Chapter 2: Quantum Dynamics and Geometric Phases

2.1 Evolution Operator & Heisenberg Picture

Evolution Operator

For an infinitesimal time evolution, the state changes as: \(\ket{\psi(t + \Delta t)} = e^{- \frac{i}{\hbar} \hat{H}(t) \Delta t} \ket{\psi(t)}\)

So for a finite time evolution, we have:

\[ \ket{\psi(t + T)} = \lim_{N \to \infty} \left( e^{- \frac{i}{\hbar} \hat{H}(t + (N-1) \Delta t) \Delta t} \cdots e^{- \frac{i}{\hbar} \hat{H}(t + \Delta t) \Delta t} e^{- \frac{i}{\hbar} \hat{H}(t) \Delta t} \right) \ket{\psi(t)} \]

Special cases:

  1. if \(\hat{H}\) is time-independent, then we have:

    \[ \ket{\psi(t + T)} = e^{- \frac{i}{\hbar} \hat{H} T} \ket{\psi(t)} \]

  2. if \([\hat{H}(t_1), \hat{H}(t_2)] = 0\) for any \(t_1\) and \(t_2\), then we have:

    \[ \ket{\psi(t + T)} = e^{- \frac{i}{\hbar} \int_t^{t+T} \hat{H}(t') dt'} \ket{\psi(t)} \]

Dyson series: for the common case, we have:

\[ \ket{\psi(t + T)} = \left( 1 + \sum_{n=1}^{\infty} \left( -\frac{i}{\hbar} \right)^n \int_t^{t+T} dt_1 \int_t^{t_1} dt_2 \cdots \int_t^{t_{n-1}} dt_n \hat{H}(t_1) \hat{H}(t_2) \cdots \hat{H}(t_n) \right) \ket{\psi(t)} \]

which is the expansion of the time-ordered exponential: \(\mathcal{T} e^{- \frac{i}{\hbar} \int_t^{t+T} \hat{H}(t') dt'}\).

Heisenberg Picture

What is reality? In classical mechanics, we always think that the “mass point” is real, and its position and momentum are well-defined. But in quantum mechanics, the mass point having certain position and momentum is not real, instead, we can only measure the expectation values of a operator \(\hat{A}\) in a state \(\ket{\psi}\), which is \(\langle \hat{A} \rangle = \bra{\psi} \hat{A} \ket{\psi}\).

Based on this idea, we have two equivalent pictures to describe the time evolution of a quantum system: Schrödinger picture and Heisenberg picture. In Schrödinger picture, the state \(\ket{\psi(t)}\) evolves with time, while the operator \(\hat{A}\) is time-independent. In Heisenberg picture, we let the operator evolve with time, while the state is time-independent.

Base on this idea, we can get the Heisenberg equations:

\[ i \hbar \partial_t \ket{\psi^H} = 0 \]

\[ i \hbar \partial_t \hat{A}^H (t) = i \hbar \partial_t \left(e^{\frac{i}{\hbar} \hat{H} t} \hat{A} e^{-\frac{i}{\hbar} \hat{H} t}\right) = [\hat{H}, \hat{A}^H(t)] \]

In the Heisenberg picture, the state is frozen: \(\ket{\psi^H} = \ket{\psi(0)}\), and operators carry the time dependence:

\[ \hat{A}^H(t) = e^{\frac{i}{\hbar}\hat{H}t}\, \hat{A}_S\, e^{-\frac{i}{\hbar}\hat{H}t} \]

where \(\hat{A}_S\) is the Schrödinger-picture operator (time-independent). Differentiate using the product rule:

\[ \begin{aligned} i\hbar \frac{d}{dt}\hat{A}^H(t) &= i\hbar \left[\frac{i}{\hbar}\hat{H}\, e^{\frac{i}{\hbar}\hat{H}t}\hat{A}_S e^{-\frac{i}{\hbar}\hat{H}t} + e^{\frac{i}{\hbar}\hat{H}t}\hat{A}_S \left(-\frac{i}{\hbar}\hat{H}\right) e^{-\frac{i}{\hbar}\hat{H}t}\right] \\ &= -\hat{H}\, e^{\frac{i}{\hbar}\hat{H}t}\hat{A}_S e^{-\frac{i}{\hbar}\hat{H}t} + e^{\frac{i}{\hbar}\hat{H}t}\hat{A}_S e^{-\frac{i}{\hbar}\hat{H}t}\,\hat{H} \\ &= -\hat{H}\,\hat{A}^H(t) + \hat{A}^H(t)\,\hat{H} \\ &= [\hat{A}^H(t),\, \hat{H}] \end{aligned} \]

where we used \([\hat{H}, e^{\pm \frac{i}{\hbar}\hat{H}t}] = 0\). Rearranging:

\[ \boxed{i\hbar \frac{d}{dt}\hat{A}^H(t) = [\hat{A}^H(t),\, \hat{H}]} \]

If the Schrödinger-picture operator has explicit time dependence \(\hat{A}_S(t)\), one must add an extra term:

\[ i\hbar \frac{d}{dt}\hat{A}^H(t) = [\hat{A}^H(t), \hat{H}] + i\hbar \left(\frac{\partial \hat{A}_S}{\partial t}\right)^H \]

Ehrenfest Theorem:

\[ \partial_t \langle \hat{A} \rangle = \frac{1}{i \hbar} \langle [\hat{A}, \hat{H}] \rangle \]

For a free particle, \(\hat{H} = \frac{\hat{p}^2}{2m}\).

Momentum: Since \([\hat{p}, \hat{H}] = [\hat{p}, \frac{\hat{p}^2}{2m}] = 0\), the Heisenberg equation gives

\[ \frac{d}{dt}\hat{p}(t) = \frac{1}{i\hbar}[\hat{p}, \hat{H}] = 0 \quad \Rightarrow \quad \hat{p}(t) = \hat{p}(0) \]

Momentum is conserved, as expected from translational symmetry.

Position: We have \([\hat{x}, \hat{H}] = [\hat{x}, \frac{\hat{p}^2}{2m}] = \frac{1}{2m}[\hat{x}, \hat{p}^2]\). Using the identity \([\hat{A}, \hat{B}^2] = [\hat{A}, \hat{B}]\hat{B} + \hat{B}[\hat{A}, \hat{B}]\):

\[ [\hat{x}, \hat{p}^2] = i\hbar \hat{p} + \hat{p}(i\hbar) = 2i\hbar \hat{p} \]

So

\[ \frac{d}{dt}\hat{x}(t) = \frac{1}{i\hbar} \cdot \frac{2i\hbar \hat{p}}{2m} = \frac{\hat{p}(t)}{m} = \frac{\hat{p}(0)}{m} \]

Integrating:

\[ \hat{x}(t) = \hat{x}(0) + \frac{\hat{p}(0)}{m}t \]

Taking expectation values:

\[ \langle \hat{x}(t) \rangle = \langle \hat{x}(0) \rangle + \frac{\langle \hat{p}(0) \rangle}{m}t, \quad \langle \hat{p}(t) \rangle = \langle \hat{p}(0) \rangle \]

These are exactly Newton’s equations for a free particle — this is the content of Ehrenfest’s theorem: quantum expectation values obey classical equations of motion.

