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Homework Help: Basic wave-function question

  1. Apr 28, 2013 #1
    1. The problem statement, all variables and given/known data

    A one-dimensional time-independent wave-function can be written as:

    [itex] \Psi = 5e^{2ix} – e^{-2ix}, (x ≤ 0) [/itex]
    [itex] \Psi = 4e^{3ix}, (x > 0)[/itex]

    Show that this wave-function obeys the boundary conditions, find the probability that an incident particle represented by the wave-function is reflected, and sketch the potential energy function, indicating where the total energy of the particle lies.

    The attempt at a solution

    First part is easy enough, proving boundary condition are obeyed means both equations will equal each other at x=0, which would give, 5-1 = 4, which is correct.

    I think the probability of the incident particle reflecting concentrates on the first equation which would be the superposition of the incident wave( [itex] 5e^{2ix}[/itex] ) and reflected wave ( [itex] -e^{-2ix}[/itex] ).

    The probability would be;

    [itex] R = \dfrac{|B|^{2}}{|A|^{2}} = \dfrac{1^{2}}{5^{2}} = \dfrac{1}{25}[/itex]

    I think the potential energy function would be the wave function approaching a barrier with higher energy, with some transmitted part of lesser amplitude and a even smaller amplitude reflecting back. Where the total energy lies, I am not really sure?

    If someone would be kind enough to confirm my attempt so far and help on the last bit.
    Last edited by a moderator: Apr 30, 2013
  2. jcsd
  3. Apr 28, 2013 #2


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    There's an additional boundary condition that you need to show is satisfied.

    Your reflection coefficient looks correct.

    To see the nature of the potential energy function, can you use the numbers 2 and 3 appearing in the exponents to tell you something about the momentum and/or kinetic energy of the particle in each region?
  4. Apr 30, 2013 #3
    The additional boundary condition being that the first derivative of each function equal each other at x=0?


    [itex] \dfrac{\delta}{\delta x}5e^{2ix}-e^{-2ix} = 10ie^{2ix} + 2ie^{-2ix}[/itex]

    [itex] \dfrac{\delta}{\delta x}4e^{3ix} = 12ie^{3ix}[/itex]

    Again at x = 0 cancels out exponential function so just 10 + 2 = 12 and is satisfied.

    The numbers in the exponent correspond to values of 'k' I think where,

    [itex] k = \dfrac{\sqrt{2mE}}{\hbar}[/itex]

    The momentum would equal the top of the fraction, the numerator, and E within that value is the kinetic energy.

    I am not sure what else if anything I am to take from this?
    Thanks for your help thus far.
  5. Apr 30, 2013 #4


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    Good, the derivative of the wavefunction is continuous.

    OK, it might be best to let E be the total energy and K the kinetic energy. So

    [itex] k = \dfrac{\sqrt{2mK}}{\hbar}[/itex]

    So, what happens to the kinetic energy as k switches from 2 to 3? What does that tell you about the shape of the potential function? Does it step up or step down as you go from x<0 to x>0?
  6. Apr 30, 2013 #5
    As 'k' changes from 2 to 3, [itex]\hbar[/itex] is a constant so the numerator must rise as well. So the momentum rises as does the kinetic energy when crossing from x<0 to x>o. It steps up out a well?
  7. Apr 30, 2013 #6


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    What happens to kinetic energy if potential energy increases? Decreases?

    Therefore, what would happen to k if potential energy increases? Decreases?
  8. Apr 30, 2013 #7
    Ahhhh sorry wrong energy! I see why you wanted to change notation from start now. Soo, k is increasing, as the incident wave transmits at the boundary a wave with 1 higher k. And if this increases that means the kinetic energy will increase, energy is the sum total of potential and kinetic energy and must be conserved, so potential must come down. If I were to plot a graph of potential energy versus distance along x-axis then it would drop at the boundary. Goes down a step?
  9. Apr 30, 2013 #8


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    Yes, that's right.
  10. Apr 30, 2013 #9
    Thanks very much for your help.
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