Undergrad How to Find the Berry Phase in Slowly Varying Systems

[ergospherical]

I was trying to follow this page:
https://phyx.readthedocs.io/en/latest/TI/Lecture notes/2.html

With $|n(\mathbf{R})\rangle$ being the eigenstates of $H|\mathbf{R} \rangle$ of eigenvalue $E_n(\mathbf{R})$, the task is to solve the Schroedinger equation $i | \dot{\psi}(t) \rangle = H(\mathbf{R}(t)) |\psi(t)\rangle$ when the parameter vector $\mathbf{R}(t)$ varies slowly with time. They suggest the trial solution\begin{align*}
|\psi(t)\rangle = e^{i\gamma_n(t)} e^{-i\phi_n(t)} |n(\mathbf{R}(t))\rangle
\end{align*}with $\phi_n(t) = \displaystyle{\int}_{0}^t E_n(\mathbf{R}(t')) dt'$. I obtain (omitting functional dependencies for clarity)\begin{align*}
|\dot{\psi}\rangle &= e^{i\gamma_n } e^{-i\phi_n}(i\dot{\gamma}_n |n\rangle - i \dot{\phi}_n |n\rangle+ |\dot{n}\rangle ) \\
&= e^{i\gamma_n} e^{-i\phi_n }(i\dot{\gamma}_n |n\rangle - i E_n |n\rangle+ |\dot{n}\rangle )
\end{align*}so putting $|\dot{\psi}\rangle = -iH |\psi \rangle = -iE_n e^{i\gamma} e^{-i\phi} |n\rangle$ gives\begin{align*}
|\dot{n}\rangle = -i\dot{\gamma}_n |n\rangle
\end{align*}which looks like what they got. Now I'm trying to figure out how they solved for $\gamma_n$, viz:\begin{align*}
\gamma_n(\mathcal{C}) = \int_{\mathcal{C}} i \langle n(\mathbf{R}) | \nabla_{\mathbf{R}} n(\mathbf{R}) \rangle d\mathbf{R} \ \ \ (\dagger)
\end{align*}I assume that $\mathcal{C}$ is a path in $\mathbf{R}$-space, parameterised by time $t$? In other words,\begin{align*}
i\gamma_n(t) &= -\int_0^t \langle n(\mathbf{R}(t') | \nabla_{\mathbf{R}(t')} n(\mathbf{R}(t')) \rangle \dot{\mathbf{R}}(t') dt' \\
i\dot{\gamma}_n(t) &= - \langle n(\mathbf{R}(t) | \nabla_{\mathbf{R}(t)} n(\mathbf{R}(t)) \rangle \dot{\mathbf{R}}(t)
\end{align*}From here (again omitting functional depencies for clarity), I'm not totally sure how to show that $i\dot{\gamma}_n |n \rangle = -|\dot{n}\rangle$, so that $(\dagger)$ is indeed a solution of the differential equation. I had the idea to re-write\begin{align*}
| \nabla_{\mathbf{R}} n \rangle \dot{\mathbf{R}} = |\dot{n} \rangle
\end{align*}so that\begin{align*}
i\dot{\gamma}_n |n \rangle &= - \langle n |\dot{n} \rangle | n \rangle = -| n \rangle \langle n |\dot{n} \rangle
\end{align*}but the RHS doesn't look quite like $|\dot{n}\rangle$, since the identity operator is rather a sum $\displaystyle{\sum_n} | n \rangle \langle n |$ over $n$. Where did I go wrong?

 

[HomogenousCow]

So you've got $$|\dot{n}\rangle = -i\dot{\gamma}_n |n\rangle ,$$ which as you've correctly pointed out can be re-written as
$$\nabla_{\mathbf{R}} | n \rangle \cdot \dot{\mathbf{R}} = -i\dot{\gamma}_n |n\rangle, $$ now apply $\langle n |$ to both sides and you get $$\langle n |\nabla_{\mathbf{R}} | n \rangle \cdot \dot{\mathbf{R}} = -i\dot{\gamma}_n, $$ then just take the integral.

 

[ergospherical]

Got it, thanks! I hadn't realized that the eigenstates $|n\rangle$ are normalised :)