Let $\hat{A}^i (i = 1, . . . , n)$ be a set of hermitian observables and $F_i$ a corresponding set of external fields that are linearly coupled to $\hat{A}^i$. Starting from the ground-state at $F_i = 0$ imagine a process in which the $F_i$ ’s are slowly changed (always remaining small) around the closed loop $$0\rightarrow \delta F_i^{(1)}\rightarrow \delta F_i^{(2)}\rightarrow \delta F_i^{(1)}\rightarrow 0$$ where $\delta F_i^{(1)}$ and $\delta F_i^{(2)}$ are two sets of small increments. At the end of the cycle the system will return to the ground-state having acquired a 'Berry phase' $e^{i\gamma}$ where $$ \gamma=\sum_{ij}\delta F_{i}^{(1)}\Omega_{ij}\delta F_{j}^{(2)} $$ and $$ \Omega_{ij}\equiv 2\Im m \langle \frac{\partial \psi}{\partial F_i} | \frac{\partial \psi}{\partial F_j} \rangle $$ is the Berry curvature. Here $$| \frac{\partial \psi}{\partial F_j} \rangle =\frac{\partial }{\partial F_j} |\psi\rangle,$$ where $| \psi \rangle$ is the state of the system regarded as a functional of the $F_i$. By evaluating $\frac{\partial }{\partial F_0} |\psi\rangle$ by first-order perturbation theory and comparing the resulting expression for $\Omega_{ij}$ with the Lehmann representation enter image description here

for the linear response function $\chi_{A_i A_j}(\omega)$ (at $T = 0$) show that the Berry curvature can be expressed in the following manner:

enter image description here

  • $\begingroup$ Ref: Gabriele Giuliani,"QUANTUM THEORY OF THE ELECTRON LIQUID" $\endgroup$
    – Abolfazl
    Dec 10, 2014 at 17:45
  • $\begingroup$ the point and my problem mostly is $| \frac{\partial \psi}{\partial F_j} \rangle $ $\endgroup$
    – Abolfazl
    Dec 11, 2014 at 19:00


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