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Potential Step Scattering

  1. Mar 28, 2012 #1
    A particle of energy E is incident from the left on a potential step. If the energy of the particle is greater than the height of the step, it is acceptable to discard:
    [tex]
    Fe^{-ikx}
    [/tex]
    saying that there is nothing incident from the right.

    If E is less than the step, it is not acceptable to discard:
    [tex]
    Fe^{-kx}
    [/tex]

    When combined with time dependence, both are waves moving right to left. The only difference is that one is a sine/cose wave and the other exponential. Why does this matter to the argument?
     
  2. jcsd
  3. Mar 28, 2012 #2
    I'm not sure that I have understood your question; anyway, if I have understood the right thing, one possible answer is the following (by convention, the step is located at x=0, V=potential step); I tell you in advance that the point that I want to reach is that , in my opinion, the final answer to your question is just a matter of convention, since the physics predicts a very precise thing.

    1) E> V

    in this case the solution of the Schroedinger equation we have that for x<0

    [itex]\psi(x)=e^{ipx}+Be^{-ipx}[/itex]

    while for x>0

    [itex]\psi(x)=Ce^{ip'x}+De^{-ip'x}[/itex]

    with [itex]p=\sqrt{2mE}[/itex], [itex]p'=\sqrt{2m(E-V)}[/itex] and where I have "normalized" the solution by imposing that the coefficient of [itex]e^{ipx}[/itex] is 1.
    In order to find B,C,D one has to impose the continuity of the wave function and its first derivative with respect to x (two relations in three unknowns); let us say that we can find B and C as a function of D and E.

    Of course we have not found all the eigenfunctions for E>V: we are left to find those whose [itex]e^{ipx}[/itex] term coefficient is 0 (and so they cannot be normalizaed in such a way that the [itex]e^{ipx}[/itex] term coefficient is 1); they can be easily find by posing for x<0
    [itex]\psi(x)=Be^{-ipx}[/itex]
    and for x>0

    [itex]\psi(x)=Ce^{ip'x}+e^{-ip'x}[/itex]
    where I have normalized in such a way to pose the [itex]e^{-ip'x}[/itex] coefficient equal to 1; B and C are determined by the border conditions.

    Anyway, one notes, if I didn't make any mistake, that for D different from zero in the first set of solutions, one can simply find that it is a superposition of a solution of the first type with D=0 and a solution of the second type.


    2) 0< E< V

    one can proceed as above, finds the exponential solution for x>0 and takes only the non divergent solution for x->infinity; the eigenfunction is determined once we have normalized to 1 the [itex]e^{ipx}[/itex] term coefficient and, if I didn't make any mistake, there are no solution if we impose that the [itex]e^{ipx}[/itex] term coefficient is 0.

    So we have found a whole set of ortho"normal" eigenfunctions [itex]\psi_E[/itex].


    The important point of the discussion, in my opinion, is the following: let us consider the physical situation in which a particle goes from left to right; from a quantum point of view this has a meaningful interpretation just in terms of wave packets; let us impose that at a time [itex]\tau[/itex] "very negative" we have a given wave packet; in particluar we can construct, say, a gaussian packet [itex]\psi_G[/itex] that is centered at x0<0 and with some dispersion \delta x; then one finds the Fourier transformed function to see for momenta distribution and finds that the momentum distribution is located at p0>0 with a particular dispersion (which of course automatically satisfy the Heisenberg uncertainty principle); now, in order to find the expansion of [itex]\psi_G[/itex] in terms of our eigenfunctions you should, as usual,

    [itex]\psi_G(x)=∫dE \psi_E(x)∫dy \psi_E(y)^*\psi_G(y)[/itex].

    This is what makes sense, in my opinion; sentences like "not incident from the right" are just a short way to express the fact that a particular coefficient of the solution is zero; moreover, I note that also in a solution with D=0, for x>0 the term C is in general different from zero; this means that, if we take the energy eigenvalue then we have a nonzero probability to find the particle in the region x>0.
    Best,
    Francesco
     
  4. Mar 28, 2012 #3
    Thanks for the help. Sorry about not defining the problem first. I was thinking about it so long that I forgot to put it in context. I'll just stick to your definitions.

    I'm not sure I completely understand your full argument. Is is basically this:
    For E>V, applying the boundary conditions gives two sets of solutions: one set has D=0 and the other set has A=0.
    For E<V, applying the boundary conditions gives one set of solutions, where C=0.

    Then you claim "incident from the left" just means a solution for which D=0.

    First, I see this works for E>V. But it doesn't work for E<V because D is not zero. Why does "incident from the left" mean different things for E>V and E<V?

    Second, this seems like a circular argument if you just define "incident from the left" as setting the coefficients equal to the correct solution. I was thinking there must be a physical reason certain coefficients are set to zero. Namely, they represent waves traveling in a certain direction. For me, "incident from the left" means "no wave traveling left from the right side from the right side of the step". But this is clearly wrong when E<V.
     
  5. Mar 28, 2012 #4
    Hi once again! Sorry for me not to have been clear; I will try to be clearer now.

    Yes, this is my opinion.


    Yes, I claim that "incident from the left just means a solution for which D=0" ("by definition", as you have pointed out); the point is that, in my opinion, one can claim that a wave function represents a particle incident from the left (and, for example, located at x0<0) only if the wavefunction that represents it is a packet centered at x0 and its Fourier transform is a packet centered at p0>0. Since a generic particle represented by an energy eigenfunction is not centered at a certain point (and, in particular, it is not localized) I think that it's meaningless to give to an energy eigenfunction a physical attribute as "incident from the left" (in this sense I have remarked that in an energy eigenstate the probability to find the particle at x>0 is different from zero); I think, then, that "incident from the left" can only have a mathematical meaning (and, reasonably, this can be identified, by definition, with the requirement D=0).
    The whole point of the discussion is not to give a meaning to "incident from the left" for E<V, but make the sentence "incident from the left" physically meaningless for E>V.

    Maybe (I must warn you that this is just an idea that has just occured to my mind) one can make the notion of wave travelling from the left in scattering theory, where there is a formulation in terms of plane waves; but I also know that this formulation is meaningful when one can identify a kinetic term and a potential term (which is the case), but, as far as I know, the potential is required to satisfy some conditions (for example, it should be short range, otherwise problems arise). Let me know what you think.
    Best,
    Francesco

    PS I think anyway that in order to make the scattering theory apparatus consistent and meaningful, one should alway interpretate his operations in terms of wave packet, see Goldberger and Watson "Collision Theory"
     
    Last edited: Mar 28, 2012
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