring points meet.

Now try to identify parts of the result as weight diagrams belonging to the irreps $c_i$ and scrape them off your drawing. This will give you

$$ a \otimes b = c_1 \oplus \dots \oplus c_k. $$

I hope I described the process clearly enough, as I don't like drawing things (sorry about that).

As to your example, which is $3/2 \otimes 1$, try doing what I just described with the $3/2$ weight diagram (4 points connected by 3 lines) and $1$ diagram (3 points with 2 lines), you will get something like this (apologies for the poor artwork):

enter image description here

Here I started with the (black) $3/2$ weight diagram and for each weight drew a (green) $1$ weight diagram in its place.

Then I looked at the resulting mess and decomposed it into three weight diagrams for irreps $5/2$, $3/2$ and $1/2$, which means that

$$ 3/2 \otimes 1 = 1/2 \oplus 3/2 \oplus 5/2, $$

in agreement with the first formula in my answer :)

Hope this helps, and please let me know if you need any clarifications.

This post imported from StackExchange Physics at 2020-10-29 20:03 (UTC), posted by SE-user Prof. Legolasov
answered Sep 30, 2017 by Prof. Legolasov (20 points) [ no revision ]

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