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Two-State Hamiltonians

A two-state Hamiltonian is a Hermitian two-by-two matrix acting on a two-dimensional Hilbert space. Its diagonalization is the reusable calculation behind level splitting, coherent oscillation, avoided crossings, spin precession, and many qubit models.

In a chosen orthonormal basis {∣1⟩,∣2⟩}\{\lvert1\rangle,\lvert2\rangle\}, the general closed-system Hamiltonian is

H=(E1ΔΔ∗E2).H= \begin{pmatrix} E_1 & \Delta\\ \Delta^* & E_2 \end{pmatrix}.

Here E1E_1 and E2E_2 are real, while Δ\Delta may be complex. The off-diagonal term mixes the basis states.

Separate the average energy

c0=E1+E22c_0=\frac{E_1+E_2}{2}

from the traceless part. Define

ϵ=E1−E22.\epsilon=\frac{E_1-E_2}{2}.

Then

H=c0I+(ϵΔΔ∗−ϵ).H=c_0I+ \begin{pmatrix} \epsilon & \Delta\\ \Delta^* & -\epsilon \end{pmatrix}.

The term c0Ic_0I shifts both energies by the same amount. It contributes an overall phase to time evolution and usually does not affect transition probabilities.

Using Pauli matrices,

H=c0I+b⋅σ,H=c_0I+\mathbf b\cdot\boldsymbol\sigma,

with

b=(Re⁡Δ, −Im⁡Δ, ϵ).\mathbf b =\left( \operatorname{Re}\Delta,\, -\operatorname{Im}\Delta,\, \epsilon \right).

This representation turns a two-state Hamiltonian into the geometry of an effective vector b\mathbf b. Its magnitude is

Ω=∣b∣=ϵ2+∣Δ∣2.\Omega=\lvert\mathbf b\rvert =\sqrt{\epsilon^2+\lvert\Delta\rvert^2}.

The energies are

E±=c0±Ω.E_\pm=c_0\pm\Omega.

Thus the energy splitting is

E+−E−=2Ω.E_+-E_-=2\Omega.

If Δ\Delta is complex, a basis phase choice can often make it real and nonnegative for a single static two-level Hamiltonian. Assume that has been done, so Δ≥0\Delta\ge0.

Define a mixing angle θ\theta by

tan⁡(2θ)=Δϵ.\tan(2\theta)=\frac{\Delta}{\epsilon}.

Equivalently,

cos⁡(2θ)=ϵΩ,sin⁡(2θ)=ΔΩ.\cos(2\theta)=\frac{\epsilon}{\Omega}, \qquad \sin(2\theta)=\frac{\Delta}{\Omega}.

One convenient choice of normalized eigenstates is

∣+⟩=cos⁡θ ∣1⟩+sin⁡θ ∣2⟩,\lvert+\rangle =\cos\theta\,\lvert1\rangle +\sin\theta\,\lvert2\rangle,

and

∣−⟩=−sin⁡θ ∣1⟩+cos⁡θ ∣2⟩.\lvert-\rangle =-\sin\theta\,\lvert1\rangle +\cos\theta\,\lvert2\rangle.

When Δ=0\Delta=0, the basis states are already energy eigenstates. When ϵ=0\epsilon=0, the mixing is maximal:

θ=π4.\theta=\frac{\pi}{4}.

For time-independent HH, suppose the system starts in ∣1⟩\lvert1\rangle. After removing the physically irrelevant overall phase from c0Ic_0I, the probability to find the system in ∣2⟩\lvert2\rangle is

P1→2(t)=∣Δ∣2Ω2sin⁡2(Ωtℏ).P_{1\to2}(t) =\frac{\lvert\Delta\rvert^2}{\Omega^2} \sin^2\left(\frac{\Omega t}{\hbar}\right).

This formula contains two important effects:

  • the oscillation frequency is set by Ω/ℏ\Omega/\hbar;
  • detuning suppresses the maximum transition probability.

On resonance, ϵ=0\epsilon=0, so Ω=∣Δ∣\Omega=\lvert\Delta\rvert and the oscillation can reach unit probability:

P1→2(t)=sin⁡2(∣Δ∣tℏ).P_{1\to2}(t) =\sin^2\left(\frac{\lvert\Delta\rvert t}{\hbar}\right).

