# Momentum space representation for finite lattices - continued

1. Mar 19, 2012

### QuantumLeak

I have been banned, maybe my nickname was not so kind. I let the topic continue here. I report my last comment:

"Ok, I got the point. thanks for replying!
It's just a change of basis that under boundary condition diagonalize the Hamiltonian. But then a subtle point:

In order for k-representation to be a good basis change (i.e. orthogonality and completeness properties) I guess that one has to choose the k values AS if there were periodic boundary conditions (i.e. k=2pi*n/L). Right?

Then another last point:
how do you really model a finite size chain? A finite size chain has zero boundary condition. In this case I use the k representation with k chosen as if there were periodic boundary condition and just then reconstruct the physical wavefunction by making a wavepacket that is zero on the borders?
"

2. Mar 19, 2012

### M Quack

A finite object in real space will need an infinite number of Fourier components in reciprocal space to fully describe it.

When you have nanometer scale objects with a few 100s or so atoms in each direction, then it is not so obvious which representation will give you a more accurate picture.