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Step potential. What if E=U?

  1. Jun 25, 2012 #1

    CAF123

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    I understand the two cases dealt with when we consider the step potential. The two cases being when E>U (i.e the total energy is less than the max potential of the potential barrier) and when E<U. What is the physics of the situation when E=U? Is it a sort of a hybrid situation between the two cases? What would this look like?

    Classically, the situation is quite clear. If we imagine a marble in a bowl. If E>U, the marble will fall over the lid of the bowl and vice versa if E<U, the marble simply staying put, confined in the bowl. If E=U, the marble just reaches the tip of the bowl, but no more.
    Quantum mechanically, what does the resulting wavefunction look like?
     
    Last edited: Jun 25, 2012
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  3. Jun 25, 2012 #2

    Ken G

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    If E=U, we know the second spatial derivative of the wavefunction is zero, so the wavefunction is linear in x. That means it must be zero, or else it would blow up. So ironically, E=U has no chance of being found in the "impossible" regime, so ends up being "more classical" than the E<U situation. But note we would need to choose the potential very specially to even allow an energy eigenvalue that is equal to U.
     
    Last edited: Jun 25, 2012
  4. Jun 25, 2012 #3

    jtbell

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    It can also be constant but nonzero, and that's what it turns out to be if you solve the problem in the usual way by applying the boundary conditions at the step boundary to the solutions on both sides. Of course this isn't normalizable, but we usually analyze the step potential using monochromatic plane waves which are non-normalizable idealizations anyway.

    Consider the attached (very crude) graphs which have the step boundary at x = 0. The potential is zero to the left and U0 to the right. Focus on the wave function to the right of the step boundary:

    E << U0: the wave function decays very "rapidly"

    E slightly < U0: the wave function decays very "slowly"

    E = U0: The wave function is constant (neither decays nor oscillates)

    E slightly > U0: the wave function oscillates with a long wavelength.

    E >> U0: the wave function oscillates with a wavelength nearly equal to the wavelength on the left side of the step.

    On the left side, the wavelength should be the same in all cases. I haven't tried to show the relative amplitudes on the left and right for the cases where E > U0.
     

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  5. Jun 26, 2012 #4

    Ken G

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    That's an interesting point about the non-normalizability, but I think the wavefunction being unnormalizable is only not a problem when we have a propagating wavefunction and can interpret it as a transmission vs. reflection probability, which involves a ratio of the unnormalizable wavefunction on the left and right. In the E=U case, we don't have propagation on the right, so we can't interpret that ratio as a transmission amplitude, and we can't be happy with a constant non-propagating unnormalizable wavefunction. Put differently, we could imagine the U jump is placed at L/2 of a hugely wide infinite square well of width L, with a particle of definite energy E (suitably chosen) bouncing around in there, and see what happens in the limit as the width L gets arbitrarily large. We might then ask, what is the probability, after many crossing times, of finding the particle on the right side? In the E < U case, that probabliity goes to 0 as L gets large. In the E > U case, that probability goes to something between 0 and 1/2 (depending on the phase of sine function when it arrives at L/2). But in the E=U case, since we have a zero constraint on the wavefunction at L, the wavefunction on the right has to always be zero-- regardless of how large L gets. So in the limit that L is infinite, it must still be zero over there-- the probability of finding the particle on the right side is zero for all L, so should also be zero in the transmission/reflection problem.

    This also means that the width L must be specially chosen to allow E=U as an eigenvalue.
    (Edit-- scratch that, if it has to have a continuous first derivative at L/2, and has to be zero, then it is just plain not possible. There is no L that allows E=U to be an energy eigenvalue, so it's also not an energy eigenstate in the limit as L goes to infinity. It's an inherently time variable problem, but it seems the transmission has to go to zero when E=U, if E is for a free particle incident from the left.)
     
    Last edited: Jun 26, 2012
  6. Jun 26, 2012 #5

    mfb

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    In the case of an exact energy E=U of the particle, the hand-wavy evaluation with incoming and outgoing waves does not work. As this energy would have to be an eigenvalue, the state would be stationary and there would be no moving particles at all.
    In the case of an incoming wave packet with <E>=U, some parts can be transmitted, some parts can be reflected.
     
  7. Jun 26, 2012 #6

    Ken G

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    But the question is, when the uncertainty in E gets arbitrarily small, is there any transmission when <E>=U? I argued above that there must not be, the transmission should go to zero as the uncertainty does, even though it is impossible to make the uncertainty formally zero because E=U cannot be an eigenvalue.
     
  8. Jun 26, 2012 #7

    jtbell

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    The transmission coefficient goes to zero as E approaches U from above. The wave function "beyond" the barrier is nonzero for E < U, but it does not represent a propagating wave. The transmission coefficient is defined in terms of the amplitudes of propagating waves.
     
  9. Jun 26, 2012 #8

    Ken G

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    What is its ratio of the magnitude of the amplitude on the right with the magnitude of the amplitude on the left? If it is nonzero but nonnormalizable, it must still have a meaningful value for that ratio that is different from zero. But it can't-- it has to be zero. This is shown, I would argue, but placing the system in a giant square well of size L, letting the system bounce around for a long time, and ask what is the probability of finding the particle on the right side. At the very least, that device proves that we cannot be solving the TISE here.
     
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