--- title: "Berry phase" firstLetter: "B" publishDate: 2021-11-29 categories: - Physics - Quantum mechanics date: 2021-11-25T20:42:45+01:00 draft: false markup: pandoc --- # 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](/know/concept/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 $\ket{\nabla_\vb{R} \psi_n}$. ## 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.