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Solving Schrodinger equation in two dimensions

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Solve the time independent Schrodinger equation for an electron in a 2-D potential well having dimensions Lx and Ly in the x and y directions respectively.


    2. Relevant equations

    d^2Y/dx^2 + d^2Y/dy^2 + 2m/h^2*EY = 0


    3. The attempt at a solution

    Y(x) = A exp(jkx) + B exp(-jkx)

    I'm not even sure that is correct as a starting point, as that is what is given for the example in the book for the Schrodinger equation in one dimension, not two dimensions.
     
  2. jcsd
  3. Oct 14, 2009 #2

    gabbagabbahey

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    Is this for a finite well or an infinite well?
     
  4. Oct 14, 2009 #3
    Infinite potential well.
     
  5. Oct 14, 2009 #4

    gabbagabbahey

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    Okay, so for [itex]|x|\leq L_x[/itex] and [itex]|y|\leq L_y[/itex] the schroedinger equation reduces to:

    [tex]\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{2mE}{\hbar^2}\right)\Psi(x,y)=0[/tex]

    right?

    Now use separation of variables; assume that [itex]\Psi(x,y)=X(x)Y(y)[/itex]....what do you get when you plug that into Schroedinger's equation?
     
  6. Oct 14, 2009 #5
    I forgot how to do separation of variables. How do you do that part?
     
  7. Oct 14, 2009 #6

    gabbagabbahey

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    You assume the wavefunction is separable (i.e. assume [itex]\Psi(x,y)=X(x)Y(y)[/itex])....substitute that into the Schrodinger equation and simplify....what do you get?
     
  8. Oct 14, 2009 #7
    I get d^2Y/dx^2 + d^2Y/dy^2 + 2m/h^2*EX(x)Y(y) = 0

    But doesn't the substitution also have to be done for the second derivatives? How would that be done?
     
  9. Oct 14, 2009 #8

    gabbagabbahey

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    No, you don't get that. And yes, the second derivatives act on [itex]\psi(x,y)=X(x)Y(y)[/itex]
     
  10. Oct 14, 2009 #9
    When you plug in for each of the second derivatives, do you treat the other variable as a constant?
     
  11. Oct 14, 2009 #10

    gabbagabbahey

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    You tell me, how does one usually take the partial derivative of a multivariable function?
     
  12. Oct 14, 2009 #11
    I googled it and found out that for multivariable partial derivatives the other variables are held constant.

    So is this the answer?

    Y(y)X(x)'' + X(x)Y(y)'' + 2m/h^2*E*X(x)Y(y) = 0
     
  13. Oct 14, 2009 #12

    gabbagabbahey

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    Yes.:approve:

    Now, what do you get when you divide both sides of the equation by [itex]X(x)Y(y)[/itex]?
     
  14. Oct 14, 2009 #13
    X(x)''/X(x) + Y(y)''/Y(y) + 2m/h^2*E = 0
     
  15. Oct 14, 2009 #14

    gabbagabbahey

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    Right, now look carefully at that equation...what variable(s) is the quantity X''(x)/X(x) a function of ? What variable(s) is the quantity Y''(y)/Y(y) a function of? Under what conditions can the add together to get a constant value of [itex]\frac{2mE}{\hbar^2}[/itex]?
     
  16. Oct 14, 2009 #15
    X''(x)/X(x) is a function of x and Y''(y)/Y(y) is a function of y.

    Are you asking when

    X(x)''/X(x) + Y(y)''/Y(y) + 2m/h^2*E = 0

    will be equal to 2m/h^2*E = 0 ?

    If that's the case, then it will happen either when both X(x)''/X(x) and Y(y)''/Y(y) are equal to 0, or when X(x)''/X(x) = -Y(y)''/Y(y).
     
  17. Oct 14, 2009 #16

    gabbagabbahey

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    Right, and 'x' and 'y' are independent variables, right?

    No, I'm asking when X(x)''/X(x) + Y(y)''/Y(y) can equal -2m/h^2*E? How can a function of 'x' added to a function of 'y' result in a constant?
     
  18. Oct 14, 2009 #17
    Yes, x and y are independent. A function of x and a function of y added together will result in a constant when the variables cancel each other out or become 0.
     
  19. Oct 14, 2009 #18

    gabbagabbahey

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    Well, the variables can't cancel eachother out, since they are independent.

    What if both functions are constants?
     
  20. Oct 14, 2009 #19
    If both functions are constant, then adding them together would result in a constant, but I don't see how they can be constants since they have variables in them.
     
  21. Oct 14, 2009 #20

    gabbagabbahey

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    Are you telling me that you don't consider f(x)=1 to be a function of x? But you do consider f(x)=0 to be a function of x?
     
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