4.2.1 Berry Phase#

Worked solutions for the homework problems in the 4.2.1 Berry Phase lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

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.

Solution.

(a) Act with \(\hat{H}\) inside the matrix element \(\langle m\vert\hat{H}\vert n\rangle\) in the two possible directions. Letting \(\hat{H}\) act to the right on its eigenstate \(\vert n\rangle\),

Letting it act to the left: \(\hat{H}\) is Hermitian, \(\hat{H}^\dagger = \hat{H}\), so \(\langle m\vert\hat{H} = (\hat{H}\vert m\rangle)^\dagger = (E_m\vert m\rangle)^\dagger = E_m^*\langle m\vert\), and a Hermitian operator has real eigenvalues, \(E_m^* = E_m\). Hence

The matrix element is a single number, so the two evaluations must agree:

so \((E_m - E_n)\langle m\vert n\rangle = 0\).

Since the levels are distinct, \(E_m - E_n \ne 0\), the only way the product can vanish is \(\langle m\vert n\rangle = 0\). Eigenstates of a Hermitian operator belonging to different eigenvalues are orthogonal.

(b) Prepared in \(\vert n\rangle\), the state at time \(t\) is \(\vert\psi(t)\rangle = \mathrm{e}^{-\mathrm{i}\hat{H}t/\hbar}\vert n\rangle\). Because \(\vert n\rangle\) is an eigenstate of \(\hat{H}\), it is also an eigenstate of any function of \(\hat{H}\), in particular of the evolution operator:

The evolution multiplies \(\vert n\rangle\) by a pure phase and nothing else — this is the property used: \(\mathrm{e}^{-\mathrm{i}\hat{H}t/\hbar}\) is diagonal in the energy eigenbasis, so it acts on an energy eigenstate as a scalar phase (\(\mathrm{e}^{-\mathrm{i}\hat{H}t/\hbar}\) commutes with \(\hat{H}\) and cannot induce transitions between its eigenspaces). The probability of finding \(\vert m\rangle\) is therefore

for all \(t\), using \(\langle m\vert n\rangle = 0\) from part (a). An energy eigenstate is stationary: it never develops overlap with a different eigenstate.

(c) Once \(\hat{H}(\boldsymbol{R})\) depends on parameters, the eigenstates \(\vert n(\boldsymbol{R})\rangle\) themselves move as \(\boldsymbol{R}\) moves, and the clean argument of (b) no longer applies verbatim — the state the system follows is the instantaneous eigenstate, which is changing. Two effects must be kept apart.

Leakage between levels. When \(\boldsymbol{R}\) shifts by \(\delta\boldsymbol{R}\), the Hamiltonian changes by \(\delta\hat{H} = (\nabla_{\boldsymbol{R}}\hat{H})\cdot\delta\boldsymbol{R}\). First-order perturbation theory gives the admixture of a different level \(m\) into the evolving state an amplitude of order

This is suppressed by the energy gap \(E_n - E_m\) in the denominator. If \(\boldsymbol{R}\) is varied slowly (the adiabatic limit) and the gap never closes, the leakage probability — quadratic in the small ratio \(\hbar\dot{\boldsymbol{R}}/(\text{gap})\) — stays small, and the system continues to occupy level \(n\) with probability close to one. This is a dynamical, gap-protected effect; it is exactly the residue of the strict orthogonality of (a)–(b).

Motion of the in-level eigenstate. The statement “the system stays in level \(n\)” constrains only the occupation probability, not the eigenstate itself. The vector \(\vert n(\boldsymbol{R})\rangle\) is an \(O(1)\) function of \(\boldsymbol{R}\): as \(\boldsymbol{R}\) traverses a finite loop, \(\vert n(\boldsymbol{R})\rangle\) can rotate substantially within its level — a spin eigenstate, for instance, follows the field direction all the way around the Bloch sphere. Nothing forbids this; orthogonality between distinct levels says nothing about how a single level’s eigenvector is parametrised.

The geometric content of the Berry phase lives precisely in this second effect. The system never leaves level \(n\), yet the in-level eigenstate is dragged around parameter space, and the phase it accumulates relative to itself after a closed loop is the gauge-invariant Berry phase. “Staying in the level” and “the eigenstate moving a lot” are perfectly compatible — the first is about probability, the second about geometry.

