Two state system

A two state system is the simplest to exhibit quantum behavior. For such a system, we can write the Hamiltonian as a 2 by 2 matrix in some basis $\{ \ket 1, \ket 2 \}$

Here $\braket{1 | H | 1} = E_1$ is the energy of being in state $\ket 1$, $E_2$ is the energy of being in state $\ket 2$, and $\Delta$ is the energy associated with transitioning between the two states (i.e. tunneling). In some basis, $H$ will be diagonal, but this is often not the most natural basis to write it in.

To find the eigenstates and diagonalize $H$ we can take a similar approach to what we did with spin. We defined the three Pauli matrices $\sigma_{xyz}$ and the spin operators $S_{xyz}$, and then used these to define a spin operator $S_{\vec n} = n_x S_x + n_y S_y + n_z S_z$ in some arbitrary direction $\vec n = (n_x, n_y, n_z)$. We found the eigenvalues were $\pm \frac\hbar2$ and found the eigenvectors.

Let’s write $H$ in this form

We can write the $g_n$ constants in terms of $E_1,E_2$ and $\Delta$

The last term of the new expression $g \, \vec n \cdot \vec \sigma$ is like the spin operator $S_{\vec n}$, but with a $g$ instead of $\frac\hbar2$. We can use the same eigenvectors and eigenvalues we already found for $S_n$ here. The $g_0 \hat 1$ term is just the identity, and so will add $g_0$ to each eigenvalue without affecting the eigenvector.

We write the eigenstates as

The eigenvalues $g_0 \pm g$ represent a ground state and an excited state. When $\Delta = 0$, $g_1$ and $g_2$ disappear, and the eigenvectors become $(1,0)$ and $(0,1)$ as expected, with eigenvalues $E_1$ and $E_2$.