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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 $S_z = \sum _i S_z^{(i)}$. $\endgroup$
    – cxx
    Mar 9, 2019 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, 2019 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. $\endgroup$ Mar 9, 2019 at 21:14

2 Answers 2

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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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Set $\hbar=1$ for simplicity. The overall normalization is irrelevant when it comes to eigenvalues, so consider \begin{equation} |\alpha\rangle= {\sqrt{6}} |\psi\rangle = \left(2 \uparrow \uparrow \downarrow - \downarrow \uparrow \uparrow - \uparrow \downarrow \uparrow \right). \end{equation}

You also know for one doublet, $$ S_z|\uparrow \rangle = \frac{1}{2} |\uparrow \rangle , \qquad S_z|\downarrow \rangle = -\frac{1}{2} |\downarrow \rangle ,\\ S_-|\uparrow \rangle = |\downarrow \rangle , \qquad 0 ~~\hbox{otherwise} ,\\ S_+|\downarrow \rangle = |\uparrow \rangle , \qquad 0 ~~\hbox{otherwise} . $$

The "total" spin operators acting on your triple tensor product state are the triple coproducts, $$ {\cal S}_i=S_i \otimes 1\!\!1 \otimes 1\!\!1 + 1\!\!1 \otimes S_i \otimes 1\!\!1 + 1\!\!1 \otimes 1\!\!1 \otimes S_i, $$ easy to check to satisfy the same Lie algebra as your $S_i$.

So, trivially compute $$ {\cal S}_z |\alpha \rangle = {1\over 2} |\alpha \rangle, $$ as you noticed, since you count the arrows, and each term of that state has the same eigenvalue--not coincidentally.

The quadratic Casimir operator is $$ {\cal S}\cdot {\cal S}= {\cal S}_z {\cal S}_z + {1\over 2} ({\cal S}_+ {\cal S}_- +{\cal S}_- {\cal S}_+) . $$ The first term acts on your state trivially, as above, $$ {\cal S}_z^2 |\alpha \rangle = {1\over 4} |\alpha \rangle, $$ while the 3rd one, the second term inside the parenthesis, kills that state. Do you see it? It has to, since ${\cal S}_+$ would raise the spin and this is not possible if you are to have spin 1/2 doublets and not spin 3/2 quartets!

Equally straightforwardly, see that $$ {\cal S}_- |\alpha \rangle = \left(2 \downarrow \uparrow \downarrow +2 \uparrow \downarrow \downarrow- \downarrow \downarrow \uparrow -\downarrow \uparrow \downarrow - \downarrow \downarrow \uparrow - \uparrow \downarrow \downarrow \right)\\ = \left( \downarrow \uparrow \downarrow + \uparrow \downarrow \downarrow- 2 \downarrow \downarrow \uparrow \right ), $$ a spin down reflection of α, predictably.

Finally, unsurprisingly, $$ {\cal S}_+ {\cal S}_- |\alpha \rangle = |\alpha \rangle ~~~~~~\leadsto ~~~~~ {\cal S}\cdot {\cal S} |\alpha \rangle= {3\over 4} |\alpha \rangle =(1/2) (1+1/2) |\alpha \rangle, $$ as it had no choice but to...

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