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Homework Help: Quantum mechanics

  1. Apr 5, 2014 #1
    [STRIKE][/STRIKE]1. The problem statement, all variables and given/known data

    A particle coming from +∞ with energy E colides with a potential of the form:

    V = ∞ , x<0 (III)
    V = -V0 , 0<x<a (II)
    V = 0, x>a (I)

    a) Determine the wave function of the particle considering that the amplitude of the incident wave is A. Writting the amplitude of the reflected wave at x=a in the form

    [itex] \frac{B}{A} = e^{i\delta} [/itex]

    determine [itex]\delta[/itex] . What is the value of [itex]\delta[/itex] in the limit where V0 = 0 ?

    b) Determine the probability density current in x>a

    3. The attempt at a solution


    For region I :

    [itex] \Psi_{I} (x) = Ae^{-ikx} + Be^{ikx} , k=\frac{\sqrt{2mE}}{\hbar} [/itex]


    [itex] \Psi_{II} (x) = Csin(lx) + Dcos(lx) , l=\frac{\sqrt{2(mE+V_{0}}}{\hbar} [/itex]


    [itex] \Psi_{III} (x) = 0 [/itex]

    Boundary conditions:

    at x=0:

    [itex] 0 = Csin(lx) + Dcos(lx) (=) D = 0 [/itex]

    So [itex] \Psi_{II} (x) = Csin(lx) [/itex]

    at x=a:

    [itex] Ae^{-ika} + Be^{ika} = Csin(la) [/itex] (1)


    [itex] -ikAe^{-ika} + ikBe^{ika} = lCcos(la) [/itex] (2)

    Dividing (1) for (2) I got:

    [itex] \frac{B}{A} = e^{-2ika} \frac{(-\frac{1}{l}tan(la)ik-1)}{(1-\frac{1}{l}tan(la)ik)} [/itex]

    How can I get rid of that constant that is multiplying by the exponential? Is this even right?
    Last edited: Apr 5, 2014
  2. jcsd
  3. Apr 6, 2014 #2

    Simon Bridge

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    B/A appears to be a complex number - is that right?

    Note: did you try writing ##\psi_{II}## as a sum of complex exponentials?
  4. Apr 6, 2014 #3

    Hi Simon Bridge, thank you for your reply.

    I did write psiII as a sum of complex exponentials and I got the following:

    [itex] \frac{B}{A} = \frac{\frac{k}{l}e^(ika-ila)-\frac{k}{l}e^(-ika + ila) - e^(-ika -ila) - e^(-ika + ila)}{\frac{k}{l}e^(ika-ila) - \frac{k}{l}e^(ika+ila) +e^(ika+ila) + e^(ika-ila)} [/itex]

    is there any way to simplify this ? This problem is killing me, because I'm not even sure what "writing the amplitude of the reflected wave at x=a in the form [itex] \frac{B}{A} [/itex] " means.

    I mean, I assumed the amplitude of the reflected wave at x=a is B, if I write B/A it isn't the amplitude of the reflected wave anymore.
  5. Apr 6, 2014 #4

    Simon Bridge

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    B/A is the proportion of the incident amplitude that is reflected.
    If you multiplied it by 100, you's be able to say, "100B/A percent got reflected".

    You basically have to look for common factors and cancel things off a bit at a time.
    It is not going to be easy. But that, I am afraid, is the exercise.

    Note: ##e^{a-b}+e^{-a-b}=(e^a+e^{-a})e^{-b}## ... stuff like that. maybe ##e^{-b}## cancels something in the denominator? If not - look for something else.

    Of course, since they are all constant terms, it may be possible to add them up geometrically.
    Separate the real and imaginary components.
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