# 4.2.1 Berry Phase
Worked solutions for the homework problems in the [4.2.1 Berry Phase](../ch4_phase-and-gauge/4-2-1-berry-phase) lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

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**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 $\vert n\rangle$,

$$
\langle m\vert\hat{H}\vert n\rangle = \langle m\vert\bigl(\hat{H}\vert n\rangle\bigr) = E_n\langle m\vert n\rangle .
$$

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

$$
\langle m\vert\hat{H}\vert n\rangle = \bigl(\langle m\vert\hat{H}\bigr)\vert n\rangle = E_m\langle m\vert n\rangle .
$$

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

$$
E_n\langle m\vert n\rangle = E_m\langle m\vert n\rangle,
$$

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:

$$
\mathrm{e}^{-\mathrm{i}\hat{H}t/\hbar}\vert n\rangle = \mathrm{e}^{-\mathrm{i}E_n t/\hbar}\vert n\rangle .
$$

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

$$
\bigl\vert\langle m\vert\psi(t)\rangle\bigr\vert^2
= \bigl\vert\mathrm{e}^{-\mathrm{i}E_n t/\hbar}\bigr\vert^2\,\bigl\vert\langle m\vert n\rangle\bigr\vert^2
= 1\cdot 0 = 0 ,
$$

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

$$
\frac{\langle m\vert(\nabla_{\boldsymbol{R}}\hat{H})\vert n\rangle\cdot\delta\boldsymbol{R}}{E_n - E_m} .
$$

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.

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**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,

$$
0 = \nabla_{\boldsymbol{R}}\langle n\vert n\rangle = \langle\nabla n\vert n\rangle + \langle n\vert\nabla n\rangle = z^{*} + z = 2\,\mathrm{Re}\,z,
$$

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

$$
\mathcal{K}\vert\nabla n(\boldsymbol{R})\rangle = \vert\nabla n(\boldsymbol{R})\rangle.
$$

(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 $\psi = n$, $\phi = \nabla n$:

$$
\langle n\vert\nabla n\rangle = \langle\mathcal{K}n\vert\mathcal{K}\nabla n\rangle = \overline{\langle n\vert\nabla n\rangle},
$$

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:

$$
\langle n\vert\nabla n\rangle = 0
\quad\Longrightarrow\quad
\boldsymbol{A}_n(\boldsymbol{R}) = 0,
\qquad
\Phi_\mathrm{Berry} = \oint\boldsymbol{A}_n\cdot\mathrm{d}\boldsymbol{R} = 0
$$

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

$$
\hat{H}\,\mathcal{T}\vert n\rangle = \mathcal{T}\hat{H}\vert n\rangle = E_n\,\mathcal{T}\vert n\rangle,
$$

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

$$
\mathcal{T}\bigl(\mathrm{e}^{\mathrm{i}\beta/2}\vert n\rangle\bigr) = \mathrm{e}^{-\mathrm{i}\beta/2}\,\mathcal{T}\vert n\rangle = \mathrm{e}^{-\mathrm{i}\beta/2}\,\mathrm{e}^{\mathrm{i}\beta}\vert n\rangle = \mathrm{e}^{\mathrm{i}\beta/2}\vert n\rangle,
$$

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 $\mathcal{T} = \mathcal{K}$:

$$
\mathcal{K}\hat{\sigma}^x\mathcal{K} = \hat{\sigma}^x,
\qquad
\mathcal{K}\hat{\sigma}^z\mathcal{K} = \hat{\sigma}^z,
\qquad
\mathcal{K}\hat{\sigma}^y\mathcal{K} = -\hat{\sigma}^y,
$$

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.

