I've come across a problem involving a system with three spin-1/2 particles in a given state, for which the total (spin) angular momentum can be calculated using the $\hat{S}^2$ operator in the representation $$\hat{S}^2 = \hat{S}_-\hat{S}_+ + \hbar\hat{S}_z + \hat{S}^2_z$$ The three particles are in the state $$|\psi\rangle = \frac{1}{\sqrt{6}}\left(-2|\downarrow\downarrow\uparrow\rangle + |\downarrow\uparrow\downarrow\rangle + |\uparrow\downarrow\downarrow\rangle\right)$$ I am seeing that the solution to this gives a total spin angular momentum of $1/2\hbar$, but working through the problem myself does not give me the same result. I can see that applying each operator to the state gives a value corresponding to $S(S+1)\hbar$ which should then give a value for $S$ but, for example, the worked solution I have gives the result of the $\hbar\hat{S}_z$ operator on the state to be $-1/2\hbar^2$, which is not what I'm seeing.

How do I operate the constituents of $\hat{S}^2$ on this state?

  • $\begingroup$ Which book are you using? For one, $S^2$ and $S_z$ don't operate on the system "nicely," and won't give the result $\hbar s(s+1)$. You can apply the total angular momentum operator after using the Clebsch–Gordan coefficients to decompose your coupled states. (or couple your decomposed states? I'm not sure how the wording goes.) $\endgroup$
    – cxx
    Jan 23, 2019 at 20:11

1 Answer 1


Consider each of the states in the superposition. Each of them is an eigenvector of the collective operator $$S_z = S_z^{(1)}\otimes 1\otimes 1 + 1\otimes S_z^{(2)}\otimes 1+ 1\otimes 1\otimes S_z^{(3)} $$ since they each have two spins down and one spin up. Thus, they have a net eigenvalue of spin $1/2$ down. Explicitly, for the third element (taking $\hbar = 1$), $$ S_z |\uparrow\downarrow\downarrow\rangle = \frac12|\uparrow\downarrow\downarrow\rangle + (-\frac12)|\uparrow\downarrow\downarrow\rangle + (-\frac12)|\uparrow\downarrow\downarrow\rangle = -\frac12 |\uparrow\downarrow\downarrow\rangle.$$

  • $\begingroup$ That makes sense but does the factor (-2) at the first element not give a result of 0 overall? i.e. -2(-1/2) -1/2 -1/2 = 0 $\endgroup$ Jan 23, 2019 at 15:32
  • 1
    $\begingroup$ No, because each of those coefficients is multiplying a ket (in the case of the -2 it is the $|\uparrow \downarrow \downarrow\rangle$ ket. You then factor out the factor of 1/2 and see that the entire combination of vectors is an eigenvector with eigenvalue 1/2, which tells you what the total $S_z$ is. $\endgroup$ Jan 26, 2019 at 21:39

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