I'm wondering if this idea is as un-natural as it feels in my head. My understanding of CGC's is that I have various fractional spin states like:

$$ \text{particle 1:} \ |j_1=\frac12, m_2=\frac12,-\frac12\rangle, $$$$ \text{particle 2:} \ |j_2=\frac12, m_2=\frac12,-\frac12\rangle, $$

where the lower case $j$'s represent the magnitude of the maximum eigen value and I can make measurements between that value and the negative of that value, which are represented by the possibilities of $m$'s. Then, I could also think about the coupled states. The coupled states can have some maximum $J$ and measurable between $J,J-1,\cdots,-J+1, -J$. Then I find the overlap of the states.

I'm curious what the relationship is between these ideas and composing quarks together in Mesons and Baryons. This is for a couple of reasons, the spin number we associate with a Quark feels different than what we associate with an electron for instance. With an electron I understand that it is measurable to see $\pm\frac12$ whereas with a quark (which I am unsure if they are individually observable), would we see for an up quark measurables of $\frac13,-\frac23$ so that I move by increments of $1$ or would I see $\frac13,0,-\frac13$ so that I keep the same magnitude of spin. Or, is it a possible 3rd option where a quark has a fixed spin number which is just a magnitude and not a vector, meaning we just add the sums together when forming baryons. If this is the case, I would be a little curious why we get CGC like things when making mesons. For example, the Pion: $$ \text{Pion: } \frac{u\bar{u}-d\bar{d}}{\sqrt{2}} $$ feels very similar to the $|J=0,M=0\rangle$ state: $$ |J=0,M=0\rangle = \frac{ |+\frac12 -\frac12\rangle - | -\frac12 +\frac12 \rangle}{\sqrt{2}} $$

Thank you for any response. I'm still trying to sort out what I think my question is at the moment so if you ask for clarification in the comments I will try to respond in a timely manner.


1 Answer 1


I suspect you are learning about the quark model and you want the formal analogy between the spin in two particle composites (A), like the positron (which it is always assumed you have studied first), and the isospin of simple mesons (B), quark composites like the pion, fully fleshed out.

In the case of (A), you "add" the spins of the two constituents to a total spin of the composite J and, indeed the CG coupling coefficients parse out how the spins in the coproduct ("spin sum") are distributed.

The analogy to (B) is quite close, with two small conceptual detours.

  1. The quarks have spin 1/2, and by focussing on pseudoscalar mesons like the pions you are choosing the J=0 state in the spin sum you seem to accept. The quarks are "observable" in some sense, indirectly, although they are not asymptotic states, but in the last sixty years they have been studied thoroughly inside their hadrons by scattering experiments. (I have no idea about what you call 1/3, and -2/3 measurable... looks like the charges of the antiquarks, and I don't know what you are trying to say there.) Both terms of the pion wavefunction have spin zero, and it is a symmetric superposition of two different spin 0 states, $\frac{1}{\sqrt 2}(\uparrow \downarrow+\downarrow\uparrow )$ normally implied and omitted from the wavefunction! Anyway, let's move on to the isospin analogy for the isospin part.

  2. The neutral pion flavor wavefunction (we omit the spin and color parts), $$ \pi^0=\frac{u\bar{u}-d\bar{d}}{\sqrt{2}}, $$ is the formal $I=1$, $I_z=0$ analog of a spin vector $|J=1,M=0\rangle $ state, to which the spin state tensor product $$\frac{1}{\sqrt 2}(|j_1=1/2,m_1=1/2\rangle \otimes |j_2=1/2,m_2=-1/2 \rangle +|j_1=1/2,m_1=-1/2\rangle \otimes |j_2=1/2,m_2=1/2 \rangle )$$ projects; you probably represented it as $\frac{1}{\sqrt 2}(\uparrow \downarrow+\downarrow\uparrow )$ in college. Alert: The difference in sign between the + of the symmetric spin triplet state and the minus of the isospin triplet state above is due to the peculiarity of the conjugate su(2) representation for antiquarks. That is, $\bar d$ is isospin up, but $-\bar u$ is isospin down, and we shrugged off an overall minus sign in the pion wavefunction. This way, the formal analogy between spin and isospin is complete, and the same CG distributions of the composite wavefunctions apply elegantly and without a hitch.

  • Of course, for baryon wavefunctions, you apply similar arguments to composing three spin 1/2 light fermions to get a spin 1/2 or spin 3/2 baryon. This time the isospin will be 1/2 or 3/2 too, as the conjugate representation flip for antiquarks above is absent! Time to master the $\Delta$ baryon wavefunctions!
  • $\begingroup$ Thank you! Sorry for the mistake about the charge as spin. I was getting a little confused and tired when I decided to post the question. Regardless, you discussion in point 2) was very enlightening! Am I correct in understanding then that when we are writting in 2) $u\bar{u}$ we are actually secretly writing $[u(\text{spin up})\bar{u}(\text{spin down}) + u(\text{spin down})\bar{u}(\text{spin up})]/\sqrt2$? If yes, I think I may fully understand. Thanks again :) $\endgroup$
    – akozi
    Feb 13, 2023 at 14:22
  • $\begingroup$ Yes; absolutely yes. This is covered in most introductions to the quark model. $\endgroup$ Feb 13, 2023 at 14:29

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