2.2 Simple Harmonic Oscillator & Coherent States

Ladder Operators & Number States

Let us derive the coherent states from scratch. We start with the Hamiltonian of a simple harmonic oscillator (SHO):

\[ \hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2} m \omega^2 \hat{x}^2 \]

We can express the position and momentum operators in terms of the ladder operators \(\hat{a}\) and \(\hat{a}^\dagger\):

\[ \hat{a} = \sqrt{\frac{m \omega}{2 \hbar}} \left( \hat{x} + \frac{i}{m \omega} \hat{p} \right), \quad \hat{a}^\dagger = \sqrt{\frac{m \omega}{2 \hbar}} \left( \hat{x} - \frac{i}{m \omega} \hat{p} \right) \]

Or \(a^\dagger\) the creation operator, and \(a\) the annihilation operator.

Then, the Hamiltonian can be rewritten as:

\[ \hat{H} = \hbar \omega \left( \hat{a}^\dagger \hat{a} + \frac{1}{2} \right) \]

and the commutation relation is:

\[ [\hat{a}, \hat{a}^\dagger] = 1 \]

Now we consider the number states \(\ket{n}\), which are eigenstates of the number operator \(\hat{N} = \hat{a}^\dagger \hat{a} \equiv \frac{\hat{H}}{\hbar \omega} - \frac{1}{2}\). We have another commutation relation:

\[ [\hat{N}, \hat{a}] = -\hat{a}, \quad [\hat{N}, \hat{a}^\dagger] = \hat{a}^\dagger \]

Put \(\hat{N}\) on the \(\ket{n}\), we have:

\[ \hat{N} \ket{n} = \hat{a}^\dagger \hat{a} \ket{n} = \left(\frac{E_n}{\hbar \omega} - \frac{1}{2} \right) \ket{n} \equiv n \ket{n} \]

add another \(\hat{a}\) on above equation, we get:

\[ \hat{a} \hat{N} \ket{n} = n \hat{a} \ket{n} = (\hat{a}^\dagger \hat{a} + [\hat{a}, \hat{a}^\dagger]) \hat{a} \ket{n} = \hat{N} \hat{a}\ket{n} + \hat{a} \ket{n} \]

So

\[ \hat{N} \hat{a} \ket{n} = (n-1) \hat{a} \ket{n} \]

Similarly, we can show that:

\[ \hat{a} \ket{n} = C \ket{n-1}, \quad \hat{a}^\dagger \ket{n} = C' \ket{n+1} \]

If we want to find the coefficients \(C\) and \(C'\), we can use the normalization condition:

\[ \bra{n}\ket{\hat{a}^\dagger \hat{a} \ket{n} = n \bra{n} n } = n \]

so

\[ C = \sqrt{n}, \quad C' = \sqrt{n+1} \]

Using induction relation, we can find the explicit form of the number states:

\[ \ket{n} = \frac{(\hat{a}^\dagger)^n}{\sqrt{n!}} \ket{0} \]

Dynamics of Oscillator

By Heisenberg equation of motion, we have:

\[ \frac{d}{dt} \hat{a}(t) = \frac{1}{i \hbar} [\hat{a}(t), \hat{H}] = -i \omega \hat{a}(t) \]

so \(\hat{a}(t) = \hat{a}(0) e^{-i \omega t}\), and similarly \(\hat{a}^\dagger(t) = \hat{a}^\dagger(0) e^{i \omega t}\).

So we have the momentum and position operators as:

\[ \hat{x}(t) = \sqrt{\frac{\hbar}{2 m \omega}} \left( \hat{a}(0) e^{-i \omega t} + \hat{a}^\dagger(0) e^{i \omega t} \right), \quad \hat{p}(t) = -i \sqrt{\frac{\hbar m \omega}{2}} \left( \hat{a}(0) e^{-i \omega t} - \hat{a}^\dagger(0) e^{i \omega t} \right) \]

Or in real form:

\[ \begin{aligned} \hat{x}(t) &= \hat{x}(0) \cos(\omega t) + \frac{\hat{p}(0)}{m \omega} \sin(\omega t) \\ \hat{p}(t) &= \hat{p}(0) \cos(\omega t) - m \omega \hat{x}(0) \sin(\omega t) \end{aligned} \]

While the space and momentum operators oscillate with time, the expectation values of \(x\) and \(p\) in an energy eigenstate \(\ket{n}\) do not oscillate:

\[ \langle n| \hat{x}(t) |n\rangle = \sqrt{\frac{\hbar}{2 m \omega}} \left( \bra{n} \hat{a}(0) \ket{n} e^{-i \omega t} + \bra{n} \hat{a}^\dagger(0) \ket{n} e^{i \omega t} \right) = 0 \]

since \(\bra{n}\hat{a}\ket{n} = \sqrt{n}\braket{n|n-1} = 0\) and likewise \(\bra{n}\hat{a}^\dagger\ket{n} = 0\). Similarly, \(\langle n|\hat{p}(t)|n\rangle = 0\) for all \(t\).

If we are on an eigenstate of the Hamiltonian, then we can’t observe the oscillation of the particle — the expectation values are static.

Coherent states resolve this. For a coherent state \(\ket{\alpha}\) (eigenstate of \(\hat{a}\) with eigenvalue \(\alpha\)), we have \(\bra{\alpha}\hat{a}\ket{\alpha} = \alpha \neq 0\). Writing \(\alpha = |\alpha|e^{i\phi}\):

\[ \langle \alpha | \hat{x}(t) | \alpha \rangle = \sqrt{\frac{\hbar}{2m\omega}}\left(\alpha e^{-i\omega t} + \alpha^* e^{i\omega t}\right) = \sqrt{\frac{2\hbar}{m\omega}}\,|\alpha|\cos(\omega t - \phi) \]

\[ \langle \alpha | \hat{p}(t) | \alpha \rangle = -i\sqrt{\frac{\hbar m \omega}{2}}\left(\alpha e^{-i\omega t} - \alpha^* e^{i\omega t}\right) = -\sqrt{2\hbar m \omega}\,|\alpha|\sin(\omega t - \phi) \]

These are exactly the classical equations of motion for a harmonic oscillator with amplitude \(x_0 = \sqrt{\frac{2\hbar}{m\omega}}|\alpha|\). Moreover, the uncertainty in position and momentum of a coherent state saturates the minimum uncertainty relation:

\[ \Delta x = \sqrt{\frac{\hbar}{2m\omega}}, \quad \Delta p = \sqrt{\frac{\hbar m\omega}{2}}, \quad \Delta x \cdot \Delta p = \frac{\hbar}{2} \]

independent of \(\alpha\) and \(t\). The coherent state is a minimum-uncertainty Gaussian wave packet that oscillates without spreading — it is the most classical state in quantum mechanics.

