4.2.1 Berry Phase#
Prompts
If a Hamiltonian does not change in time, why do transitions between different energy levels have exactly zero probability? What does that say about long-time evolution at fixed \(\hat{H}\)?
When the Hamiltonian shifts by a tiny amount, how do its eigenstates respond, and why does the system stay in approximately the same energy level? Where does the room for any nontrivial geometric effect come from?
Why is the eigenstate \(\vert n(\boldsymbol{R})\rangle\) defined only up to a local phase \(\mathrm{e}^{\mathrm{i}\alpha(\boldsymbol{R})}\), and how is this the same kind of gauge freedom that a charged particle’s wavefunction has in real space?
What does the Berry connection \(\boldsymbol{A}_n(\boldsymbol{R}) = \mathrm{i}\langle n(\boldsymbol{R})\vert\nabla_{\boldsymbol{R}}\vert n(\boldsymbol{R})\rangle\) track, why is its open-path integral unphysical, and why is the closed-loop integral (the Berry phase) physical?
Why is the Berry curvature of a spin-1/2 eigenstate a magnetic-monopole field at the centre of the Bloch sphere, and how does that picture explain precession as a Lorentz-type response?
Lecture Notes#
Overview#
The eigenstates of a Hamiltonian \(\hat{H}(\boldsymbol{R})\) depend on the parameters \(\boldsymbol{R}\) that define \(\hat{H}\). Tracking how the eigenstate moves through parameter space is a purely geometric problem: each diagonalisation leaves a local phase undetermined, and absorbing that ambiguity requires a gauge connection on parameter space. The line integral of that connection around a closed loop is the Berry phase, a gauge-invariant observable that mirrors the electromagnetic phase in real space, with parameter space playing the role of position.
Parameter-dependent eigenstates#
Let \(\hat{H}(\boldsymbol{R})\) depend smoothly on a set of parameters \(\boldsymbol{R}\) — a magnetic-field direction, a lattice momentum, the position of a slow source, or any other variable. At each \(\boldsymbol{R}\) the eigenvalue problem reads
The question is how the eigenstate \(\vert n(\boldsymbol{R})\rangle\) in a chosen level \(n\) moves as \(\boldsymbol{R}\) moves.
Why levels do not mix
If \(\boldsymbol{R}\) is held fixed, eigenstates of distinct energies are orthogonal,
so the transition probability \(\vert\langle m\vert n\rangle\vert^{2}\) between different levels is exactly zero. The system stays in level \(n\) indefinitely.
When \(\boldsymbol{R}\) shifts by a small amount, \(\hat{H}\) changes by a small amount, and each eigenstate is modified by a small amount. The overlap with the original level \(n\) remains close to one, and the overlap with any other level \(m\) remains small. The system continues to occupy level \(n\) — only the in-level eigenstate \(\vert n(\boldsymbol{R})\rangle\) itself moves with \(\boldsymbol{R}\).
This picture fails only when the gap \(E_m - E_n\) closes at some \(\boldsymbol{R}\); the level under study then mixes resonantly with another level and is no longer well-defined on its own.
The interesting question is therefore not whether the system leaves level \(n\) — generically it does not — but how the in-level eigenstate \(\vert n(\boldsymbol{R})\rangle\) itself is parametrised across \(\boldsymbol{R}\).
Phase ambiguity in parameter space#
If \(\vert n(\boldsymbol{R})\rangle\) is an eigenstate of \(\hat{H}(\boldsymbol{R})\) in a non-degenerate level, then \(\mathrm{e}^{\mathrm{i}\alpha(\boldsymbol{R})}\,\vert n(\boldsymbol{R})\rangle\) is also an eigenstate for any real \(\alpha(\boldsymbol{R})\). Diagonalising \(\hat{H}\) independently at each \(\boldsymbol{R}\) leaves an uncontrolled local phase choice at every point:
This is a gauge transformation — but on parameter space, not real space. A charged particle’s wavefunction \(\psi(\boldsymbol{r}) \to \mathrm{e}^{\mathrm{i}\alpha(\boldsymbol{r})}\psi(\boldsymbol{r})\) has the same local phase ambiguity over position. In that case the resolution was to introduce a gauge connection (the vector potential \(\boldsymbol{A}\)) that keeps track of how the phase reference rotates between nearby points. The same recipe applies here.
