# Rabi Hamiltonian with three instead of two Pauli matrices

This question was motivated by a question on Mathoverflow, in which a Hamiltonian is considered that looks like the Rabi Hamiltonian, but with three instead of two Pauli matrices: $$H=\omega a^\dagger a+\Delta\sigma_z+g_x\sigma_x(a+a^\dagger)+ig_y\sigma_y(a-a^\dagger).$$ In the literature I have only found the Rabi Hamiltonian with $$g_y=0$$, so with two Pauli matrices. The case $$g_y\neq 0$$ with three Pauli matrices seems essentially different, at least if $$g_x\neq g_y$$ and one does not make the rotating wave approximation. Has it been considered before?

Since this model is so fundamental, I am a bit surprised at not having found this generalization, I may be missing something obvious.

• If you start with a general Hamiltonian for an atom coupled to an electromagnetic field, you may come up with sll kinda of terms. It gets even worse, when one deals with solid state, quantum dots, etc. Sep 4, 2021 at 19:09

In my answer to the Mathoverflow question I made a conjecture for the ground state energy in the special case that $$\Delta=2g_xg_y$$:
The generalised Rabi Hamiltonian $$H_0=\begin{pmatrix} a^\dagger a+\alpha^2&\alpha a+\beta a^\dagger\\ \alpha a^\dagger+ \beta a&a^\dagger a+ \beta^2 \end{pmatrix}$$ has a doubly degenerate ground state at zero energy for any $$\alpha,\beta\in\mathbb{R}$$.
I now see that this conjecture follows from the Bethe Ansatz solution of the generalized Rabi Hamiltonian given in the Tomka et al. paper, section II.B.1. The case $$\Delta=2g_xg_y$$ corresponds to $$\lambda_-=\delta$$ in their notation, which is the "exceptional case"$$^\ast$$ in which the Rabi Hamiltonian has an eigenvalue $$E_0=-\lambda_+=-(g_x^2+g_y^2)$$. This corresponds to a zero eigenvalue for $$H_0$$, because the spectrum of $$H_0$$ is shifted relative to the Rabi spectrum by $$(\alpha^2+\beta^2)/2=g_x^2+g_y^2$$.
The zero eigenvalue is the ground state, because $$H_0$$ is positive definite, $$H_0=Q^\dagger Q,\;\;\text{for}\;\;Q=\begin{pmatrix} \alpha&a\\ a&\beta \end{pmatrix}.$$ The two ground state wave functions, such that $$Q\Psi_\pm=0$$, are $$\Psi_\pm={{\sqrt\beta}\choose{\mp\sqrt\alpha}}|\!\pm\!\!\sqrt{\alpha\beta}\rangle$$, in terms of the coherent state $$|\gamma\rangle$$ with $$a|\gamma\rangle=\gamma|\gamma\rangle$$.
$$^\ast$$ In arXiv:1407.5213 the symmetry is identified that enforces the doubly degenerate ground state.