Coherent States

Definition: A coherent state \(\ket{\alpha}\) is defined as the eigenstate of the annihilation operator \(\hat{a}\):

\[ \hat{a} \ket{\alpha} = \alpha \ket{\alpha} \]

where \(\alpha\) is a complex number.

Now we will derive the explicit form of the coherent state. We can express \(\ket{\alpha}\) as a superposition of number states:

\[ \ket{\alpha} = \sum_{n=0}^{\infty} c_n \ket{n} \]

add the annihilation operator on both sides, we have:

\[ \hat{a} \ket{\lambda} = \sum_{n=0}^{\infty} c_n \hat{a} \ket{n} = \sum_{n=1}^{\infty} c_n \sqrt{n} \ket{n-1} = \sum_{n=0}^{\infty} c_{n+1} \sqrt{n+1} \ket{n} \]

So we have the recurrence relation:

\[ \alpha c_n = c_{n+1} \sqrt{n+1} \Rightarrow c_{n+1} = \frac{\alpha}{\sqrt{n+1}} c_n \Rightarrow c_n = \frac{\alpha^n}{\sqrt{n!}} c_0 \]

By the normalization condition:

\[ \begin{aligned} \sum_{n=0}^{\infty} |c_n|^2 &= 1 \\ \sum_{n=0}^{\infty} |c_0|^2 \frac{|\alpha|^{2n}}{n!} &= 1 \Rightarrow |c_0|^2 e^{|\alpha|^2} = 1 \Rightarrow c_0 = e^{-|\alpha|^2/2} \end{aligned} \]

So we have the explicit form of the coherent state:

\[ \ket{\alpha} = e^{-|\alpha|^2/2} \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} \ket{n} \]

Discussions:

  1. In space representation:

    \[ \ket{\alpha} = e^{-|\alpha|^2/2} \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} \frac{(\hat{a}^\dagger)^n}{\sqrt{n!}} \ket{0} = e^{-|\alpha|^2/2} e^{\alpha \hat{a}^\dagger} \ket{0} \]

    So we have \(\phi_\alpha (x) = \bra{x}\ket{\alpha}\):

    \[ \phi_\alpha (x) = e^{-|\alpha|^2/2} \bra{x} e^{\alpha \hat{a}^\dagger} \ket{0} \]

    Utilizing the Baker-Campbell-Hausdorff formula, we have:

    \[ e^{\alpha \hat{a}^\dagger} = e^{-\frac{|\alpha|^2}{2}} e^{\alpha \hat{a}^\dagger - \alpha^* \hat{a}} e^{\alpha^* \hat{a}} \]

    BCH formula: \(e^{\hat{A} + \hat{B}} = e^{-\frac{1}{2} [\hat{A}, \hat{B}]} e^{\hat{A}} e^{\hat{B}}\), if \([\hat{A}, [\hat{A}, \hat{B}]] = [\hat{B}, [\hat{A}, \hat{B}]] = 0\).

    So finally we have:

    \[ \phi_\alpha (x) = \left( \frac{m \omega}{\pi \hbar} \right)^{1/4} \exp \left( -\frac{m \omega}{2 \hbar} (x - x_0)^2 + \frac{i p_0}{\hbar} (x - x_0/2) \right) \]

2.3 Landau Level

Landau levels are the quantized energy levels of a charged particle in a uniform magnetic field.

MOE in Magnetic Field

Consider the magnetic field \(\mathbf{B} = B \hat{z}\), we choose the symmetric gauge:

\[ \vec{A} = (A_x, A_y) = \frac{B}{2} (-y, x) \]

Because we import the magnetic field, so the Hamiltonian should be rewrited by canonical momentum \(\vec{\pi}\):

\[ H_0 = \frac{1}{2m} \vec{\pi}^2 = \frac{1}{2m} (\vec{p} - e \vec{A})^2, \quad \text{minus for the electron} \]

Written in x-y coordinates:

\[ \hat{H} = \frac{1}{2m}(\hat{\pi}_x^2 + \hat{\pi}_y^2) \]

In quantum mechanics, \(\pi_x\) and \(\pi_y\) do not commute:

\[ [\hat{\pi}_x, \hat{\pi}_y] = [\hat{p}_x - e A_x, \hat{p}_y - e A_y] = -i \hbar (\partial_x A_y - \partial_y A_x) = i \hbar e B \]

Similar to the previous ladder operators, we can define the ladder operators for the Landau levels:

\[ \begin{aligned} \hat{a} = \sqrt{\frac{1}{2e\hbar B}} (\hat{\pi}_x + i \hat{\pi}_y) \\ \hat{a}^\dagger = \sqrt{\frac{1}{2e \hbar B}} (\hat{\pi}_x - i \hat{\pi}_y) \end{aligned} \]

which satisfy: \([\hat{a}, \hat{a}^\dagger] = 1\). We can also define the number operator: \(\hat{N} = \hat{a}^\dagger \hat{a}\) and write the hamiltonian as \(\hat{H} = \hbar \omega (\hat{N} + 1/2)\).

Conclusion: The energy of electron in the universal magnetic field is quantized and the energy gap is \(\hbar \omega\).

Ground Eigenstates

Similarly, we have

\[ \hat{a} \ket{0} = -i \sqrt{\frac{\hbar}{2eB}} (2 \frac{\partial}{\partial \overline{z}} + \frac{eB}{2\hbar} z) \ket{0} = 0 \]

So we have

\[ \psi (z, \overline{z}) = f(z) e^{- \frac{eB}{4\hbar} \overline{z}z} \]

and \(f(z)\) is arbitrary. We choose a good form:

\[ f_n(z) \approx N_n z^n \]

Thus,

\[ \psi_n (z, \overline{z}) = N_n z^n e^{- \frac{eB}{4 \hbar} \overline{z} z} \]

where the normalization factor is:

\[ \int d^2 x |\phi_n|^2 = N_n^2 \int 2 \pi r dr \, r^{2n} e^{- \frac{eBr^2}{2 \hbar}} = N_n^2 \pi \hbar \int_0^\infty r^{2n+1} e^{- \frac{eBr^2}{2 \hbar}} dr = N_n^2 \pi (\frac{2 \hbar}{eB})^{n+1} n! = 1 \]

Corollary:

\[ <r^2> = \bra{\phi_n} r^2 \ket{\phi_n} = (n+1) \frac{2\hbar}{eB} \]

Ground State Degeneracy

For an infinite system, \(n\) can be any non-negative integer, so the ground state degeneracy is infinite. For a finite system of area \(A\), the degeneracy becomes finite.

