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?

  • $\begingroup$ For L=2, you found the ground state is the unique antisymmetric singlet, no? $\endgroup$ Jan 16 '21 at 15:35
  • $\begingroup$ Looks like H should have a minus before the sum, otherwise |up,up> is not a ground state $\endgroup$ Jan 16 '21 at 16:01
  • $\begingroup$ 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. $\endgroup$
    – Zachary
    Jan 16 '21 at 16:09
  • $\begingroup$ 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? $\endgroup$
    – Zachary
    Jan 16 '21 at 16:10
  • 2
    $\begingroup$ 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? $\endgroup$
    – mike stone
    Jan 16 '21 at 18:46

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