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**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:

$$
\nabla_{\boldsymbol{R}}\vert\tilde{n}\rangle
= \mathrm{i}\,(\nabla_{\boldsymbol{R}}\alpha)\,\mathrm{e}^{\mathrm{i}\alpha}\vert n\rangle
+ \mathrm{e}^{\mathrm{i}\alpha}\,\nabla_{\boldsymbol{R}}\vert n\rangle .
$$

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:

$$
\langle\tilde{n}\vert\nabla_{\boldsymbol{R}}\vert\tilde{n}\rangle
= \mathrm{i}\,(\nabla_{\boldsymbol{R}}\alpha)\,\langle n\vert n\rangle
+ \langle n\vert\nabla_{\boldsymbol{R}}\vert n\rangle
= \mathrm{i}\,\nabla_{\boldsymbol{R}}\alpha + \langle n\vert\nabla_{\boldsymbol{R}}\vert n\rangle ,
$$

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

$$
\tilde{\boldsymbol{A}}_n
= \mathrm{i}\,\langle\tilde{n}\vert\nabla_{\boldsymbol{R}}\vert\tilde{n}\rangle
= \mathrm{i}\,(\mathrm{i}\,\nabla_{\boldsymbol{R}}\alpha)
+ \mathrm{i}\,\langle n\vert\nabla_{\boldsymbol{R}}\vert n\rangle
= -\nabla_{\boldsymbol{R}}\alpha + \boldsymbol{A}_n .
$$

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

$$
\tilde{\boldsymbol{\Omega}}_n
= \nabla_{\boldsymbol{R}}\times\tilde{\boldsymbol{A}}_n
= \nabla_{\boldsymbol{R}}\times\bigl(\boldsymbol{A}_n - \nabla_{\boldsymbol{R}}\alpha\bigr)
= \nabla_{\boldsymbol{R}}\times\boldsymbol{A}_n - \nabla_{\boldsymbol{R}}\times\nabla_{\boldsymbol{R}}\alpha .
$$

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

$$
\tilde{\boldsymbol{\Omega}}_n = \nabla_{\boldsymbol{R}}\times\boldsymbol{A}_n = \boldsymbol{\Omega}_n .
$$

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 $\boldsymbol{R}_b$:

$$
\int_{\boldsymbol{R}_a}^{\boldsymbol{R}_b}\tilde{\boldsymbol{A}}_n\cdot\mathrm{d}\boldsymbol{R}
= \int_{\boldsymbol{R}_a}^{\boldsymbol{R}_b}\boldsymbol{A}_n\cdot\mathrm{d}\boldsymbol{R}
- \int_{\boldsymbol{R}_a}^{\boldsymbol{R}_b}\nabla_{\boldsymbol{R}}\alpha\cdot\mathrm{d}\boldsymbol{R} .
$$

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

$$
\int_{\boldsymbol{R}_a}^{\boldsymbol{R}_b}\nabla_{\boldsymbol{R}}\alpha\cdot\mathrm{d}\boldsymbol{R}
= \alpha(\boldsymbol{R}_b) - \alpha(\boldsymbol{R}_a) ,
$$

so

$$
\int_{\boldsymbol{R}_a}^{\boldsymbol{R}_b}\tilde{\boldsymbol{A}}_n\cdot\mathrm{d}\boldsymbol{R}
= \int_{\boldsymbol{R}_a}^{\boldsymbol{R}_b}\boldsymbol{A}_n\cdot\mathrm{d}\boldsymbol{R}
+ \alpha(\boldsymbol{R}_a) - \alpha(\boldsymbol{R}_b) .
$$

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:

$$
\oint_{\mathcal{C}}\tilde{\boldsymbol{A}}_n\cdot\mathrm{d}\boldsymbol{R}
= \oint_{\mathcal{C}}\boldsymbol{A}_n\cdot\mathrm{d}\boldsymbol{R}
+ \alpha(\boldsymbol{R}_a) - \alpha(\boldsymbol{R}_a)
= \oint_{\mathcal{C}}\boldsymbol{A}_n\cdot\mathrm{d}\boldsymbol{R} ,
$$

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.)

