Dynamical Phase versus Geometric Phase

Quantum phases have two different sources that are easy to confuse. A dynamical phase is accumulated because the state has energy over time. A geometric phase is accumulated because the state, or an eigenspace, is carried around a path in a space of rays or parameters.

In the standard Berry-phase setting, a Hamiltonian depends slowly on parameters $R(t)$ and an instantaneous eigenstate follows a closed path $C$; the minimal eigenstate-following assumption is summarized in Adiabatic Theorem Reminder:

$$H(R)\lvert n(R)\rangle = E_n(R)\lvert n(R)\rangle.$$

After one adiabatic cycle, the state returns to the same ray:

$$\lvert\psi(T)\rangle = e^{i\alpha_n} \lvert n(R(0))\rangle.$$

The total phase separates as

$$\alpha_n = -\frac{1}{\hbar} \int_0^T E_n(t)\,dt + \gamma_n[C].$$

The first term is dynamical. The second term is geometric.

Dynamical Phase

For a stationary energy eigenstate,

$$\lvert\psi(t)\rangle = e^{-iEt/\hbar} \lvert E\rangle.$$

The phase

$$\alpha_{\mathrm{dyn}} = -\frac{Et}{\hbar}$$

is dynamical. It depends on the energy and on elapsed time. If the energy changes during an adiabatic process, the eigenstate component accumulates

$$\alpha_{\mathrm{dyn}} = -\frac{1}{\hbar} \int_0^T E_n(t)\,dt.$$

Changing the speed of traversal changes this phase. If the same parameter-space path is traversed more slowly, the integral usually changes because the state spends more time at each energy.

For a single isolated state, an overall dynamical phase is not directly observable. It becomes observable when compared with another amplitude, another internal component, or another path.

Geometric Phase

The geometric phase is not determined by energy times duration. In Berry’s adiabatic setting it is

$$\gamma_n[C] = \oint_C \mathbf A_n(R)\cdot dR,$$

where

$$\mathbf A_n(R) = i\langle n(R)|\nabla_R n(R)\rangle$$

is the Berry connection in a smooth eigenvector gauge that returns to the same vector after the loop. Otherwise add the endpoint phase $\arg\langle n(0)|n(T)\rangle$ to the connection integral; this prevents a nonperiodic representative from changing the answer.

For a closed loop, $\gamma_n[C]$ is invariant modulo $2\pi$ under smooth single-valued changes of eigenvector phase. It depends on the path taken through parameter space and on how the eigenstate line twists along that path.

If the same closed path is traversed at a different speed while the adiabatic approximation remains valid, the Berry phase is unchanged. That speed-insensitivity is a useful diagnostic, but it is not the definition. The definition is geometric holonomy of the state ray or eigenstate line.

Side-by-Side Comparison

Feature	Dynamical phase	Geometric phase
Source	energy over time	path through rays or parameters
Typical formula	$-\hbar^{-1}\int E(t)\,dt$	$\oint \mathbf A\cdot dR$
Depends on duration?	yes	not directly, under adiabatic conditions
Needs a closed path?	no	closed paths give the clean gauge-invariant case
Physical access	interference with a reference	interference after path-dependent transport

Both are phases of a quantum amplitude. The distinction is not “physical versus unphysical.” The distinction is what information the phase records.

Example: Same Path, Different Timing

Suppose a nondegenerate eigenstate is transported adiabatically around the same closed loop $C$ twice. In the first run the traversal time is $T$; in the second it is $2T$, with the same path and a uniformly stretched schedule, $R_{2T}(t)=R_T(t/2)$.

If the energy profile as a function of path position is the same, the adiabatic dynamical-phase integral doubles:

$$\alpha_{\mathrm{dyn}}(2T) = 2\alpha_{\mathrm{dyn}}(T).$$

The Berry phase is the same:

$$\gamma[C]_{2T} = \gamma[C]_T.$$

This is why experiments often try to cancel or subtract dynamical phases when measuring geometric phases.

Example: Spin Following a Field Direction

For a spin-$1/2$ in a slowly changing magnetic field, the instantaneous eigenstate depends on the direction $\hat{\mathbf n}$ of the field. If $\hat{\mathbf n}$ traces a closed loop on the unit sphere, the Berry phase is proportional to the solid angle $\Omega$ enclosed by the loop:

$$\gamma = -\frac{\Omega}{2}$$

for the $+1$ eigenstate of $\hat{\mathbf n}\cdot\boldsymbol\sigma$, using $A=i\langle n|dn\rangle$ and the oriented solid angle. The opposite eigenstate has the opposite sign.

The dynamical phase still depends on the Zeeman energy and the time spent along the path. The geometric phase depends on the loop on the sphere. The worked example is Berry Phase for Spin-1/2.

