The energy for nonzero total spin of 1-dimensional XY model

Question

I want to compute the energies and eigenstates for non-zero total spin of the 1-dimensional XY model.

The Hamiltonian for the 1-dimensional XY model is given by: \begin{align*} H = -J \sum_{i=1}^{N} (S_i^x S_{i+1}^x + S_i^y S_{i+1}^y) \end{align*} where $ S_i^x $ and $ S_i^y $ are the spin-1/2 components at site $ i $. We assume $N$ is even and periodic boundary conditions ($ S_{N+1} = S_1 $).

For convenience, We often transform the spin operators into fermion operators. We use the Jordan-Wigner transformation. This transforms the Hamiltonian into: \begin{align*} H = -\frac{J}{2} \sum_{i=1}^{N} (c_i^\dagger c_{i+1} + c_{i+1}^\dagger c_i) \end{align*} where $ c_i $ and $ c_i^\dagger $ are fermion annihilation and creation operators, respectively.

The total spin $S^z_{tot}$ is related with the fermion number $M=\sum_i c_i^\dagger c_{i}$ as \begin{align*} S^z_{tot}=M-\frac{N}{2}. \end{align*} The ground state is given by the $S^z=0$ sector and this corresponds to the half-filled state $M=N/2$ in the language of fermions. The ground state energy is given by $ E_{\text{ground}} = -J \sum_{k \in \text{half-filled}} \cos(k) $.

What I want to find is the (minimum) energy and eigenstate for non-zero $S^z$ sectors. As an example, we consider the $N=4$ case. For $S^z_{tot}=2\ (M=4)$, the state is $|M=4\rangle=|1111\rangle$ and the energy is 0 where $|1\rangle /|0\rangle$ is the fermions' occupied/unoccupied state. For $S^z_{tot}=1\ (M=3)$, I think the eigenstate is given by: \begin{align*} |M=3\rangle=\frac{1}{2}\left[|0111\rangle + |1011 \rangle+| 1101 \rangle + | 1110 \rangle \right] \end{align*} and the energy is $-J/2$. For $S^z_{tot}=0\ (M=2)$, the energy is $-J/\sqrt{2}$. How do you find the energy and eigenstate in general?

Answer 1

The fermionic Hamiltonian \begin{align} H = −\frac{J}{2} \sum _{i=1}^N (c^\dagger _i c_{i+1} + c^\dagger _{i+1} c_i) \end{align} can be immediately diagonalized by going into momentum space, \begin{align} c_j = \frac{1}{\sqrt{L}} \sum _k e^{ikj} c_k, \end{align} which then leads to the diagonal Hamiltonian \begin{align} H = \sum _k \varepsilon _k c^\dagger _k c_k, \end{align} with eigenenergy \begin{align} \varepsilon _k = - J \cos (k). \end{align} Then one can in principle obtain all the eigenstates and corresponding eigenenergies by adding to the fermionic vacuum the momentum fermion operators (the vacuum is $|0\rangle$ such that $c_q |0\rangle = 0$ $\forall q$). Then if you are interested in the unnormalized state with a given number of fermions $M = |\mathcal{K}|$ you construct them as \begin{align} | \psi _\mathcal{K} \rangle = \prod _{k \in \mathcal{K}} c^\dagger _k | 0 \rangle , \end{align} where the energy is given by \begin{align} E_\mathcal{K} = \frac{\langle \psi _\mathcal{K} | H | \psi _\mathcal{K} \rangle}{ \langle \psi _\mathcal{K} | \psi _\mathcal{K} \rangle } = \sum _{k \in \mathcal{K}} \varepsilon _k . \end{align} In other words if you want to find the state with the smallest amount of energy for $M$ fermion you just occupy the $M$ lowest eigenmodes of the system.

I want to point out that there are some caveats to translating the spins to fermions using the Jordan-Wigner transformation. For an overview of dealing with the XY chain rigorously I can recommend the beginner-friendly notes: The quantum Ising chain for beginners. In short, the XY model with open boundary conditions is not exactly $H = −\frac{J}{2} \sum _{i=1}^N (c^\dagger _i c_{i+1} + c^\dagger _{i+1} c_i)$, but a slightly modified version with boundary conditions for the fermions depending on the parity of fermions in the system (for fixed boundary conditions for the spins).