# Geometric phase in magnetic field

$$\newcommand{\ket}[1]{\left|#1\right>}$$ $$\newcommand{\bra}[1]{\left<#1\right|}$$ $$\newcommand{\dv}[2]{\frac{d #1}{d#2}}$$ $$\newcommand{\braket}[2]{\left<#1\middle|#2\right>}$$

Following Griffiths "Introduction to Quantum Mechanics" [2nd edition] example 10.2, I'm trying to get result (10.55):

We have an electron with charge $$-e$$ and mass $$m$$ at rest at the origin, in the presence of a magnetic field whose magnitude $$B_0$$ is constant, but whose direction sweeps out a cone, of opening angle $$\alpha$$, at constant angular velocity $$\omega$$.

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In the adiabatic regime, the eigenspinors are given by the approximation \begin{align} \chi(t)\simeq e^{-i\omega_1 t/2}e^{i(\omega \cos\alpha) t/2}e^{-i\omega t/2}\chi_+(t), \end{align} where $$\omega_1=eB/m$$ is the "cyclotron frequency", $$\omega$$ is the constant precession angular velocity of the magnetic field in the angle $$\alpha$$, and \begin{align} \chi_+(t)=\begin{pmatrix} \cos(\alpha/2)\\ e^{i\omega t}\sin(\alpha/2) \end{pmatrix} \end{align} is the spin up eigenvector.

Geometric phase is given by \begin{align} \gamma(t)=i\int_0^{t}\braket{\chi(t')}{\dv{\chi(t')}{t'}} dt'. \end{align}

My attempt:

It's only algebraic steps.

\begin{align} \dv{\chi(t)}{t}&=e^{-i(\omega_1+\omega(1-\cos\alpha))t/2}\left(-\frac{i}{2}\left(\omega_1+\omega(1-\cos\alpha\right))\chi_+(t)+\dot{\chi}_+(t)\right)\\ &=e^{-i(\omega_1+\omega(1-\cos\alpha))t/2}\left[-\frac{i}{2}\left(\omega_1+\omega(1-\cos\alpha)\right)\begin{pmatrix} \cos(\alpha/2)\\ e^{i\omega t}\sin(\alpha/2) \end{pmatrix}+\begin{pmatrix} 0\\ i\omega e^{i\omega t}\sin(\alpha/2) \end{pmatrix}\right]\\ &=-\frac{i}{2}e^{-i(\omega_1+\omega(1-\cos\alpha))t/2}\begin{pmatrix} (\omega_1+\omega(1-\cos\alpha))\cos(\alpha/2)\\ (\omega_1-\omega(1+\cos\alpha))e^{i\omega t}\sin(\alpha/2). \end{pmatrix} \end{align} Also \begin{align} \chi(t)=e^{-i(\omega_1+\omega(1-\cos\alpha))t/2}\begin{pmatrix} \cos(\alpha/2)\\ e^{i\omega t}\sin(\alpha/2) \end{pmatrix}, \end{align} thus \begin{align} \braket{\chi(t)}{\dv{\chi(t)}{t}}&=-\frac{i}{2}\begin{pmatrix} \cos(\alpha/2) & e^{-i\omega t}\sin(\alpha/2) \end{pmatrix} \begin{pmatrix} (\omega_1+\omega(1-\cos\alpha))\cos(\alpha/2)\\ (\omega_1-\omega(1+\cos\alpha))e^{i\omega t}\sin(\alpha/2). \end{pmatrix}\\ &=-\frac{i}{2}\left\{\left[\omega_1+\omega(1-\cos\alpha)\right]\cos^2(\alpha/2)+\left[\omega_1-\omega(1+\cos\alpha)\right]\sin^2(\alpha/2)\right\}\\ &=-\frac{i}{2}\left\{\omega_1+\omega\left[\cos^2(\alpha/2)-\sin^2(\alpha/2)-\cos\alpha\right]\right\}\\ &=-\frac{i}{2}\omega_1, \end{align} where I used in the third to fourth line the fact that $$\cos^2\theta-\sin^2\theta=\cos(2\theta)$$. Then the geometric phase is readily \begin{align} \gamma_+(t)=\frac{\omega_1}{2}t, \end{align} which clearly differs from (10.55) (even in the $$\omega_1$$-dependency). What I am doing wrong? Maybe I should derive wrt the "inner parameter" $$\omega_1$$?