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**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 $\hbar = 1$)

$$
\vert s=1, m=+1; \boldsymbol{n}\rangle = \cos^{2}(\theta/2)\,\vert +1\rangle + \tfrac{1}{\sqrt 2}\sin\theta\,\mathrm{e}^{\mathrm{i}\varphi}\vert 0\rangle + \sin^{2}(\theta/2)\,\mathrm{e}^{2\mathrm{i}\varphi}\vert -1\rangle
$$

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

$$
\Phi^{(1)}_\mathrm{Berry}(\theta_0) = -2\pi\,(1 - \cos\theta_0) = -\Omega_\text{solid},
$$

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

$$
\vert s=1, m=+1; \boldsymbol{n}\rangle = \vert\uparrow;\boldsymbol{n}\rangle^{(1)}\otimes\vert\uparrow;\boldsymbol{n}\rangle^{(2)}.
$$

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 $\sin^{2}\theta = 4\sin^{2}(\theta/2)\cos^{2}(\theta/2)$,

$$
\vert c_+\vert^{2} + \vert c_0\vert^{2} + \vert c_-\vert^{2} = \cos^{4}(\theta/2) + 2\sin^{2}(\theta/2)\cos^{2}(\theta/2) + \sin^{4}(\theta/2) = \bigl[\cos^{2}(\theta/2) + \sin^{2}(\theta/2)\bigr]^{2} = 1.
$$

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 $\hat{S}_\pm\vert 0\rangle = \sqrt 2\vert\pm 1\rangle$, $\hat{S}_+\vert -1\rangle = \sqrt 2\vert 0\rangle$, $\hat{S}_-\vert +1\rangle = \sqrt 2\vert 0\rangle$,

$$
\begin{split}
\hat{S}_z\vert\Psi\rangle &= c_+\vert+1\rangle - c_-\vert -1\rangle,\\
\hat{S}_+\vert\Psi\rangle &= \sqrt 2\,c_0\vert+1\rangle + \sqrt 2\,c_-\vert 0\rangle,\\
\hat{S}_-\vert\Psi\rangle &= \sqrt 2\,c_+\vert 0\rangle + \sqrt 2\,c_0\vert -1\rangle.
\end{split}
$$

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.*

$$
\cos\theta\,c_+ + \tfrac{\sin\theta}{2}\sqrt 2\,c_0\,\mathrm{e}^{-\mathrm{i}\varphi}
= \cos^{2}(\theta/2)\cos\theta + \tfrac{\sin^{2}\theta}{2}
= \cos^{2}(\theta/2)\bigl[2\cos^{2}(\theta/2)-1 + 2\sin^{2}(\theta/2)\bigr]
= c_+. \quad\checkmark
$$

*The $\vert 0\rangle$ component.*

$$
\tfrac{\sin\theta}{2}\sqrt 2\bigl(c_-\mathrm{e}^{-\mathrm{i}\varphi} + c_+\mathrm{e}^{\mathrm{i}\varphi}\bigr)
= \tfrac{\sin\theta}{\sqrt 2}\,\mathrm{e}^{\mathrm{i}\varphi}\bigl[\sin^{2}(\theta/2) + \cos^{2}(\theta/2)\bigr]
= c_0. \quad\checkmark
$$

*The $\vert -1\rangle$ component.*

$$
-\cos\theta\,c_- + \tfrac{\sin\theta}{2}\sqrt 2\,c_0\,\mathrm{e}^{\mathrm{i}\varphi}
= \mathrm{e}^{2\mathrm{i}\varphi}\bigl[-\sin^{2}(\theta/2)\cos\theta + \tfrac{\sin^{2}\theta}{2}\bigr]
= \mathrm{e}^{2\mathrm{i}\varphi}\sin^{2}(\theta/2)\bigl[2\cos^{2}(\theta/2) - (2\cos^{2}(\theta/2)-1)\bigr]
= c_-. \quad\checkmark
$$

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

$$
\langle\Psi\vert\partial_\varphi\Psi\rangle = \mathrm{i}\vert c_0\vert^{2} + 2\mathrm{i}\vert c_-\vert^{2} = \mathrm{i}\bigl[2\sin^{2}(\theta/2)\cos^{2}(\theta/2) + 2\sin^{4}(\theta/2)\bigr] = 2\mathrm{i}\sin^{2}(\theta/2) = \mathrm{i}(1-\cos\theta),
$$

so

$$
A_\varphi = \mathrm{i}\langle\Psi\vert\partial_\varphi\Psi\rangle = -(1-\cos\theta).
$$