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 \(\langle n(\boldsymbol{R})\vert n(\boldsymbol{R})\rangle = 1\) to show that \(\langle n\vert\nabla_{\boldsymbol{R}} n\rangle\) is purely imaginary, hence \(\boldsymbol{A}_n\) is a real-valued vector field on parameter space (the factor \(\mathrm{i}\) in the definition is precisely there to absorb the \(\mathrm{i}\) of the overlap).

(b) Suppose there exists a fixed orthonormal basis \(\{\vert i\rangle\}\) (independent of \(\boldsymbol{R}\)) in which the eigenstate has real components — equivalently, an antiunitary operator \(\mathcal{K}\) acting as \(\mathcal{K}\vert i\rangle = \vert i\rangle\) (extended antilinearly, \(\mathcal{K}(c\vert\psi\rangle) = c^{*}\mathcal{K}\vert\psi\rangle\)) satisfies \(\mathcal{K}\vert n(\boldsymbol{R})\rangle = \vert n(\boldsymbol{R})\rangle\). Using antiunitarity \(\langle\mathcal{K}\psi\vert\mathcal{K}\phi\rangle = \overline{\langle\psi\vert\phi\rangle}\) together with \(\boldsymbol{R}\)-independence of \(\mathcal{K}\), show that \(\langle n\vert\nabla n\rangle\) is real. Combine with (a) to conclude \(\boldsymbol{A}_n(\boldsymbol{R}) = 0\) — real eigenstates carry no Berry phase.

(c) The existence of such a real eigenstate is protected by time-reversal symmetry. For a spinless particle, time reversal is the antiunitary operator \(\mathcal{T} = \mathcal{K}\) above. Show that if \([\hat{H}(\boldsymbol{R}),\mathcal{T}] = 0\) at every \(\boldsymbol{R}\), then each non-degenerate eigenstate can be rephased so that \(\mathcal{T}\vert n(\boldsymbol{R})\rangle = \vert n(\boldsymbol{R})\rangle\) — automatically inheriting the hypothesis of (b). Conclude that a nontrivial Berry phase requires time-reversal symmetry breaking along the loop. For \(\hat{H}(\boldsymbol{R}) = \boldsymbol{h}(\boldsymbol{R})\cdot\hat{\boldsymbol{\sigma}}\), identify the Pauli matrix that breaks \(\mathcal{T}\) and use this to characterise the TRS-protected loops on the Bloch sphere \(\boldsymbol{n} = (\sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta)\) — the closed loops whose Berry phase is forced to vanish.

Solution.

(a) Differentiating \(\langle n\vert n\rangle = 1\) at every \(\boldsymbol{R}\) and applying the Leibniz rule,

where \(z \equiv \langle n\vert\nabla_{\boldsymbol{R}} n\rangle\) and we used \(\langle\nabla n\vert n\rangle = z^{*}\). Hence \(z\) is purely imaginary; write \(z = \mathrm{i}\boldsymbol{f}(\boldsymbol{R})\) with \(\boldsymbol{f}\) real. Then \(\boldsymbol{A}_n = \mathrm{i}z = -\boldsymbol{f}\) is a real-valued vector field — the factor of \(\mathrm{i}\) in the definition converts the imaginary overlap into a real connection, letting \(\boldsymbol{A}_n\) play the role of a genuine gauge potential.

(b) Because \(\{\vert i\rangle\}\) is fixed (independent of \(\boldsymbol{R}\)), differentiating \(\mathcal{K}\vert n(\boldsymbol{R})\rangle = \vert n(\boldsymbol{R})\rangle\) gives

(Explicitly: \(\vert n\rangle = \sum_i c_i(\boldsymbol{R})\vert i\rangle\) has \(c_i\in\mathbb{R}\), so \(\nabla\vert n\rangle = \sum_i (\nabla c_i)\vert i\rangle\) also has real components.) Apply antiunitarity \(\langle\mathcal{K}\psi\vert\mathcal{K}\phi\rangle = \overline{\langle\psi\vert\phi\rangle}\) to \(\psi = n\), \(\phi = \nabla n\):

so \(\langle n\vert\nabla n\rangle\) is real. Combined with (a), which forces it to be purely imaginary, the only consistent value is zero:

around every closed loop: real eigenstates carry no Berry phase.

