Categories: Physics, Quantum mechanics.

Berry phase

Consider a Hamiltonian $$\hat{H}$$ that does not explicitly depend on time, but does depend on a given parameter $$\vb{R}$$. The Schrödinger equations then read:

\begin{aligned} i \hbar \dv{t} \ket{\Psi_n(t)} &= \hat{H}(\vb{R}) \ket{\Psi_n(t)} \\ \hat{H}(\vb{R}) \ket{\psi_n(\vb{R})} &= E_n(\vb{R}) \ket{\psi_n(\vb{R})} \end{aligned}

The general full solution $$\ket{\Psi_n}$$ has the following form, where we allow $$\vb{R}$$ to evolve in time, and we have abbreviated the traditional phase of the “wiggle factor” as $$L_n$$:

\begin{aligned} \ket{\Psi_n(t)} = \exp\!(i \gamma_n(t)) \exp\!(-i L_n(t) / \hbar) \: \ket{\psi_n(\vb{R}(t))} \qquad L_n(t) \equiv \int_0^t E_n(\vb{R}(t')) \dd{t'} \end{aligned}

The geometric phase $$\gamma_n(t)$$ is more interesting. It is not included in $$\ket{\psi_n}$$, because it depends on the path $$\vb{R}(t)$$ rather than only the present $$\vb{R}$$ and $$t$$. Its dynamics can be found by inserting the above $$\ket{\Psi_n}$$ into the time-dependent Schrödinger equation:

\begin{aligned} \dv{t} \ket{\Psi_n} &= i \dv{\gamma_n}{t} \ket{\Psi_n} - \frac{i}{\hbar} \dv{L_n}{t} \ket{\Psi_n} + \exp\!(i \gamma_n) \exp\!(-i L_n / \hbar) \dv{t} \ket{\psi_n} \\ &= i \dv{\gamma_n}{t} \ket{\Psi_n} + \frac{1}{i \hbar} E_n \ket{\Psi_n} + \exp\!(i \gamma_n) \exp\!(-i L_n / \hbar) \: \ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t} \\ &= i \dv{\gamma_n}{t} \ket{\Psi_n} + \frac{1}{i \hbar} \hat{H} \ket{\Psi_n} + \exp\!(i \gamma_n) \exp\!(-i L_n / \hbar) \: \ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t} \end{aligned}

Here we recognize the Schrödinger equation, so those terms cancel. We are then left with:

\begin{aligned} - i \dv{\gamma_n}{t} \ket{\Psi_n} &= \exp\!(i \gamma_n) \exp\!(-i L_n / \hbar) \: \ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t} \end{aligned}

Front-multiplying by $$i \bra{\Psi_n}$$ gives us the equation of motion of the geometric phase $$\gamma_n$$:

\begin{aligned} \boxed{ \dv{\gamma_n}{t} = - \vb{A}_n(\vb{R}) \cdot \dv{\vb{R}}{t} } \end{aligned}

Where we have defined the so-called Berry connection $$\vb{A}_n$$ as follows:

\begin{aligned} \boxed{ \vb{A}_n(\vb{R}) \equiv -i \braket{\psi_n(\vb{R})}{\nabla_\vb{R} \psi_n(\vb{R})} } \end{aligned}

Importantly, note that $$\vb{A}_n$$ is real, provided that $$\ket{\psi_n}$$ is always normalized for all $$\vb{R}$$. To prove this, we start from the fact that $$\nabla_\vb{R} 1 = 0$$:

\begin{aligned} 0 &= \nabla_\vb{R} \braket{\psi_n}{\psi_n} = \braket{\nabla_\vb{R} \psi_n}{\psi_n} + \braket{\psi_n}{\nabla_\vb{R} \psi_n} \\ &= \braket{\psi_n}{\nabla_\vb{R} \psi_n}^* + \braket{\psi_n}{\nabla_\vb{R} \psi_n} = 2 \Re\{ - i \vb{A}_n \} = 2 \Im\{ \vb{A}_n \} \end{aligned}

Consequently, $$\vb{A}_n = \Im \braket{\psi_n}{\nabla_\vb{R} \psi_n}$$ is always real, because $$\braket{\psi_n}{\nabla_\vb{R} \psi_n}$$ is imaginary.