*$\theta$ 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

$$
\langle\Psi\vert\partial_\theta\Psi\rangle = -\tfrac{1}{2}\sin\theta\cos^{2}(\theta/2) + \tfrac{1}{2}\sin\theta\cos\theta + \tfrac{1}{2}\sin\theta\sin^{2}(\theta/2),
$$

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,

$$
\langle\Psi\vert\partial_\theta\Psi\rangle = \tfrac{1}{2}\sin\theta\bigl[\sin^{2}(\theta/2) - \cos^{2}(\theta/2) + \cos\theta\bigr] = \tfrac{1}{2}\sin\theta\bigl[-\cos\theta + \cos\theta\bigr] = 0,
$$

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,

$$
\Phi^{(1)}_\mathrm{Berry}(\theta_0) = \oint A_\varphi\,\mathrm{d}\varphi = -(1-\cos\theta_0)\!\int_{0}^{2\pi}\mathrm{d}\varphi = -2\pi(1-\cos\theta_0) = -\Omega_\text{solid}.
$$

**(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

$$
\vert\Psi\rangle = \vert s=1, m=+1; \boldsymbol{n}\rangle = \vert\uparrow;\boldsymbol{n}\rangle^{(1)}\otimes\vert\uparrow;\boldsymbol{n}\rangle^{(2)}.
$$

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:

$$
\nabla\vert\Psi\rangle = \vert\nabla\uparrow\rangle^{(1)}\otimes\vert\uparrow\rangle^{(2)} + \vert\uparrow\rangle^{(1)}\otimes\vert\nabla\uparrow\rangle^{(2)}.
$$

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,

$$
\langle\Psi\vert\nabla\Psi\rangle = \langle\uparrow\vert\nabla\uparrow\rangle^{(1)} + \langle\uparrow\vert\nabla\uparrow\rangle^{(2)} = 2\langle\uparrow;\boldsymbol{n}\vert\nabla\vert\uparrow;\boldsymbol{n}\rangle.
$$

Multiplying by $\mathrm{i}$,

$$
\boldsymbol{A}^{(1)}(\boldsymbol{n}) = \mathrm{i}\langle\Psi\vert\nabla\Psi\rangle = 2\,\boldsymbol{A}^{(1/2)}(\boldsymbol{n}).
$$

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),

$$
\Phi^{(1/2)}_\mathrm{Berry}(\theta_0) = \tfrac{1}{2}\Phi^{(1)}_\mathrm{Berry}(\theta_0) = -\tfrac{1}{2}\Omega_\text{solid} = -\pi(1-\cos\theta_0).
$$

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,

$$
\vert s,+s;\boldsymbol{n}\rangle = \bigl(\vert\uparrow;\boldsymbol{n}\rangle\bigr)^{\otimes 2s}.
$$

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

$$
\boldsymbol{A}^{(s)} = 2s\,\boldsymbol{A}^{(1/2)},
\qquad
\Phi^{(s)}_\mathrm{Berry} = 2s\cdot\bigl(-\tfrac{1}{2}\Omega_\text{solid}\bigr) = -s\,\Omega_\text{solid}.
$$

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.

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**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):

$$
\vert\uparrow_S\rangle
= \mathrm{e}^{-\mathrm{i}\varphi}\cos(\theta/2)\vert\uparrow\rangle + \sin(\theta/2)\vert\downarrow\rangle
= \mathrm{e}^{-\mathrm{i}\varphi}\Bigl[\cos(\theta/2)\vert\uparrow\rangle + \mathrm{e}^{\mathrm{i}\varphi}\sin(\theta/2)\vert\downarrow\rangle\Bigr] .
$$

The bracket is exactly $\vert\uparrow_N\rangle$, so

$$
\vert\uparrow_S\rangle = \mathrm{e}^{-\mathrm{i}\varphi}\,\vert\uparrow_N\rangle .
$$

