--- title: "Berry phase" sort_title: "Berry phase" date: 2021-11-29 categories: - Physics - Quantum mechanics layout: "concept" --- 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](/know/concept/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](https://doi.org/10.1098/rspa.1984.0023), 1984, Royal Society. 2. G. Grosso, G.P. Parravicini, *Solid state physics*, 2nd edition, Elsevier.