Define the magnetic length \(\ell_B = \sqrt{\frac{\hbar}{eB}}\). The ground state wavefunctions \(\psi_n(z, \bar{z}) = N_n z^n e^{-|z|^2/(4\ell_B^2)}\) are localized at radius \(r_n = \sqrt{2(n+1)}\,\ell_B\). If the system is a disk of area \(A\), we require \(r_n^2 \lesssim A\), so the maximum angular momentum quantum number is:

\[ n_{\max} \approx \frac{A}{2\pi \ell_B^2} - 1 = \frac{eBA}{2\pi\hbar} - 1 \]

The total number of degenerate states in each Landau level is therefore:

\[ \boxed{ N_\phi = \frac{eBA}{2\pi\hbar} = \frac{\Phi}{\Phi_0} } \]

where \(\Phi = BA\) is the total magnetic flux through the sample and \(\Phi_0 = \frac{2\pi\hbar}{e} = \frac{h}{e}\) is the magnetic flux quantum. Each Landau level can accommodate exactly one state per flux quantum threading the sample.

Quantum Hall Effect

Consider a 2D electron gas (2DEG) in a strong perpendicular magnetic field \(B\). The energy spectrum consists of discrete Landau levels \(E_n = \hbar\omega_c(n + 1/2)\) with \(\omega_c = eB/m\). Each level has degeneracy \(N_\phi = \Phi/\Phi_0\). The filling factor is defined as:

\[ \nu = \frac{N_e}{N_\phi} = \frac{N_e h}{eBA} = \frac{n_e h}{eB} \]

where \(N_e\) is the total number of electrons and \(n_e\) is the 2D electron density.

Integer Quantum Hall Effect

When \(\nu = n\) is an integer (i.e., exactly \(n\) Landau levels are completely filled), the system has a spectral gap \(\hbar\omega_c\) and exhibits:

\[ \boxed{ \sigma_{xy} = \nu \frac{e^2}{h}, \quad \sigma_{xx} = 0 } \]

The Hall conductance is quantized to extraordinary precision (\(\sim 10^{-9}\)), independent of sample details such as disorder, geometry, or material parameters.

Laughlin’s gauge argument: Consider the 2DEG on an annular (Corbino) geometry threaded by an Aharonov-Bohm flux \(\Phi_{AB}\). Adiabatically inserting one flux quantum \(\Phi_0\) maps each single-particle state to the next angular momentum state. For \(\nu\) filled Landau levels, this transfers exactly \(\nu\) electrons from the inner edge to the outer edge. The resulting charge transport gives:

\[ I = \nu e \cdot \frac{1}{T}, \quad V_H = \frac{\Phi_0}{T} \quad \Rightarrow \quad \sigma_{xy} = \frac{I}{V_H} = \nu \frac{e^2}{h} \]

where \(T\) is the time to insert one flux quantum.

Topologically, the IQHE can be understood via the TKNN (Thouless-Kohmoto-Nightingale-den Nijs) invariant: the Hall conductance of the \(n\)-th filled band equals its first Chern number \(C_n\), and \(\sigma_{xy} = \frac{e^2}{h}\sum_{n\,\text{filled}} C_n\).

Fractional Quantum Hall Effect

At fractional filling factors such as \(\nu = 1/3, 2/5, 3/7, \dots\), the Hall conductance is again quantized:

\[ \sigma_{xy} = \nu \frac{e^2}{h} \]

This cannot be explained by non-interacting electrons — it is an intrinsically many-body phenomenon driven by electron-electron Coulomb interactions within a partially filled Landau level.

Laughlin wavefunction (for \(\nu = 1/m\), \(m\) odd):

\[ \Psi_m(z_1, \dots, z_N) = \prod_{i < j}(z_i - z_j)^m \exp\!\left(-\frac{1}{4\ell_B^2}\sum_k |z_k|^2\right) \]

The Jastrow factor \((z_i - z_j)^m\) keeps electrons apart, minimizing Coulomb energy. The exponent \(m\) must be odd for fermions (antisymmetry). The quasiparticle excitations above this state carry fractional charge \(e^* = e/m\) and obey fractional (anyonic) statistics.

2.4 Gauge transformation & Aharonov-Bohm Effect

Gauge Transform in Classical Mechanics

Let’s review the GT in EM first. We can write the Maxwell Eqs. in potential’s form:

\[ \begin{aligned} - \nabla^2 \phi - \frac{\partial}{\partial t} (\nabla \cdot \vec{A}) &= \frac{\rho}{\epsilon_0} \\ \nabla \times (\nabla \times \vec{A}) &= \mu_0 \vec{J} - \frac{1}{c^2} \frac{\partial^2 \vec{A}}{\partial t^2} - \frac{1}{c^2} \nabla \frac{\partial \phi}{\partial t} \end{aligned} \]

and the counterpart GT is:

\[ \begin{aligned} \vec{A} \to \vec{A} + \nabla \Lambda \\ \phi \to \phi - \frac{\partial \Lambda}{\partial t} \end{aligned} \]

This kind of GT is called local, which doesn’t change the physical E&M fields. In covariant 4-dimensional notation, the gauge transformation takes a unified form.

Define the four-potential \(A^\mu = (\phi/c,\, \vec{A})\) (or in natural units \(A^\mu = (\phi,\, \vec{A})\)). The gauge transformation becomes:

\[ \boxed{A_\mu \to A_\mu + \partial_\mu \Lambda} \]

The physical (gauge-invariant) content of the electromagnetic field is encoded in the field strength tensor:

\[ F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu \]

which is manifestly invariant under \(A_\mu \to A_\mu + \partial_\mu \Lambda\) since \(\partial_\mu \partial_\nu \Lambda - \partial_\nu \partial_\mu \Lambda = 0\). The components of \(F_{\mu\nu}\) are:

\[ F_{\mu\nu} = \begin{pmatrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{pmatrix} \]

Maxwell’s equations then take the compact form:

\[ \partial_\mu F^{\mu\nu} = \mu_0 J^\nu, \quad \partial_{[\mu} F_{\nu\rho]} = 0 \]

where the second equation (Bianchi identity) is automatically satisfied by \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\).

Gauge Transform in Quantum Mechanics

In Quantum Mechanics, to ensure the local gauge transformation:

\[ \psi(\vec{x}, t) \to \psi'(\vec{x}, t) = e^{i \frac{e}{\hbar} \Lambda(\vec{x}, t)} \psi(\vec{x}, t) \]

For the free system:

\[ \hat{H}_0 = \frac{1}{2m} (-i \hbar \nabla)^2 + e A_0 \]

Add the vector potential \(\vec{A}\) to the Hamiltonian, we have:

\[ \hat{H} = \frac{1}{2m} (-i \hbar \nabla - e \vec{A})^2 + e A_0 \]

which eigenfunction can be written as:

\[ \psi(\vec{x}) = \psi_0 (\vec{x}) \exp \left[ i \frac{e}{\hbar} \int_{\vec{x}_0}^{\vec{x}} \vec{A}(\vec{x}') \cdot d\vec{x}' \right] \]

AB Effect by GT

In vanilla case, there is no magnetic field:

\[ \psi^{(0)} (\vec{x}) = \psi_1^{(0)} (\vec{x}) + \psi_2^{(0)} (\vec{x}) \] where \(\psi_1^{(0)}\) and \(\psi_2^{(0)}\) are the wavefunctions of the two slits.

