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The surprisingly simple linear potential solution

  1. Jan 21, 2009 #1
    Once upon a time I was taking a quantum class and I asked the instructor why we did not cover the linear potential. What happened to F=ma in quantum mechanics? He said it was quite non-trivial and had to be done with Airy functions; not suitable for an introductory class.

    Ok.

    Given the Schrodinger equation

    [tex]\{ -\frac{1}{2m}\frac{\partial^2}{\partial x^2}-Fx\}\Psi=i \frac{\partial\Psi}{\partial t} [/tex]

    how about

    [tex]\Psi\propto exp\{i[(k+Ft)x-\frac{k^2 t}{2m}-\frac{kFt^2}{2m}-\frac{F^2t^3}{6m}]\}[/tex]

    where k is arbitrary?

    Why isn't this well known? I spent some time digging through quantum texts in a science library and found only the barest mention of it, just a quick reference in a chapter exercise.

    Seems like since it is so very simple it should get mentioned. I presume that it is because it is uninteresting. But why?
     
  2. jcsd
  3. Jan 21, 2009 #2
    what is the normalization constant?
     
  4. Jan 21, 2009 #3

    Avodyne

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    The linear potential is easy to solve in momentum space, where the time-indepdent Schrodinger equation is (with [itex]\hbar=1[/itex])

    [tex]{p^2\over 2m}\widetilde\psi_E(p) - iF{d\over dp}\widetilde\psi_E(p) = E\widetilde\psi_E(p)[/tex]

    The (unnormalizable) solution is

    [tex]\widetilde\psi_E(p) = \exp\!\left[{i\over F}\left(Ep-{p^3\over 6m\right)\right][/tex]

    To get to position space, we have to Fourier transform from [itex]p[/itex] to [itex]x[/itex], and this yields an Airy function.

    However, the general solution of the time-dependent Schrodinger equation (in momentum space) is given by

    [tex]\widetilde\psi(p,t)=\int_{-\infty}^{+\infty}dE\;c(E)\,e^{-iEt}\;\widetilde\psi_E(p) [/tex]

    where the coefficient [itex]c(E)[/itex] is an arbitary function of [itex]E[/itex]. If we choose [itex]c(E)=\exp(-ikE/F)[/itex], where [itex]k[/itex] is arbitrary, we get

    [tex]\widetilde\psi(p,t)\propto \delta(p-k-Ft)\exp(-ip^3/6mF) [/tex]

    The time-dependent wave function in position space is then

    [tex]\psi(x,t) = {1\over\sqrt{2\pi}}\int_{-\infty}^{+\infty}dp\;e^{ipx}\;\widetilde\psi(p,t)[/tex]

    which works out to be

    [tex]\psi(x,t) \propto\exp[i(k+Ft)x-i(k+Ft)^3/6mF][/tex]

    This is your solution with an extra constant phase factor of [itex]\exp(-ik^3/6mF)[/itex]. It is nice that it can be expressed exactly, but physically it corresponds to a superposition of energy eigenstates with equal probability for every energy (and a particular energy-dependent phase factor). This is not a particularly useful wave function to know.
     
  5. Jan 22, 2009 #4
    In 1-D, a normalization constant of [tex]\frac{1}{\sqrt{2\pi\hbar}}[/tex] gives [tex]\langle\Psi_k|\Psi_{k'}\rangle=\delta(k-k')[/tex] same as for free momentum states. Notice that this solution reduces to the plane wave for F=0.
     
    Last edited: Jan 22, 2009
  6. Jan 22, 2009 #5
    I expect you're right. I myself have no sense of what is useful or not.

    However, you can also look at this solution as an accelerating plane wave. At any particular time t its shape is a pure plane wave corresponding to momentum k+Ft, that is, a plane wave whose wavelength shortens with time. So if you are dealing with a system in which momentum eigenstates are more important than energy eigenstates, wouldn't you prefer this solution?
     
  7. Jan 24, 2009 #6

    Dr Transport

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    The solution for the linear potential is an Airy function. The details are found in many tests on solid state optics.
     
  8. Jan 24, 2009 #7
    Yes, that is true. The solutions in terms of Airy functions are the energy eigenstates.

    The expression in the OP is also an exact solution ... but one that is much simpler.
     
  9. Jan 24, 2009 #8
    I'm sorry pellman I was trying to point out that your solution has a problem with normalisation.

    for one the nomalization depends on time, which means that a normalised solution would no longer be a solution to the schrodinger equation.


    although your response puzzles me, are you trying to solve the eigenvalue problem for the linear potential (which are the airy functions) or are you trying to write the full solution. if its the latter that would mean your solution satisfies <Y||Y>=1
     
  10. Jan 25, 2009 #9
    Check again. Unless I'm mistaken, if

    [tex]\Psi_k(x,t) = \frac{1}{\sqrt{2\pi}} exp{\{i[(k+Ft)x-\frac{k^2 t}{2m}-\frac{kFt^2}{2m}-\frac{F^2t^3}{6m}]\}}[/tex]

    then

    [tex]\int{\Psi_{k'}^*(x,t)\Psi_k(x,t)dx}=\delta(k-k')[/tex]

    The solution is an eigenstate of momentum. The eigenvalue k+Ft depends on time, true, but since time is a parameter here and not a dynamical variable, I don't think that is a problem.
     
  11. Jan 26, 2009 #10
    my mistake, you are correct
     
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