Counting operators with group theory

[Luca_Mantani]

Homework Statement

I have an exercise that I do not know how to solve. $N$ is a nucleon field, in the fundamental representation of $SU(4)$. We want to classify operators by their $SU(4)$ transformation properties, bearing in mind that the nucleon is a fermion and we need antisymmetric products of fields in our operators. The problem asks to combine the representations for two $N$'s and two $N^\dagger$'s and count the number of operators (with no derivatives) from counting the representations. The operator to consider is:
$$(N^\dagger N)(N^\dagger N)$$

Homework Equations

The problem provide these two helpful equations:
$$4\otimes \bar{4}=1\oplus 15$$
$$4\otimes 4 = 6\oplus 10$$

The Attempt at a Solution

I don't know where to start, I only know that the solution has to be of the form $x\oplus y \oplus z$. Also as a follow up question, it asks to find out if each of these representations has a $SU(2)$ singlet subgroup or not.

It would be helpful if someone know hot to tackle this kind of problem and is willing to explain to me the procedure.
Thanks in advance!

 

[mark726]

Hello,
To solve this problem, we need to use the equations provided to break down the product of nucleon fields into different representations of $SU(4)$. Let's start by expanding the product:
$$(N^\dagger N)(N^\dagger N) = N^\dagger N N^\dagger N$$
Using the first equation, we can break down the product of two $N$'s into $1$ and $15$ representations:
$$N^\dagger N = (N^\dagger N)\otimes 1 \oplus (N^\dagger N)\otimes 15$$
Similarly, we can break down the product of two $N^\dagger$'s into $1$ and $15$ representations. Therefore, our original product can be written as:
$$(N^\dagger N)(N^\dagger N) = (N^\dagger N)\otimes (N^\dagger N) = [(N^\dagger N)\otimes 1 \oplus (N^\dagger N)\otimes 15] \otimes [(N^\dagger N)\otimes 1 \oplus (N^\dagger N)\otimes 15]$$
Now, using the second equation, we can break down the product of two $4$ representations into $6$ and $10$ representations. Therefore, our product now becomes:
$$(N^\dagger N)(N^\dagger N) = [(N^\dagger N)\otimes 1 \oplus (N^\dagger N)\otimes 15] \otimes [(N^\dagger N)\otimes 1 \oplus (N^\dagger N)\otimes 15] = [(N^\dagger N)\otimes 1 \oplus (N^\dagger N)\otimes 15] \otimes [(N^\dagger N)\otimes 6 \oplus (N^\dagger N)\otimes 10]$$
Now, we need to combine these representations to get the final answer. Remember that we need to take into account the antisymmetry of the product of fields. For example, the product of two $N^\dagger$'s in the $15$ representation will give us an antisymmetric combination, while the product of two $N$'s in the $15$