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 $\{\lvert1\rangle,\lvert2\rangle\}$, the general closed-system Hamiltonian is

$$H= \begin{pmatrix} E_1 & \Delta\\ \Delta^* & E_2 \end{pmatrix}.$$

Here $E_1$ and $E_2$ are real, while $\Delta$ may be complex. The off-diagonal term mixes the basis states.

Helpful background. Two-Level Systems introduces the physical basis and the validity of a two-state description; Pauli matrices supplies the matrix algebra used below.

Average Energy And Detuning

Separate the average energy

$$c_0=\frac{E_1+E_2}{2}$$

from the traceless part. Define

$$\epsilon=\frac{E_1-E_2}{2}.$$

Then

$$H=c_0I+ \begin{pmatrix} \epsilon & \Delta\\ \Delta^* & -\epsilon \end{pmatrix}.$$

The term $c_0I$ shifts both energies by the same amount. Within this isolated two-state system it contributes only an overall phase, which cancels from transition probabilities. Retain it when comparing absolute energies.

Pauli-Matrix Form

Using Pauli matrices,

$$H=c_0I+\mathbf b\cdot\boldsymbol\sigma,$$

with

$$\mathbf b =\left( \operatorname{Re}\Delta,\, -\operatorname{Im}\Delta,\, \epsilon \right).$$

The minus sign in $b_y=-\operatorname{Im}\Delta$ follows from the standard convention $(\sigma_y)_{12}=-i$. Conversely,

$$E_1=c_0+b_z,\qquad E_2=c_0-b_z,\qquad \Delta=b_x-ib_y.$$

The components depend on the chosen basis. Under a unitary basis change the Pauli coordinates rotate, while $c_0$, $|\mathbf b|$, and probabilities between specified physical states are invariant. A phase redefinition of one basis state mixes the $x$ and $y$ components.

The magnitude of this effective vector is

$$\Omega=\lvert\mathbf b\rvert =\sqrt{\epsilon^2+\lvert\Delta\rvert^2}.$$

The Pauli identity $(\mathbf b\cdot\boldsymbol\sigma)^2=|\mathbf b|^2I$, together with zero trace, gives energies

$$E_\pm=c_0\pm\Omega.$$

Thus the energy splitting is

$$E_+-E_-=2\Omega.$$

Here $\Omega$ has units of energy; $\Omega/\hbar$ is an angular frequency. If $\Omega=0$, then $\epsilon=\Delta=0$ and $H=c_0I$: every state is an energy eigenstate, and there is no distinguished energy axis.

Projectors onto energy eigenstates

For $\Omega>0$, set $\hat{\mathbf b}=\mathbf b/\Omega$. The energy projectors are

$$P_\pm=\frac12\left(I\pm\hat{\mathbf b}\cdot\boldsymbol\sigma\right), \qquad H=E_+P_++E_-P_-.$$

They obey $P_\pm^2=P_\pm$, $P_+P_-=0$, and $P_++P_-=I$. For a normalized state $|\psi\rangle$, the energy probabilities are $\langle\psi|P_\pm|\psi\rangle$. This avoids choosing eigenvector phases. At $\Omega=0$, the single energy outcome $c_0$ has projector $I$; the displayed rank-one projectors are undefined because the eigenbasis is arbitrary.

Eigenvectors And Mixing Angle

If $\Delta$ is complex, a constant basis phase choice can make it real and nonnegative for a single static two-level Hamiltonian. Assume that has been done, so $\Delta\ge0$. States and other operators must be expressed in the same basis. A time-dependent rephasing also changes the Hamiltonian through the moving basis; the static argument does not discard that additional term.

For $\Omega>0$, define $0\leq\theta\leq\pi/2$ by

$$\cos(2\theta)=\frac{\epsilon}{\Omega}, \qquad \sin(2\theta)=\frac{\Delta}{\Omega}.$$

The relation $\tan(2\theta)=\Delta/\epsilon$ follows when $\epsilon\ne0$, but alone does not fix the quadrant. The sine and cosine specify the branch even when the detuning changes sign.

