# Finding the Eigenvalues and Eigenvectors of the Hamiltonian for three spin-1/2 particles coupled antiferromagnetically

## Problem

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 $$S_i^2$$.

## Work

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 $$\vec{S}_i \cdot \vec{S}_{i+1} = \frac{1}{2}\Big[\Big(\vec{S}_i + \vec{S}_{i+1}\Big)^2 - \Big(\vec{S}_i^2 + \vec{S}_{i+1}^2\Big) \Big].$$

## Issue

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.

• 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. – my2cts Oct 30 '18 at 20:47
• 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 – ZeroTheHero Oct 30 '18 at 21:55
• 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 – Grant Cates Oct 31 '18 at 7:00
• physicsforums.com/threads/… – Boris Valderrama Oct 11 '20 at 21:19

$$S_{1}S_{2}+S_{2}S_{3}+S_{3}S_{1}=\frac{1}{2}\left(S^{2}-S_{1}^{2}-S_{2}^{2}-S_{3}^{2}\right)$$
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.