(c) Suppose \([\hat{H}(\boldsymbol{R}),\mathcal{T}] = 0\) for an antiunitary symmetry \(\mathcal{T}\) at every \(\boldsymbol{R}\). Acting on \(\hat{H}\vert n\rangle = E_n\vert n\rangle\) with \(\mathcal{T}\) — antiunitary leaves the real eigenvalue \(E_n\) untouched — gives

so \(\mathcal{T}\vert n\rangle\) is an eigenstate with the same eigenvalue. If \(E_n\) is non-degenerate, \(\mathcal{T}\vert n\rangle = \mathrm{e}^{\mathrm{i}\beta(\boldsymbol{R})}\vert n(\boldsymbol{R})\rangle\) for some phase \(\beta\). Rephase the eigenstate \(\vert n\rangle \to \mathrm{e}^{\mathrm{i}\beta/2}\vert n\rangle\); antiunitarity then gives

i.e. the rephased state satisfies \(\mathcal{T}\vert n\rangle = \vert n\rangle\) — automatically the hypothesis of (b). By (b), \(\boldsymbol{A}_n \equiv 0\) and the Berry phase vanishes on every loop.

Therefore whenever the Hamiltonian respects time-reversal symmetry along the loop, the Berry phase of a non-degenerate eigenstate is zero. A nontrivial Berry phase requires \(\mathcal{T}\)-symmetry breaking somewhere on the loop.

For the two-level family \(\hat{H} = \boldsymbol{h}\cdot\hat{\boldsymbol{\sigma}}\) with spinless \(\mathcal{T} = \mathcal{K}\):

since \(\hat{\sigma}^y = \begin{pmatrix} 0 & -\mathrm{i}\\ \mathrm{i} & 0\end{pmatrix}\) alone has imaginary entries. So \([\hat{H},\mathcal{K}] = 0\) iff \(h_2 = 0\); the \(\hat{\sigma}^y\) component is the TRS-odd Pauli. For the Bloch-sphere parametrisation, \(h_2 = h\sin\theta\sin\varphi\), which vanishes precisely on the great circle \(n_y = 0\) — the \(\sigma^x\)-\(\sigma^z\) plane through both poles (the union of the meridians \(\varphi = 0\) and \(\varphi = \pi\)). Any closed loop confined to this great circle has \(h_2 \equiv 0\) pointwise, so \([\hat{H},\mathcal{T}] = 0\) along the loop and the Berry phase is forced to vanish by TRS protection. (More trivially, a Hamiltonian with \(h_2 \equiv 0\) everywhere on parameter space has zero Berry phase on every loop, since it is real-symmetric throughout.) Loops that leave the \(n_y = 0\) great circle, in contrast, are no longer protected — TRS is broken at points where \(\sin\varphi \neq 0\), and the Berry phase is generically nonzero.

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.

Solution.

(a) Apply the product rule to \(\vert\tilde{n}\rangle = \mathrm{e}^{\mathrm{i}\alpha(\boldsymbol{R})}\vert n\rangle\). The gradient hits the phase factor and the ket in turn:

The corresponding bra is \(\langle\tilde{n}\vert = \mathrm{e}^{-\mathrm{i}\alpha}\langle n\vert\). Form the overlap; the phase factors \(\mathrm{e}^{-\mathrm{i}\alpha}\mathrm{e}^{\mathrm{i}\alpha} = 1\) cancel in both terms:

using \(\langle n\vert n\rangle = 1\). Multiply by \(\mathrm{i}\) to form the new connection:

So \(\tilde{\boldsymbol{A}}_n = \boldsymbol{A}_n - \nabla_{\boldsymbol{R}}\alpha\). The rephasing of the eigenstate shifts the Berry connection by a gradient — exactly the transformation law \(\boldsymbol{A}\to\boldsymbol{A}+\nabla\alpha\) of an electromagnetic vector potential, with a sign fixed by the convention that \(\alpha\) rephases the state.

