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Solving TISE in 3D

  1. Mar 20, 2012 #1
    1. The problem statement, all variables and given/known data

    A particle in a 3D cubical potential well. The walls are Lx, Ly, Lz long.
    Inside the well, V(x,y,z)=0 when 0<x<Lx, 0<y<Ly, 0<z<Lz. V= ∞ elsewhere.
    Solve the TISE to find the eigenfunctions and eigenvalues of this potential. And to normalise the wavefunctions.

    (Hint: look for separable solutions in the form ψ(x,y,z)=X(x)Y(y)Z(z))


    2. Relevant equations

    (-hbar/2m)(∇2)ψ(x,y,z)+V(x,y,z)ψ(x,y,z)=Eψ(x,y,z)


    3. The attempt at a solution

    I'm confused about which equation to use for this question. I know that for a potential well, the V terms are zero, and i can write -hbar/2m and E in terms of k, so the equation reduces to ∇2ψ(x,y,z)=-k2ψ(x,y,z).

    we did the one dimensional well in class. my teacher used ψ(x)=Asin(kx)+Bsin(kx) as a general solution, then subbed in the boundary conditions like when x=L, ψ(x)=0 and so on.

    However, I know that ψ(x)=Aeikx is also a general solution to the differential equation. And this is used for free particles. I did some research online, and apparently I can treat a particle in the well as a free particle.

    So i don't know which equation i should use, and what are the differences between the two equations? Some sites i've seen used the trig, some used the complex.

    I tried to use the complex one, I got (after normalisation) ψ(x,y,z)=(1/sqrt3L)ei(kxx+kyy+kzz). Then I tried to sub in the boundary conditions, only to realize no matter what i did i couldn't make ψ=0.
    So I tried to use the trig, but the equation wasn't separable.

    I really don't know what to do here, some help would be appreciated! Thanks!
     
    Last edited: Mar 20, 2012
  2. jcsd
  3. Mar 20, 2012 #2
    There are two independent solutions, eikx and e-ikx. You will need both in order to satisfy your boundary conditions.
     
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