One convenient choice of normalized eigenstates is

$$\lvert+\rangle =\cos\theta\,\lvert1\rangle +\sin\theta\,\lvert2\rangle,$$

and

$$\lvert-\rangle =-\sin\theta\,\lvert1\rangle +\cos\theta\,\lvert2\rangle.$$

When $\Delta=0$ and $\epsilon\ne0$, the basis states are already energy eigenstates; their upper/lower labels depend on the sign of $\epsilon$. When $\epsilon=0$ and $\Delta>0$, the mixing is maximal:

$$\theta=\frac{\pi}{4}.$$

Exact static time evolution

For a time-independent Hamiltonian, separate the common phase:

$$U(t)=e^{-ic_0t/\hbar}U_{\mathrm{rel}}(t).$$

When $\Omega>0$, the identity $(\hat{\mathbf b}\cdot\boldsymbol\sigma)^2=I$ lets the even and odd powers of the exponential sum to

$$U_{\mathrm{rel}}(t) =\cos\left(\frac{\Omega t}{\hbar}\right)I -i\sin\left(\frac{\Omega t}{\hbar}\right) \hat{\mathbf b}\cdot\boldsymbol\sigma.$$

Equivalently, $U(t)=e^{-iE_+t/\hbar}P_++e^{-iE_-t/\hbar}P_-$. At $\Omega=0$, use $U_{\mathrm{rel}}(t)=I$: only the common phase evolves. These formulas do not apply to a general changing field by substituting $\mathbf b(t)$; time-dependent Hamiltonians can fail to commute at different times.

Coherent Oscillation

For time-independent $H$ with $\Omega>0$, suppose the system starts in $\lvert1\rangle$. In the original basis the off-diagonal propagator element is $-i e^{-ic_0t/\hbar}(\Delta^*/\Omega)\sin(\Omega t/\hbar)$. Its squared magnitude gives

$$P_{1\to2}(t) =\frac{\lvert\Delta\rvert^2}{\Omega^2} \sin^2\left(\frac{\Omega t}{\hbar}\right).$$

This formula contains two important effects:

• the population oscillates with angular frequency $2\Omega/\hbar$: $\sin^2(\Omega t/\hbar)=[1-\cos(2\Omega t/\hbar)]/2$;
• detuning suppresses the maximum transition probability.

On resonance, $\epsilon=0$ with $\Delta\ne0$, so $\Omega=\lvert\Delta\rvert$ and the oscillation can reach unit probability:

$$P_{1\to2}(t) =\sin^2\left(\frac{\lvert\Delta\rvert t}{\hbar}\right).$$

Far off resonance, $\lvert\epsilon\rvert\gg\lvert\Delta\rvert$, the maximum transition probability is small. With $\Delta=0$ there is no transfer, including the fully degenerate case $\Omega=0$.

Avoided Crossings

Suppose a control parameter changes the detuning $\epsilon$, while $\Delta$ is a fixed nonzero coupling. The two energies are

$$E_\pm(\epsilon) =c_0\pm\sqrt{\epsilon^2+\lvert\Delta\rvert^2}.$$

At $\epsilon=0$, the minimum gap is

$$E_+-E_-=2\lvert\Delta\rvert.$$

If $\Delta=0$, the levels cross. If $\Delta\ne0$, 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.

Effective field and Bloch-sphere motion

For $\Omega>0$, the static propagator rotates the Bloch vector about $\hat{\mathbf b}$ through angle

$$\phi(t)=\frac{2\Omega t}{\hbar}.$$

The factor of two relates the spinor’s half-angle to the Bloch-vector rotation. An energy eigenstate lies on the rotation axis and has a stationary Bloch vector. When $\mathbf b=0$, every Bloch vector is stationary and no axis is selected.

For a spin-$1/2$ magnetic moment with gyromagnetic ratio $\gamma$,

$$H=-\gamma\mathbf S\cdot\mathbf B =-\frac{\gamma\hbar}{2}\mathbf B\cdot\boldsymbol\sigma, \qquad \mathbf b=-\frac{\gamma\hbar}{2}\mathbf B.$$

The sign of $\gamma$ fixes the orientation of the energy axis relative to the magnetic field. In a double well, atomic subspace, or dimer, $\mathbf b$ instead encodes detuning and complex coupling; it need not represent a magnetic field.

The wave-mechanics visualization is developed in Bloch Sphere: Wave-Mechanics Perspective. The spinor interpretation is developed in Bloch Sphere.

Common Mistakes

• Forgetting that the off-diagonal element can be complex while the Hamiltonian remains Hermitian.
• Treating $E_1$ and $E_2$ as exact energies when $\Delta\ne0$.
• Keeping the common shift $c_0I$ in transition-probability calculations where it cancels.
• Confusing detuning $\epsilon$ with the full energy splitting $2\Omega$.
• Assuming an avoided crossing occurs without coupling.
• Forgetting that phase conventions can change the apparent phase of $\Delta$.
• Using the static propagator for a changing field without checking time ordering.

