Eigenstate and real space representation

[Arya_]

Hi All,

I was going through a paper on quantum simulations. Below is an extract from the paper; I would be obliged if anyone 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

 

[DrClaude]

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.

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)

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 $\chi_\mathbf{k}(y,z)$.

k and ρ are both 2D vectors.

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