Far off resonance, ∣ϵ∣≫∣Δ∣\lvert\epsilon\rvert\gg\lvert\Delta\rvert, the maximum transition probability is small.

Suppose a control parameter changes the detuning ϵ\epsilon, while Δ\Delta stays nonzero. The two energies are

E±(ϵ)=c0±ϵ2+∣Δ∣2.E_\pm(\epsilon) =c_0\pm\sqrt{\epsilon^2+\lvert\Delta\rvert^2}.

At ϵ=0\epsilon=0, the minimum gap is

E+−E−=2∣Δ∣.E_+-E_-=2\lvert\Delta\rvert.

If Δ=0\Delta=0, the levels cross. If Δ≠0\Delta\ne0, they repel and form an avoided crossing. This is the standard two-level explanation of why coupling turns a crossing of bare basis energies into a nonzero spectral gap.

Time-dependent passage through an avoided crossing is introduced in Landau–Zener Problem: First Encounter, with the advanced treatment in Landau–Zener Transition.

The traceless Hamiltonian b⋅σ\mathbf b\cdot\boldsymbol\sigma generates rotations of the state vector on the Bloch sphere. The direction of b\mathbf b sets the rotation axis, and its magnitude sets the angular frequency.

The wave-mechanics visualization is developed in Bloch Sphere: Wave-Mechanics Perspective. The spinor interpretation is developed in Bloch Sphere. This page keeps the Hamiltonian algebra as the canonical wave-mechanics first encounter.

  • Forgetting that the off-diagonal element can be complex while the Hamiltonian remains Hermitian.
  • Treating E1E_1 and E2E_2 as exact energies when Δ≠0\Delta\ne0.
  • Keeping the common shift c0Ic_0I in transition-probability calculations where it cancels.
  • Confusing detuning ϵ\epsilon with the full energy splitting 2Ω2\Omega.
  • Assuming an avoided crossing occurs without coupling.
  • Forgetting that phase conventions can change the apparent phase of Δ\Delta.
  • J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
  • R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
  • D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
  • L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
  1. Derive the eigenvalues of the general two-state Hamiltonian
H=(E1ΔΔ∗E2).H= \begin{pmatrix} E_1 & \Delta\\ \Delta^* & E_2 \end{pmatrix}.
Solution

Write

H=c0I+(ϵΔΔ∗−ϵ),H=c_0I+ \begin{pmatrix} \epsilon & \Delta\\ \Delta^* & -\epsilon \end{pmatrix},

where c0=(E1+E2)/2c_0=(E_1+E_2)/2 and ϵ=(E1−E2)/2\epsilon=(E_1-E_2)/2. The traceless part has determinant

−ϵ2−∣Δ∣2,-\epsilon^2-\lvert\Delta\rvert^2,

and trace zero. Its eigenvalues are therefore

±ϵ2+∣Δ∣2.\pm\sqrt{\epsilon^2+\lvert\Delta\rvert^2}.

Adding back c0c_0 gives

E±=c0±ϵ2+∣Δ∣2.E_\pm=c_0\pm\sqrt{\epsilon^2+\lvert\Delta\rvert^2}.
  1. For ϵ=0\epsilon=0, show that an initial ∣1⟩\lvert1\rangle state reaches ∣2⟩\lvert2\rangle with unit probability.
Solution

When ϵ=0\epsilon=0,

Ω=∣Δ∣.\Omega=\lvert\Delta\rvert.

The transition probability formula becomes

P1→2(t)=sin⁡2(∣Δ∣tℏ).P_{1\to2}(t) =\sin^2\left(\frac{\lvert\Delta\rvert t}{\hbar}\right).

At

t=πℏ2∣Δ∣,t=\frac{\pi\hbar}{2\lvert\Delta\rvert},

the sine equals 11, so P1→2=1P_{1\to2}=1.

  1. What is the minimum gap in an avoided crossing with energies E±=c0±ϵ2+∣Δ∣2E_\pm=c_0\pm\sqrt{\epsilon^2+\lvert\Delta\rvert^2}?
Solution

The gap is

E+−E−=2ϵ2+∣Δ∣2.E_+-E_- =2\sqrt{\epsilon^2+\lvert\Delta\rvert^2}.

This is minimized at ϵ=0\epsilon=0, giving

ΔEmin=2∣Δ∣.\Delta E_{\mathrm{min}}=2\lvert\Delta\rvert.