Given three spin-1/2 particles with the total spin operator $\vec{S}=\sum\limits_{i=1}^3 \vec{S}_i$ and its $z$ projection $S_z=\sum\limits_{i=1}^3 S_{z,i}$, and the Hamiltonian

$$H = J\sum\limits_{i=1}^3 \vec{S}_i \cdot \vec{S}_{i+1} $$ (assuming for $i=3$ that $i+1=1$), calculate the eigenstates and the eigenenergies.

Hint: Rewrite $H$ as a function of $S^2$ and $S_i^2$.


I've already calculated the basis for $\vec{S}^2$ and $S_z$

$$ \vert 3/2,3/2\rangle \equiv \vert\uparrow\uparrow\uparrow \rangle \\ \vert3/2,1/2\rangle \equiv \frac{1}{\sqrt{3}}\Big( \vert\downarrow\uparrow\uparrow \rangle + \vert\uparrow\uparrow\downarrow \rangle + \vert\uparrow\downarrow\uparrow \rangle \Big)\\ \vert3/2,-1/2\rangle \equiv \sqrt{\frac{2}{3}}\Big( \vert\downarrow\downarrow\uparrow \rangle + \vert\downarrow\uparrow\downarrow \rangle + \vert\uparrow\downarrow\downarrow \rangle \Big)\\ |3/2,-3/2\rangle \equiv \vert\downarrow\downarrow\downarrow \rangle \\ $$ with eigenvalues according to

$$S^2\vert s,m \rangle = \hbar^2s(s+1)\vert s,m \rangle = \frac{15\hbar^2}{4}\vert 3/2,m \rangle \\ S_z\vert s,m \rangle = \hbar m\vert 3/2,m \rangle \text{ for } m=-\frac{3}{2},-\frac{1}{2},\frac{1}{2}\frac{3}{2}. $$

I'm now attempting to rewrite the Hamiltonian according to the hint, using $$\vec{S}_i \cdot \vec{S}_{i+1} = \frac{1}{2}\Big[\Big(\vec{S}_i + \vec{S}_{i+1}\Big)^2 - \Big(\vec{S}_i^2 + \vec{S}_{i+1}^2\Big) \Big].$$


I'm not certain I'm applying the hint correctly. With the above, $$H = \frac{J}{2}\Big[ \Big(\vec{S}_1 + \vec{S}_2 \Big)^2 + \Big(\vec{S}_1 + \vec{S}_3 \Big)^2 + \Big(\vec{S}_2 + \vec{S}_3 \Big)^2 -2\Big( \vec{S}_1^2 + \vec{S}_2^2 +\vec{S}_3^2\Big) \Big],$$

which once expanded and using the expansion of $\vec{S}^2 = \Big(\vec{S}_1 + \vec{S}_2 +\vec{S}_3\Big)^2$ gets me to

$$H = J\sum\limits_{i=1}^2 S_i^2,$$

which just seems wrong to me, since it's not written as a function of $S^2$ and $S_i^2$, as the hint suggests.

  • $\begingroup$ This is a frustration problem. Magnetic frustration that is! You can construct all spin states by starting from the easiest one and applying spin operators to it. $\endgroup$
    – my2cts
    Oct 30, 2018 at 20:47
  • $\begingroup$ Note that your $\vert 3/2,-1/2\rangle$ state is certainly incorrect as it is obviously not normalized and should be closely related to the $\vert 3/2,1/2\rangle$ state $\endgroup$ Oct 30, 2018 at 21:55
  • $\begingroup$ I just realized the normalization issue when I was looking over my work this morning. I'll go back through and see what might have gone wrong. My instinct tells me it should have the same prefctor as the $|3/2,1/2\rangle$ state $\endgroup$ Oct 31, 2018 at 7:00
  • $\begingroup$ physicsforums.com/threads/… $\endgroup$ Oct 11, 2020 at 21:19

2 Answers 2


Note that


and that's what you need. Your calculation is almost correct (your last Hamiltonian is wrong), but longer than it should be.


Just for fun, there's a second way to approach this problem. A generic basis state is of the form $| a_1 a_2 a_3 \rangle$ where $a_i$ is an up- or down-arrow. Let $P$ be the operator that exchanges the different spins: $P| a_1 a_2 a_3 \rangle = | a_2 a_3 a_1 \rangle$, so it's like a discrete momentum. It's easy to write down all of the eigenstates of $P$ (there are three possible eigenvalues). Now notice that $H$ commutes with $P$. This means that $H$ only mixes states that can have the same eigenvalue of $P$, so it's a much "smaller" problem to diagonalize $H$ in this basis.

The above technique is really common in condensed matter physics, so it's cool to master (Google the term "magnon").


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