# Berry phase and an emergent gauge field

In Nakahara 10.6.2 the case of system with fast $$r$$ and slow $$R$$ degrees (might be more than one of each) of freedom is discussed. The Hamiltonian is -

$$H=\frac{p^2}{2m}+\frac{P^2}{2M} + V(r;R)$$ Where $$p(P)$$ is the momentum associated with $$r(R)$$.

In describing the Schrodinger equation for the entire system, the total wavefunction of both dofs $$\Psi(r,R)=\Phi(R)\left|R\right\rangle$$ where $$\Phi(R)$$ is the slow dof wavefunction, $$\left|R \right\rangle$$ is the fast dofs wavefunction for fixed values of $$R$$.

In the derivation Nakahara apears to use this equality -

$$\left\langle R|\nabla^2_R|R \right\rangle=\left\langle R|\nabla_R|R \right\rangle)^2$$

However, I do not understand how it is reached. When starting with the LHS, and using integration by parts, I find a minus sign which I cannot wave off.

I do not have a copy of Nakahara to hand, but I don't think it is an equality, but rather an approximation. Your notation is a bit confusing as $$\Psi(r,R)= \Phi(R)|R\rangle$$ does not make sense as written. The LHS is a function and $$|R\rangle$$ is a vector in the Hilbert space, so the two sides are quite different things. I think what is meant is that we can factor the combined wavefunction for the slow nucleaus and the fast electron as $$\Psi(r,R)= \psi(r,R)\Phi(R)$$ where the wavefunction for the fast moving electron is $$\langle r|R\rangle= \psi(r,R).$$ Here $$R$$ is a parameter. Berry connection is $$i {\bf A}(R)= \langle R| \nabla_R |R\rangle=\int \psi^*(r,R)\nabla_R \psi(r,R) d^3{\bf r}.$$ We want an equation for the $$\Phi(R)$$ wavefunction to include the gauge covariant derivative $$(\nabla_R+i{\bf A}(R))^2.$$ However, when we plug the factored wavefunction into the full Shroedinger equation, we get $$\langle R|\nabla^2_R|R\rangle$$ instead of the $$(iA)^2$$. We therefore insert a complete set of states between the two $$\nabla_R$$'s but it is pecisely at this point that we make the adiabatic approximation by only keeping the single intermediate state $$|R\rangle \langle R|$$.
• When you make a small change in $R$ the $\psi(r,R)$ ony changes a little. By far its largest overlap is still with the unchanged $\psi(r,R)$. The essence of the adiabatic approx is that we ignore the overlap with all states orthogonal to the original $\psi(r,R)$. – mike stone Feb 22 at 17:40