(b) The Berry curvature of the rephased state is

The curl of a gradient vanishes identically: for a single-valued \(\alpha\) the mixed second partials commute, \(\partial_i\partial_j\alpha = \partial_j\partial_i\alpha\), so \((\nabla\times\nabla\alpha)_k = \epsilon_{kij}\partial_i\partial_j\alpha = 0\). Therefore

The Berry curvature is gauge-invariant — a local, observable field on parameter space, the analogue of the magnetic field \(\boldsymbol{B} = \nabla\times\boldsymbol{A}\).

(c) Open path. Integrate the transformation law along a path from \(\boldsymbol{R}_a\) to \(\boldsymbol{R}_b\):

The integral of a gradient is the change of the function between the endpoints (the fundamental theorem of calculus for line integrals),

so

The open-path integral shifts by the endpoint terms \(\alpha(\boldsymbol{R}_a) - \alpha(\boldsymbol{R}_b)\). It depends on the arbitrary phase convention chosen at the two ends and is therefore not a physical observable on its own.

Closed loop. For a closed loop \(\mathcal{C}\) the start and end points coincide, \(\boldsymbol{R}_a = \boldsymbol{R}_b\), so the shift collapses:

provided \(\alpha\) is single-valued so that it returns to the same value after one circuit. The closed-loop integral — the Berry phase — is gauge-invariant. The same conclusion follows from (b) via Stokes’ theorem: \(\oint_{\mathcal{C}}\boldsymbol{A}_n\cdot\mathrm{d}\boldsymbol{R} = \int_{\Sigma}\boldsymbol{\Omega}_n\cdot\mathrm{d}\boldsymbol{S}\), and the right-hand side is built from the gauge-invariant curvature. (If \(\alpha\) is allowed to wind — to be multi-valued, changing by \(2\pi\times\)integer around the loop — the Berry phase shifts by an integer multiple of \(2\pi\) and so is well-defined only modulo \(2\pi\); this is exactly the situation met in Problem 5.)

4. Spin-1 Berry phase. Consider a spin-1 parametrised by a unit vector \(\boldsymbol{n}(\theta,\varphi) = (\sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta)\) on the Bloch sphere, with the highest-weight state \(\vert s=1, m=+1;\boldsymbol{n}\rangle\) identified as the eigenstate of \(\hat{\boldsymbol{S}}\cdot\hat{\boldsymbol{n}}\) with the maximum eigenvalue (the overall energy scale plays no role in the Berry phase, so we omit it). Compute the Berry phase of this state around the latitude \(\theta = \theta_0\).

(a) Highest-weight state. Verify that (in units \(\hbar = 1\))

is normalised, and is the eigenstate of \(\hat{\boldsymbol{S}}\cdot\hat{\boldsymbol{n}}\) with eigenvalue \(+1\). Use the spin-1 ladder action \(\hat{S}_\pm\vert m\rangle = \sqrt{2 - m(m\pm 1)}\,\vert m\pm 1\rangle\) and the rotation decomposition \(\hat{\boldsymbol{S}}\cdot\hat{\boldsymbol{n}} = \hat{S}_z\cos\theta + \tfrac{1}{2}\sin\theta\,(\hat{S}_+\mathrm{e}^{-\mathrm{i}\varphi} + \hat{S}_-\mathrm{e}^{\mathrm{i}\varphi})\).

(b) Berry connection. Differentiate the state in (a) and compute the connection components \(A_\varphi = \mathrm{i}\langle\cdot\vert\partial_\varphi\cdot\rangle\) and \(A_\theta = \mathrm{i}\langle\cdot\vert\partial_\theta\cdot\rangle\). Show that \(A_\varphi = -(1-\cos\theta)\) and \(A_\theta = 0\).

(c) Berry phase along a latitude. Integrate \(A_\varphi\) around \(\varphi\in[0,2\pi)\) at fixed \(\theta = \theta_0\). Show that

the full solid angle \(\Omega_\text{solid} = 2\pi(1-\cos\theta_0)\) of the enclosed spherical cap.

(d) SU(2) shortcut. Reproduce the result without the 3-component algebra. The spin-1 representation lives inside \(\mathbf{2}\otimes\mathbf{2}\) as the symmetric subspace; for \(\hat{\boldsymbol{J}} = \hat{\boldsymbol{S}}^{(1)} + \hat{\boldsymbol{S}}^{(2)}\) with each \(\hat{\boldsymbol{S}}^{(i)}\) a spin-1/2, the maximum projection \(J_n = +1\) along \(\boldsymbol{n}\) forces \(S^{(1)}_n = S^{(2)}_n = +\tfrac{1}{2}\), so

Use Leibniz on the product to derive \(\boldsymbol{A}^{(1)} = 2\,\boldsymbol{A}^{(1/2)}\) at the level of Berry connections, and conclude that the spin-1/2 Berry phase along the same latitude is half the spin-1 result, \(\Phi^{(1/2)}_\mathrm{Berry} = -\Omega_\text{solid}/2\). Generalise to spin-\(s\) (the symmetric subspace of \(2s\) spin-1/2’s): \(\Phi^{(s)}_\mathrm{Berry} = -s\,\Omega_\text{solid}\), identifying the spin-\(s\) “Berry monopole” at the centre of the Bloch sphere as a Dirac monopole of charge \(s\).

Solution.

Abbreviate the components as \(c_+ = \cos^{2}(\theta/2)\), \(c_0 = \tfrac{1}{\sqrt 2}\sin\theta\,\mathrm{e}^{\mathrm{i}\varphi}\), \(c_- = \sin^{2}(\theta/2)\,\mathrm{e}^{2\mathrm{i}\varphi}\).

(a) Normalisation and eigenvalue. Using \(\sin^{2}\theta = 4\sin^{2}(\theta/2)\cos^{2}(\theta/2)\),

For the eigenvalue, act on \(\vert\Psi\rangle = c_+\vert+1\rangle + c_0\vert 0\rangle + c_-\vert -1\rangle\) with \(\hat{\boldsymbol{S}}\cdot\hat{\boldsymbol{n}}\). Using \(\hat{S}_z\vert m\rangle = m\vert m\rangle\) and \(\hat{S}_\pm\vert 0\rangle = \sqrt 2\vert\pm 1\rangle\), \(\hat{S}_+\vert -1\rangle = \sqrt 2\vert 0\rangle\), \(\hat{S}_-\vert +1\rangle = \sqrt 2\vert 0\rangle\),

Assemble component by component, using \(c_0\mathrm{e}^{-\mathrm{i}\varphi} = \sin\theta/\sqrt 2\) and \(c_-\mathrm{e}^{-\mathrm{i}\varphi} = \sin^{2}(\theta/2)\mathrm{e}^{\mathrm{i}\varphi}\):

The \(\vert +1\rangle\) component.

The \(\vert 0\rangle\) component.

The \(\vert -1\rangle\) component.

Hence \(\hat{\boldsymbol{S}}\cdot\hat{\boldsymbol{n}}\,\vert\Psi\rangle = (+1)\,\vert\Psi\rangle\) — the state is the highest-weight eigenstate along \(\boldsymbol{n}\).

(b) Berry connection.

\(\varphi\) component. Only \(c_0\) and \(c_-\) carry \(\varphi\) dependence; \(\partial_\varphi c_0 = \mathrm{i}c_0\) and \(\partial_\varphi c_- = 2\mathrm{i}c_-\). Therefore

so

\(\theta\) component. Derivatives: \(\partial_\theta c_+ = -\tfrac{1}{2}\sin\theta\), \(\partial_\theta c_0 = \tfrac{1}{\sqrt 2}\cos\theta\,\mathrm{e}^{\mathrm{i}\varphi}\), \(\partial_\theta c_- = \tfrac{1}{2}\sin\theta\,\mathrm{e}^{2\mathrm{i}\varphi}\). Then

where the cross-phase factors \(\mathrm{e}^{-\mathrm{i}\varphi}\mathrm{e}^{\mathrm{i}\varphi}\) and \(\mathrm{e}^{-2\mathrm{i}\varphi}\mathrm{e}^{2\mathrm{i}\varphi}\) both collapse to \(1\). Grouping,

using \(\sin^{2}(\theta/2) - \cos^{2}(\theta/2) = -\cos\theta\). Hence \(A_\theta = 0\).