Berry connection#
Berry connection
The Berry connection on parameter space is the real-valued vector field
Under a rephasing \(\vert n(\boldsymbol{R})\rangle \to \mathrm{e}^{\mathrm{i}\alpha(\boldsymbol{R})}\vert n(\boldsymbol{R})\rangle\), it transforms as
exactly the gauge form of a vector potential.
The reality of \(\boldsymbol{A}_n\) follows from normalisation: \(\langle n\vert n\rangle = 1\) forces \(\langle n\vert\nabla_{\boldsymbol{R}}\vert n\rangle\) to be purely imaginary, so the factor of \(\mathrm{i}\) makes \(\boldsymbol{A}_n\) real.
The gauge transformation rule mirrors the electromagnetic gauge transformation \(\boldsymbol{A}\to\boldsymbol{A}+\nabla\alpha\) of 4.1.3 Gauge Invariance, with a sign that absorbs the rephasing of the eigenstate.
Derivation: reality and gauge transformation
Reality. Differentiate \(\langle n(\boldsymbol{R})\vert n(\boldsymbol{R})\rangle = 1\) with respect to \(\boldsymbol{R}\):
The first term is the complex conjugate of the second, so \(\langle n\vert\nabla_{\boldsymbol{R}} n\rangle = -\langle n\vert\nabla_{\boldsymbol{R}} n\rangle^{*}\) is purely imaginary. Multiplying by \(\mathrm{i}\) gives a real field, \(\boldsymbol{A}_n = \mathrm{i}\langle n\vert\nabla_{\boldsymbol{R}}\vert n\rangle \in \mathbb{R}\).
Gauge transformation. Under \(\vert n\rangle \to \vert\tilde{n}\rangle = \mathrm{e}^{\mathrm{i}\alpha(\boldsymbol{R})}\vert n\rangle\), the product rule gives
Taking the inner product with \(\langle\tilde{n}\vert = \langle n\vert\mathrm{e}^{-\mathrm{i}\alpha}\) kills the phase factors:
Multiplying by \(\mathrm{i}\),
Berry phase along a path#
The Berry connection lets one compare the phase reference at two parameter points. The accumulated rephasing along a path \(\mathcal{C}\) from \(\boldsymbol{R}_a\) to \(\boldsymbol{R}_b\) in parameter space is the line integral
Berry phase
Along an open path \(\mathcal{C}\), the line integral shifts under a gauge transformation by
\[ \Phi_{\mathrm{Berry}}(\mathcal{C}) \;\longrightarrow\; \Phi_{\mathrm{Berry}}(\mathcal{C}) + \alpha(\boldsymbol{R}_a) - \alpha(\boldsymbol{R}_b), \]so it depends on the phase convention at the endpoints and is not a physical observable on its own.
Along a closed loop \(\mathcal{C} = \partial\Sigma\), the boundary terms cancel and the loop integral is gauge-invariant. The Berry phase
\[ \Phi_{\mathrm{Berry}}(\mathcal{C}) = \oint_{\mathcal{C}} \boldsymbol{A}_n(\boldsymbol{R})\cdot\mathrm{d}\boldsymbol{R} \]is a genuine physical observable: a phase the eigenstate picks up relative to itself after being parallel-transported once around \(\mathcal{C}\).
This is the same open-vs-closed dichotomy as the electromagnetic holonomy of 4.1.3 Gauge Invariance § Open path vs closed loop, lifted from real space to parameter space.
Berry curvature#
A local gauge-invariant version of the Berry phase is obtained by taking the curl of the Berry connection.
