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Whereas I can calculate the Chern number of a quantum state (or band) from the integration of the Berry curvature in all space. How can I infer the topology of the quantum state from this result? What is the physical meaning of a quantum state with non-zero Chern number? |
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I am in part trying to understand this myself. The Berry phase is computed from differential forms, such as the one-forms $\omega$ constructed from states $$ \omega~=~\langle\psi|d\psi\rangle $$ and with the covariant differential $D~=~d~+~\omega$ the two-forms $$ \Omega~=~D\omega~=~d\omega~+~\omega\wedge\omega $$ The tensor components of the 2-form $F$ are elements of a self-adjoint principal bundle $P$. The determinant of these elements $$ det\Big|1~+~\frac{ixF}{2\pi}\Big|~=~\sum_nc_jx^n $$ which is a characteristic polynomial which represents the Chern class. Each $c_n(P)$ is an element of $H^{2n}(M)$. So the curvature form for the Berry phase, or the Fubini-Study metric $\Omega=~dz\wedge d{\bar z}/(1~+~|z|^2)^2$ is evaluated $\int\Omega~=~2\pi i$ and gives $c_1~=~1$ So there is a nontrivial cocycle on the “2-level. For this projective geometry there are alternating Betti numbers $1,~0$ for even and odd. If you had some product of states $\prod_n |\psi_n\rangle$, say in an entangled state etc, you could apply the differential $d$ up to $n$ times and form and $n$-form. For instance the product $|\psi_1,~\psi_2\rangle$ $=~|\psi_1\rangle|\psi_2\rangle$ defines the one-form $$ \omega~=~d|\psi_1,~\psi_2\rangle~=~d|\psi_1\rangle|\psi_2\rangle~+~|\psi_1\rangle d|\psi_2\rangle $$ and one could then build up a system of differential forms on various chains. The analogue of the projective geometry for this is a $G_2(V)$ Grassmannian and this continues up for n-product spaces. |
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