# Physical Realization of Three-Level System

I have come across the Hamiltonian (where $\varepsilon,\Delta\geq0$) in one of my problem sets:

$$H= \left(\begin{array}{c c c} 0&0&0\\ 0&\varepsilon-\Delta&0\\ 0&0&\varepsilon+\Delta\\ \end{array}\right),$$

Which has a very interesting statistical mechanics, showing a quantum phase transition at $\Delta=\varepsilon$ for the parameter

$$M=\frac{2\sinh(\beta\Delta)}{e^{\beta\varepsilon}+2\cosh(\beta\Delta)}.$$

But the thing is, I only thought to compute that parameter because of the following physical argument.

I thought of this Hamiltonian as a model for the following system: some particle, in the absense of a magnetic field, can have three states, a ground state with zero energy and two excited states with energy $\varepsilon$. The excited state degeneracy is due to the (possibly unphysical) half-integer spin of the particle, which is nonexistant in the ground state and only manifests itself on the excited state.

As such, in the excited state we can introduce a magnetic field, which would give rise to an energy difference $2\Delta$ between the once degenerate excited states. In this physical system, $M$ is proportional to the expected magnetization of the system.

The problem is, I don't think such a system exists, one that has no spin in a ground state and half-integer spin in the excited state.

Does it exist after all? And if not, is there a system that actually is modeled by that Hamiltonian and gives us a physical, measurable interpretation of the $M$ parameter?

• what is the phase transition that you are modelling? – wcc Aug 16 '18 at 4:48
• Why not just talk about a spin-1 particle? – Ryan Thorngren Aug 16 '18 at 7:02
• @IamAStudent At zero temperature ($\beta\rightarrow\infty$) the magnetization can have two different limits. Zero if $\Delta<\varepsilon$ and 1 if $\Delta>\varepsilon.$ The zero temperature magnetization is discontinuous at $\Delta=\varepsilon$, a second order transition. – Gabriel Golfetti Aug 16 '18 at 11:09
• @RyanThorngren an ordinary spin 1 particle would have three energy eigenstates in a magnetic field, for sure. However they would be equally distributed at energies $-\Delta, 0,\Delta$, not what I'm looking for. – Gabriel Golfetti Aug 16 '18 at 11:11
• So why would the magnetization you described be interesting? FYI, in practice experimenters optically pump the NV centers to polarize them into the $|m_s = 0\rangle$ state. – wcc Aug 16 '18 at 20:45

Nitrogen vacancy (NV) centers in diamond has a ground state structure with $|m_s = 0\rangle$ and $|m_s = \pm 1\rangle$ that are ~ $h \cdot 2.87\,\text{GHz}$ above the $|m_s = 0\rangle$. The zero-field splitting is from spin-spin interaction. I grabbed the numbers from this paper. So the total spin is not 1/2 as you desired but at least it has the structure of zero-field splitting.