Question about a step in Griffith's QM book on Berry's phase for an electron in time-varying magnetic field

Question

Some Context

I am going through David Griffith's Introduction to Quantum Mechanics (Second Edition), Chapter 10, Section 10.2 on Berry's phase.

Suppose we have a system with a time-dependent Hamiltonian $\hat{H}(t)$ that is non-degenerate for all $t$ with discrete energies $E_{1}<E_{2}<\cdots$. Suppose further that the system starts in one of the energy eigenstate $|\psi(0)\rangle = |\psi_{n}(0)\rangle$. In the book, we conclude that if a system is undergoing adiabatic change, then the state of the system is approximately $$ |\psi(t)\rangle = e^{i\theta_{n}(t)}e^{i\gamma_{n}(t)}|\psi_{n}(t)\rangle $$ where $$ \theta_{n}(t) = -\frac{1}{\hbar}\int_{0}^{t} E_{n}(\tau)\, d\tau \qquad\text{and}\qquad \gamma_{n}(t) = i\int_{C} \langle\psi_{n} | \nabla_{R}\psi_{n} \rangle\cdot d\vec{R}. $$ See (10.23) and (10.45) in the book. Here $\vec{R} = (R_{1}, \ldots, R_{N})$ is a vector of time-varying parameters that are responsible for the time-dependence of the Hamiltonian: $$ \hat{H}(t) = \hat{H}(R_{1}(t), \ldots, R_{N}(t)). $$

Now in the book, equation (10.43), we took $$ \nabla_{R}\psi_{n}\cdot \frac{d\vec{R}}{dt} = \frac{\partial\psi_{n}}{\partial R_{1}}\frac{dR_{1}}{dt} + \cdots + \frac{\partial\psi_{n}}{\partial R_{N}}\frac{dR_{N}}{dt} $$ in the derivation of (10.45).

Now my questions concerns a specific step in Example 10.2. In the example, we are considering an electron in a magnetic field of constant magnitude $B_{0}$ and time-varying direction given by $(\theta(t), \phi(t))$ in spherical coordinates. The spin-up eigenstate along $\vec{B}(t)$ is given by $$ |\chi_{+}(t)\rangle = \begin{pmatrix} \cos(\theta/2) \\ e^{i\phi}\sin(\theta/2) \end{pmatrix}. $$ See (10.57). Now afterwards, to calculate the Berry phase the book takes the gradient as \begin{align*} \nabla|\chi_{+}(t)\rangle = \frac{\partial |\chi_{+}\rangle}{\partial r} \vec{e}_{r} + \frac{1}{r}\frac{\partial |\chi_{+}\rangle}{\partial\theta}\vec{e}_{\theta} + \frac{1}{r\sin\theta}\frac{\partial |\chi_{+}\rangle}{\partial\phi}\vec{e}_{\phi}. \end{align*}

Question

Now what I don't understand here is, if $\theta(t)$ and $\phi(t)$ are the time-varying parameters $R_{1}$ and $R_{2}$, shouldn't we simply have \begin{align*} \nabla|\chi_{+}(t)\rangle = \frac{\partial |\chi_{+}\rangle}{\partial R_{1}} \vec{e}_{1} + \frac{\partial |\chi_{+}\rangle}{\partial R_{2}}\vec{e}_{2} = \frac{\partial |\chi_{+}\rangle}{\partial \theta} \vec{e}_{\theta} + \frac{\partial |\chi_{+}\rangle}{\partial\phi}\vec{e}_{\phi}? \end{align*} Griffiths offers no reason why the parameters should be interpreted as physical coordinates, and the equation (10.43) involves the gradient only in Cartesian form. What justifies us taking the gradient in terms of spherical coordinates in that scenario?

I think more broadly, my question is, the formula for Berry's curvature (10.45) involves a gradient with respect to parameters $R_{1}, \ldots, R_{N}$. But a gradient is something that depends on a metric? What metric are we presupposing in this equation?

An answer to either of my questions would be greatly appreciated.

Answer 1

The object you are integrating is the differential form $$ d\gamma = A_\mu dx^\mu.$$ We could either write it as $d\gamma = A_\theta d\theta + A_\phi d\phi$ which would use the standard definition $ A_\mu = i\langle \psi| \partial_\mu |\psi\rangle $ or we could equivalently use gradients and write it as $ d\gamma = \mathbf{A} \cdot d\mathbf{R}$, where now $\mathbf{A} = i\langle \psi | \nabla |\psi\rangle$, where the gradient is defined in polar coordinates. I believe the author uses polar coordinates for the gradient because there is a relationship between the Euclidean space the spin is situated in and the abstract parameter space, where the parameter space is the orientation of the spin in Euclidean space.

In terms of metrics:

$$ A_\mu = \langle \psi | \partial_\mu |\psi\rangle = g_{\mu\nu} \langle \psi | \nabla^\nu |\psi\rangle \quad \Rightarrow \quad A_\mu dx^\mu = g_{\mu \nu} \langle \psi | \nabla^\nu |\psi\rangle dx^\mu = \mathbf{A} \cdot d \mathbf{x}$$

The additional factors from using the gradient in the polar coordinate system will be cancelled as the dot product and line element $d \mathbf{R}$ will be expressed in polar coordinates, so we get the same integrand with either definition.

Answer 2

The integral for $\gamma_{n}(t)$ was derived from an application of the chain rule in (10.43). If you consider that part, you will find that it doesn't matter what metric you use. By the chain rule we have $$ \frac{\partial\psi}{\partial t} = (\nabla_{1}\psi)\cdot_{1}\frac{d\vec{R}}{dt} = (\nabla_{2}\psi)\cdot_{2}\frac{d\vec{R}}{dt} $$ where the gradient and dot-product are taken with respect to any metrics $g_{1}, g_{2}$. As long as the dot-product and gradient use the same metric and we are consistent, it doesn't matter what metric is used.

I find it unfortunate that Griffiths doesn't make this explicit, so it seems like there is a gap in reasoning in the book.