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Closed-form solutions to the wave equation

  1. Nov 4, 2008 #1
    Hi all,

    I'm interested to find a solution to the wave equation corresponding to
    Gaussian initial conditions
    [tex] \psi(0,x) = e^{-x^2/2} [/tex]
    A solution which satisfies these initial conditions is (up to some constant factor)
    [tex] \psi(t,x) = \int \frac{d^3k}{(2\pi)^3} e^{-k^2/2 + i(k \cdot x - \omega t)} [/tex]
    where \omega = |k|. If we use spherical coordinates ( so that k.x = |k| r cos \theta )
    then the angular dependence can be integrated out ... I get
    [tex] \psi(t,x) = \int_{0}^{\inf} \frac{dk}{(2\pi)^2} \frac{2k \sin(kr)}{r} e^{-k^2/2 - \omega t} [/tex]
    but can't get any further. Any ideas? Maybe this isn't the right direction to go anyway.

    The reason I'm looking at this is because I'm interested to see the viability of
    Hermite functions as a basis for state space in QM: the above function e^-x^2/2 is the
    zeroth Hermite function. Hermite functions form a countable orthonormal basis, they
    can be defined to have unit normalization, and all of their moments are finite. None
    of these are true for the usual plane wave basis... so I feel the Hermite functions provide
    a more concrete realisation of the Hilbert space of states. Of course, they won't be much
    use unless I get them in closed form for t>0.

  2. jcsd
  3. Nov 6, 2008 #2
    I have an idea, but I'm not sure whether it will be helpful.. I mean, if you rewrite sin(kr) as (exp(ikr)-exp(-ikr))/2i, then you have something on the form k*exp(f(k)). Then maybe you can use integration by parts to integrate? Maybe it won't work...
  4. Nov 7, 2008 #3
    The exponent of the initial solution is a quadratic equation. Complete the square, do a change of variables, done.
  5. Nov 10, 2008 #4

    Hans de Vries

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    Science Advisor
    Gold Member

    Do a plane wave decomposition from the x to the p domain. Then presume positive
    energy and thus :

    [tex] \omega = \sqrt{k^2+m^2}[/tex]

    This gives you the time-evolution of the independent plane waves. The last step is
    the 3d inverse Fourier from the (t,p) domain to the (t,x) domain. The result comes
    down to a convolution between the Gaussian and a second order Bessel K function
    with imaginary argument. (A Hankel function)

    Regards, Hans.
  6. Nov 11, 2008 #5
    Thanks for the three responses, they were a lot of help, and in the end helped me piece together an answer for the massless case... it's pretty epic.

    The problem:
    Massless propagation of free scalar field with Gaussian initial conditions
    \Box^2 \psi(t,x) = 0
    \psi(0,x) = \frac{1}{ a^{3/2} \pi^{3/4} }\exp \left( -\frac{x^2}{2a^2} \right)
    where x is shorthand for the three space dimensions (x,y,z). 'a' is a constant.

    The solution is
    \psi(t,x) = \frac{a^{1/2}}{ \pi^{3/4} 2r } \left( F(t+r) - F(t-r) \right)
    where r = \sqrt(x^2 + y^2 + z^2) and F is given by
    F(s) = s \exp \left( -\frac{s^2}{2a^2} \right) \left( 1- i \text{Erfi} \frac{s}{\sqrt{2}a} \right)
    The error functions arise because, as pointed out in the responses, the second integral in the original post is of the form k*exp(quadratic), and this can be done by substitution and it gives rise to an error function. Erfi is notation used by Mathematica; it's basically just Erf rotated by 90 degrees in the complex plane.

    I've attached two plots which show respectively the real and imaginary parts of the solution \psi(t,r), plotted on the (t,r) plane. Gaussian initial conditions can be seen along the axis t=0. It's kind of nifty, the light-like propagation is clearly visible in each plot, as the packets propagate along the lightcone t=+/-r.

    If there's any interest to see the details, I guess I can try write them up somehow.

    Thanks again for the tips,


    Attached Files:

  7. Nov 11, 2008 #6
    Sorry, that should be
    \psi(t,x) = \frac{1}{a^{3/2} \pi^{3/4} } \frac{1}{2r} \left( F(t+r) - F(t-r) \right)
    ... the answer in the previous post is out by a factor of a^2. For which I earn the 'wrong dimension' hat.
  8. Nov 12, 2008 #7

    Hans de Vries

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    Science Advisor
    Gold Member

    At first glance that looks to be what it should be. With the two terms coming
    from the sine used in the radial Fourier transform. Nice images BTW.

    Regards, Hans
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