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

$$
\alpha(\theta,\varphi) = -\varphi .
$$

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:

$$
A_\varphi^{(N)} = -\sin^{2}(\theta/2) = -\frac{1-\cos\theta}{2} .
$$

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

$$
\vert\uparrow_S\rangle = \begin{pmatrix}\mathrm{e}^{-\mathrm{i}\varphi}\cos(\theta/2)\\ \sin(\theta/2)\end{pmatrix},
\qquad
\langle\uparrow_S\vert = \begin{pmatrix}\mathrm{e}^{\mathrm{i}\varphi}\cos(\theta/2) & \sin(\theta/2)\end{pmatrix}.
$$

Now only the upper entry carries $\varphi$,

$$
\partial_\varphi\vert\uparrow_S\rangle = \begin{pmatrix}-\mathrm{i}\,\mathrm{e}^{-\mathrm{i}\varphi}\cos(\theta/2)\\ 0\end{pmatrix},
$$

and the overlap is

$$
\langle\uparrow_S\vert\partial_\varphi\vert\uparrow_S\rangle
= \mathrm{e}^{\mathrm{i}\varphi}\cos(\theta/2)\,\bigl(-\mathrm{i}\,\mathrm{e}^{-\mathrm{i}\varphi}\cos(\theta/2)\bigr)
= -\mathrm{i}\cos^{2}(\theta/2) .
$$

Hence

$$
A_\varphi^{(S)} = \mathrm{i}\langle\uparrow_S\vert\partial_\varphi\vert\uparrow_S\rangle
= \mathrm{i}\,(-\mathrm{i})\cos^{2}(\theta/2)
= \cos^{2}(\theta/2)
= \frac{1+\cos\theta}{2} .
$$

The two values differ by

$$
A_\varphi^{(S)} - A_\varphi^{(N)}
= \cos^{2}(\theta/2) - \bigl(-\sin^{2}(\theta/2)\bigr)
= \cos^{2}(\theta/2) + \sin^{2}(\theta/2) = 1 .
$$

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 $\alpha = -\varphi$,

$$
-\partial_\varphi\alpha = -\partial_\varphi(-\varphi) = +1 ,
$$

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:

$$
\oint A_\varphi^{(N)}\,\mathrm{d}\varphi = -2\pi\sin^{2}(\theta_0/2) = -\pi(1-\cos\theta_0),
\qquad
\oint A_\varphi^{(S)}\,\mathrm{d}\varphi = 2\pi\cos^{2}(\theta_0/2) = \pi(1+\cos\theta_0) .
$$

These two *numbers* are not equal — they differ by

$$
\pi(1+\cos\theta_0) - \bigl(-\pi(1-\cos\theta_0)\bigr) = 2\pi .
$$

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

$$
-\oint\nabla\alpha\cdot\mathrm{d}\boldsymbol{R}
= -\bigl[\alpha(\varphi{=}2\pi) - \alpha(\varphi{=}0)\bigr]
= -(-2\pi) = +2\pi ,
$$

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

$$
\mathrm{e}^{\mathrm{i}(\Phi_{\mathrm{Berry}} + 2\pi)} = \mathrm{e}^{\mathrm{i}\Phi_{\mathrm{Berry}}} ,
$$

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.

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**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
$\boldsymbol{R}\to\boldsymbol{k}$,

$$
\boldsymbol{A}_n(\boldsymbol{k}) = \mathrm{i}\langle u_{n\boldsymbol{k}}\vert\nabla_{\boldsymbol{k}}\vert u_{n\boldsymbol{k}}\rangle,
\qquad
\boldsymbol{\Omega}_n(\boldsymbol{k}) = \nabla_{\boldsymbol{k}}\times\boldsymbol{A}_n(\boldsymbol{k}) .
$$

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**

$$
c_1 = \frac{1}{2\pi}\int_{\mathrm{BZ}}\boldsymbol{\Omega}_n\cdot\mathrm{d}\boldsymbol{S}
= \frac{1}{2\pi}\int_{\mathrm{BZ}}\Omega_n(\boldsymbol{k})\,\mathrm{d}^2k .
$$

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:

$$
\boldsymbol{\Omega}_n(-\boldsymbol{k}) = -\boldsymbol{\Omega}_n(\boldsymbol{k}) .
$$

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.)