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Wave Equation in 2 Dimensions - Basic

  1. Jun 25, 2009 #1
    Hey Everyone,

    So i've been working on some very basic QM mathematics. Basically I've worked out the wave equation for a particle in one dimension (briefly) like so:

    Code (Text):

    -[tex]\frac{\hbar [SUP]2[/SUP]}{2m}[/tex][tex]\psi[/tex]"(x) + V(x)[tex]\psi[/tex](x) = E[tex]\psi[/tex](x)

    V = 0 for 0 < x < L  ; (L = "Length" of the Boundary)

    => [tex]\psi[/tex](x) = A sin([tex]\frac{n \pi x}{L}[/tex])

    => A = [tex]\frac{L}{2}[/tex]
    The trouble I'm having is trying to extrapolate this to two spatial dimensions (if that can be done in the fashion I'm trying).

    I follow the same process except my solution to the Schrodinger equation (solution to the differential equation) is

    Code (Text):

    [tex]\psi[/tex](x,y) = A sin([tex]\frac{n \pi x}{L}[/tex]) + B sin([tex]\frac{n \pi y}{L}[/tex])

    A[SUP]2[/SUP]([tex]\frac{L}{2}[/tex])y + B[SUP]2[/SUP]([tex]\frac{L}{2}[/tex])x = 1

    ^ From Normalizing the Solution with Limits of integration for the double integral of 0 < (x,y) < L
    The problem is that instead of finding the 'constants' A and B i've now got a relationship between them.

    If someone could point out what I did wrong in my process (determine the wave equation differential equation, normalize and solve) and if that's alright where I go from here in writing the full wave-function of the model, I'd greatly appreciate it.

  2. jcsd
  3. Jun 25, 2009 #2


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    Why do you think you can simply add solutions? You can use the method of separation of variables to find solutions of the form [tex]\psi (x,y) = X(x)Y(y)[/tex] however. To be sure, plug your solution into the equation and you'll see that it isn't a solution.
  4. Jun 26, 2009 #3
    Thank you, I'll work that one out
  5. Jun 28, 2009 #4
    I'm in need of some direction... I have differential equation experience and partial integrals/derivatives but I believe this is a partial differential equation now that both the x and y variables have been introduced and I have not worked on partial diff eqs. Where do I start to tackle this one? I tried a few solutions with a guess and check and none of them have worked and I'm a bit stuck with this one. Any help would be appreciated, thank you
  6. Jun 28, 2009 #5


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    Well, you should find a section on Separation of Variables in your QM text if you have one or a book on PDE's. You seem like you have an understanding enough for me to just say the following: Look for solutions of the form [tex]\psi (x,y) = X(x)Y(y)[/tex] . Your Hamiltonian now is the 2-D Laplacian so your DE looks like [tex]\frac{{ - h^2 }}{{2m}}(\frac{{\partial ^2 }}{{\partial x^2 }} + \frac{{\partial ^2 }}{{\partial y^2 }})\psi (x,y) = E\psi (x,y)[/tex]. There's some math that shows the solution form we have is valid so you can check that out on your own.

    Now simple arranging shows this is also [tex]\frac{{\partial ^2 \psi (x,y)}}{{\partial x^2 }} + \frac{{\partial ^2 \psi (x,y)}}{{\partial y^2 }} = \frac{{ - 2mE}}{{h^2 }}\psi (x,y)[/tex]. At this point, you plug in your solution [tex]\psi (x,y) = X(x)Y(y)[/tex]. At this point you can do some slight manipulations and what you'll get is basically 2 seperated ODE's summing to 0. For PDE's, the only way this is possible is if both ODE's are equal to a constant (otherwise you couldn't have independent variations between x and y which is a requirement from your DE). Make up a set of constants, say m and -m and from there, solve the ODE's and look back on your solution form and construct your wavefunction.
  7. Jun 29, 2009 #6
    I don't have a QM text but I know of seperation of variables from diff. eq. I have to take the time to read through your explanation but from skimming it it looks like I can follow your steps. Thank you!
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