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Leaving out numerical factors, we have that $$\mathrm{d}j_A = \mathrm{Tr}(F \wedge F)$$ This already shows that we are dealing with a topological quantity, since the RHS is the second Chern character of the gauge field (or rather, the principal bundle associated to it). Now, there is also the (3D) Chern-Simons form  \omega = \mathrm{Tr}(F \wedge A - ...

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If you assume separability of the wave function, i.e., $\psi(\mathbf x)=u(x)v(y)w(z)$, you can solve the individual components separately: \begin{align} -\frac{\hbar^2}{2\mu}\frac{d^2u(x)}{dx^2}+V_1(x)u(x)&=E_1u(x)\\ -\frac{\hbar^2}{2\mu}\frac{d^2v(y)}{dy^2}+V_2(y)v(y)&=E_2v(y)\tag{1}\\ ...

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No. Feynman diagrams are made by summing over the perturbative contributions of quantum amplitudes. They cannot hold non-perturbative information.

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