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How to work in the momentum representation?

  1. Nov 15, 2007 #1
    In plain old QM, we can in principle take any state and expand it either in terms of momentum eigenstates or position eigentstates (i.e., the wave function). But in practice this usually means solving the Schrodinger equation, i.e., working in the postion representation from the start and then going over to the momentum representation later, if needed.

    For instance for a free particle we start with

    [tex]\frac{\hat{p}^2}{2m}|\psi(t)\rangle=i\frac{\partial}{\partial t}|\psi(t)\rangle[/tex]

    and go over to

    [tex]-\frac{1}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t)=i\frac{\partial}{\partial t}\psi(x,t)[/tex]

    This has the set of solutions [tex]\psi(x,t)\propto e^{i(kx-\frac{k^2}{2m}t)}[/tex] where k is an integration constant. Then we apply specific boundary conditions depending whether this is particle in a box, a free particle, the 2-slit experiment, etc.

    But why not start in the momentum representation? Well the trouble is that in the momentum representation the momentum operator is represented by multiply-by-p. So we have simply

    [tex]\frac{p^2}{2m}\phi(p,t)=i\frac{\partial}{\partial t}\phi(p,t)[/tex]

    the solution to which is

    [tex]\phi(p,t)\propto g(p)e^{-i\frac{p^2}{2m}t}[/tex]

    where g(p) is any old function of p. How do we get that g must take the form [tex]A e^{i\beta p}+Be^{-i\beta p}[/tex], where beta is a constant (later to be identified with position)?

    I was thinking that, hey, maybe some problems are easier to solve in the momentum representation from the start instead of the position rep, but I can't even figure out how to do the free particle.
     
    Last edited: Nov 15, 2007
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  3. Nov 15, 2007 #2

    Demystifier

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    g does NOT need to take this form. Instead, g(p) may be any function that can be represented by a Fourier series/transform in p. A similar statement is true for the original psi(x).

    And yes, many problems are easier to solve in the momentum space.
     
  4. Nov 15, 2007 #3
    Doh!

    You're right, of course, Demystifier. i see it now. Thanks.
     
  5. Nov 15, 2007 #4

    Demystifier

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    I am glad that you have understood that so fast. Once you realize that you can also expand in other bases as well, including the basis consisting of hermite polinomials, you will realize that the answer to your question on non-integer numbers of particles is not more complicated than this.
     
  6. Nov 15, 2007 #5

    reilly

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    pellman:
    In fact, g(p) must be a delta function (Kroneker delta in a finite system) to maintain orthogonality and completeness of the set of solutions of the free H, and to conform to common practice of describing particles with momentum k. (Yes, you have to play a few games with delta function products.)

    Properly used, the Hermite functions lead to creation and annihilation operators. In some instances, a configuration space or momentum space approach works with angular momentum states -- the multipole expansion of classical E&M --. Typically this approach is used for working with partial wave cross sections in scattering problems.

    Momentum representation is widely used in particle physics and in many-body physics. In fact, Feynman diagrams represent momentum space computations.

    You ask very good questions.

    Regards, Reilly Atkinson
     
  7. Nov 16, 2007 #6
    Thanks, reilly, but I think Demystifier is right. g can be any function normalized to [tex]\int |g(p)|^2dp=1[/tex]. Then the position representation is

    [tex]\psi(x)=\frac{1}{\sqrt{2\pi}}\int g(p)e^{i(px-\frac{p^2}{2m}t)}dp[/tex]
     
  8. Nov 16, 2007 #7

    reilly

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    pellman -- Try config space to see that your wave function is not an eigenfunction of H = -d/dx(d/dx)-- you get an extra p*p inside the integral, and you can't take it outside the integral. So your Wavefunction satisfies the time-dependent Schrodinger EQ. , but not the time independent Schrodinger Eq-- unless you go with the delta functions. Your wave function is a superposition of eigenstates of the free Hamiltonian.

    Regards,
    Reilly Atkinson
     
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