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The hamiltonian of a collection of noninteracting quantum spin-$1/2$ operators $\hat{S}_i$ fixed at each lattice site $i(=1,2,..., N)$ in presence of magnetic field ${\bf B}=B\hat{{\bf z}}$ $$\hat{H}=-B\sum\limits_{i=1}^{N}\hat{S}_i^z.\tag{1}$$ Clearly, since the spins are noninteracting, the quantum mechanical ground state of $\hat{H}$ is a simple product $$|0\rangle=\prod\limits_{i=1}^{N}|\uparrow_z\rangle_i.\tag{2}$$

Now consider the quantum Ising model, described by the hamiltonian, $$\hat{H}^\prime=-J\sum\limits_{\langle i,j\rangle}\hat{S}_i^z\hat{S}_j^z,~~(J>0)\tag{3}$$ Since in this case there are (nearest-neighbour) interactions, the Hamiltonian $\hat{H}^\prime$ is not a sum of one-particle hamiltonians, and hence, one does not expect the ground state to be a simple product of the form $(2)$.

Question How does one find the exact ground state wavefunction of the hamiltonian $\hat{H}^\prime$ (in Eq.$(3)$) and what does it look like?

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  • $\begingroup$ Are they spin-1/2 operators? $\endgroup$ – Clara Diaz Sanchez May 8 '20 at 17:18
  • $\begingroup$ Yes, added. Thanks! $\endgroup$ – SRS May 8 '20 at 17:20
  • $\begingroup$ This problem doesn't have an easy general solution unless you somehow restrict the graph of neighbors. I would expect in dimensions 1 and 2 in a square lattice probably enough is known; but if you want to go beyond that you either have to use computational methods or a simplification like a mean field approximation. $\endgroup$ – Ege Erdil May 8 '20 at 17:28
  • $\begingroup$ This looks like the usual ferromagnet. The ground state is a manifold $\sim SU(2)/U(1)$, which also contains the state (2) in your text... $\endgroup$ – Vivek May 8 '20 at 17:41
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The system is a 1D lattice of spin-half particles. And the Hamiltonian considers nearest neighbour interaction with a coupling that favours alignment. Intuitively we can see that the ground state at temperature $T=0$ is when all of the spins are aligned (along z).

Firstly we need to find the eigenstates of the system. At each lattice point, we can have a spin. In general a superposition of up and down. And the total state is a direct product of all these spins. But the form of the Hamiltonian restricts the form of the eigenstates to be: each neighbouring pairs being up or down. Then all possible eigenstates are just binary strings of the form up/down of length equalling the number of sites.

It’s easy to see that this state (all aligned along z) is an eigenstate of the Hamiltonian. All we have to show is that the this is the least energy configuration. Consider a lattice with all spins pointing up except one pointing down at site-k. This state is an eigenstate of the Hamiltonian and has an excess energy $J/2$ as compared to when all spins are aligned.

Further notice that the lattice with the highest energy is the one where the neighbours are anti parallel (evens up and odds down).

Thus the ground state is the direct product of all ups or all downs or any superposition of the two.

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For positive J the solution is a product of parallel spins. It has spin S=N/2 and N+1 degeneracy.

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  • $\begingroup$ J is positive. So alternating spins would be the highest energy state. $\endgroup$ – Superfast Jellyfish May 8 '20 at 20:17
  • $\begingroup$ Oops, true. It simplifies the answer. I just edited it. $\endgroup$ – my2cts May 8 '20 at 20:22

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