Now add a \(\Phi\) between two slits, so wavefunctions gain extra phase of vector potential:

\[ \psi^{(1)} (\vec{x}) = \psi_1^{(0)} (\vec{x}) \exp \left[ i \frac{e}{\hbar} \int_{\vec{x}_0}^{\vec{x}} \vec{A}(\vec{x}') \cdot d\vec{x}' \right] + \psi_2^{(0)} (\vec{x}) \exp \left[ i \frac{e}{\hbar} \int_{\vec{x}_0}^{\vec{x}} \vec{A}(\vec{x}') \cdot d\vec{x}' \right] \]

Rewrite the format:

\[ \psi(\vec{x}) = \psi_1^{(0)} (\vec{x}) + e^{i \frac{2\pi \Phi}{\Phi_0}} \psi_2^{(0)} (\vec{x}) \] where \(\Phi_0 = \frac{h}{e}\) is the magnetic flux quantum.

AB Effect by Path Integral

If voltages of two slits are different (\(V_1 \neq V_2\)), then we can gain extra phase:

\[ \phi_j = - \frac{1}{\hbar} \int_0^t (p^2 / 2m + V_j) dt \]

So the wavefunctions gain extra phase:

\[ \Delta \phi = \frac{1}{\hbar} \int_0^t (V_2 - V_1) dt \]

After all, we can write the wave function in 4-dim format:

\[ \psi(x, t) \to \psi_0(x, t) \exp \left[ i \frac{e}{\hbar} \int_0^x A_\mu dx^\mu \right] \]

2.5 Geometric Phase

Usually, there are two types of phase: dynamic phase and geometric phase.

  • Dynamic phase: To describe the evolution of a system over time.
  • Geometric phase: To describe the evolution of a system over some paths in parameters’ space.

Adiabatic Theorem

Adiabatic Process

Consider a \(H(\lambda(t))\), which satisfies

\[ i\hbar \frac{d}{dt} |\psi(t)\rangle = H(\lambda(t)) |\psi(t)\rangle \]

We assume that \(\ket{n(t)}\) and \(E_n(t)\) are the eigenstates and eigenenergies of the Hamiltonian \(H(t)\) at time \(t\), respectively, and \(\{\ket{n(t)}\}\) forms a complete set. Then the system can be written as:

\[ \ket{\psi(t)} = \sum_n C_n (t) \ket{n(t)} \]

Substitute into the Schrödinger equation:

\[ i\hbar \sum_n \dot{C}_n \ket{n(t)} + i\hbar \sum_n C_n \dot{\ket{n(t)}} = \sum_n C_n H(t) \ket{n(t)} \]

Act on the both sides by \(\bra{m(t)}\):

\[ i \hbar \dot{C}_m + i\hbar C_m \bra{m}\ket{\dot{m}} + i \hbar \sum_{n \neq m} C_n \bra{m} \ket{\dot{n}} = E_m C_m \]

For the term \(\sum_{n \neq m} C_n \bra{m} \ket{\dot{n}}\):

Act \(\partial_t\) on Schrödinger equation \[\dot{H} \ket{n} + H \ket{\dot{n}} = \dot{E}_n \ket{n} + E_n \ket{\dot{n}}\] then act \(\bra{m}\) on both sides \[ \bra{m} \dot{H} \ket{n} + E_m \bra{m} \ket{\dot{n}} = E_n \bra{m} \ket{\dot{n}} \to \bra{m} \ket{\dot{n}} = \frac{\bra{m} \dot{H} \ket{n}}{E_n - E_m} \]

Adiabatic condition: If \(|{\bra{m} \dot{H} \ket{n}}| \ll |E_n - E_m|\), then it is called an adiabatic process, with \(|\bra{m}\ket{\dot{n}}| \ll 1\).

Adiabatic Theorem

Theorem: When a quantum system changes sufficiently slowly with external parameters, the system will always remain in its initial eigenstate. After a period of time, the quantum state may accumulate an additional geometric phase beyond the dynamical phase.

Using the adiabatic condition mentioned above, we can rewrite the Schrödinger equation as:

\[ i \hbar \dot{C}_m(t) + i\hbar C_m \bra{m}\ket{\dot{m}} = E_m C_m(t) \]

Then get \[ \frac{1}{C_m} \frac{d}{dt} C_m (t) = - \frac{i}{\hbar} E_m (t) - \bra{m}\ket{\dot{m}} \] \[ \to C_m (t) = C_m(t) \exp \left[ - \frac{i}{\hbar} \int_0^t E_m (t') dt' \right] e^{i \gamma(t)} \] where \[ \boxed{ \begin{aligned} \gamma(t) &= i \int_0^t \bra{m(t')} \frac{d}{dt} \ket{m(t')} dt' \\ &= i \int_{\lambda_i}^{\lambda_f} \bra{m(\lambda_i)} \frac{\partial}{\partial\lambda_i} \ket{m(\lambda_i)} d\lambda_i \\ &= i \int_{\lambda_i}^{\lambda_f} \bra{m(\lambda_i)} \vec\nabla_\lambda \ket{m(\lambda_i)} \cdot d\vec\lambda \end{aligned} } \]

Thus, we can see that the quantum state accumulates a new phase related to the path in parameter space—known as the geometric phase.

Berry Phase

When the Hamiltonian of the system returns to its initial value after a period of time, the total accumulated geometric phase is \[ \gamma_c = i \oint_c\bra{m(\lambda)}\vec\nabla_\lambda \ket{m(\lambda)} \cdot d \lambda \equiv \oint_c \vec A(\lambda) \cdot d \vec \lambda \] which is the Berry phase. Also, we often call \(\vec A(\lambda)\) the Berry connection.

