Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Result of measuring H_x

  1. Nov 16, 2007 #1
    [tex]H=H_{x}+H_{y}[/tex] and

    [tex]\Phi(x,y) = cos (\pi x/a) cos (\pi y/a) sin(2 \pi x/a) sin(2 \pi y/a)[/tex]

    if we measure [tex]H_{x}[/tex] what do we get?

    the energy for this an infinit potential well is [tex]E_{n} propotional to n^{2}[/tex]

    so do we just say that [tex]H_{x} |\Phi>_{x,y} = E_{n} |\Phi>_{x,y}[/tex] ?
    and the n here is only 1 and 2 ie E1 and E2? what about measuring H? do we get E1+E2 or am I misunderstanding all this? Any help is appreciated
    Last edited: Nov 16, 2007
  2. jcsd
  3. Nov 17, 2007 #2
    I guess for one thing it is important to not forget that the only physical result you can really measure is the full Hamiltonian( the Sum of both the x and y part).
    This is important because you can´t just measure only H_x independently the eigenstates of this measurement e.g. of part of the hamiltonian hold no physical significane the only thing that measures are the eigenstates of the full Hamiltonian.
    An Eigenstate of H_x doesn´t automatically have to be one of the whole hamiltonian ( Does not give any physical Energy measurement at all!!, even though one could take it as the kinetic energy contained in the movement in x direction ).

    So in conclusion the H_x doesn´t correspond to the E_n´s of your system, and so H_x does not correspond to an E1 or something.
    The n´s can run from 1 to infinity the correspond to the energy levels of your whole system.
    But i guess in this 2D square well you need 2 quantum numbers to fully take care of the degeneracy of the system don´t you because i guess the complete solution of your system is :

    [tex]\Phi(x,y) = cos (\pi nx/a) cos (\pi ky/a) sin(2 \pi nx/a) sin(2 \pi ky/a)[/tex]

    so the k´s correspond to eigenstates of the H_y and the n´s correspond to eigenstates of the H_x whereby the sum of both maybe n+k = l corresponds to the energy quantum number of the system and could be used to enumerate the eigenstates of H.
    I guess this is a good way to see the 2 fold degeneracy because most l could be made up of different combinations of k and n for example l= 3 could correspond to n=1 k=2 or k=2 and n=2. e.g there are 2 states of energy l=3.
    Or for l=4 you have even more states.

    hope i could help you :)
  4. Nov 19, 2007 #3
    That was very helpful. I was so confused, thanks alot.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook