Berry Phase for Spin-1/2

The standard worked example of Berry phase is a spin-$1/2$ eigenstate that adiabatically follows a magnetic-field direction around a closed loop. The result is geometric:

$$\gamma_+ = -\frac{\Omega}{2}$$

for the convention used below, where $\Omega$ is the oriented solid angle enclosed by the loop on the unit sphere of field directions.

This example is useful because everything is visible at once:

• the instantaneous eigenstate is a point on the Bloch sphere;
• the Berry connection is a local one-form on the sphere;
• the Berry curvature is the field of an effective monopole at the degeneracy;
• the final phase depends on solid angle, not on traversal time, once the adiabatic approximation is valid.

Hamiltonian Convention

Take the two-level Hamiltonian

$$H(\hat{\mathbf n}) = -\frac{\Delta}{2}\, \hat{\mathbf n}\cdot\boldsymbol\sigma, \qquad \Delta>0,$$

where $\hat{\mathbf n}$ is a unit vector specifying the direction of the magnetic field or effective field. In terms of a physical Zeeman coupling, $\Delta$ is the positive level splitting.

The eigenstate of $\hat{\mathbf n}\cdot\boldsymbol\sigma$ with eigenvalue $+1$ has energy

$$E_+ = -\frac{\Delta}{2}.$$

The eigenstate with eigenvalue $-1$ has energy

$$E_- = \frac{\Delta}{2}.$$

Thus, for this Hamiltonian, $\lvert +;\hat{\mathbf n}\rangle$ is the lower-energy state. Reversing the sign of the Hamiltonian or following the other eigenstate reverses the sign of the Berry phase.

Parameter Sphere

Write the field direction as

$$\hat{\mathbf n} = (\sin\theta\cos\phi,\, \sin\theta\sin\phi,\, \cos\theta).$$

The parameter space at fixed field magnitude is a sphere $S^2$. The point $\Delta=0$, where the two levels become degenerate, is excluded. This excluded degeneracy is why the curvature picture looks like a monopole.

For this simple Hamiltonian, the parameter sphere of $\hat{\mathbf n}$ also matches the Bloch sphere of the lower eigenstate. That identification is special to this two-level model. In a general Berry-phase problem, parameter space need not be the same as projective Hilbert space.

Instantaneous Eigenstate

A convenient local eigenvector for the $+1$ eigenvalue of $\hat{\mathbf n}\cdot\boldsymbol\sigma$ is

$$\lvert +;\theta,\phi\rangle = \begin{pmatrix} \cos(\theta/2)\\ e^{i\phi}\sin(\theta/2) \end{pmatrix}.$$

This gauge is smooth away from one pole of the sphere. The singularity is not a physical singularity of the state ray; it is a sign that no single smooth phase convention covers the full sphere.

The corresponding Bloch vector is

$$\langle +;\theta,\phi| \boldsymbol\sigma |+;\theta,\phi\rangle = \hat{\mathbf n}.$$

Thus this state really is spin-up along the instantaneous field direction.

Berry Connection

Using

$$A_+ = i\langle +;\theta,\phi\rvert d\lvert +;\theta,\phi\rangle,$$

one obtains

$$A_+ = -\frac{1-\cos\theta}{2}\,d\phi.$$

Equivalently,

$$A_\theta=0, \qquad A_\phi = -\frac{1-\cos\theta}{2}.$$

The connection depends on the gauge. A different phase choice changes $A_+$ by $-d\chi$. The closed-loop phase factor, after the usual patching and modulo $2\pi$, is gauge invariant.

Berry Curvature

The curvature is

$$F_+ = dA_+ = -\frac12\sin\theta\,d\theta\wedge d\phi.$$

For a surface $\Sigma$ on the field-direction sphere,

$$\int_\Sigma F_+ = -\frac12 \int_\Sigma \sin\theta\,d\theta\,d\phi = -\frac{\Omega(\Sigma)}{2}.$$

If the closed path $C$ bounds $\Sigma$, then

$$\gamma_+[C] = \oint_C A_+ = \int_\Sigma F_+ = -\frac{\Omega}{2},$$

with the usual orientation convention. The same loop with the opposite orientation gives the opposite Berry phase.

The solid angle $\Omega$ is defined modulo $4\pi$ for a closed loop on a sphere. Since $\gamma_+$ is defined modulo $2\pi$, the phase factor $e^{i\gamma_+}$ is unambiguous.

Constant-Polar-Angle Loop

For a loop at fixed $\theta$ with $\phi:0\to2\pi$,

$$\gamma_+ = \int_0^{2\pi} -\frac{1-\cos\theta}{2}\,d\phi.$$

Therefore

$$\gamma_+ = -\pi(1-\cos\theta).$$

The enclosed solid angle is

$$\Omega = 2\pi(1-\cos\theta),$$

so this is exactly $-\Omega/2$.

Special cases are useful checks:

• If $\theta=0$, the loop shrinks to the north pole and $\gamma_+=0$ in this gauge.
• If $\theta=\pi/2$, the loop is the equator and $\gamma_+=-\pi$.
• If the loop orientation is reversed, $\gamma_+$ changes sign.

Dynamical Phase Is Separate

If the system stays in $\lvert +;\hat{\mathbf n}(t)\rangle$ for time $T$, the total phase is

$$\alpha_+ = -\frac{1}{\hbar} \int_0^T E_+(t)\,dt + \gamma_+[C].$$

For constant splitting $\Delta$,

$$-\frac{1}{\hbar} \int_0^T E_+\,dt = \frac{\Delta T}{2\hbar}.$$

Thus

$$\alpha_+ = \frac{\Delta T}{2\hbar} - \frac{\Omega}{2}.$$

The first term changes if the loop is traversed more slowly. The second does not, provided the evolution remains adiabatic and follows the same oriented loop.

