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Schrödinger equation in momentum space and number of solutions

  1. Nov 15, 2011 #1

    Note: I will be sloppy with constant factors in this post. Only the general structure of the equations matters.

    Consider a particle in a linear potential,

    [itex]\frac{\mathrm d^2}{\mathrm d x^2} \psi(x) + x \psi(x) - E \psi(x) = 0.[/itex]​

    Mathematically, this is a second-order ODE, and there are two solutions, related to the Airy functions Ai and Bi. Physically, only the Ai solution has the correct boundary condition; Ai(y) drops exponentially to zero for y > 0, whereas Bi(y) diverges exponentially. So one is led to select the Ai solution as the physical one and discard the Bi solution.

    Now, let's go to momentum space, where

    [itex]\frac{\mathrm d}{\mathrm d k} \psi(k) + k^2 \psi(k) - E \psi(k) = 0.[/itex]​

    By the familiar properties of the Fourier transform, d2/dx2 became a k2, and x became a d/dk. We are left with a first-order ODE, which has one linearly independent solution, namely the physical one corresponding to Ai.

    The question that occurs to me is, what happened to the second solution? Why did it disappear when going to momentum space, and how was the Fourier transform able to pick out the physical solution? Where is the Airy of yesteryear?

    Additional notes:
    1. Both Ai and Bi seem to have well-defined Fourier transforms, see http://functions.wolfram.com/Bessel-TypeFunctions/AiryAi/22/01/ and http://functions.wolfram.com/Bessel-TypeFunctions/AiryBi/22/01/.
    2. This stated problem is the one that brought up the question, but I could ask a very similar one about Dirac's treatment of the harmonic oscillator.
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Nov 15, 2011 #2


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    Well, there's only square integrable solution of the SE either way, so it's ok. The boundary conditions for such ODE's are imposed to assure 2 conditions:
    * solutions are within the Hilbert space of normalizable states.
    * the formal differential operators such as the Hamiltonian (2nd order in coordinate representation) and the momentum (1st order in coordinate representations) are self-adjoint.

    These 2 restrictions should lead you to acceptable solutions for the ODE's. That's why we're rejecting Bi(x). Going to momentum space, it may happen that some ODE in the coordinate representations becomes either another ODE, or an integral equation, or intego-differential equation, or neither, for example for the free spinless Galilean particle. Sometimes it's easier to find solutions in the coordinate representation, sometimes in the momentum representations, depends on the model and its space-time symmetries.

    The Fourier transformation has the merit of converting physical solutions (normalizable wavefunctions) into physical solutions, since it's an automorphism of the Hilbert space.

    Bottom line, a spectral equation for an operator in coordinate representation leads to PDE's or ODE's for which we always have boundary conditions which select either normalizable states, or scattering states. The Fourier transformation takes normalizable into normalizable, or scatteting into scattering, nothing is really lost due to it. The loss (if there is one) is only due to the various boundary conditions which may occur.
  4. Nov 15, 2011 #3

    Physics Monkey

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    Without thinking in detail, the potential you're considering is already pretty sick since it runs away to negative infinity. The physical nature of solutions can be quite delicate in such a situation.

    On the other hand, if you cutoff the potential, say by considering only x > 0, then your second equation does not contain the correct potential.
  5. Nov 17, 2011 #4
    Thank you for your replies. Also, I apologize for a mistake in my question where I wrote that Bi possesses a "well-defined" Fourier transform; I only saw that Wolfram Functions had a (rather complicated-looking) formula for [itex]\mathscr F[\operatorname{Bi}](z)[/itex], but looking more closely, I see that it is proportional to [itex](z - \sqrt{z^2})[/itex], so it would be 0 for real z. So there is my answer, I guess. See also: calclab.math.tamu.edu/~fulling/m412/f07/airywkb.pdf.

    In another sense, I still think it's an interesting thing that the Fourier transform picks out exactly what we want. (Is this in some sense a general property? Is it possible to construct potentials where Fourier discards acceptable solutions, or even picks wrong ones?)

    I do want to comment that neither Ai nor Bi are square integrable. So the reason for rejecting Bi has more to do with its divergent nature.

    After reading up a little bit on my Hilbert space theory [http://arxiv.org/abs/quant-ph/9907069] [Broken], I would conjecture that Ai is in the dual space (meaning that [itex]\int \text d x Ai(x) \psi(x) < \infty[/itex] for [itex]\psi \in L^2[/itex]) which Bi surely is not, and that this is the deeper reason.
    Last edited by a moderator: May 5, 2017
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