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diff --git a/content/know/concept/berry-phase/index.pdc b/content/know/concept/berry-phase/index.pdc new file mode 100644 index 0000000..339599b --- /dev/null +++ b/content/know/concept/berry-phase/index.pdc @@ -0,0 +1,219 @@ +--- +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. |