Expectation Values for momentum and a particle in a square well

In summary, the expectation values of p and p2 for a particle in state n=2 in a one dimensional square well potential can be calculated using the wave equation derived through separation of variables. The meaning of n=2 in this context is unclear as it is a two dimensional problem. It is important to apply the momentum operator in both the x and y directions during integration.
  • #1
muffins08
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Homework Statement


Calculate the expectation values of p and p2 for a particle in state n=2 in a square well potential.



Homework 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


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?
 
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  • #2
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.
 
  • #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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