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 \Inprod{\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} \Inprod{\psi_n}{\psi_n} = \Inprod{\nabla_\vb{R} \psi_n}{\psi_n} + \Inprod{\psi_n}{\nabla_\vb{R} \psi_n} \\ &= \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}^* + \Inprod{\psi_n}{\nabla_\vb{R} \psi_n} = 2 \Real\{ - i \vb{A}_n \} = 2 \Imag\{ \vb{A}_n \} \end{aligned}

Consequently, $\vb{A}_n = \Imag \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ is always real, because $\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ 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 $\inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}$:

\begin{aligned} \inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n} &= \exp(i f) \Big( i \nabla_\vb{R} f \: \inprod{\tilde{\psi}_n}{\psi_n} + \inprod{\tilde{\psi}_n}{\nabla_\vb{R} \psi_n} \Big) \\ &= i \nabla_\vb{R} f \: \inprod{\psi_n}{\psi_n} + \inprod{\psi_n}{\nabla_\vb{R} \psi_n} \\ &= i \nabla_\vb{R} f + \inprod{\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}) = \Imag\!\Big\{ \nabla_\vb{R} \cross \Inprod{\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 \Inprod{\psi_n}{\nabla_\vb{R} \psi_n} &= \Inprod{\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} \Inprod{\nabla_\vb{R} \psi_n}{\psi_m} \cross \Inprod{\psi_m}{\nabla_\vb{R} \psi_n} \end{aligned}

The fact that $\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ 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} \Inprod{\nabla_\vb{R} \psi_n}{\psi_m} \cross \Inprod{\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} \Inprod{\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 = \Imag \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 $\Ket{\nabla_\vb{R} \psi_n}$.

## References

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.