# Eigenstate and real space representation

1. Dec 8, 2013

### Arya_

Hi All,

I was going through a paper on quantum simulations. Below is an extract from the paper; I would be obliged if any one can help me to understand it:

We will use eigenstate representation for transverse direction(HT) and real space for longitudinal direction(HL) Hamiltonians.

HL= h^2/2m * d^2/dx^2 + U (makes sense because x is longitudinal direction)

HT= h^2/2m *( d^2/dy^2 + d^2/dz^2) + Uy.z (writing in terms of y,z isn't a real space representation again? This was meant to be eigenstate representaion)

then,

Xk(ρ) = exp(i.k.ρ)/area
where HT.Xk = εk.Xk

k and ρ are both 2D vectors.

Now, k is wave vector, Xk(ρ) is function of ρ, what is ρ? What does it physically corresponds to?

Using finite difference approximation for HL:

HLψ= -tψn-1 +2t+Un.ψn +....

here I can interpret ψ is function of n, which is discretized real space. However I am not able to figure out what is ρ?

Thanks,
Arya

2. Dec 18, 2013

### Staff: Mentor

I was able to find the article by Googling. Posting a reference would have been helpful. It seems to be written by someone that has little experience in quantum mechanical numerical simulations and uses a very non-standard notation.

The Hamiltonian doesn't change because of the representation, the wave function (or rather, its representation) does. The wave function is sampled on a series of grid points $x_n$ along with a finite basis of eigenfunctions $\chi_\mathbf{k}(y,z)$.

You truncated the quotation: "both 2D vectors in the y–z plane." I take it that $\rho = (y,z)$.