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This is probably a trivial question and I am missing something conceptually simple here.

I have the spin part of the total wave function of a baryon consisting of three light quarks:

\begin{equation} \psi = \frac{1}{\sqrt{6}}\left(2\big\uparrow\big\uparrow\big\downarrow - \big\downarrow\big\uparrow\big\uparrow - \big\uparrow\big\downarrow\big\uparrow \right) \end{equation}

My task is to compute the value of the spin ($S$) and its projection ($S_z$) of the state described by this wave function.

Naturally, I would get $S$ by applying the operator $\textbf{S}^2$ on my $\psi$, which would allow me to get the value of $S(S+1)$. But here my question comes: I am confused, how should I apply this operator on such a 'complicated' wave function consisting of three terms? And how do I take into account properly the (Clebsch–Gordan) coefficients entering before each term?

When I try to do that, I get an answer which is definitely wrong, however (I think) I understand how to perform such an exercise on a simple wave function consisting of only one term, such as $\psi'=\big\downarrow\big\downarrow\big\downarrow$, where a simple 'arrow counting' gives you a correct answer $3/2$.

Regarding the $S_z$, it is quite easy to guess to be $+1/2$ just counting the arrows, but again, the CG coefficients confuse me as I am not sure whether they should be taken into account or can be ignored.

I am pretty sure there's a ridiculously simple way to get the answer without any complicated math, but apparently I am missing something, so would be grateful for some catchy explanations.

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  • $\begingroup$ How did you use the CB coefficients? I'm not sure if there's a quick way to find $S^2$, but $S_z$ can always just be added together, since the total $S_z = \sum _i S_z^{(i)}$. $\endgroup$ – Hanting Zhang Mar 9 '19 at 19:22
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    $\begingroup$ $S^2$ is something like $S_z^2 + \left(S_+S_- + S_-S_+\right)/2$. So check this definition for yourself, as it looks a lot like a homework assignment, and try it. Another approach is to construct all the spin states of three spins. Then you will encounter this one soon enough. Note that 3 spins easily give frustration ... I mean the spins. $\endgroup$ – my2cts Mar 9 '19 at 19:35
  • $\begingroup$ @Hanting Zhang. Yes, there is a shortcut: Three spin 1/2s give you a spin 3/2 and two spin 1/2s. Raising the state ψ with $S_+$ yields 0, so this is definitely not in the spin 3/2 quartet, so, inevitably, it it is in one of the two spin 1/2 doublets or a linear combination thereof! my2cts already indicates the straight way to verify this is, indeed, a pure doublet. $\endgroup$ – Cosmas Zachos Mar 9 '19 at 21:14
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Your equation for $| \psi \rangle $ is

\begin{equation} |\psi\rangle = \frac{1}{\sqrt{6}}\left(2\big\uparrow\big\uparrow\big\downarrow - \big\downarrow\big\uparrow\big\uparrow - \big\uparrow\big\downarrow\big\uparrow \right). \end{equation}

If you want to measure the total spin of the state, you indeed act with $S^2 = (S^{(1)} + S^{(2)} + S^{(2)})^2$, where the superscripts denote the fact that the spin operators only act on their respective particles.


To get the total spin of your state you just expand out $S^2 = (S^{(1)} + S^{(2)} + S^{(2)})^2$ and act on each state in the sum with the operator. Note that you know that, for example, $[S^{(1)}]^2 \uparrow\uparrow\downarrow = \hbar^2(\frac{1}{2})(1+\frac{1}{2})$. You will also get terms like $S^{(1)} \cdot S^{(2)}$ which you will have to expand into components, but you can nevertheless calculate.

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