# Ground state degeneracy of $\hat{H}= \sum_{i=1}^{L-1} \mathbf{\hat{S}}_i\cdot \mathbf{\hat{S}}_{i+1}$, given a possible ground state

Consider $$\hat{H}= \sum_{i=1}^{L-1} \mathbf{\hat{S}}_i\cdot \mathbf{\hat{S}}_{i+1}$$ where $$\mathbf{\hat{S}}_i\cdot \mathbf{\hat{S}}_{i+1} = 1_1\otimes\cdots\otimes 1_{i-1}\otimes \mathbf{\hat{S}}_i\otimes \mathbf{\hat{S}}_{i+1} \otimes 1_{i+2}\otimes\cdots\otimes 1_L.$$ Suppose that $$\mid\uparrow\rangle ^{\otimes^L}$$ (all spin ups) is one of the possible ground state configurations. What can be said about the ground state degeneracy?

My attempt: First, I considered the simple case of $$L=2$$; then $$\hat{H} = \mathbf{\hat{S}}_1\cdot \mathbf{\hat{S}}_{2} = S_{1x}\otimes S_{2x}+S_{1y}\otimes S_{2y}+S_{1z}\otimes S_{2z}$$ has eigenvalues $$-3\hbar^2/4$$ (multiplicity 1) and $$\hbar^2/4$$ (multiplicity 3).

Now we're given that $$\mid \uparrow\uparrow\rangle = \begin{pmatrix} 1\\0\\0\\0\end{pmatrix}$$ describes a possible ground state. It is an eigenvector of $$\hat{H}$$, with corresponding eigenvalue $$\hbar^2/4$$ (can be checked manually). So, the ground state is 3-fold degenerate. The state $$\mid\uparrow\rangle ^{\otimes^L}$$ can be represented as $$(1\,0\,\dots\,0\,0)$$, conaining $$L^2$$ components. I can see how the first column of the matrix of $$\hat{H}$$ would be this unit vector as well. I need to find the eigenvalues of $$\hat{H}$$ and their multiplicities. Since $$\mathbf{\hat{S}}_i\cdot \mathbf{\hat{S}}_{i+1}$$ will always contains the factor $$\hbar^2/4$$, I would essentially have to show that this is indeed an eigenvalue of $$\hat{H}$$ and I need to find its multiplicity. I don't really know how I would go about finding its multiplicity though. Any hints for this problem?

• For L=2, you found the ground state is the unique antisymmetric singlet, no? Jan 16 '21 at 15:35
• Looks like H should have a minus before the sum, otherwise |up,up> is not a ground state Jan 16 '21 at 16:01
• It could be with an minus sign indeed. I found this problem online a few days ago, but can't find it back to check the exact expression. Jan 16 '21 at 16:09
• I'm not really looking for a complete solution, I would just like to know how to handle these kinds of problems: you're given a Hamiltonian and a possible ground state, what can be said about the ground state degeneracy? Jan 16 '21 at 16:10
• The ground state has $L$ spins up. It is a symmetric state and therefore andeigenstate of $J_{\rm tot}$ with eigenvalue $m= L/2$. It is therefore the highest spin in an $j=L/2$ multiplet with dimension $2j+1$. So the degeracy is? Jan 16 '21 at 18:46