Can the minimum energy state of an Ising-Hamiltonian be negative?

Question

I am simulating 1D many body (N>2) quantum chains, governed by Ising-Hamiltonians. More on the Ising-Hamiltonian.

My simulation generates the Hamiltonian based on the following formula:

$$\hat H = J\sum_i^N{\sigma_i^z\sigma_{i+1}^z} + \sum_i^{N-1}{\lambda(t)\sigma_i^x} $$

I am fairly certain I am generating the Hamiltonian correctly. As expected, however, extracting the eigenvalues from the Hamiltonian gives us a vector (in MATLAB) with all of the energy states of the system. To find the minimum energy state I am supposed to examine the minimum eigenvalue.

My question is: the eigenvalues often contain negative values. Do I ignore these? Take the absolute value of them? Can the minimum energy be negative?

Thanks for your time.

Answer 1

Yes the minimum energy should be negative. No you shouldn't ignore the negative eigenvalues.

Take the negative eigenvector of $\sigma^x$ at each lattice site (assuming $\lambda$ is positive, otherwise take the positive eigenvector) and call this state $|\psi\rangle$. The expected energy of $|\psi\rangle$ is $$\langle\psi|\hat{H}|\psi\rangle=0+\langle\psi|\sum_{i=0}^{N-1}{\lambda\sigma_i^x}|\psi\rangle=N\lambda \left(-1/2\right)\leq0$$

The ground state has an energy less than this, so the ground state has negative energy. You could do this same variational argument using a trial state that is an eigenvector of the first term in the Hamiltonian instead if you want, you'll again see it has to be negative.