Why don't neutrons and protons have variable half-integer spin?

Question

Protons and neutrons are made up of 3 quarks say Q1, Q2, Q3. Each quark is a 1/2 spin particle. Now, $S(proton/neutron) = S(Q1)\otimes S(Q2)\otimes S(Q3)$. So, shouldn't the proton/neutron have a range of spins from 0 to 3/2? Am I missing something basic here? I have done my quantum mechanics courses but still have to do the particle physics course.

Answer 1

As the QM courses may have explained, you can combine two of the three spin $\frac12$ particles to either spin $0$ or spin $1$. If you then add the third, you can only get spin $\frac12$ or spin $\frac32$. Actually it is a bit more complicated, since we could imagine combining them in a more general way, but this result still turns out to be true.

So the case of spin $0$ that you mention is not possible. Still, there are two possibilities, so the question remains, why do the three combine in a way that gives spin $\frac12$? A short answer would be: they also can combine into a spin $\frac32$ state, but then we call the resulting particle a $\Delta^0$ or $\Delta^+$, see [Delta baryons]. Actually it is (again) a bit more complicated: the distinction between nucleons ($n$ or $p$ particles) and $\Delta$'s is more related to the fact that the quark flavors combine in a different way.

So the question still is: why can't the quark flavors combine in the way they do in the nucleons, but with their spins combined to spin $\frac32$? This is related to earlier questions like [Why isn't there a second baryon octet] and [Isospin doublet and quark content from contraction of quarks]. To answer your question we need to look at all the quantum properties of the quarks we have to combine:

1. The spatial wave function.
2. Their spin, described by the group $SU(2)$.
3. Color charge, described by color-$SU(3)$.
4. Their flavor, described by flavor-$SU(N)$.

We then assume that what we find in nature is the lowest energy state in which they can be combined following the rules of the groups mentioned, and under the restriction that the combined state is totally anti-symmetric in the three particles.

This simplifies matters a bit: The spatial wave function is in the lowest $s$-state, totally symmetric under permutation. And we take only the lightest "up" and "down" quark flavors, described by isospin $SU(2)$ (if we would also consider strange quarks that would already be flavor-$SU(3)$, not to mention the other, even heavier quark flavors).

Still, with spin $SU(2)$, color-$SU(3)$ and flavor-$SU(2)$ (isospin), we have $2\times 3\times2=12$ states per quark, so $12^3=1728$ states for a 3-quark combination even if we assume the spatial wave function to be the ground state. Selecting the $4$ states only that we need for the $p$ and $n$ with each two spin states, is still a large job.

Fortunately, color confinement does help us. Only color-neutral combinations are allowed, and only the totally antisymmetric combining of the three color states gives this result (as is usually quickly stated! Actually proving this would of course be, again, more complicated). So then we only have to create a totally symmetric combined spin and isospin state. Still $(2\times2)^3=64$ states to work with, but it is becoming manageable. So the rest can be left as an exercise for the reader.

But let's give an outline anyway:

1. Take the totally symmetric spin combinations, that would include ${\small |+++\rangle}$ and ${\small |---\rangle}$, with $S=\frac32, S_z=\pm\frac32$, but also two in between with $S_z=\pm\frac12$ that we can obtain with the raising and lowering operators, giving $\frac1{\sqrt3}\,({\small |-++\rangle}+{\small |+-+\rangle}+{\small |++-\rangle})$ and $\frac1{\sqrt3}\,({\small |+--\rangle}+{\small |-+-\rangle}+{\small |--+\rangle})$. In the same way create totally symmetric isospin combinations and together with the already chosen antisymmetric color and symmetric spatial state we have 16 states. The $\Delta^{++},\Delta^{+},\Delta^{0}$, and $\Delta^{-}$, each with four $S_z$ values.
2. Try to create totally antisymmetric spin combinations. Combined with totally antisymmetric isospin combinations that would give the required totally symmetric spin-isospin combination. But three spin-$\frac12$ states (or isospin-$\frac12$ states) cannot be totally antisymmetric! So this does not work...
3. Work with "mixed symmetry" spin states. Combined with mixed-symmetry isospin states that still can give totally symmetric spin-isospin combinations. This does work. We need spin states orthogonal to the earlier symmetric ones, e.g. for $S=\frac12,S_z=+\frac12$ that would be $\frac1{\sqrt2}({\small |-++\rangle}-{\small|+-+\rangle})$ or $\frac1{\sqrt2}({\small|+-+\rangle}-{\small|++-\rangle})$. Likewise for isospin we create $\frac1{\sqrt2}(|duu\rangle-|udu\rangle),$ and $\frac1{\sqrt2}(|udu\rangle-|uud\rangle)$. Then some simple math will show that there is one linear combination of these to give the $p$ with $S_z=+\frac12$ (that's the exercise for the reader!) In a similar way we get the other $S_z$ and $I_z$ values, together $4$ states for $p$ and $n$.
4. Convince ourselves that we never can combine a mixed state spin with totally symmetric isospin, or vice versa, so we are done with our search: the $16$ $\Delta$ states and $4$ nucleon states are the only totally symmetric ones of those $64$ spin-isospin combinations we were left with.

