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Time-independent Schrödinger equation, normalizing

  1. Jun 20, 2017 #1
    1. The problem statement, all variables and given/known data
    An electron coming from the left encounters/is trapped the following potential:
    -a<x<0; V=0
    0<x<a; V=V0
    infinity elsewhere
    the electron has energy V0
    a)Write out the wave function
    b)normalize th wave function
    2. Relevant equations


    3. The attempt at a solution
    for -a<x<0
    $$Ψ(x)=Acos(kx)+Bsin(kx)$$
    $$k^2=\frac{2mV_0}{ħ^2}$$
    and for 0<x<a
    $$Ψ(x)=Cx+D$$
    and 0 elsewhere
    i used the sine and cosine because it seemed it would be better for continuity condition in x=0, if you would use exponential form please do explain why.
    so this is what my teacher expects for a).
    for b)
    applying continuity conditions on x=0 i get:
    A=D
    B=C
    and so:$$\int_{-a}^{0}|Ψ(x)|^2=1$$
    im a bit confused here, is this the norm or the module? i think its the norm and if so ot might have been worth it to write the wave function in exponential form, so before i transcribe this big integral please clarify this for me.

    Furthermore this should look like a particle traped in a box correct? i dont really understand what happens when E=V, i understand the probabiity part, it decays linearly further inside the step, correct?
    And what about if E>V0 is it a particle traped in a box, but in the 0-a area the amplitude decreses? And the allowed energy levels for that area start at V0? what about penetration? and when E is smaller what happens?
    Thank you!

    Edit:would it be easier if i shifted the potential by -a so that it is in the range [0;2a]?
     
    Last edited: Jun 20, 2017
  2. jcsd
  3. Jun 24, 2017 #2

    vela

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    ##B=C## isn't quite correct. You should also apply continuity conditions for ##\psi(x)## at ##x=-a## and ##x=a##.

    The normalization requirement is
    $$\int_{-\infty}^\infty \lvert \psi(x) \rvert^2\,dx = 1.$$ In this problem, since the wave function vanishes for ##|x|>a##, you have
    $$\int_{-a}^a \lvert \psi(x) \rvert^2\,dx = 1.$$
     
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