Berry curvature
The Berry curvature is
Under a gauge transformation \(\boldsymbol{A}_n \to \boldsymbol{A}_n - \nabla_{\boldsymbol{R}}\alpha\), the curvature is unchanged because the curl of a gradient vanishes. By Stokes’ theorem the closed-loop Berry phase equals the curvature flux through any surface \(\Sigma\) bounded by \(\mathcal{C}\),
The analogy is direct: \(\boldsymbol{A}_n\) plays the role of a vector potential and \(\boldsymbol{\Omega}_n\) the role of a magnetic field — both living on parameter space rather than real space.
Same gauge structure, different parameter space#
In §4.1, the local phase ambiguity was \(\psi(\boldsymbol{r}) \to \mathrm{e}^{\mathrm{i}\alpha(\boldsymbol{r})}\psi(\boldsymbol{r})\) on real space, and the gauge connection that absorbed it was the electromagnetic vector potential \(q\boldsymbol{A}/\hbar\). Here the local ambiguity \(\vert n(\boldsymbol{R})\rangle \to \mathrm{e}^{\mathrm{i}\alpha(\boldsymbol{R})}\vert n(\boldsymbol{R})\rangle\) lives on parameter space, and the connection is the Berry connection \(\boldsymbol{A}_n(\boldsymbol{R})\). The structure — connection, curvature, open-path gauge dependence, closed-loop physical observable — is identical. The electromagnetic case is the special instance in which the parameter happens to be the position of the particle.
Example: spin-1/2 on the Bloch sphere#
Consider a spin-1/2 whose eigenstate is parametrised by a unit vector \(\hat{\boldsymbol{n}}\) on the Bloch sphere (the field direction, the easy axis, or any other unit-vector parameter). Use spherical angles
The eigenstate pointing along \(\hat{\boldsymbol{n}}\) in the standard “north-pole” gauge is
Derivation: Berry connection and curvature for spin-1/2
Use the eigenstate and its bra:
Berry connection components in parameter space are
First the derivatives:
Then the inner products:
so \(A_\theta = 0\), and
hence
The curl in \((\theta,\varphi)\) coordinates is
To convert the coordinate-component curvature to a vector field on the unit sphere, match surface-flux integrals. The sphere’s vector area element is \(\mathrm{d}\boldsymbol{S} = \hat{\boldsymbol{n}}\,\sin\theta\,\mathrm{d}\theta\,\mathrm{d}\varphi\), so for any patch \(\Sigma\)
which forces \(\boldsymbol{\Omega}\cdot\hat{\boldsymbol{n}} = \Omega_{\theta\varphi}/\sin\theta = -\tfrac12\). The eigenstate depends only on \(\hat{\boldsymbol{n}}\), so the curvature is purely radial,
This is exactly the field of a magnetic monopole of strength \(-1/2\) sitting at the centre of the Bloch sphere.
Spin-1/2 Berry phase
For a closed loop \(\mathcal{C}\) that subtends solid angle \(\Omega_{\text{solid}}\) on the Bloch sphere, the spin-up eigenstate \(\vert\uparrow(\hat{\boldsymbol{n}})\rangle\) acquires Berry phase
The Berry phase of a spin-1/2 equals minus half the solid angle subtended by the path on the Bloch sphere.
Monopole-Lorentz picture of precession
The Berry curvature \(\boldsymbol{\Omega} = -\hat{\boldsymbol{n}}/2\) on the Bloch sphere is a radial field with no source on the sphere itself: it is the field of a magnetic monopole at the centre. The quantum spin therefore lives on a parameter sphere threaded by monopole flux.
A torque from an external field pushes the spin direction \(\hat{\boldsymbol{n}}\) tangentially on the Bloch sphere — analogous to an “electric” force on a charged particle constrained to that sphere. On a sphere threaded by a monopole, the Lorentz response to such a tangential force is perpendicular to it: instead of relaxing along the torque, the particle moves sideways, tracing a small circle around the force axis. That sideways motion is precession. The same dictionary — the gyroscopic term as a monopole Lorentz force — appears at the classical level in 4.4.1 Classical Spin.
Example: spin along the equator
Problem. A spin-1/2 eigenstate \(\vert\uparrow(\hat{\boldsymbol{n}})\rangle\) is carried once around a great circle of the Bloch sphere (the equator). What Berry phase does it acquire?
Solution. The equator bounds a hemisphere of solid angle \(\Omega_{\text{solid}} = 2\pi\), so
and the eigenstate returns to itself up to a geometric sign flip \(\mathrm{e}^{-\mathrm{i}\pi} = -1\). This is the geometric content of the statement that a spin-1/2 requires a \(4\pi\) rotation to return to itself.
Poll: Berry phase on the Bloch sphere
A spin-1/2 eigenstate \(\vert\uparrow(\hat{\boldsymbol{n}})\rangle\) is parallel-transported around a closed loop \(\mathcal{C}\) on the Bloch sphere subtending solid angle \(\Omega_{\text{solid}}\). Which statement is correct?
(A) The Berry phase equals the loop’s enclosed area divided by the total surface area of the Bloch sphere.
(B) The Berry phase is \(-\Omega_{\text{solid}}/2\) — a gauge-invariant geometric quantity unrelated to the eigenenergy.
(C) The Berry phase vanishes because the eigenstate returns to itself; any nonzero phase would violate single-valuedness.
(D) The Berry phase depends on how fast the loop is traversed: slower traversal gives a smaller Berry phase.
Summary#
A Hamiltonian \(\hat{H}(\boldsymbol{R})\) that depends on parameters \(\boldsymbol{R}\) defines an eigenstate \(\vert n(\boldsymbol{R})\rangle\) at each \(\boldsymbol{R}\); a system in level \(n\) stays in level \(n\) as \(\boldsymbol{R}\) moves, but the in-level eigenstate itself moves with \(\boldsymbol{R}\).
The eigenstate is defined only up to a local phase \(\mathrm{e}^{\mathrm{i}\alpha(\boldsymbol{R})}\), a gauge freedom on parameter space.
The Berry connection \(\boldsymbol{A}_n = \mathrm{i}\langle n\vert\nabla_{\boldsymbol{R}}\vert n\rangle\) absorbs this ambiguity; it is real and transforms like a vector potential, \(\boldsymbol{A}_n \to \boldsymbol{A}_n - \nabla_{\boldsymbol{R}}\alpha\).
The open-path integral of \(\boldsymbol{A}_n\) shifts by boundary terms under gauge transformations and is not physical; the closed-loop integral, the Berry phase, is gauge-invariant and physical.
The Berry curvature \(\boldsymbol{\Omega}_n = \nabla_{\boldsymbol{R}}\times\boldsymbol{A}_n\) is locally gauge-invariant; the Berry phase is the flux of \(\boldsymbol{\Omega}_n\) through the enclosed surface.
For a spin-1/2 the Berry curvature is the field of a unit monopole at the centre of the Bloch sphere, \(\boldsymbol{\Omega} = -\hat{\boldsymbol{n}}/2\), and the closed-loop Berry phase is \(-\Omega_{\text{solid}}/2\); precession is the Lorentz-type response of the spin in this monopole field.
See Also
4.1.3 Gauge Invariance: the same connection, curvature, and holonomy structure on real space, with the electromagnetic vector potential playing the role of \(\boldsymbol{A}_n\).
4.2.2 Aharonov-Bohm Effect: Berry phase in the special case where the parameter is the position of a charged particle around a flux-threaded loop.
4.4.1 Classical Spin: classical version of the spin-monopole dictionary; the gyroscopic term in the spin-axis equation is precisely a monopole Lorentz force.
Homework#
1. Static orthogonality. Let \(\hat{H}\) be a parameter-independent Hermitian operator with \(\hat{H}\vert n\rangle = E_n\vert n\rangle\) and \(\hat{H}\vert m\rangle = E_m\vert m\rangle\), \(E_m \ne E_n\).
