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.


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

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

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

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

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

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

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


1 Answer 1


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 $$(U^\dagger\dot{U})_{ia} = -\langle n_a|\dot{n}_i\rangle.$$


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