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Free electron or empty lattice schrodinger equation solution

  1. Apr 3, 2015 #1
    in the solution for free electron we start with

    X(x) = A exp (ikx) + B exp (-ikx)

    then using boundary conditions we eliminate B if the wave is traveling in the positive direction and vice versa
    my questions are:
    1. what is the boundary condition used?
    2. is it X(-inf) = 0? because this would cancel out B and leave A
    3. But then if we also assume that X(inf)=0 then A=0 so we have no wave function and cancel out the electron which is a fallacy

    please help i have my midterm on sunday
     
  2. jcsd
  3. Apr 3, 2015 #2
    As this is a free electron, remember that if we were to take the integral ##\int_{-\infty}^{\infty} \mathrm{X}^{*}\mathrm{X} = 1##
     
  4. Apr 3, 2015 #3
    X=Aexp(ikx)+Bexp(-ikx)
    X*=Aexp(-ikx)+Bexp(ikx)
    X.X*=A^2+B^2+AB[exp(-2ikx)+exp(2ikx)]
    <XIX>=1
    X(A^2+B^2) +[AB/2ik]*[exp(2ikx)-exp(-2ikx)] = 1

    how can I solve that?!!

    please help me i am losing time
     
  5. Apr 3, 2015 #4
    really guys I need help here
     
  6. Apr 4, 2015 #5

    DrClaude

    User Avatar

    Staff: Mentor

    It's not so much a question of boundary conditions as one of initial condition. Taking k to be positive, then one case corresponds to a particle of wave number k traveling towards +x, the other to a particle of wave number k traveling towards -x. You then keep the solution that corresponds to the physical situation you want to describe, for instance a particle coming in from ##x = -\infty##.

    As for the normalization, I don't know what superluminal had in mind, but a plane wave is not square integrable, and hence can't describe a physical state. For that, you need to build a wave packet by summing over a few plane waves.
     
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