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Solving for the wave function

  1. Oct 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Assume a free particle, V=0, in a infinite potential square well between -L/2 and L/2
    solve for the wave function for this particle.

    2. Relevant equations

    Time independent schrodinger equation


    3. The attempt at a solution

    After arriving at the second order differential equation, I get a general solution of
    ψ=Acos(kx) + Bsin(kx)

    How do I go on to solve for A, B and k?

    I've plugged in the boundary conditions, but the solution is still quite indeterminate.

    I have to assume A=0, then ψ=Bsin(2n∏x/L)
    however, if I assume B=0, I get ψ=Acos([2n+1]∏x/L)

    which is correct?
    I do need to pick one before going onto normalising right?
     
  2. jcsd
  3. Oct 21, 2013 #2

    DrClaude

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    Staff: Mentor

    Can you find a trigonometric relation between these two solutions?
     
  4. Oct 21, 2013 #3
    You can rewrite the solution to ψ=Aexp(ik)+Bexp(-ik) (k=2pi/λ)

    when x→+∞,ψ=0.so A=0.
    when x→-∞,ψ=0, so B=0

    that means ψ is a segmented function.

    your solution is right,and hope my answer can help you to comprenhen it in other way.
     
  5. Oct 21, 2013 #4
    er, nope.
    some pointers pls?
     
  6. Oct 22, 2013 #5

    DrClaude

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    Staff: Mentor

    The wave function is limited to the range -L/2, L/2 because of the infinite wall.
     
  7. Oct 22, 2013 #6

    DrClaude

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    Staff: Mentor

    $$\cos(x + \pi/2) = -\sin(x)$$
    Your two solutions are the same
     
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