Adiabatic Condition

The state follows the instantaneous eigenstate only when transitions to the other level are negligible. A useful schematic condition is

$$\hbar |\dot{\hat{\mathbf n}}| \ll \Delta,$$

up to angular factors and matrix elements. More invariantly, the off-diagonal nonadiabatic coupling must be small compared with the energy gap. The path must also avoid the degeneracy $\Delta=0$.

The solid-angle formula is not a statement about arbitrary fast spin motion. Fast driving can produce nonadiabatic transitions, Rabi dynamics, or Aharonov–Anandan phases rather than the simple Berry phase derived here.

Opposite Eigenstate

For the upper state $\lvert -;\hat{\mathbf n}\rangle$ in the same Hamiltonian convention, the Berry phase has the opposite sign:

$$\gamma_-[C] = +\frac{\Omega}{2}.$$

This is consistent with the curvature picture: the two eigenline bundles carry opposite Berry curvature. It is also why sign conventions must be stated carefully. A formula copied without the Hamiltonian convention, eigenstate label, and orientation convention is incomplete.

Monopole Interpretation

The full parameter space before fixing field magnitude is $\mathbb R^3$ with the degeneracy at the origin removed. The curvature through a sphere around the origin has total flux

$$\int_{S^2}F_+ = -2\pi.$$

The corresponding Chern number is

$$C_+ = \frac{1}{2\pi} \int_{S^2}F_+ = -1$$

for the convention used here. This is the geometric meaning of the phrase “Berry monopole”: the degeneracy acts as a source of curvature flux in parameter space.

The physical Berry phase around a loop is the flux through any spanning surface that avoids the degeneracy, computed with the correct patching if a single gauge is not available.

Common Mistakes

• Quoting $\gamma=-\Omega/2$ without specifying which eigenstate and Hamiltonian sign are being used.
• Confusing the field-direction sphere with ordinary physical space.
• Forgetting the dynamical phase when predicting the total phase in an experiment.
• Applying the formula when the field passes through zero and the gap closes.
• Treating the local gauge singularity of the spinor as a physical singularity of the state.
• Assuming that all two-level cyclic phases are Berry phases; the adiabatic and eigenstate-following assumptions matter.
• Forgetting that reversing the loop orientation reverses the sign of the Berry phase.

Cross-Links

• Berry Phase
• Adiabatic Theorem Reminder
• Berry Connection
• Berry Curvature
• Berry Phase Problems
• Dirac Monopole Preview
• Chern Numbers
• Bloch Sphere
• Spin in Magnetic Fields
• Larmor Precession
• Spin Rotations
• Spin in Magnetic Field Hamiltonian
• Berry Connection as a Mathematical Object
• Chern Numbers

References

• M. V. Berry, “Quantal phase factors accompanying adiabatic changes,” Proceedings of the Royal Society A 392, 45-57, 1984.
• B. Simon, “Holonomy, the quantum adiabatic theorem, and Berry’s phase,” Physical Review Letters 51, 2167-2170, 1983.
• A. Shapere and F. Wilczek, eds., Geometric Phases in Physics, World Scientific, 1989.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.

Exercises

1. Compute $A_+=i\langle +;\theta,\phi\rvert d\lvert +;\theta,\phi\rangle$ for the spinor used on this page.

Solution

The spinor is

$$\lvert +;\theta,\phi\rangle = \begin{pmatrix} \cos(\theta/2)\\ e^{i\phi}\sin(\theta/2) \end{pmatrix}.$$

The $\phi$ derivative gives

$$\partial_\phi \lvert +;\theta,\phi\rangle = \begin{pmatrix} 0\\ i e^{i\phi}\sin(\theta/2) \end{pmatrix}.$$

Therefore

$$\langle +|\partial_\phi +\rangle = i\sin^2(\theta/2),$$

so

$$A_\phi = i\langle +|\partial_\phi +\rangle = -\sin^2(\theta/2) = -\frac{1-\cos\theta}{2}.$$

The $\theta$ component vanishes, so

$$A_+ = -\frac{1-\cos\theta}{2}\,d\phi.$$

2. Find the Berry phase for a constant-$\theta$ loop.

Solution

Use

$$A_+ = -\frac{1-\cos\theta}{2}\,d\phi.$$

For $\phi:0\to2\pi$,

$$\gamma_+ = \int_0^{2\pi} -\frac{1-\cos\theta}{2}\,d\phi = -\pi(1-\cos\theta).$$

Since $\Omega=2\pi(1-\cos\theta)$, this is $-\Omega/2$.

3. For constant splitting $\Delta$ and loop duration $T$, compute the total phase of the lower state.

Solution

The lower state has

$$E_+ = -\frac{\Delta}{2}.$$

The dynamical phase is

$$-\frac{1}{\hbar} \int_0^T E_+\,dt = \frac{\Delta T}{2\hbar}.$$

The Berry phase is $-\Omega/2$, so

$$\alpha_+ = \frac{\Delta T}{2\hbar} - \frac{\Omega}{2}.$$

4. What happens to the Berry phase if the same loop is traversed in the opposite direction?

Solution

The oriented line integral changes sign:

$$\oint_{-C} A_+ = -\oint_C A_+.$$

Equivalently, the oriented solid angle changes sign. Therefore

$$\gamma_+[-C] = -\gamma_+[C].$$