(a) Evaluate \(\langle m\vert\hat{H}\vert n\rangle\) two ways and conclude \((E_m - E_n)\langle m\vert n\rangle = 0\), hence \(\langle m\vert n\rangle = 0\).
(b) Argue that if the system is prepared in \(\vert n\rangle\) and evolves under unitary time evolution generated by \(\hat{H}\), then the probability of being found in \(\vert m\rangle\) at any later time is exactly zero. State which property of \(\mathrm{e}^{-\mathrm{i}\hat{H}t/\hbar}\) you used.
(c) Now let \(\hat{H}(\boldsymbol{R})\) depend on parameters and let \(\boldsymbol{R}\) shift by a small amount. Explain qualitatively why the leakage probability between distinct levels remains small, while the in-level eigenstate \(\vert n(\boldsymbol{R})\rangle\) itself can change noticeably.
2. Reality of the Berry connection. The Berry connection is \(\boldsymbol{A}_n(\boldsymbol{R}) = \mathrm{i}\langle n(\boldsymbol{R})\vert\nabla_{\boldsymbol{R}}\vert n(\boldsymbol{R})\rangle\) for a normalised eigenstate.
(a) Differentiate \(\langle n(\boldsymbol{R})\vert n(\boldsymbol{R})\rangle = 1\) with respect to \(\boldsymbol{R}\) and show \(\langle n\vert\nabla_{\boldsymbol{R}} n\rangle\) is purely imaginary.
(b) Conclude that \(\boldsymbol{A}_n(\boldsymbol{R})\) is a real-valued vector field on parameter space.
(c) Without the factor of \(\mathrm{i}\), identify \(\mathrm{Re}\langle n\vert\nabla_{\boldsymbol{R}} n\rangle\) as a total derivative of a quantity that is fixed by normalisation, and explain why this real part carries no geometric information.
3. Gauge transformation of the connection. Under the rephasing \(\vert n(\boldsymbol{R})\rangle \to \vert\tilde{n}(\boldsymbol{R})\rangle = \mathrm{e}^{\mathrm{i}\alpha(\boldsymbol{R})}\vert n(\boldsymbol{R})\rangle\):
(a) Compute \(\nabla_{\boldsymbol{R}}\vert\tilde{n}\rangle\) via the product rule and show \(\tilde{\boldsymbol{A}}_n = \boldsymbol{A}_n - \nabla_{\boldsymbol{R}}\alpha\).
(b) Show that the Berry curvature \(\boldsymbol{\Omega}_n = \nabla_{\boldsymbol{R}}\times\boldsymbol{A}_n\) is unchanged by the gauge transformation.
(c) Show that the open-path line integral \(\int_{\boldsymbol{R}_a}^{\boldsymbol{R}_b}\boldsymbol{A}_n\cdot\mathrm{d}\boldsymbol{R}\) shifts by \(\alpha(\boldsymbol{R}_a) - \alpha(\boldsymbol{R}_b)\), while the closed-loop integral \(\oint_{\mathcal{C}}\boldsymbol{A}_n\cdot\mathrm{d}\boldsymbol{R}\) is gauge-invariant.
4. Spin-1/2 Berry connection. The spin-up eigenstate along \(\hat{\boldsymbol{n}}(\theta,\varphi) = (\sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta)\) is \(\vert\uparrow(\hat{\boldsymbol{n}})\rangle = \cos(\theta/2)\vert\uparrow\rangle + \mathrm{e}^{\mathrm{i}\varphi}\sin(\theta/2)\vert\downarrow\rangle\).
(a) Compute \(A_\theta = \mathrm{i}\langle\uparrow\vert\partial_\theta\vert\uparrow\rangle\) and \(A_\varphi = \mathrm{i}\langle\uparrow\vert\partial_\varphi\vert\uparrow\rangle\).
(b) Compute the Berry phase along a latitude circle at fixed polar angle \(\theta = \theta_0\), \(\varphi \in [0, 2\pi)\).
