Categories: Physics, Quantum mechanics.

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 \(\ket{\nabla_\vb{R} \psi_n}\).

- M.V. Berry, Quantal phase factors accompanying adiabatic changes, 1984, Royal Society.
- 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.