Berry phase for effective gauge potential

Question

On page 290 of Wens QFT he says that for the adiabatic motion of a single quasiparticle, for small t,

$$ \left\langle\Psi^{h}\left(\xi(t+\Delta t), \xi^{*}(t+\Delta t)\right) | \Psi^{h}\left(\xi(t), \xi^{*}(t)\right)\right\rangle = e^{\mathrm{i} \Delta t \mathbf{a} \cdot \dot{x}} $$

And from this he concludes that

$$\mathrm{i}\left\langle\Psi^{h}\left(\xi(t), \xi^{*}(t)\right)\left|\frac{\mathrm{d}}{\mathrm{d} t}\right| \Psi^{h}\left(\xi(t), \xi^{*}(t)\right)\right\rangle=\boldsymbol{a} \cdot \dot{x}$$

I’m wondering how he gets the right side of the first equation and then how he concludes with this second equation. This doesn’t seem straightforward. Any help appreciated!

Answer 1

I do not know what "a" is here, but I think the solution can be as follows:

1/ For the first relation, you take the time derivative of the lhs and, then, use the Schroedinger's equation in ket and bra notations (remember the adiabatic condition). After that, you use the first-order expansion: $$H(t+\Delta t)=H(t)+\Delta tH^\prime(t)+O(t^2)$$ to simplify the expression. I guess that "a" is encoded in the Hamiltonian. You will end up with a simple (?!) first-order differential equation. Solve it and get the result!

2/ You just expand to first order in $\Delta t$ for both sides of the first relation, and quickly you will see the point.