The Berry phase

Intuition

Carry a Hamiltonian slowly round a closed loop of its parameters, back to where it started. The state returns to the same eigenstate, with two phases: the dynamical one, which depends on how long the journey took, and a geometric one, which depends only on the loop. Berry showed in 1984 that the geometric phase cannot be removed by choosing the phases of the eigenstates differently, and that interference measures it. For a spin that follows a magnetic field whose direction goes round a loop, it is minus half the solid angle the loop encloses.

Walk a pointer over a globe from the North Pole down to the equator, a quarter of the way along it, and back to the pole, never turning it. It comes back turned by a right angle, the solid angle enclosed. The turn records the shape of the loop, not the speed of the walk; so does the Berry phase.

A field $\mathbf{B}$ whose direction is carried once round a cone of half-angle $\theta$ about the $z$ axis. Its tip traces a loop on the unit sphere enclosing the solid angle $\Omega=2\pi(1-\cos\theta)$ ; a spin that follows the field returns with the geometric phase $-\Omega/2$ , however fast or slow the trip.

Let the Hamiltonian depend on parameters $\mathbf{R}$ , carried slowly round a closed loop $C$ . The eigenstate being followed returns with its dynamical phase and the geometric phase

\gamma_{n}(C)=i\oint_{C}\braket{n(\mathbf{R})|\nabla_{\mathbf{R}}\,n(\mathbf{R})}\cdot d\mathbf{R}

Properties

Time drops out of $\int\braket{n|\dot n}\,dt=\int\braket{n|\nabla_{\mathbf{R}}n}\cdot d\mathbf{R}$ : the phase depends on the loop, not on the speed.

The phase of a spin carried round a cone

\begin{aligned}&\ket{\mathbf{n}}=\cos\tfrac{\theta}{2}\ket{\uparrow}+e^{i\varphi}\sin\tfrac{\theta}{2}\ket{\downarrow},\ \ \varphi:0\to2\pi\\ &\Longrightarrow\ \gamma=-\tfrac12\Omega,\qquad \Omega=2\pi(1-\cos\theta)\end{aligned}

As the field goes round the cone, $\varphi$ runs from $0$ to $2\pi$ at fixed $\theta$ and the spin state along the field follows. Differentiate the state along the loop, take its overlap with the state itself and integrate: the result is minus half the solid angle of the cone.

Proof steps

\ket{\mathbf{n}}=\cos\tfrac{\theta}{2}\ket{\uparrow}+e^{i\varphi}\sin\tfrac{\theta}{2}\ket{\downarrow}

The state along the field, single-valued in $\varphi$ .

\partial_{\varphi}\ket{\mathbf{n}}=i\,e^{i\varphi}\sin\tfrac{\theta}{2}\ket{\downarrow}

Only the second component depends on $\varphi$ .

\braket{\mathbf{n}|\partial_{\varphi}\mathbf{n}}=i\sin^{2}\tfrac{\theta}{2}

Its overlap with the state.

\begin{aligned}\gamma&=i\oint\braket{\mathbf{n}|\partial_{\varphi}\mathbf{n}}\,d\varphi=-2\pi\sin^{2}\tfrac{\theta}{2}\\ &=-\pi(1-\cos\theta)\end{aligned}

Integrate round the loop, with $2\sin^{2}\tfrac{\theta}{2}=1-\cos\theta$ .

\Omega=2\pi(1-\cos\theta)\ \Longrightarrow\ \gamma=-\tfrac12\Omega

The solid angle of a cone of half-angle $\theta$ .

Applications

Practice

Geometry, Not Speed

The geometric phase is an integral along the loop in parameter space, with time nowhere in it. Going round the same loop faster or slower, but still slowly enough, gives the same phase.

\gamma_{n}=i\oint_{C}\braket{n|\nabla_{\mathbf{R}}n}\cdot d\mathbf{R}

Try it

On what does the Berry phase of a slow loop depend?