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Fourier Analysis in QM

  1. Jun 4, 2004 #1
    Myself and a user on another board have come to the following hurdle we can't overcome:


    Mathematically we represent an arbiarty matter wave as a superposition of plane waves; using the theory of Fourier analysis. We can write an arbitary wavepacked as a Fourier integral of the form:
    [tex]$\psi (x,t) = \frac{1}{{\sqrt {2\pi } }}\int\limits_{ - \infty }^\infty {A(p_x )e^{\frac{i}{\hbar }(p_x x - Et)} dP_x } $
    [/tex]

    What this means is we can mathematically constuct a localised complace wavefunction by adding up (integrating) a large number of plane waves each with different momentum. The relaive amplitde of the component waves is determined by the function A(p[x]) which is given by:

    [tex]$A(p_x ) = \frac{1}{{\sqrt {2\pi } }}\int\limits_{ - \infty }^\infty {\psi (x,t)e^{\frac{i}{\hbar }(p_x x - Et)} dx} $
    [/tex]

    The functions psi(x,t) and A(p[x])) are called Fourier transofrms of one another.

    How could we use this practically? Isn't there a circular argument there? To get teh amplitudes, we need the wavefunction, but to get teh wavefunction we need the amplitudes. The only answer I could give was that the amplitudes may fall naturally out of the system being studied. However having not 'offically' studied Fourier series before, is there a simpiler explanation?
     
  2. jcsd
  3. Jun 4, 2004 #2
    Well my knowledge of fourier series being better than my knowledge of quantum mechanics formalisms (though still not that great), i would say to remember that these are general cases. We can use experimental data to determine what we want the wavefunction to be. and from there the component amplitudes are fairly simple(EDIT:this is a relative term, they may not be imple to actually obtain, but the process should be obvious). Also it might be easiser to think in terms of the non-complex fourier series, rather than the complex forms of the transforms.
     
  4. Jun 4, 2004 #3

    turin

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    Homework Helper

    The wavefunction must satisfy the dynamical equation in theory. If you know your physical system, then you can solve for the theoretical wavefunction, and this may be most convenient in the momentum basis. Even if the momentum basis doesn't make things convenient for you, you may still want to know what momenta to expect in an experiment once you find what the theory says the wavefunction should be (in the configuration basis). In other words: sometimes you start with ψ(x,t) and for whatever reason you want ψk.

    Another thing you can do is make a bunch of momentum measurements. This would give you the Fourier amplitudes (technically, I think it would have to be a fourier series). Then, you would use the transform if you wanted to find the wavefunction as a spatial probability distribution. In other words: sometimes you start with ψk and for whatever reason you want ψ(x,t).
     
  5. Jun 5, 2004 #4

    reilly

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    FT's and QM

    Your question could well be articulated for Electricity and Magnetism, heat conduction, mechanical vibrations. The use of Fourier Transforms -- or series-- is often a useful tool for the solution of differential equations., ordinary and partial.The wave equation, for the Schrodinger Eq. or Dirac Eq. for a free particle, is transformed by the Fourier Transform into an algebraic equation, one that's easy to solve.

    One approach to proof of the Uncertainty Princ. uses a Gaussian wave function to describe a wave packet. And the FT of a Gaussian is a Gaussian. As it turns out, a narrow spatial wave packet leads to a broad momentum space wave packet. So FTs are useful for basic theoretical work.

    Note that Fourier - like transforms and series can be developed with many sets of functions -- Legendre Polynomials for angular momentum, Hermite Functions for oscillators, and others as well.

    Many books on mathematical physics, QM, E&M discuss all of this in great detail. Like everything else, FTs are sometime useful and appropriate, and sometimes not.

    Regards,
    R. Atkinson
     
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