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Expectation Values for momentum and a particle in a square well

  1. Aug 31, 2011 #1
    1. The problem statement, all variables and given/known data
    Calculate the expectation values of p and p2 for a particle in state n=2 in a square well potential.



    2. Relevant equations


    [itex]\Psi[/itex](x,y) = (2/L)*sin(n1[itex]\pi[/itex]x/L)*sin(n2[itex]\pi[/itex]y/L)

    p= -i[itex]\hbar[/itex][itex]\partial[/itex]/[itex]\partial[/itex]x


    3. The attempt at a solution

    [itex]\int[/itex][itex]\Psi[/itex]p[itex]\Psi[/itex]dxdy limits being from 0 to L for both.

    I derived the two dimension wave equation using separation of variables. This is where I had some questions. For the state n=2, does that mean both n1 and n2 equal 2 or does one of them equal 1 and the other 2 since from what I understand 1,2 would be the next highest energy level. Also as I was integrating, the x portion integrated nicely due to the momentum operator while the y portion stayed rather "unclean". Was I suppose to apply the momentum operator in both the x and y direction?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 31, 2011 #2

    Dick

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    I don't know what n=2 is supposed to mean for a two dimensional square well where you have two quantum numbers. Are you sure they don't mean a one dimensional problem? That's also called a 'square well' referring to the shape of the potential. Square doesn't necessarily mean two dimensions.
     
  4. Aug 31, 2011 #3
    Completely forgot I posted here, and yes I just realized that it's just a one dimensional well after lots of thinking...it's been a long a day heh. Thanks for the input though!
     
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