Suppose now that the parameter $$\vb{R}(t)$$ is changed adiabatically (i.e. so slow that the system stays in the same eigenstate) for $$t \in [0, T]$$, along a circuit $$C$$ with $$\vb{R}(0) \!=\! \vb{R}(T)$$. Integrating the phase $$\gamma_n(t)$$ over this contour $$C$$ then yields the Berry phase $$\gamma_n(C)$$:

\begin{aligned} \boxed{ \gamma_n(C) = - \oint_C \vb{A}_n(\vb{R}) \cdot \dd{\vb{R}} } \end{aligned}

But we have a problem: $$\vb{A}_n$$ is not unique! Due to the Schrödinger equation’s gauge invariance, any function $$f(\vb{R}(t))$$ can be added to $$\gamma_n(t)$$ without making an immediate physical difference to the state. Consider the following general gauge transformation:

\begin{aligned} \ket*{\tilde{\psi}_n(\vb{R})} \equiv \exp\!(i f(\vb{R})) \: \ket{\psi_n(\vb{R})} \end{aligned}

To find $$\vb{A}_n$$ for a particular choice of $$f$$, we need to evaluate the inner product $$\braket*{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}$$:

\begin{aligned} \braket*{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n} &= \exp\!(i f) \Big( i \nabla_\vb{R} f \: \braket*{\tilde{\psi}_n}{\psi_n} + \braket*{\tilde{\psi}_n}{\nabla_\vb{R} \psi_n} \Big) \\ &= i \nabla_\vb{R} f \: \braket*{\psi_n}{\psi_n} + \braket*{\psi_n}{\nabla_\vb{R} \psi_n} \\ &= i \nabla_\vb{R} f + \braket*{\psi_n}{\nabla_\vb{R} \psi_n} \end{aligned}

Unfortunately, $$f$$ does not vanish as we would have liked, so $$\vb{A}_n$$ depends on our choice of $$f$$.

However, the curl of a gradient is always zero, so although $$\vb{A}_n$$ is not unique, its curl $$\nabla_\vb{R} \cross \vb{A}_n$$ is guaranteed to be. Conveniently, we can introduce a curl in the definition of $$\gamma_n(C)$$ by applying Stokes’ theorem, under the assumption that $$\vb{A}_n$$ has no singularities in the area enclosed by $$C$$ (fortunately, $$\vb{A}_n$$ can always be chosen to satisfy this):

\begin{aligned} \boxed{ \gamma_n(C) = - \iint_{S(C)} \vb{B}_n(\vb{R}) \cdot \dd{\vb{S}} } \end{aligned}

Where we defined $$\vb{B}_n$$ as the curl of $$\vb{A}_n$$. Now $$\gamma_n(C)$$ is guaranteed to be unique. Note that $$\vb{B}_n$$ is analogous to a magnetic field, and $$\vb{A}_n$$ to a magnetic vector potential:

\begin{aligned} \vb{B}_n(\vb{R}) \equiv \nabla_\vb{R} \cross \vb{A}_n(\vb{R}) = \Im\Big\{ \nabla_\vb{R} \cross \braket{\psi_n(\vb{R})}{\nabla_\vb{R} \psi_n(\vb{R})} \Big\} \end{aligned}

Unfortunately, $$\nabla_\vb{R} \psi_n$$ is difficult to evaluate explicitly, so we would like to rewrite $$\vb{B}_n$$ such that it does not enter. We do this as follows, inserting $$1 = \sum_{m} \ket{\psi_m} \bra{\psi_m}$$ along the way:

\begin{aligned} i \vb{B}_n = \nabla_\vb{R} \cross \braket{\psi_n}{\nabla_\vb{R} \psi_n} &= \braket{\psi_n}{\nabla_\vb{R} \cross \nabla_\vb{R} \psi_n} + \bra{\nabla_\vb{R} \psi_n} \cross \ket{\nabla_\vb{R} \psi_n} \\ &= \sum_{m} \braket{\nabla_\vb{R} \psi_n}{\psi_m} \cross \braket{\psi_m}{\nabla_\vb{R} \psi_n} \end{aligned}

The fact that $$\braket{\psi_n}{\nabla_\vb{R} \psi_n}$$ is imaginary means it is parallel to its complex conjugate, and thus the cross product vanishes, so we exclude $$n$$ from the sum:

\begin{aligned} \vb{B}_n &= \sum_{m \neq n} \braket{\nabla_\vb{R} \psi_n}{\psi_m} \cross \braket{\psi_m}{\nabla_\vb{R} \psi_n} \end{aligned}

From the Hellmann-Feynman theorem, we know that the inner products can be rewritten:

\begin{aligned} \braket{\psi_m}{\nabla_\vb{R} \psi_n} = \frac{\matrixel{\psi_n}{\nabla_\vb{R} \hat{H}}{\psi_m}}{E_n - E_m} \end{aligned}

Where we have assumed that there is no degeneracy. This leads to the following result:

\begin{aligned} \boxed{ \vb{B}_n = \Im \sum_{m \neq n} \frac{\matrixel{\psi_n}{\nabla_\vb{R} \hat{H}}{\psi_m} \cross \matrixel{\psi_m}{\nabla_\vb{R} \hat{H}}{\psi_n}}{(E_n - E_m)^2} } \end{aligned}

Which only involves $$\nabla_\vb{R} \hat{H}$$, and is therefore easier to evaluate than any $$\ket{\nabla_\vb{R} \psi_n}$$.

1. M.V. Berry, Quantal phase factors accompanying adiabatic changes, 1984, Royal Society.
2. G. Grosso, G.P. Parravicini, Solid state physics, 2nd edition, Elsevier.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.