# Definition Consider the [[Berry phase]] of a system that depends on a [[vector]] $\pmb{x}(t)$ that varies [[adiabatic|adiabatically]] in time. Assume the vector evolves cyclically along a closed [[Contour]] $C$, then the [[Berry phase]] is: $ \begin{align} \gamma[C] &= i \oint_C \text{d} \pmb{x} \cdot \bra{n, \pmb{x}(t)} \nabla_{\pmb{x}} \ket{n, \pmb{x}(t)},\\ &\equiv \oint_C \text{d} \pmb{x} \cdot \pmb{A}_n(\pmb{x}),\\ \end{align} $ Where we have defined the Berry connection as follows: $ \begin{align} \pmb{A}_n(\pmb{x}) &\equiv i \bra{n,\pmb{x}} \nabla_{\pmb{x}} \ket{n,\pmb{x}} \end{align} $ This assumes the system is non-degenerate, and the Berry connection is *abelian*. Note that the definition sometimes has a minus sign, and this article uses the minus sign convention when discussing the nonabelian Berry connection. ==I should clean up that section to avoid the minus sign== # Gauge Invariance The [[Berry Connection]] must be integrated along a closed path to become physical (giving the [[Berry phase]]. This is in contrast to the [[Berry Curvature]]. This can be seen easily: $ \begin{align} \ket{n, \pmb{x}} &\rightarrow e^{-i \beta(\pmb{x})} \ket{n, \pmb{x}}, \\ \gamma[C] &\rightarrow \gamma[C] + 2 \pi n, \quad n \in \mathbb{Z}, \\ \pmb{A}_n(\pmb{x}) &\rightarrow \pmb{A}_n(\pmb{x}) + \nabla_{\pmb{x}} \beta(\pmb{x}),\\ \end{align} $ # $U(2N)$ Berry Connection Matrix For a system with $2N$ bands, the *nonabelian* $U(2N)$ Berry connection matrices are defined as: $ \begin{align} \pmb{a}_{mn}(\pmb{k}) = - i \bra{u_{m,\pmb{k}}}\nabla_{\pmb{k}}\ket{u_{n,\pmb{k}}} .\end{align} $ We can relate $\pmb{a}(-\pmb{k})$ and $\pmb{a}(\pmb{k})$ as follows: $ \begin{align} \pmb{a}_{mn}(-\pmb{k}) &= - i \bra{u_{m,-\pmb{k}}}\nabla_{-\pmb{k}}\ket{u_{n,-\pmb{k}}} ,\\ &= i \bra{u_{m,-\pmb{k}}}\nabla_{\pmb{k}}\ket{u_{n,-\pmb{k}}} , \\ &= i \sum_{l,l'} \bra{u_{l',\pmb{k}}}\Theta^\dagger (w_{ml'}^*(\pmb{k}))^\dagger \nabla_{\pmb{k}} \times(w_{nl}^*(\pmb{k}))\Theta\ket{u_{l,-\pmb{k}}} , \\ &= -i \sum_{l,l'} \bra{u_{l',\pmb{k}}}\Theta w_{l'm}(\pmb{k}) \times ([\nabla_{\pmb{k}} w_{nl}^*(\pmb{k})]\Theta\ket{u_{l,-\pmb{k}}} +w_{nl}^*(\pmb{k})\nabla_{\pmb{k}}\Theta\ket{u_{l,-\pmb{k}}}) ,\\ %&= -i \sum_{l,l'} \bra{u_{l',\pmb{k}}}\Theta w_{l'm}(\pmb{k}) [\nabla_{\pmb{k}} w_{nl}^*(\pmb{k})]\Theta\ket{u_{l,-\pmb{k}}} + \bra{u_{l',\pmb{k}}}\Theta w_{l'm}(\pmb{k}) w_{nl}^*(\pmb{k})\nabla_{\pmb{k}}\Theta\ket{u_{l,-\pmb{k}}} , \\ \end{align} $ $ \begin{align} \pmb{a}_{mn}(-\pmb{k})&= -i \sum_{l,l'} w_{l'm}(\pmb{k}) \bra{u_{l',\pmb{k}}}\Theta^2\ket{u_{l,-\pmb{k}}} \times \nabla_{\pmb{k}} (w^\dagger(\pmb{k}))_{ln} \\ &+ \bra{u_{l',\pmb{k}}}\Theta w_{l'm}(\pmb{k}) w_{nl}^*(\pmb{k})\nabla_{\pmb{k}}\Theta\ket{u_{l,-\pmb{k}}} , %&= i \sum_l w_{lm}(\pmb{k})\nabla_{\pmb{k}} (w^\dagger(\pmb{k}))_{ln} \end{align} $ Here, $w_{mn}$ is the [[sewing matrix]]. Thus, we get: $ \begin{align} \pmb{a}(-\pmb{k}) = w(\pmb{k}) \pmb{a}^*(\pmb{k}) w^\dagger(\pmb{k}) + i w(\pmb{k}) \nabla_{\pmb{k}} w^\dagger(\pmb{k}) \end{align} $ Observe that: $ \begin{align} \pmb{a}^*_{mn}(\pmb{k}) &= (- i \bra{u_{m,\pmb{k}}}\nabla_{\pmb{k}}\ket{u_{n,\pmb{k}}} )^* \\ %&= i \bra{u_{n,\pmb{k}}}\nabla_{\pmb{k}}\ket{u_{m,\pmb{k}}}\\ &= \pmb{a}_{nm} .\end{align} $ And thus: $ \begin{align} \text{tr}[\pmb{a^*}(\pmb{k})] = \sum_n \pmb{a^*}_{nn}(\pmb{k}) = \sum_n \pmb{a}_{nn}(\pmb{k}) = \text{tr}[\pmb{a}(\pmb{k})] .\end{align} $ Moreover: $ \begin{align} \nabla_{\pmb{k}} (w w^\dagger) &= \nabla_{\pmb{k}}(1) = 0 ,\\ &= (\nabla_{\pmb{k}} w) w^\dagger + w (\nabla_{\pmb{k}} w^\dagger) = 0 \end{align} $ Thus: $ \begin{align} w \nabla_{\pmb{k}} w^\dagger = -(\nabla_{\pmb{k}} w) w^\dagger .\end{align} $ Taking the [[trace]] of the equation for $\pmb{a}(-\pmb{k})$ above and using the invariance of the trace under [[Unitary Operator|unitary transformations]]: $ \begin{align} \text{tr}[\pmb{a}(-\pmb{k})] &=\text{tr}[w(\pmb{k}) \pmb{a}^*(\pmb{k}) w^\dagger(\pmb{k})]\\ &+ i \text{tr}[w(\pmb{k}) \nabla_{\pmb{k}} w^\dagger(\pmb{k})]\nonumber ,\\ &=\text{tr}[\pmb{a}^*(\pmb{k})] + i \text{tr}[w(\pmb{k}) \nabla_{\pmb{k}} w^\dagger(\pmb{k})] .\end{align} $ Now replacing $\pmb{k} \to -\pmb{k}$ and using what we have derived so far, the cyclic property of the [[trace]], as well as $\text{tr}[AB] = \text{tr}[(AB)^T] = \text{tr}[B^T A^T] = \text{tr}[A^T B^T]$, we get: $ \begin{align} \text{tr}[\pmb{a}(\pmb{k})] &=\text{tr}[\pmb{a}^*(-\pmb{k})] + i \text{tr}[w(-\pmb{k}) \nabla_{-\pmb{k}} w^\dagger(-\pmb{k})] ,\\ &=\text{tr}[\pmb{a}(-\pmb{k})] - i \text{tr}[w^T(\pmb{k}) \nabla_{\pmb{k}} (w^\dagger(\pmb{k}))^T] ,\\ &=\text{tr}[\pmb{a}(-\pmb{k})] - i \text{tr}[w(\pmb{k}) \nabla_{\pmb{k}} (w^\dagger(\pmb{k}))] ,\\ &=\text{tr}[\pmb{a}(-\pmb{k})] + i \text{tr}[(\nabla_{\pmb{k}} w(\pmb{k})) w^\dagger(\pmb{k})] ,\\ &=\text{tr}[\pmb{a}(-\pmb{k})] + i \text{tr}[w^\dagger(\pmb{k}) \nabla_{\pmb{k}} w(\pmb{k})] .\end{align} $