# Proving that the Berry curvature is always zero

If we have an energy eigenstate $|\psi(\boldsymbol \lambda )\rangle$ which is a function of some external parameter $\boldsymbol \lambda = (a,b)$, then the associated Berry connection is defined to be $$\mathcal A_n(\boldsymbol \lambda) = i \langle \psi(\boldsymbol \lambda) | \partial_n | \psi(\boldsymbol \lambda)\rangle \quad (n = a,b)$$ and the Berry curvature is $$\Omega(\boldsymbol \lambda) = \partial_a \mathcal A_b(\boldsymbol \lambda) - \partial_b \mathcal A_a(\boldsymbol \lambda).$$

However, it seems I can show that $\partial_a \mathcal A_b(\boldsymbol \lambda) = 0$. Clearly this must be wrong, but I cannot find the mistake. Can anyone spot it? The argument goes as follows:

\begin{align} \boxed{-i\partial_a \mathcal A_b(\boldsymbol \lambda)} &= \partial_a \left( \langle \psi(\boldsymbol \lambda) | \partial_b | \psi(\boldsymbol \lambda)\rangle \right) \\ &= \left( \partial_a\langle \psi(\boldsymbol \lambda) | \right) \partial_b | \psi(\boldsymbol \lambda)\rangle + \langle \psi(\boldsymbol \lambda) | \partial_a \partial_b | \psi(\boldsymbol \lambda)\rangle \\ &= - \langle \psi(\boldsymbol \lambda) | \partial_a \partial_b | \psi(\boldsymbol \lambda)\rangle +\langle \psi(\boldsymbol \lambda) | \partial_a \partial_b | \psi(\boldsymbol \lambda)\rangle \\ &= \boxed{ 0}. \end{align}

For going from the second line to the third line, I used that $$\boxed{ \left( \partial_a \langle \psi(\boldsymbol \lambda) | \right) = - \langle \psi(\boldsymbol \lambda) | \partial_a }.$$ To prove this, extend a given $|\psi(\boldsymbol \lambda)\rangle$ to a complete, orthornormal basis $$\{ |\varphi_0(\boldsymbol \lambda)\rangle, |\varphi_1(\boldsymbol \lambda)\rangle, |\varphi_2(\boldsymbol \lambda)\rangle, \dots\} \quad \textrm{where } |\varphi_0(\boldsymbol \lambda)\rangle = |\psi(\boldsymbol \lambda)\rangle.$$ Moreover, choose this such that for a local neighborhood of $\boldsymbol \lambda$, the above states are smooth as a function of $\boldsymbol \lambda$. Surely there can be global obstructions against this, but locally this should be doable. Then we have \begin{align} \langle \psi(\boldsymbol \lambda) | \varphi_l(\boldsymbol \lambda ) \rangle &= \delta_{0,l}\\ & \Downarrow \\ \partial_a \left( \langle \psi(\boldsymbol \lambda) | \varphi_l(\boldsymbol \lambda ) \rangle \right) &= 0 \\ & \Downarrow \\ \left( \partial_a \langle \psi(\boldsymbol \lambda) | \right) | \varphi_l(\boldsymbol \lambda ) \rangle &= - \langle \psi(\boldsymbol \lambda) | \partial_a | \varphi_l(\boldsymbol \lambda ) \rangle \end{align} such that the claimed result follows from the completeness of the basis.

NB: I am aware of subtleties where you can seemingly prove that, e.g., the Chern number has to be zero etc. But all those arguments use global assumptions, which are then violated by the fact that the Berry connection has to be singular at some point (for the Chern number to be non-zero). What puzzles me about the above argument, however, is that it seems entirely local... I.e. the above conclusion should seemingly apply wherever $|\psi(\boldsymbol \lambda)\rangle$ is non-singular, but that does not seem right.

It is incorrect to generalize from $$\left( \partial_a \langle \psi(\boldsymbol \lambda) | \right) | \varphi_l(\boldsymbol \lambda ) \rangle = - \langle \psi(\boldsymbol \lambda) | \partial_a | \varphi_l(\boldsymbol \lambda ) \rangle$$ to $$\left( \partial_a \langle \psi(\boldsymbol \lambda) | \right) = - \langle \psi(\boldsymbol \lambda) | \partial_a . \tag{incorrect!}$$ The first formula is explicitly referring to the $\varphi_l$, but that's it. You cannot willy-nilly substitute them for other functions (like the $\partial_b \varphi_l$, as you've done here).
• But it is a complete basis... If $\{|e_n \rangle \}$ is a complete basis, and if $\langle \alpha | e_n \rangle = \langle \beta | e_n \rangle$ (for any $n$), then we know that $| \alpha \rangle = | \beta\rangle$, right? One can prove this by expanding out in the basis. – Ruben Verresen May 22 '18 at 14:20
• You've got different bases. If you want to do that then you need to extend your description to a tensor product with $L_2$ over the space of the $\boldsymbol\lambda$s. – Emilio Pisanty May 22 '18 at 14:26
• What do you mean by 'you've got different bases'? $\boldsymbol \lambda$ can be considered to be fixed in the above manipulation. For conceptual convenience, one might rewrite $\partial_a |\varphi_l(\boldsymbol \lambda)\rangle = \sum_m A_{lm}(\boldsymbol \lambda) |\varphi_m(\boldsymbol \lambda) \rangle$ to convince oneself that it does make sense to imagine fixing $\boldsymbol \lambda$. – Ruben Verresen May 22 '18 at 14:29
• @RubenVerresen Precisely. And then if you differentiate $\partial_a |\varphi_l(\boldsymbol \lambda)⟩ = \sum_m A_{lm}^{(a)}(\boldsymbol \lambda) |\varphi_m(\boldsymbol \lambda)⟩$ again you get $$\partial_b\partial_a |\varphi_l(\boldsymbol \lambda)⟩ = \sum_m \partial_bA_{lm}^{(a)}(\boldsymbol \lambda) |\varphi_m(\boldsymbol \lambda)⟩+\sum_m A_{lm}^{(a)}(\boldsymbol \lambda)\partial_b |\varphi_m(\boldsymbol \lambda)⟩,$$ where the latter gives you back the doubly-applied linear operator but the former doesn't. – Emilio Pisanty May 22 '18 at 15:34