(c) Berry phase along the latitude. With \(\mathrm{d}\theta = 0\) along the latitude,

(d) SU(2) shortcut. The total spin operator \(\hat{\boldsymbol{J}} = \hat{\boldsymbol{S}}^{(1)} + \hat{\boldsymbol{S}}^{(2)}\) has projection \(\hat{J}_n = \hat{S}^{(1)}_n + \hat{S}^{(2)}_n\) along \(\boldsymbol{n}\). Each individual \(\hat{S}^{(i)}_n\) has spectrum \(\{+\tfrac{1}{2},-\tfrac{1}{2}\}\), so \(\hat{J}_n\) has spectrum \(\{+1,0,-1\}\) — the spin-1 multiplet. The maximum eigenvalue \(J_n = +1\) is achieved only when both individual projections are maximal, giving the unique product state

The product is automatically symmetric under exchange of the two spin-1/2’s, placing it inside the spin-1 (symmetric) irrep of \(\mathbf{2}\otimes\mathbf{2} = \mathbf{3}_\text{sym}\oplus\mathbf{1}_\text{antisym}\).

Apply Leibniz to the gradient of the product:

Contracting with \(\langle\Psi\vert = \langle\uparrow;\boldsymbol{n}\vert^{(1)}\otimes\langle\uparrow;\boldsymbol{n}\vert^{(2)}\) and using \(\langle\uparrow\vert\uparrow\rangle = 1\) on each factor,

Multiplying by \(\mathrm{i}\),

The spin-1 connection is exactly twice the spin-1/2 connection — no Wigner-\(d\)-matrix manipulation needed. Integrating around the latitude, \(\Phi^{(1)}_\mathrm{Berry} = 2\Phi^{(1/2)}_\mathrm{Berry}\), and using the value from (c),

The spin-1/2 Berry phase is half the solid angle — the famous geometric result here predicted entirely from the spin-1 calculation, with no separate spin-1/2 derivation needed.

Spin-\(s\) generalisation. The highest-weight state of total spin \(J = s\) lives in the fully symmetric subspace of \(2s\) spin-1/2’s,

The same Leibniz argument as above gives \(\langle\Psi\vert\nabla\Psi\rangle = 2s\,\langle\uparrow\vert\nabla\uparrow\rangle\), hence

The “Berry monopole” at the centre of the Bloch sphere has charge \(s\): spin-1/2 is a half-monopole, spin-1 is a unit monopole, and so on. The half-integer/integer distinction of angular momentum is the same geometric fact as the Dirac quantisation of monopole charge on \(S^{2}\) — a deep connection that emerges for free from the tensor-product reduction.

5. Two gauges on Bloch sphere. Consider two phase conventions for the spin-up eigenstate along \(\boldsymbol{n}\):

• gauge (N): \(\vert\uparrow_N(\boldsymbol{n})\rangle = \cos(\theta/2)\vert\uparrow\rangle + \mathrm{e}^{\mathrm{i}\varphi}\sin(\theta/2)\vert\downarrow\rangle\),

• gauge (S): \(\vert\uparrow_S(\boldsymbol{n})\rangle = \mathrm{e}^{-\mathrm{i}\varphi}\cos(\theta/2)\vert\uparrow\rangle + \sin(\theta/2)\vert\downarrow\rangle\).

(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.

Solution.

(a) Factor a common phase out of gauge (S):

The bracket is exactly \(\vert\uparrow_N\rangle\), so

Matching to \(\vert\uparrow_S\rangle = \mathrm{e}^{\mathrm{i}\alpha}\vert\uparrow_N\rangle\) identifies the rephasing function

Both kets describe the same physical spin direction \(\boldsymbol{n}\); they differ only by the bookkeeping phase \(\mathrm{e}^{-\mathrm{i}\varphi}\). The two conventions are the Bloch-sphere analogue of two electromagnetic gauges. Note that gauge (N) is smooth at the north pole (\(\theta=0\) gives \(\vert\uparrow\rangle\), independent of \(\varphi\)) but ill-defined at the south pole (\(\theta=\pi\) gives \(\mathrm{e}^{\mathrm{i}\varphi}\vert\downarrow\rangle\), \(\varphi\)-dependent), whereas gauge (S) is smooth at the south pole and singular at the north — they are the two patches of the Wu-Yang construction.