(c) Verify \(\Phi_{\mathrm{Berry}} = -\Omega_{\text{solid}}/2\), where \(\Omega_{\text{solid}} = 2\pi(1 - \cos\theta_0)\) is the solid angle of the spherical cap above the latitude.
5. Two gauges on Bloch sphere. Consider two phase conventions for the spin-up eigenstate along \(\hat{\boldsymbol{n}}\):
gauge (N): \(\vert\uparrow_N(\hat{\boldsymbol{n}})\rangle = \cos(\theta/2)\vert\uparrow\rangle + \mathrm{e}^{\mathrm{i}\varphi}\sin(\theta/2)\vert\downarrow\rangle\),
gauge (S): \(\vert\uparrow_S(\hat{\boldsymbol{n}})\rangle = \mathrm{e}^{-\mathrm{i}\varphi}\cos(\theta/2)\vert\uparrow\rangle + \sin(\theta/2)\vert\downarrow\rangle\).
(a) Find the rephasing function \(\alpha(\theta,\varphi)\) that relates the two gauges and verify \(\vert\uparrow_S\rangle = \mathrm{e}^{\mathrm{i}\alpha}\vert\uparrow_N\rangle\).
(b) Compute \(A_\varphi\) in both gauges and check the two values differ by \(-\partial_\varphi\alpha\), consistent with the Berry-connection transformation rule.
(c) Show that the closed-loop Berry phase around any latitude \(\theta = \theta_0\) is the same in both gauges, even though \(A_\varphi\) itself is not.
6. Precession from monopole field. Take the Berry curvature on the Bloch sphere of unit-vector parameters to be \(\boldsymbol{\Omega} = -\hat{\boldsymbol{n}}/2\) — a unit monopole at the centre.
(a) For a fictitious charged “particle” whose position is \(\hat{\boldsymbol{n}}\) on the unit sphere, write down the equation of motion that balances a tangential applied force \(\boldsymbol{F}\) against the Lorentz force from the monopole field \(\boldsymbol{\Omega}\) (use the analogy \(\boldsymbol{F}_{\mathrm{Lorentz}} \propto \dot{\hat{\boldsymbol{n}}}\times\boldsymbol{\Omega}\)).
(b) Show that the force-balanced trajectory traces a small circle (a latitude) about the axis along \(\boldsymbol{F}\), rather than moving along \(\boldsymbol{F}\).
(c) Identify this small-circle motion with Larmor precession of the spin, and explain why precession is the geometric signature of a Berry curvature concentrated as a monopole field on the parameter sphere.
7. Berry phase in Bloch band. Consider a Bloch electron in a crystalline solid, with the wavevector \(\boldsymbol{k}\) in the Brillouin zone (BZ) as the parameter and the periodic part \(\vert u_{n\boldsymbol{k}}\rangle\) of the Bloch function as the eigenstate of the Bloch Hamiltonian \(\hat{H}(\boldsymbol{k})\).
(a) Write the Berry connection \(\boldsymbol{A}_n(\boldsymbol{k}) = \mathrm{i}\langle u_{n\boldsymbol{k}}\vert\nabla_{\boldsymbol{k}}\vert u_{n\boldsymbol{k}}\rangle\) and the Berry curvature \(\boldsymbol{\Omega}_n(\boldsymbol{k})\). Explain why the BZ being a torus (a closed manifold) makes the total curvature flux well-defined and independent of gauge.
(b) Define the Chern number \(c_1 = \frac{1}{2\pi}\int_{\mathrm{BZ}}\boldsymbol{\Omega}_n\cdot\mathrm{d}\boldsymbol{S}\) for a filled band, and argue that it cannot change under smooth deformations of \(\hat{H}(\boldsymbol{k})\) that keep the band gap open.
(c) Time-reversal symmetry forces \(c_1 = 0\) for every band. Without proving this, explain qualitatively how time reversal constrains \(\boldsymbol{\Omega}_n(\boldsymbol{k})\), and what kind of physical mechanism is required to obtain a nonzero \(c_1\).