Exercises

1. Derive the eigenvalues of the general two-state Hamiltonian

$$H= \begin{pmatrix} E_1 & \Delta\\ \Delta^* & E_2 \end{pmatrix}.$$

Solution

Write

$$H=c_0I+ \begin{pmatrix} \epsilon & \Delta\\ \Delta^* & -\epsilon \end{pmatrix},$$

where $c_0=(E_1+E_2)/2$ and $\epsilon=(E_1-E_2)/2$. The traceless part has determinant

$$-\epsilon^2-\lvert\Delta\rvert^2,$$

and trace zero. Its eigenvalues are therefore

$$\pm\sqrt{\epsilon^2+\lvert\Delta\rvert^2}.$$

Adding back $c_0$ gives

$$E_\pm=c_0\pm\sqrt{\epsilon^2+\lvert\Delta\rvert^2}.$$

2. For $\epsilon=0$ and $\Delta\ne0$, show that an initial $\lvert1\rangle$ state reaches $\lvert2\rangle$ with unit probability. What changes if $\Delta=0$?

Solution

When $\epsilon=0$,

$$\Omega=\lvert\Delta\rvert.$$

The transition probability formula becomes

$$P_{1\to2}(t) =\sin^2\left(\frac{\lvert\Delta\rvert t}{\hbar}\right).$$

At

$$t=\frac{\pi\hbar}{2\lvert\Delta\rvert},$$

the sine equals $1$, so $P_{1\to2}=1$. If $\Delta=0$ as well, $H=c_0I$ and the initial state never transfers; the displayed transfer time is not defined.

3. At fixed $\Delta$, what is the minimum gap when $E_\pm=c_0\pm\sqrt{\epsilon^2+\lvert\Delta\rvert^2}$ and $\epsilon$ is varied through zero?

Solution

The gap is

$$E_+-E_- =2\sqrt{\epsilon^2+\lvert\Delta\rvert^2}.$$

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

$$\Delta E_{\mathrm{min}}=2\lvert\Delta\rvert.$$

It is an avoided crossing for $\Delta\ne0$ and an exact crossing for $\Delta=0$.

4. Decompose $H=\begin{pmatrix}3&1-i\\1+i&1\end{pmatrix}$ into Pauli coordinates and find its energies. All entries are measured in the same energy unit.

Solution

The average is $c_0=2$, the half-detuning is $b_z=1$, and $\Delta=1-i$ gives $b_x=1$, $b_y=1$. Thus

$$H=2I+\sigma_x+\sigma_y+\sigma_z, \qquad E_\pm=2\pm\sqrt3.$$

Using $b_y=-1$ would conjugate the off-diagonal entries and describe a different matrix in the specified basis.

5. For $\Omega>0$, prove that $P_+=(I+\hat{\mathbf b}\cdot\boldsymbol\sigma)/2$ is an orthogonal projector. Why does this formula not select an energy eigenvector at $\Omega=0$?

Solution

Write $Q=\hat{\mathbf b}\cdot\boldsymbol\sigma$. Since $Q^\dagger=Q$ and $Q^2=I$,

$$P_+^\dagger=P_+, \qquad P_+^2=\frac14(I+2Q+Q^2)=P_+.$$

The complementary projector is $I-P_+$. At $\Omega=0$, $\hat{\mathbf b}$ is undefined and the Hamiltonian gives no preferred one-dimensional eigenspaces; the single energy projector is $I$.

6. For $H=K\sigma_x$ with $K>0$, find the energy eigenstates and time-evolution operator. Starting in $|1\rangle$, when does the state first reach $|2\rangle$ up to phase?

Solution

The symmetric and antisymmetric states $(|1\rangle\pm|2\rangle)/\sqrt2$ have energies $\pm K$. Their spectral phases give

$$U(t)=\cos(Kt/\hbar)I-i\sin(Kt/\hbar)\sigma_x,$$

so $U(t)|1\rangle=\cos(Kt/\hbar)|1\rangle-i\sin(Kt/\hbar)|2\rangle$. The first full transfer is at $t=\pi\hbar/(2K)$, where the state is $-i|2\rangle$. The energy gap is $2K$, not $K$.

References

• 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.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010 — Pauli coordinates, projectors, and two-state unitary evolution.
• 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.