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I am confused by the wavevector in spectral function $A(\mathbf k,\omega)$. How to understand this $\mathbf k$ for a periodic structure? And how is it related to the $\mathbf k$ (in first Brillouin Zone) we use in the band structure? If there are multibands, then in band structure for the same k we have several energies. To roughly reproduce spectral function, do I expect several peaks for the same $\mathbf k$ in $A(\mathbf k,\omega)$ or I need to unfold the multibands, and the multi energies for same $\mathbf k$ in band structure appear at different $\mathbf k$ in $A(\mathbf k,\omega)$?

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You should try to do the calculation for free electrons for a two-band model and see what happens (and tell us the resul !). – Adam Apr 15 '14 at 0:54

The spectral function has a peak function at the energy and momentum of long-lived excitations. As with all observables in a crystal the spectral function is fully defined by the first Brillouin Zone. In the case of the spectral function corresponding to the physical electron Green's function, you would see at fixed momentum a peak at the energy for every band. This is what is measured in ARPES experiments for example.

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Thanks. Then I have an another question. If we have a system with simple unit cell. In its first Brillouin Zone there is only one band. If we add to the system a potential with large periodicity, then we have a supercell, corresponding to a mini first BZ. Now band structure are folded back to mini BZ and for same k in the mini BZ we have multi bands. Are we going to observe multiple peaks at fixed momentum in ARPES? If so, it will be confusing since the perturbation potential can be infinitesimally small and ARPES results should be continuous to unperturbed case. – jinchenhao Jun 19 '13 at 21:11
If you have a supercell, the second band will show up, but with a very weak intensity. The intensity in that sense is continuous. If the perturbation is small, the intensity will be weak, almost invisible, whereas if the perturbation is strong, the intensity will be strong and additional bands will definitely be visible. – Xcheckr Jul 30 '14 at 19:43

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