Two-level Hamiltonians

Intuition

Two states that matter and a coupling between them: a molecule of ammonia with its nitrogen above or below, an electron in a double well, an atom with two levels, a superconducting qubit. The Hamiltonian is a Hermitian two-by-two matrix, and up to a constant it is always $\tfrac12(\epsilon\sigma_{z}+\Delta\sigma_{x})$ : a bias $\epsilon$ that makes one state lower than the other, and a coupling $\Delta$ that lets the system pass between them. The energies are $\pm\tfrac12\sqrt{\epsilon^{2}+\Delta^{2}}$ , and the eigenstates are superpositions set by one angle, $\tan\theta=\Delta/\epsilon$ . With no bias the eigenstates are the equal mixtures; with a large bias they are close to the two states themselves.

Two identical pendulums joined by a weak spring swing together or against each other; make one heavier and each normal mode is mostly one pendulum. Bias and coupling play the same roles for two quantum states.

The weights of $\ket{a}$ and $\ket{b}$ in the ground state of $\tfrac12(\epsilon\sigma_{z}+\Delta\sigma_{x})$ , against the bias . With a large negative bias the ground state is nearly , with a large positive bias nearly , and at it is an equal mixture.

In a basis $\ket{a},\ket{b}$ , with the constant part dropped, every two-level Hamiltonian with real couplings can be written

\begin{aligned}\hat{H}&=\tfrac12\bigl(\epsilon\,\sigma_{z}+\Delta\,\sigma_{x}\bigr)=\tfrac12\begin{pmatrix}\epsilon&\Delta\\ \Delta&-\epsilon\end{pmatrix}\\ E_{\pm}&=\pm\tfrac12\sqrt{\epsilon^{2}+\Delta^{2}},\qquad \tan\theta=\frac{\Delta}{\epsilon}\end{aligned}

Properties

$\epsilon$ is the bias, the energy difference of $\ket{a}$ and $\ket{b}$ without coupling; $\Delta$ is the coupling, which lets the system pass between them.
The eigenstates are $∣ E_{+} ⟩ =$ and , with the mixing angle .

Energies and states of a two-level Hamiltonian

\hat{H}=\tfrac12(\epsilon\sigma_{z}+\Delta\sigma_{x})\ \Longrightarrow\ E_{\pm}=\pm\tfrac12\sqrt{\epsilon^{2}+\Delta^{2}}

Square the Hamiltonian: the Pauli matrices square to one and anticommute, so the square is a number times the identity. The Hamiltonian has zero trace, so its two eigenvalues are opposite, and their square is that number. Writing the field as a length times a direction in the $x$ – $z$ plane gives the eigenstates as spin up and down along it.

Proof steps

\hat{H}^{2}=\tfrac14\bigl(\epsilon\sigma_{z}+\Delta\sigma_{x}\bigr)^{2}=\tfrac14\bigl(\epsilon^{2}+\Delta^{2}\bigr)\hat{1}

$\sigma_{z}^{2}=\sigma_{x}^{2}=\hat{1}$ and .

E_{\pm}=\pm\tfrac12\sqrt{\epsilon^{2}+\Delta^{2}}

The trace is zero, so the eigenvalues are opposite; each squares to $\tfrac14(\epsilon^{2}+\Delta^{2})$ .

\begin{aligned}\hat{H}&=\tfrac12\sqrt{\epsilon^{2}+\Delta^{2}}\,\bigl(\cos\theta\,\sigma_{z}+\sin\theta\,\sigma_{x}\bigr)\\ \tan\theta&=\Delta/\epsilon\end{aligned}

A length times a unit direction in the $x$ – $z$ plane.

\begin{aligned}\ket{E_{+}}&=\cos\tfrac{\theta}{2}\ket{a}+\sin\tfrac{\theta}{2}\ket{b}\\ \ket{E_{-}}&=-\sin\tfrac{\theta}{2}\ket{a}+\cos\tfrac{\theta}{2}\ket{b}\end{aligned}

Spin up and down along that direction, as in the spin chapter.

Applications

Practice

The Energies

A bias $\epsilon$ and a coupling $\Delta$ give two levels at plus and minus half of $\sqrt{\epsilon^{2}+\Delta^{2}}$ .

E_{\pm}=\pm\tfrac12\sqrt{\epsilon^{2}+\Delta^{2}}

Try it

A two-level system has bias $\epsilon=3$ and coupling $\Delta=4$ , in some unit of energy. What is its upper energy?