(b) Gauge (N) was done in Problem 4:

For gauge (S), write the spinor and bra,

Now only the upper entry carries \(\varphi\),

and the overlap is

Hence

The two values differ by

The transformation rule of Problem 3 predicts \(\boldsymbol{A}^{(S)} = \boldsymbol{A}^{(N)} - \nabla\alpha\), i.e. for the \(\varphi\) component \(A_\varphi^{(S)} - A_\varphi^{(N)} = -\partial_\varphi\alpha\). With \(\alpha = -\varphi\),

which matches the computed difference exactly. The Berry-connection transformation rule is confirmed.

(c) Compute the closed-loop integral around the latitude \(\theta = \theta_0\) in each gauge:

These two numbers are not equal — they differ by

The difference is exactly \(2\pi\), and the reason is precisely the caveat noted at the end of Problem 3: the rephasing function \(\alpha = -\varphi\) is not single-valued. Going once around the latitude, \(\varphi\) advances by \(2\pi\) and \(\alpha\) winds by \(-2\pi\), so the loop integral picks up

which is the \(2\pi\) mismatch found above. A single-valued gauge transformation would leave the loop integral exactly invariant (Problem 3c); a winding one shifts it by \(2\pi\times\)integer.

The resolution is that the Berry phase enters physics only through the phase factor \(\mathrm{e}^{\mathrm{i}\Phi_{\mathrm{Berry}}}\) — the amount by which the transported eigenstate is rotated. Since

the two gauges give the same physical Berry phase: it is well-defined modulo \(2\pi\), and the \(2\pi\) discrepancy between the bare integrals is invisible to any measurement. Equivalently, both numbers are valid evaluations of the curvature flux through a cap bounded by the latitude — gauge (N) integrates over the cap above (\(\Omega_{\text{solid}} = 2\pi(1-\cos\theta_0)\), giving \(-\Omega_{\text{solid}}/2\)), gauge (S) over the complementary cap below (\(\Omega'_{\text{solid}} = 2\pi(1+\cos\theta_0)\), giving \(+\Omega'_{\text{solid}}/2\)) — and the two caps differ by the whole sphere, whose total curvature flux is \(\int\boldsymbol{\Omega}\cdot\mathrm{d}\boldsymbol{S} = -\tfrac12\cdot 4\pi = -2\pi\). The closed-loop Berry phase is the same in both gauges in the only sense that matters physically: as an element of \(\mathbb{R}\bmod 2\pi\). That the two single-valued patches cannot be glued without this \(2\pi\) ambiguity is exactly what forces monopole / spin quantisation in §4.4.2.

6. 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 \(c_1\).

Solution.

(a) Here the parameter is the crystal momentum: \(\boldsymbol{R} = \boldsymbol{k}\), and the eigenstate in level \(n\) is the cell-periodic part \(\vert u_{n\boldsymbol{k}}\rangle\) of the Bloch function, satisfying \(\hat{H}(\boldsymbol{k})\vert u_{n\boldsymbol{k}}\rangle = E_n(\boldsymbol{k})\vert u_{n\boldsymbol{k}}\rangle\) with the Bloch Hamiltonian \(\hat{H}(\boldsymbol{k}) = \mathrm{e}^{-\mathrm{i}\boldsymbol{k}\cdot\hat{\boldsymbol{r}}}\hat{H}\,\mathrm{e}^{\mathrm{i}\boldsymbol{k}\cdot\hat{\boldsymbol{r}}}\). The Berry connection and curvature are the general definitions with \(\boldsymbol{R}\to\boldsymbol{k}\),

In two dimensions the curvature has a single independent component, the scalar \(\Omega_n(\boldsymbol{k}) = \partial_{k_x}A_{n,y} - \partial_{k_y}A_{n,x}\).

The Brillouin zone is periodic: crystal momenta \(\boldsymbol{k}\) and \(\boldsymbol{k}+\boldsymbol{G}\) differing by a reciprocal-lattice vector are physically identical. Identifying opposite faces of the BZ turns it into a closed manifold — in 2D a torus \(T^2\) — with no boundary. Two consequences make the total flux well-defined and gauge-independent:

• The Berry curvature is pointwise gauge-invariant (Problem 3b): under a rephasing \(\vert u_{n\boldsymbol{k}}\rangle\to\mathrm{e}^{\mathrm{i}\alpha(\boldsymbol{k})}\vert u_{n\boldsymbol{k}}\rangle\) the curvature is unchanged, \(\tilde{\boldsymbol{\Omega}}_n = \boldsymbol{\Omega}_n\). So the integrand of \(\int_{\mathrm{BZ}}\boldsymbol{\Omega}_n\cdot\mathrm{d}\boldsymbol{S}\) is gauge-invariant everywhere.