Because \(\vec A(\lambda)\) is seems like a kind of vector potential, we can also define the Berry curvature as the ‘’magnetic field’’ in parameter space: \[ \vec B(\lambda) = \nabla_\lambda \times \vec A(\lambda) \]

在电磁学中，磁场 \(\boldsymbol{B}\) 是矢量势 \(\boldsymbol{A}\) 的旋度：

\[ \boldsymbol{B}(\boldsymbol{r}) = \nabla_{\boldsymbol{r}} \times \boldsymbol{A}(\boldsymbol{r}) \]

这与 Berry 理论完全平行：

Berry 理论 电磁学
参数空间 \(\boldsymbol{R}\) 实空间 \(\boldsymbol{r}\)
Berry 联络 \(\boldsymbol{\mathcal{A}}(\boldsymbol{R})\) 矢量势 \(\boldsymbol{A}(\boldsymbol{r})\)
Berry 曲率 \(\boldsymbol{\mathcal{F}} = \nabla_R \times \boldsymbol{\mathcal{A}}\) 磁场 \(\boldsymbol{B} = \nabla_r \times \boldsymbol{A}\)
Berry 相位 \(\oint \mathcal{A} \cdot d\boldsymbol{R}\) AB 相位 \(\frac{q}{\hbar}\oint \boldsymbol{A} \cdot d\boldsymbol{r}\)

Aharonov–Bohm 相位作为 Berry 相位：带电粒子在磁场中绕闭合路径 \(C\) 获得的相位为

\[ \gamma_C = \frac{q}{\hbar} \oint_C \boldsymbol{A} \cdot d\boldsymbol{r} = \frac{q}{\hbar} \iint_S \boldsymbol{B} \cdot d\boldsymbol{S} \]

从纤维丛角度看，\(\boldsymbol{A}\) 是实空间上的 \(U(1)\) 规范联络，\(\boldsymbol{B}\) 是其曲率。因此：

磁场 = 实空间上的 Berry 曲率；
Berry 曲率 = 参数空间（如动量空间）上的”磁场”

两者本质上是同一 \(U(1)\) 规范场结构在不同底空间的体现。

spin 1/2 on Bloch Sphere

任何 \(2\times 2\) 厄米哈密顿量（去掉无关的整体能量平移）都可以写成

\[ \hat H(\boldsymbol{R}) = \frac{\hbar\Omega(\boldsymbol{R})}{2}\boldsymbol{n}(\boldsymbol{R})\cdot\boldsymbol{\sigma}, \]

其中

  • \(\boldsymbol{\sigma} = (\sigma_x,\sigma_y,\sigma_z)\) 为 Pauli 矩阵；
  • \(\Omega(\boldsymbol{R})>0\) 是某个”有效频率”；
  • \(\boldsymbol{n}(\boldsymbol{R})\) 是单位向量，代表 Bloch 球上的一点： \[ \boldsymbol{n} = (\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta). \]

典型例子：自旋 1/2 在外磁场 \(\boldsymbol{B}\) 中，

\[ \hat H = -\boldsymbol{\mu}\cdot\boldsymbol{B} = -\gamma\frac{\hbar}{2}\boldsymbol{\sigma}\cdot\boldsymbol{B} = \frac{\hbar\Omega}{2}\boldsymbol{n}\cdot\boldsymbol{\sigma}, \]

其中 \(\boldsymbol{n}\) 是 \(\boldsymbol{B}\) 的方向，\(\Omega = -\gamma |\boldsymbol{B}|\)。 这个哈密顿量的本征值为 \(E_\pm = \pm \frac{\hbar\Omega}{2}\), 对应的本征态可以取为自旋沿 \(\boldsymbol{n}\) 方向的 up/down: \[ \boldsymbol{n}\cdot\boldsymbol{\sigma}|\boldsymbol{n},\pm\rangle = (\pm 1)|\boldsymbol{n},\pm\rangle. \]

在固定的 \(\sigma_z\) 本征基 \(\{|↑\rangle,|↓\rangle\}\) 下，一种常用的参数化是

\[ |\boldsymbol{n},+\rangle = \cos\frac{\theta}{2}|↑\rangle + e^{i\phi}\sin\frac{\theta}{2}|↓\rangle, \]

\[ |\boldsymbol{n},-\rangle = -\sin\frac{\theta}{2}|↑\rangle + e^{i\phi}\cos\frac{\theta}{2}|↓\rangle. \]

下面我们就对 \(|+\rangle \equiv |\boldsymbol{n},+\rangle\) 计算几何相位；对 \(|-\rangle\) 结果只差一个符号。

Explicit Calculation: \(|+\rangle\)

回忆几何相位定义： \[ \gamma_+^{\text{geo}} = i\int_0^T \langle +(\theta(t),\phi(t))|\partial_t +(\theta(t),\phi(t))\rangle dt. \] 因为 \(|+\rangle\) 依赖于 \(\theta,\phi\)，用链式法则： \[ \partial_t |+\rangle = \dot\theta\partial_\theta |+\rangle + \dot\phi\partial_\phi |+\rangle, \] \[ \langle +|\partial_t +\rangle = \dot\theta\langle +|\partial_\theta +\rangle + \dot\phi\langle +|\partial_\phi +\rangle. \] 只要我们算出 \[ \mathcal{A}_\theta \equiv i\langle +|\partial_\theta +\rangle,\quad \mathcal{A}_\phi \equiv i\langle +|\partial_\phi +\rangle, \] 几何相位就写成 \[ \gamma_+^{\text{geo}} = \int_0^T dt\bigl[\mathcal{A}_\theta\dot\theta + \mathcal{A}_\phi\dot\phi\bigr] = \oint_C \Bigl(\mathcal{A}_\theta d\theta + \mathcal{A}_\phi d\phi\Bigr), \] 其中 \(C\) 是 Bloch 球上 \((\theta(t),\phi(t))\) 轨迹的投影。

  1. 计算 Berry 联络 \(\mathcal{A}_\theta,\mathcal{A}_\phi\)

    取 \(|+\rangle = \begin{pmatrix} \cos\frac{\theta}{2} \\ e^{i\phi}\sin\frac{\theta}{2} \end{pmatrix}\), 先算 \(\mathcal{A}_\theta\)： \[ \partial_\theta |+\rangle = \begin{pmatrix} -\frac{1}{2}\sin\frac{\theta}{2} \\ e^{i\phi}\frac{1}{2}\cos\frac{\theta}{2} \end{pmatrix}. \] \[ \langle +|\partial_\theta +\rangle = \cos\frac{\theta}{2}\left(-\frac{1}{2}\sin\frac{\theta}{2}\right) + e^{-i\phi}\sin\frac{\theta}{2}\cdot e^{i\phi}\frac{1}{2}\cos\frac{\theta}{2} = 0. \] \[ \boxed{\mathcal{A}_\theta = i\langle +|\partial_\theta +\rangle = 0.} \] 再算 \(\mathcal{A}_\phi\)： \[ \partial_\phi |+\rangle = \begin{pmatrix} 0 \\ i e^{i\phi}\sin\frac{\theta}{2} \end{pmatrix}. \] \[ \langle +|\partial_\phi +\rangle = \cos\frac{\theta}{2}\cdot 0 + e^{-i\phi}\sin\frac{\theta}{2}\cdot i e^{i\phi}\sin\frac{\theta}{2} = i\sin^2\frac{\theta}{2}. \] \[ \mathcal{A}_\phi = i\langle +|\partial_\phi +\rangle = i\cdot i\sin^2\frac{\theta}{2} = -\sin^2\frac{\theta}{2}. \] \[ \boxed{ \mathcal{A}_\phi(\theta) = -\frac{1-\cos\theta}{2},\quad \mathcal{A}_\theta = 0. } \] 这就是两能级系统中”上能级”在 Bloch 球坐标下的 Berry 联络。

