# Deriving the non-abelian Berry connection

I'm slightly confused about a manipulation in Section 1.5.4 of Tong's notes on the Quantum Hall Effect. This concerns the derivation of the non-abelian Berry phase.

Setup:

We have an $$N$$-dimensional ground state $$\mathcal{H}$$ (take its energy to be $$0$$), which is acted on by the adiabatic tuning of a parameter $$\lambda=(\lambda^{1},\ldots,\lambda^{k})$$ where $$\lambda=\lambda(t)$$. Thus, the ground state space depends on $$\lambda$$, i.e. $$\mathcal{H}=\mathcal{H}(\lambda)$$. Take an orthonormal basis of states $$|n_{a}(\lambda)\rangle_{a=1}^{N}$$ for each $$\lambda$$, such that $$n_{a}(\lambda)$$ is smooth. Now suppose that at time $$t=0$$, the system is in state $$|\psi_{a}(0)\rangle=|n_{a}(0)\rangle$$. Then at a later time $$t$$, we may write

$$$$|\psi_{a}(t)\rangle=\sum_{b} U_{ab}|n_{b}(\lambda(t))\rangle \quad \ \ \mathbf{(0)}$$$$

That is $$|\psi_{a}(t)\rangle$$ is given by the $$a^{th}$$ row of some time-dependent unitary $$U=U(\lambda(t))\in U(N)$$ (represented with respect to our choice of basis for $$\mathcal{H}(\lambda)$$).

Now we solve the Schroedinger equation $$i\hbar\partial_{t}|\psi_{a}(t)\rangle=H|\psi_{a}(t)\rangle=0$$, which using $$\mathbf{(0)}$$, apparently yields

$$$$|\dot{\psi_{a}}\rangle=\sum_{b}\left(\dot{U_{ab}}|n_{b}\rangle + U_{ab}|\dot{n_{b}}\rangle\right)=0 \quad \ \ \mathbf{(1)}$$$$

Tong (eq. 1.51) claims that this can be rearranged to give

$$$$[ U^{\dagger} \dot{U} ] \small{ba} = -\langle n_{a}|\dot{n_{b}} \rangle \quad \ \ \mathbf{(2)}$$$$

Question: How do we get from (1) to (2)?

Multiply the equation with $$U^\dagger$$: $$0 = \sum_{a,b}U^\dagger_{ia}\dot{U}_{ab}|n_b\rangle + U^\dagger_{ia}U_{ab}|\dot{n}_b\rangle$$ Carry out the sum over $$a$$ gives $$0 = \sum_{b}(U^\dagger\dot{U})_{ib}|n_b\rangle + \delta_{ib}|\dot{n}_b\rangle$$ Then you act from the left with $$\langle n_a|$$ and use the orthogonality of the basis states, which gives you the expected $$(U^\dagger\dot{U})_{ia} = -\langle n_a|\dot{n}_i\rangle.$$