1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    a)

    For region I :

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

    II:

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

    III:

    [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)

    and

    [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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Quantum mechanics
  1. Quantum mechanics (Replies: 6)

  2. Quantum Mechanics (Replies: 3)

  3. Quantum Mechanics (Replies: 3)

  4. Quantum mechanics (Replies: 0)

  5. Quantum Mechanics (Replies: 6)

Loading...