Question about calculating isospin couplings

Question

So say we have the following interaction: $$ \Sigma_c^+ \rightarrow \Lambda_c^+ + \pi^0 $$ And say that we want to find out the quark composition of $\Sigma_c^+$ through inspection of the isospins of the particles on the RHS. If $\Lambda_c^+$ is stated to be an isospin singlet $\left( J^P=\frac{1}{2}^+\right )$ with $I_3 = 0$ through the relation $\frac{1}{2}(n_u - n_d)$ where $n_U$ and $n_d$ is the number of up and down quarks respectively, I assume that it is safe to make the following assumption that $I = I_3 = 0$. Doing the same for the $\pi^0 $ and acknowledging that it is flavourless we again get $I_3 = 0$. Now, I'm not sure how to get $I$ for the pion from the data provided so I looked it up and I see that it is $I = 1$.

My question is this, how do I calculate the isospin couplings to find out what $I$ and $I_3$ are for $\Sigma_c^+$? In my notes it states we take the tensor product between the max values of the isospin ($I$) to find the total isospin $I^{tot}$ as followed:

$$ I^{tot} = 0 \otimes 1 = 1 $$ However I am quite confused by this because all the examples of tensor products I have seen involve matrices and not integers. My brain is also interpreting that as $0 * 1 = 1$ which is obviously not the case. I'm also not entirely sure of what to do with the knowledge that $J^P = \frac{1}{2}^+$ . Think I am quite confused. I understand we can use the ladder operators to find the quark composition after we find out the isospins but it's mainly just this intermedial step. Thank you for reading and I appreciate any help people can give in advance.

Answer 1

I think its more common to use the dimension of the representation ($2I+1$), rather then the quantum number, $I$. With that, the isoscalar is one-dimensional (trivial representation):

$$ {\bf 1}$$

and the isovector is 3 dimensional (adjoint representation):

$$ {\bf 3}$$

so that:

$${\bf 1}\otimes{\bf 3}={\bf 3}$$ A more active example is 2D spinor-rep (up and down, the fundamental representation):

$${\bf 2}\otimes{\bf 2}={\bf 3} \oplus {\bf 1}$$

The ${\bf 3}$ are the pions ($\pi^+,\pi^0,\pi^-$), and single combination would be eta-like, except etas have strangeness in the mix which complicates things (the group is flavor SU(3), not isospin SU(2)).

It's cleaner to first understand it using spins...where 2 electrons can be in the antisymmetric singlet state:

$$ |0,0\rangle = \frac 1 {\sqrt 2}[|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle] $$

or one of the three symmetric triplet states:

$$ |1,+1\rangle = |\uparrow\uparrow\rangle$$

$$ |1,0\rangle = \frac 1 {\sqrt 2}[|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle] $$

$$ |1,-1\rangle = |\downarrow\downarrow\rangle$$

The mathematical similarity to spin is why it was called "isospin" in nuclear physics, with the proton and neutron being the two up/down states of the nucleon.

This was naturally extended to up/down quarks, and expanded to include the strange quark. The much larger masses of charm/bottom/top quarks means flavor symmetry is so broken that SU(4), SU(5), and SU(6) isn't that useful.

Answer 2

However I am quite confused by this because all the examples of tensor products I have seen involve matrices and not integers. My brain is also interpreting that as 0*1=1, which is obviously not the case.

You are misreading your Kronecker products. A Kronecker product of two operators (here, isospin matrices) is a composition of two matrices of different dimensionalities, say n×n with m×m, in which (say, for the sake of argument) you take the right factor matrix and insert it next to each element of the left factor matrix: this will net you an nm×nm matrix, which might, or might not, be reducible (with or without subspaces transforming separately). Here, because one factor is 1-dimensional (the isosinglet is a number, a 1×1 matrix), this does not happen, and your product representation is irreducible, acting on a 3-dim vector space, 1 * 3=3.

Your brain stress results from the fact that, in such physics, one hearkens back to traditional angular momentum notation, and uses spin labels j instead of the dimensionality n= 2j+1. That is, had you used dimensionality formulas you'd have, more intuitively, the isosinglet composing with an isotriplet to an isotriplet, 1 ⊗ 3 = 3, much more intuitive, since it automatically error-checks that the components and the outcome always involve the same number of states!

Applying this to your problem, then, it asks you to compose an isosinglet $ \Lambda_c^+$ with an isotriplet $\pi^0 $, whose isospin you are indeed assumed to know, as you surmised. Appreciating one of the quarks of both baryons is a c, you are then led to an isosinglet flavor part of the wavefucntion for the $ \Lambda_c^+$, c(ud-du), antisymmetric in u and d. I am not sure how much about full baryonic wave functions you have been taught, and whether you are expected to write the full ones involving spin (and color, antisymmetrically) at your level. Likewise, the pion flavor wave function is the isotriplet $\bar u u-\bar d d$. (It is actually symmetric in the two flavors: the minus sign is a peculiarity of the conjugate su(2) representation.) The flavor wavefunction of the $\Sigma_c^+ $ then has to be c(ud+du), $$ \Sigma_c^+ \rightarrow \Lambda_c^+ + \pi^0 \\ c(ud+du) \rightarrow c(ud-du) + (\bar u u-\bar d d)/\sqrt2~ . $$