  2. 几何相位的线积分表达

    因为 \(\mathcal{A}_\theta=0\)，所以 \[ \gamma_+^{\text{geo}} = \oint_C \mathcal{A}_\phi d\phi = -\frac{1}{2}\oint_C (1-\cos\theta)d\phi. \]

    这已经是一个非常常见的公式：沿路径在 \((\theta,\phi)\) 空间中积分 \((1-\cos\theta)d\phi\)。 下一步我们要把它变成”立体角”的形式。

Curvature and Solid Angle: \(\gamma_+^{\text{geo}} = -\tfrac{1}{2}\Omega\)

在参数空间中，Berry 曲率 2-形式为 \[ \mathcal{F}_{\theta\phi} = \partial_\theta\mathcal{A}_\phi - \partial_\phi\mathcal{A}_\theta. \]

因为 \(\mathcal{A}_\theta=0\)，所以 \[ \mathcal{F}_{\theta\phi} = \partial_\theta\mathcal{A}_\phi = \partial_\theta\left[-\frac{1-\cos\theta}{2}\right] = -\frac{1}{2}\sin\theta. \]

几何上，\(\mathcal{F}_{\theta\phi} d\theta d\phi\) 正好是”磁单极子场”在球面上的通量密度：与 Bloch 球上的面积元 \(dS = \sin\theta d\theta d\phi\) 成正比。 利用 Stokes 定理（在参数空间），对闭合曲线 \(C\)： \[ \oint_C \mathcal{A}_\phi d\phi = \iint_S \mathcal{F}_{\theta\phi}d\theta d\phi = -\frac{1}{2}\iint_S \sin\theta d\theta d\phi, \] 其中 \(S\) 是由 \(C\) 围成的任意曲面（在 Bloch 球上）。 右边积分刚好是立体角 \(\Omega_S\) \(\Omega_S = \iint_S \sin\theta d\theta d\phi\). 所以 \[ \oint_C \mathcal{A}_\phi d\phi = -\frac{1}{2}\Omega_S. \] 于是得到最终结果： \[ \boxed{ \gamma_+^{\text{geo}} = -\frac{1}{2}\Omega_S, } \]

也就是：两能级系统上能级沿闭合路径做绝热演化时，其几何相位等于该路径在 Bloch 球上所围立体角的一半，并带一个负号。

GPT-5 讲几何相位

1. 从态矢到射线：几何相位真正“住”的空间

平时我们写态为 \(|\psi\rangle\)，但物理上可观的是“射线”（ray）：

  • 物理态：\(|\psi\rangle\) 与 \(e^{i\alpha}|\psi\rangle\) 完全等价；
  • 射线空间：\(\mathcal{P}(\mathcal{H}) = (\mathcal{H}\setminus\{0\})/U(1)\)，即 Hilbert 空间模掉全局相位。

几何结构：

  • 总空间：归一化态矢的集合 \(\mathcal{S}(\mathcal{H})\)；
  • 底空间：射线空间 \(\mathcal{P}(\mathcal{H})\)；
  • 纤维：每个射线上附着的 \(U(1)\) 相位自由度。

这就是一个 \(U(1)\) 主纤维丛。几何相位 = 这个纤维丛上的联络的 holonomy。

1.1 “平行移动条件”与 Berry 联络

不依赖哈密顿量，也可以直接在射线空间上选“规范”，典型做法：

  • 要求态矢的变化方向尽量“正交于自身”： \[ \langle\psi(t)|\dot\psi(t)\rangle\ \text{是纯虚数}\quad\text{或更强：}\quad \langle\psi(t)|\dot\psi(t)\rangle = 0. \]

定义 Berry 联络一形式： \[ \mathcal{A} = i\langle\psi|d\psi\rangle, \] 局部坐标： \[ A_\mu(\lambda) = i\langle\psi(\lambda)|\partial_{\lambda^\mu}\psi(\lambda)\rangle. \]

  • 规范变换 \(|\psi\rangle \to e^{i\chi(\lambda)}|\psi\rangle\)： \[ A_\mu \to A_\mu - \partial_\mu\chi, \] 完全是 \(U(1)\) 规范场的变换律。
  • 几何相位： \[ \gamma(C) = \oint_C A_\mu d\lambda^\mu. \]

要点： 这个构造完全基于射线空间几何，本身不需要绝热或本征态。Berry 相位只是这种一般几何相位在“绝热本征态路径”上的特例。

2. 非绝热几何相位：Aharonov–Anandan 相位

Berry 相位通常要求：

  • 绝热；
  • 始终在瞬时本征态；
  • 路径在参数空间中闭合。

更一般的情况是：演化可以很快、非绝热，只要求射线空间中的路径闭合： \[ |\psi(T)\rangle = e^{i\gamma_{\text{tot}}}|\psi(0)\rangle. \]

Aharonov–Anandan (AA) 相位定义如下：

  1. 总相位 \[ \gamma_{\text{tot}} = \arg\big(\langle\psi(0)|\psi(T)\rangle\big). \]

  2. 动力学相位 \[ \gamma_{\text{dyn}} = -\frac{1}{\hbar}\int_0^T \langle\psi(t)|H(t)|\psi(t)\rangle\,dt. \]

  3. 几何相位 \[ \gamma_{\text{geom}} = \gamma_{\text{tot}} - \gamma_{\text{dyn}}. \]

可以证明：\(\gamma_{\text{geom}}\) 只依赖于射线空间曲线 \([\psi(t)]\) 的“形状”，与演化快慢无关（对重新参数化 \(t\to f(t)\) 不变）。在适当规范下： \[ \gamma_{\text{geom}} = i\oint \langle\psi(t)|\dot\psi(t)\rangle\,dt. \]

与 Berry 相位关系：

  • 绝热 + 始终为瞬时本征态 \(|n(\lambda(t))\rangle\) 时，AA 相位退化为 Berry 相位；
  • AA 相位说明：几何相位的本质是“射线空间曲线的几何”，而非“绝热性”。

3. 非闭合路径：Pancharatnam 相位与 geodesic closure

课堂上多讲的是闭合路径的几何相位。但实验上常见的是态从 \(|\psi_i\rangle\) 演化到 \(|\psi_f\rangle\)，不必只差一个相位。

Pancharatnam 在偏振光干涉中提出的思想：

  • 两态 \(|\psi_1\rangle,|\psi_2\rangle\)（归一化）“相位对齐”的条件： \[ \langle\psi_1|\psi_2\rangle > 0\ \text{（实且正）}, \] 即干涉强度最大。
  • 一般两态： \[ \langle\psi_1|\psi_2\rangle = r e^{i\phi}, \] 其中 \(\phi\) 就可视为二者的“相对几何相位”。