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EDIT (spoiler) The solution for the totally symmetric spin-isospin state is:

$$\begin{align} |\,p,+\rangle=& {\small\frac1{2\sqrt3}}\Big[ \big(\,|duu\rangle-|udu\rangle\big) \big(\,|{\small -++}\rangle-|{\small +-+}\rangle\big) \\& \quad + \big(\,|udu\rangle-|uud\rangle\big) \big(\,|{\small +-+}\rangle-|{\small ++-}\rangle\big) \\& \quad + \big(\,|uud\rangle-|duu\rangle\big) \big(\,|{\small ++-}\rangle-|{\small -++}\rangle\big) \Big] \tag{1a} \\[6pt] = & \ {\small\frac1{2\sqrt3}}\Big[ \ |duu\rangle\ \big(\,2\,|{\small -++}\rangle-|{\small +-+}\rangle-|{\small ++-}\rangle\big) \\&\quad\ \ \, + |udu\rangle\ \big(-|{\small -++}\rangle+2\,|{\small +-+}\rangle-|{\small ++-}\rangle\big) \\ &\quad \ \ \, + |uud\rangle\ \big(-|{\small -++}\rangle-|{\small +-+}\rangle+2\,|{\small ++-}\rangle\big) \Big] \tag{1b} \\[6pt] = & \ {\small\frac1{2\sqrt3}}\Big[ \ \big(\,2\,|duu\rangle-|udu\rangle-|uud\rangle\big) \ |{\small -++}\rangle \\&\quad\ \, + \big(\! -|duu\rangle+2\,|udu\rangle-|uud\rangle\big) \ |{\small +-+}\rangle \\&\quad\ \, + \big(\! -|duu\rangle-|udu\rangle+2\,|uud\rangle\big) \ |{\small ++-}\rangle \Big] \tag{1c} \\[6pt] = & \ {\small\frac1{2\sqrt3}}\Big[ \ 3\,|duu\rangle\ |{\small -++}\rangle +3\,|udu\rangle\ |{\small +-+}\rangle +3\,|uud\rangle\ |{\small ++-}\rangle \\ & \quad\ -\big(\,|duu\rangle+|udu\rangle+|uud\rangle\big) \,\big(\,|{\small -++}\rangle+|{\small +-+}\rangle+|{\small ++-}\rangle\big) \Big] \tag{1d} \end{align}$$ where the four ways of writing the state are obtained by just reordering the terms. The first way explicitly uses the spin and isospin states mentioned before, constructed orthogonal to the states already used in the $\Delta$'s, the second and third are perhaps more clear as Schmidt decompositions to show how the $3$-quark flavor states are entangled with the spin states, and the last way, $(1d)$, actually shows that this is indeed the solution for a totally symmetric spin-isospin state: swapping any pair of the particles in both the isospin and spin factors immediately gives us back the same terms!

In the same way we can construct the three other nucleon states: $|\,p,-\rangle,\ |\,n,+\rangle,\ |\,n,-\rangle$, but we could also derive those from $|\,p,+\rangle$ by applying the spin lowering and isospin lowering operators.