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Homework Help: Forces in a Quantum Reflection

  1. Sep 13, 2013 #1
    Hi folks,

    So everyone at some point studying QM deals with reflections from 1D finite potential wells using the TISE, but something that I never covered was how to talk about or calculate the forces involved in turning around that particle. I tried to work something out below, but I'm not sure if it's a valid approach and was hoping for some advice.

    1. The problem statement, all variables and given/known data

    Consider a 1D finite potential well starting at position ##x=0##, length ##a##, and depth ##-V_0##. A particle is incident on the well from the left with energy ##E>0##. Calculate the expectation of the force on the particle in that region given a reflection coefficient ##R##.

    2. Relevant equations

    $$V(x) = -V_0 (θ(x)-θ(x-a)) $$
    $$F = -∇V$$
    $$ψ_{region} = e^{i k x} + R e^{-i k x}$$

    3. The attempt at a solution
    F &= V_0 ( δ(x)-δ(x-a)) \\
    <F> &= ∫_{-∞}^0 ψ^*Fψ dx \\
    &= ∫_{-∞}^0 (e^{- i k x} + R e^{i k x}) V_0 ( δ(x)-δ(x-a)) (e^{i k x} + R e^{-i k x}) dx \\
    &\text{Expanding and noting only the unshifted delta function will survive the integral} \\
    & =∫_{-∞}^0 V_0 δ(x) (1 + R(e^{2i k x} + e^{-2i k x}) + R^2) dx \\
    & =V_0 (R+1)^2

    If this is right, then at full reflection, can we expect a force of 4 times the potential strength, while at no reflection, the particle still experiences a force equal to the potential? Also, my dimensions seem weird here: after integrating out the delta function, I'm just left with energy, not force...right? It seems proportionally correct, but I can't figure out where am I going wrong...

    I'm not sure if it's a calculation error or a conceptual misunderstanding on how to deal with finding this force.

    If this is a sound line of thinking or could be massaged into one, could it also be applied to a wavefunction undergoing exponential decay as it tunnels through a finite barrier?

    Thanks for the help!
  2. jcsd
  3. Sep 13, 2013 #2


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    by calculating the expectation value of [itex]F[/itex] from -∞ to [itex]0[/itex] you calculate the average value of the force the particle is likely to experience given it is on the left side of the potential well, and is described by the wavefunction [itex]\Psi(x)[/itex]. This seems like a good way to go if you are interested in measuring the force on the particle.

    Alternatively, you could look at how the momentum probability distribution of a Gaussian wave packet traveling to the right changes with time, and by seeing how the mean value of the momentum changes in a reflection, you can infer the force on that particle with Newton's second law and Ehrenfest's theorem.
  4. Sep 14, 2013 #3
    First off, thanks for the response.

    Right, so your first comment says what I've already done verbatim, but doesn't actually address any of my questions. Like the remaining force in the case of no reflection or the dimensions of the solution which seem to be off.

    In your second comment, we're no longer in a time independent picture if we are looking at the time rate of change of momentum, right? So, not valid here? I understand the use of Newton's second law and Ehrenfest's theorem in other situations, but I don't think it can be used for this scattering state in a time independent picture...

    I can Fourier transform the state into the momentum space representation, and find the expectation value of the momentum on this side of the well, but since the representation is still time independent, taking a time derivative of that would be 0. And as I understand it, there wouldn't really be a way to discern the expectation value of the momentum before and after a reflection--there'd only be 1 expectation value.

    Also, naively, since the scattering state has the same energy going in as going out after reflection, isn't the momentum change just twice the incoming, defined by the relations $$ \frac{\hbar^2 k^2}{2m} = E$$ $$p = \hbar k$$ $$Δp = \sqrt{8 m E} $$ (Can we look at it that way?) Either way, there's no dependence on the reflection coefficient here, and I don't see how to introduce the 'per unit time' to get a force.

    I think the best way to think about the problem is in the energy-gradient picture with time independence, as I've already done, but I still don't have resolve to my questions on that approach.

    Maybe I'm wrong or am missing something with the momentum view? Anyone care to throw down some math?
    Last edited: Sep 14, 2013
  5. Sep 15, 2013 #4
    Bump to my questions?

    1) Why does my calculation of ##\langle F \rangle## have units of ##V_0##?

    2) My analysis of the limits of the force are wobbly, because I forgot that the coefficient ##R## is NOT the probability of reflection, ##|R|^2##. Therefore, the answer for ##\langle F \rangle## actually has imaginary components to it once you expand it out! For reference:

    $$R = \frac{(k^2 - q^2) \sin{a q}}{2 i k q \cos{a q} + (k^2 + q^2) \sin{a q}} $$

    What does it mean to have a complex valued ## \langle F \rangle## ? Are the imaginary components related to dispersion/dissapative forces or a force-phase?

    Plotting it all out in normalized units with respect to the experimental variables ##{E,V_0}## provides some interesting behaviors, but I'm not sure if this is even a valid technique to begin with.

    The density plot is ##| \langle F \rangle |## (lighter is larger) as ##E=[0,300]; V_0=[0,100]##;
    The line plot is a normalized plot of ##|R|^2## in BLACK, ##| \langle F \rangle |## in RED, ##\Re{ \langle F \rangle }## in GREEN and ##\Im{ \langle F \rangle }## in BLUE

    Can someone comment on whether or not you can actually calculate the reflection force this way, and shed some light on the result if valid?

    Attached Files:

  6. Sep 24, 2013 #5
    Nevermind, I solved my questions after a discussion with a friend.

    1) Units issue comes from the normalization

    2) Imaginary issue comes from the fact that I forgot to take the conjugate of R

    If anyone comes across this and wants a more detailed explaination, feel free to pm me
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