对非闭合路径 \(|\psi(0)\rangle\to|\psi(T)\rangle\)：

  • 定义 Pancharatnam 相位： \[ \phi_P = \arg\big(\langle\psi(0)|\psi(T)\rangle\big). \]
  • 再扣除动力学相位，得到： \[ \gamma_{\text{geom}} = \phi_P - \gamma_{\text{dyn}}. \]

geodesic closure 图像（特别适用于两能级 Bloch 球）：

  1. 在射线空间（对二能级系统就是 Bloch 球）上，原路径从点 \(A\) 到点 \(B\)；
  2. 用球面测地线（大圆上最短弧）把 \(B\) 接回 \(A\)；
  3. 用“原路径 + 这条测地线”围成闭合曲线；
  4. 几何相位等于该闭合曲线所围立体角的一半。

这样就把非闭合路径的相位问题，转回到闭合路径几何相位上。

4. 简并本征空间与非阿贝尔几何相位（Wilczek–Zee）

讲义里通常只讨论非简并本征态，此时每个本征态有一个 \(U(1)\) 的 Berry 相位。如果能级存在简并，情况升级为非阿贝尔几何相位。

设第 \(n\) 个能级在参数 \(\lambda\) 处的简并度为 \(d\)： \[ H(\lambda)|n,a;\lambda\rangle = E_n(\lambda)|n,a;\lambda\rangle,\quad a=1,\dots,d. \]

绝热且无跃迁到其他能级时，系统态可写为： \[ |\psi(t)\rangle = \sum_{a=1}^d c_a(t)\,|n,a;\lambda(t)\rangle. \]

代入薛定谔方程并投影到简并子空间，得到系数的演化： \[ i\hbar\,\dot{\vec c}(t) = i\hbar\,\mathbf{A}_\mu(\lambda)\dot\lambda^\mu\,\vec c(t) + E_n(\lambda)\,\vec c(t), \] 其中 \(\vec c\) 是 \(d\) 维列向量，\(\mathbf{A}_\mu\) 是 \(d\times d\) 矩阵： \[ (\mathbf{A}_\mu)_{ab} = i\langle n,a;\lambda|\partial_{\lambda^\mu} n,b;\lambda\rangle. \]

  • 对局域基变换 \[ |n,a;\lambda\rangle \to \sum_b U_{ab}(\lambda) |n,b;\lambda\rangle,\quad U(\lambda)\in U(d), \] 有 \[ \mathbf{A}_\mu \to U\mathbf{A}_\mu U^{-1} + i(\partial_\mu U) U^{-1}, \] 这就是非阿贝尔规范场的标准变换律。

  • 几何演化算符： \[ \mathcal{U}_\text{geom}(C) = \mathcal{P}\exp\left(-\int_C \mathbf{A}_\mu d\lambda^\mu\right)\in U(d), \] 即路径有序指数产生一个 \(U(d)\) 矩阵，而不只是一个相位因子。

物理意义：

  • 非阿贝尔几何相位可以在简并子空间内部“混合”不同本征态；
  • 这是 holonomic quantum computation 的理论基础：通过在参数空间做闭合回路，实现对编码在简并子空间中的量子信息的几何幺正门，实现对局部噪声相对稳定的操控。

5. 拓扑视角：Berry 曲率、Chern 数与拓扑物态

在讲义中，Berry 曲率被类比为“参数空间中的磁场”。如果参数空间本身是紧致流形（例如二维 Brillouin 区），则 Berry 曲率可以定义一个拓扑不变量：

  • Berry 曲率： \[ \Omega_n(\mathbf{k}) = \nabla_{\mathbf{k}}\times\mathbf{A}_n(\mathbf{k}), \] 对应第 \(n\) 个能带。

  • 一阶 Chern 数： \[ C_n = \frac{1}{2\pi}\iint_{\text{BZ}} \Omega_n(\mathbf{k})\,d^2k\in\mathbb{Z}. \]

Chern 数是整数，在不闭合能隙的连续形变下不变，是系统的拓扑不变量。

物理后果：

  • 整数量子霍尔效应： \[ \sigma_{xy} = \frac{e^2}{h}\sum_{n\ \text{填充}} C_n, \] 即霍尔电导由填充能带的 Chern 数决定；
  • 一维系统（如 SSH 模型）的 Zak 相位： \[ \gamma_{\text{Zak}} = \oint_{\text{BZ}} A(k)\,dk, \] 和极化、边界态密切相关；
  • 拓扑绝缘体、拓扑超导的 bulk–edge correspondence 本质上是 Berry 曲率和拓扑不变量的故事。

简言之：几何相位 → Berry 曲率 → 拓扑不变量 → 可观测的量子数（电导、边缘态数目等）。

6. 混合态与退相干：Uhlmann 相位与干涉几何相位

真实系统常是混态，密度矩阵 \(\rho\) 而不是纯态 \(|\psi\rangle\)。如何推广几何相位？

6.1 Uhlmann 相位（数学上自然）

  • 把 \(\rho\) 提纯到更大 Hilbert 空间： \[ \rho = \mathrm{Tr}_\text{anc}|\Psi\rangle\langle\Psi|. \]
  • 对“提纯态”定义平行移动条件，构造一个 Uhlmann 联络，路径结束时的 holonomy 给出 Uhlmann 相位。

Uhlmann 相位在数学上与 Berry 相位平行，但实验上不太容易直接测。

6.2 Sjöqvist 等人的干涉式混态几何相位

另一条路是从干涉实验出发：考虑一束混态系统与参考光束干涉，干涉条纹的 visibility 和相位偏移可用来定义“混态几何相位”。在无退相干的幺正演化下，这种定义与纯态 Berry 相位在结构上高度一致。

退相干的影响：

  • 射线空间上的“曲线”被模糊成“路径族”，几何信息部分丢失；
  • 可观测的几何相位会被衰减（visibility 下降），在极强退相干极限完全消失。

7. 路径积分与自旋系统：Wess–Zumino Berry 项

在路径积分形式下，几何相位常表现为作用中的一个额外“拓扑项”。

对自旋 \(S\) 系统，使用 SU(2) coherent states，可以得到有效作用： \[ S_\text{eff}[\mathbf{n}(t)] = S\int dt\,\big(1-\cos\theta\big)\dot\phi - \int dt\,H(\mathbf{n}(t)), \] 其中 \(\mathbf{n}(t)\) 是 Bloch 球上的单位矢量。这里的 \[ S\int dt\,\big(1-\cos\theta\big)\dot\phi \] 正是 Berry 相位对应的 Wess–Zumino 项（或 Wess–Zumino–Witten 项），它等价于： \[ \gamma_\text{Berry} = -S \times (\text{轨迹在球面上所围立体角}). \]

应用示例：

  • 自旋链的低能有效场论中，这个 Berry 项区分了整数自旋和半整数自旋的相行为（Haldane 猜想）；
  • 在拓扑磁性、自旋冰、skyrmion 晶体中，Berry 相位直接影响有效电磁响应、拓扑 Hall 效应等。

这条路径把几何相位和“拓扑场论”、“自旋几何”紧密联系在一起。