Example: Aharonov–Bohm Phase

The Aharonov–Bohm phase is not a Berry phase from slow change of a Hamiltonian parameter in the basic setup. It is instead a gauge holonomy in real space:

$$\exp \left( \frac{iq}{\hbar} \oint_C \mathbf A\cdot d\mathbf r \right).$$

It is geometric in the broader sense that it is a loop phase of a connection. When the magnetic field vanishes throughout the accessible deformation region, this phase depends only on the winding and enclosed flux. A hole alone is insufficient: nonzero curvature in the region can make the phase change under a smooth path deformation.

This example is useful because it shows the family resemblance between Berry phases and electromagnetic gauge phases, while also warning that not every geometric phase is a Berry phase.

A Cyclic State Need Not Follow an Eigenstate

Take the constant Hamiltonian $H=\hbar\omega\sigma_z/2$, with $\omega>0$, and initial state $(\cos(\theta/2),\sin(\theta/2))^{\mathsf T}$. Its Bloch vector precesses once in $T=2\pi/\omega$, and the state returns as $\psi(T)=-\psi(0)$. Thus one choice of total phase is $\alpha=-\pi$.

For an arbitrary cyclic Schrödinger state, the dynamical phase uses the expectation value, not an eigenvalue:

$$\alpha_{\mathrm{dyn}}=-\frac1\hbar\int_0^T\langle H\rangle\,dt =-\pi\cos\theta.$$

The geometric remainder is $\alpha-\alpha_{\mathrm{dyn}}=-\pi(1-\cos\theta)$ modulo $2\pi$. This is the Aharonov–Anandan phase of a cyclic ray path. No parameter changes slowly and, except at the poles, the state is not an energy eigenstate. The same ray geometry can therefore arise outside Berry’s adiabatic setting.

Gauge and Convention Dependence

The Berry connection is gauge dependent:

$$\lvert n(R)\rangle \mapsto e^{i\chi(R)} \lvert n(R)\rangle$$

implies, with the convention used here,

$$\mathbf A_n \mapsto \mathbf A_n-\nabla_R\chi.$$

For a closed loop,

$$\oint_C\nabla_R\chi\cdot dR = 0$$

when $\chi$ is smooth and single-valued on the loop. Therefore the closed-loop Berry phase is gauge invariant modulo $2\pi$.

For open paths, the line integral alone is not gauge invariant. One must include an endpoint comparison, such as a Pancharatnam phase convention, or specify an interferometric reference. This is one reason closed cycles are the clean first case.

Common Mistakes

• Calling any phase that survives a cycle “Berry phase” without checking the adiabatic eigenstate setup.
• Thinking geometric phase is independent of all dynamics; the path itself is produced by dynamics, and adiabaticity has conditions.
• Forgetting that the total measured phase usually contains both dynamical and geometric parts.
• Treating the Berry connection as gauge invariant rather than the closed-loop holonomy.
• Confusing “not dependent on speed” with “topological.” A geometric phase can vary continuously with the shape of a path.
• Ignoring the reference path or component needed to observe a phase.

Exercises

1. A state remains in an eigenstate with constant energy $E$ for time $T$. What is the dynamical phase? What happens if $T$ is doubled?

Solution

The dynamical phase is

$$\alpha_{\mathrm{dyn}} = -\frac{ET}{\hbar}.$$

If $T$ is doubled while $E$ is unchanged, the phase becomes

$$\alpha_{\mathrm{dyn}}(2T) = -\frac{2ET}{\hbar} = 2\alpha_{\mathrm{dyn}}(T).$$

2. Under $\mathbf A_n\mapsto\mathbf A_n-\nabla_R\chi$, show that the Berry phase around a closed loop is unchanged for single-valued $\chi$.

Solution

The transformed loop integral is

$$\oint_C \left( \mathbf A_n-\nabla_R\chi \right)\cdot dR = \oint_C\mathbf A_n\cdot dR - \oint_C\nabla_R\chi\cdot dR.$$

For a smooth single-valued $\chi$,

$$\oint_C\nabla_R\chi\cdot dR = \chi(\text{final})-\chi(\text{initial}) = 0.$$

Thus the closed-loop Berry phase is unchanged modulo the usual $2\pi$ phase convention.

3. A spin-$1/2$ eigenstate follows a closed loop enclosing solid angle $\Omega=\pi$ on the parameter sphere. What is the geometric phase in the convention $\gamma=-\Omega/2$?

Solution

Substitute $\Omega=\pi$:

$$\gamma = -\frac{\pi}{2}.$$

The physically relevant phase is defined modulo $2\pi$ and depends on the orientation and eigenstate convention.

References

• Y. Aharonov and J. Anandan, “Phase change during a cyclic quantum evolution,” Physical Review Letters 58, 1593–1596, 1987.
• M. V. Berry, “Quantal phase factors accompanying adiabatic changes,” Proceedings of the Royal Society A 392, 45–57, 1984.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• A. Shapere and F. Wilczek, eds., Geometric Phases in Physics, World Scientific, 1989.
• B. Simon, “Holonomy, the quantum adiabatic theorem, and Berry’s phase,” Physical Review Letters 51, 2167–2170, 1983.