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Calculate the rate of probability density's movement?

  1. Oct 8, 2015 #1
    Screen Shot 2015-10-08 at 12.24.30 PM.png
    1. The problem statement, all variables and given/known data

    Given the following diagram of a finite potential well, calculate the rate at which the right-going wave is bringing probability density up to the barrier. (Ignore interference with the left-going wave. ) (Hint: you can get the velocity from the energy, and the average probability density from assuming that the integral over the well must give 1, when both left and right-going parts are included.). You can think of this rate as the rate at which the particle ‘attempts’ to cross the barrier.

    2. Relevant equations

    [tex]\frac{dP}{dt} = \frac{dP}{dx} \frac{dx}{dt} [/tex]

    3. The attempt at a solution
    Using the concession given in the question -- that we can use the average probability density to calculate the answer, [tex]\frac{dP}{dx} = \frac{1}{W+L}[/tex].

    Speed is given by solving for v in [tex]E = 0.5mv^2[/tex]

    Thus, we should have [tex] \frac{\sqrt(2E/m)}{(W+L)} [/tex]

    The actual answer is [tex] \frac{\sqrt(2E/m)}{(2W)} [/tex]

    For this to be true, average probability would have to be estimated as [tex]\frac{dP}{dx} = \frac{1}{2W}[/tex]. Why?
     
  2. jcsd
  3. Oct 9, 2015 #2
    In this statement, you may ignore [itex] L [/itex] and since there are two ways of waves, left and right, you just divide it as 2.
     
  4. Oct 9, 2015 #3
    I understand why waves would travel rightwards. Why would they travel leftwards? To travel leftwards, a wave would have to first tunnel through the barrier and then tunnel back. A wave wouldn't tunnel back, because once it tunnels rightwards (ie. exits the barrier), it is unenclosed and will now move rightwards.
     
  5. Oct 9, 2015 #4
    If the time is passed, then it can be correct but in this problem, I guess the time is fixed.

    Addendum : The probability density what you get is averaged. So, for each point, the probability density is given by
    [tex] \dfrac{dP}{dx} = \dfrac{\sqrt{2E/m}}{W}. [/tex]
    However, there always exist left and right moving particles for each point and as the statement says, the particle "attempts" to cross the barrier can only be the right moving particles at the boundary. The left moving particle can be thought as the bounced particles at the wall.

    When the time passes, the particles only escape to the right side but the probability density will not change. Some particles can bounce again and again.
     
    Last edited: Oct 9, 2015
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