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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ $S^2$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.