Ordered lattice necessary for band structure?by Hyo X Tags: amorphous, band structure, conductor, solid 

#1
Aug1613, 01:45 AM

P: 37

Is it possible for a disordered or amorphous structure to have band structure?
I understand derivation of bands from KronigPenney model. E.g. does amorphous silicon have a band structure? While amorphous silicon oxide does not have a band structure? 



#2
Aug1613, 04:50 AM

P: 75

Amorphous materials do have band structures (another example: liquid metals). It's necessary to have bands in order to accommodate the electrons of the material (they can't all occupy states of the same energy, due to the exclusion principle). However, the electronic states can't be described using Brillouin zone concepts.




#3
Aug1713, 07:27 PM

P: 37

Can anyone point me to a reference to derive band structure based on disordered/amorphous solids?
All the approaches I have seen  Kronig Penney model, Bloch wavefunctions, Wannier functions, seem to all require periodic lattices... How can you predict the band structure for a disordered solid? Does it just have a uniform band structure with radial k_r rather than k_x? is the tightbinding model relevant? 



#4
Sep1713, 03:37 PM

P: 640

Ordered lattice necessary for band structure?
Amorphous materials have a band structure, but it is not a nice, clean band structure with sharp energies. Rather, it results from the fact that locally the amorphous material has more or less the same firstneighbor bonds as the crystalline material, the second neighbors are somewhat similar, and the mess starts from there on.
Amorphous materials are isotropic, so the band structure should also be isotropic, i.e. only depend on k_r as you point out. 



#5
Sep1813, 04:16 AM

P: 405

Strictly speaking, nonperiodic materials do not have a band structure. Having a band structure means that electronic states can be classified according to kvectors. kvectors are the labels of the irreducible representations of the lattice translation group. While in an amorphous system of course you can still define kvectors, if the Hamiltonian H does not commute with lattice translations T, in general H and T cannot be diagonalized by a common set of eigenvectors (with T's eigenvectors being labelled by k and H's by E(k)).



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