• Because the torus has no boundary, the integration region “the whole BZ” is canonical — there is no arbitrary choice of surface or boundary curve whose endpoints could leave a residual gauge-dependent term. The connection \(\boldsymbol{A}_n\) may be impossible to choose smooth and periodic over the entire torus at once (precisely when the flux is nonzero), but the curvature is globally defined and its integral over the closed BZ is a clean, gauge-invariant number.

(b) For a filled band \(n\), define the (first) Chern number

It is the total Berry-curvature flux through the BZ, in units of \(2\pi\). Two facts make it a robust integer:

It is an integer. Splitting the torus into patches and applying Stokes’ theorem, the total flux equals the winding of the transition functions that glue the patches’ gauge choices. Single-valuedness of the eigenstate forces that winding to be an integer multiple of \(2\pi\) — the same \(2\pi\)-ambiguity met in Problem 5(c). Hence \(c_1\in\mathbb{Z}\).

It cannot change under gap-preserving deformations. Suppose \(\hat{H}(\boldsymbol{k})\) is deformed smoothly through a family \(\hat{H}_\lambda(\boldsymbol{k})\), with the band gap above and below band \(n\) staying open for all \(\lambda\). While the gap is open the eigenstate \(\vert u_{n\boldsymbol{k}}\rangle\) is non-degenerate and depends smoothly on both \(\boldsymbol{k}\) and \(\lambda\) (the level-repulsion argument of Problem 1: no resonant mixing as long as \(E_{n\pm1}-E_n\ne 0\)). Therefore \(\boldsymbol{\Omega}_n(\boldsymbol{k})\), and with it the integral \(c_1(\lambda)\), is a continuous function of \(\lambda\). But \(c_1\) takes only integer values. A continuous function valued in \(\mathbb{Z}\) is constant. So \(c_1\) cannot change under any smooth deformation that keeps the gap open: it is a topological invariant of the gapped band. The only way to change it is to close the gap, at which point \(\vert u_{n\boldsymbol{k}}\rangle\) becomes degenerate, the curvature develops a singularity, and \(c_1\) can jump.

(c) Time reversal is implemented by an antiunitary operator \(\mathcal{T}\), and a time-reversal-symmetric crystal has \(\mathcal{T}\hat{H}(\boldsymbol{k})\mathcal{T}^{-1} = \hat{H}(-\boldsymbol{k})\). This relates the band at \(\boldsymbol{k}\) to the band at \(-\boldsymbol{k}\). Because \(\mathcal{T}\) is antiunitary — it includes complex conjugation, which flips the sign of the \(\mathrm{i}\) in the Berry connection — the curvature is odd under time reversal:

Qualitatively: time reversal maps the curvature at \(\boldsymbol{k}\) onto minus the curvature at \(-\boldsymbol{k}\), so wherever the band has positive Berry curvature it must have an equal and opposite amount at the mirror-image momentum. The curvature comes in cancelling pairs. Integrating an odd function over the \(\boldsymbol{k}\to-\boldsymbol{k}\) symmetric BZ, the contributions from \(\boldsymbol{k}\) and \(-\boldsymbol{k}\) cancel exactly, so the total flux vanishes and \(c_1 = 0\) for every band.

Consequently a nonzero Chern number is impossible without breaking time-reversal symmetry. Some physical mechanism that distinguishes the two directions of time is required: an external magnetic field (the integer quantum Hall effect — §4.3.3), spontaneous magnetic order or intrinsic magnetisation (the quantum anomalous Hall effect), or a circulating-current pattern within the unit cell. Only when \(\mathcal{T}\) is broken can the Berry curvature fail to cancel between \(\boldsymbol{k}\) and \(-\boldsymbol{k}\), leaving a net flux and a topologically nontrivial band with \(c_1\ne 0\). (Spin–orbit-coupled time-reversal-invariant insulators evade the conclusion only in a weaker sense: they still have \(c_1=0\), but support a separate \(\mathbb{Z}_2\) topological invariant — beyond the scope of this problem.)