(i,j) notation of orders of wave vector

  • Context: Undergrad 
  • Thread starter Thread starter PBJ
  • Start date Start date
  • Tags Tags
    Notation Vector Wave
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 1K views
PBJ
Messages
1
Reaction score
0
I am looking at a formula, simplified here, which says that peaks occur at wavelengths: [tex]\lambda_{ij}=...(i^{2}+j^{2})...[/tex] "where i, j correspond to the orders of the 2D wave vector"

The peak wavelengths are at [tex]\lambda_{10}[/tex],[tex]\lambda_{11}[/tex],[tex]\lambda_{20}[/tex]

(i.e. where [tex](i^{2}+j^{2})[/tex] terms are

[tex]1^{2}+0^{2}=1[/tex]
[tex]1^{2}+1^{2}=2[/tex]
[tex]2^{2}+0^{2}=4[/tex]

I don't understand what is meant by "i,j correspond to the orders of the 2D wave vector" and why the notation goes up as 10,11,20...

I expect there is a simple explanation if someone would be kind enough to help me understand this.
 
Physics news on Phys.org
In this problem it appears you are looking at two electrons (or any particles) in a potential well. Orders of the 2D wavevector, I'm assuming, are the energy states. I'm going to go out on a limb and say that you are looking for the peak amplitude of the wavefunction. The i,j notation, while somewhat confusing as this notation is generally left for complex numbers, are the wavevectors of the particles oscillating in the potential well.

Therefore the notation on the lambda symbol are the energy states of the two electrons, [10] is one electron in the first excited state and the second electron in the ground state, [11] first and second electrons in the first excited state and [20] first electron in the second excited state, second electron in the ground state.