Two-level systems · Quantum Mechanics

Intuition

Any system with two relevant states — the ammonia molecule of the dynamics chapter, an atom with two levels, a photon’s polarisation, a superconducting qubit — has a Hamiltonian that is a Hermitian two-by-two matrix. Every such matrix is a number plus $\mathbf{a}\cdot\vec{\sigma}$ . So every two-level system is a spin one-half in an effective magnetic field $\mathbf{a}$ : its energies are the number plus and minus $|\mathbf{a}|$ , and its states precess about $\mathbf{a}$ . What was learned about spin applies to them all.

Many different songs can be played on one guitar. The two-level systems of physics are different songs, but the instrument — a spin in a field — is always the same.

Every two-level system is a spin

A two-level Hamiltonian is a Hermitian $2\times2$ matrix. Writing it through the Pauli matrices gives its energies and eigenstates at once, and its dynamics is precession about the effective field $\mathbf{a}$ .

\begin{aligned}\hat{H}&=a_{0}\hat{1}+\mathbf{a}\cdot\vec{\sigma}=\begin{pmatrix}a_{0}+a_{z}&a_{x}-ia_{y}\\ a_{x}+ia_{y}&a_{0}-a_{z}\end{pmatrix}\\ E_{\pm}&=a_{0}\pm|\mathbf{a}|\end{aligned}

Consequences

The eigenstates are up and down along the direction of $\mathbf{a}$ on the Bloch sphere.
The state precesses about $\mathbf{a}$ at the angular frequency $2|\mathbf{a}|/\hbar$ , the Bohr frequency of the two levels.
Ammonia with $\hat{H} = (\begin{array}{cc} E_{0} & - A \\ - A & E_{0} \end{array})$ is a spin in a field along : energies , and the nitrogen’s flipping is precession.

The energies of a two-level system

\hat{H}=a_{0}\hat{1}+\mathbf{a}\cdot\vec{\sigma}\ \Longrightarrow\ E_{\pm}=a_{0}\pm|\mathbf{a}|

The identity part only adds $a_{0}$ . The rest squares to $|\mathbf{a}|^{2}$ times the identity, so its eigenvalues square to $|\mathbf{a}|^{2}$ ; it has trace zero, so they are and .

Proof steps

(\mathbf{a}\cdot\vec{\sigma})^{2}=\sum_{i,j}a_{i}a_{j}\sigma_{i}\sigma_{j}=|\mathbf{a}|^{2}\,\hat{1}

The cross terms cancel because different Pauli matrices anticommute.

(\mathbf{a}\cdot\vec{\sigma})\ket{v}=\lambda\ket{v}\ \Rightarrow\ \lambda^{2}=|\mathbf{a}|^{2}

Apply the square to an eigenvector.

\operatorname{tr}(\mathbf{a}\cdot\vec{\sigma})=0\ \Rightarrow\ \lambda=\pm|\mathbf{a}|

The two eigenvalues add to zero, so one of each sign.

E_{\pm}=a_{0}\pm|\mathbf{a}|

Add $a_{0}$ , which shifts every eigenvalue.

Applications

Practice

Energies at a Glance

Write a two-level Hamiltonian as $a_{0}+\mathbf{a}\cdot\vec{\sigma}$ . Its energies are .

E_{\pm}=a_{0}\pm|\mathbf{a}|

Try it

What is the higher energy of $\hat{H}=\begin{pmatrix}3&4\\4&-3\end{pmatrix}$ ?