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Fourier Transforms, Momentum and Position

  1. Jun 9, 2009 #1

    r16

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    In quantum mechanics, why does the Fourier transform

    [tex] f(x) = \int_{-\infty}^\infty F(k) e^{ikx}dk [/tex]
    represent position and

    [tex]F(k) = \int_{-\infty}^\infty f(x) e^{-ikx} dx [/tex]
    represent momentum?
     
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  3. Jun 9, 2009 #2

    malawi_glenn

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    you must have something given to you, e.g. given that f(x) is the position wavefunction, then it's fourier transform is the momenum wavefunction. It just follows since p = - i hbar d/dx

    <x'|p|p'> = -i h d/dx' <x'|p'>

    p'<x'|p'> = -i h d/dx' <x'|p'>

    solve this differential equation

    <x'|p'> = N exp { i p' x' /h }

    now look at

    psi_a(x') = <x'|a> = (insert completeness realtion ) = integral dp' <x'|p'><p'|a>

    psi_a(x') = N integral dp' exp { i p' x' /h } phi_a(p')

    determine the normalization N using the delta function:

    delta(x' - x'') = N^2 integral dp' exp { i p'( x' - x'') /h }

    we get N = 1/sqrt(2 pi h)
     
  4. Jun 10, 2009 #3

    r16

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    I'm only starting out in Quantum Mechanics (chapter 2 of the griffiths book) and I am not familiar with the notation

    I'm sure ill get to it later on in the book. Until then, could you explain it?
     
  5. Jun 10, 2009 #4

    malawi_glenn

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    you might want to look up "bra - ket" notation or "dirac notation" in Griffiths book, then I can explain if you don't understand, but basically:

    p|p'> = "p operator on p-eigenstate with momentum p' " = p'|p'> (I denote operator with the letter and eigenvalues with prime"

    since p = -i h d/dx, we can do the same operation on the right hand side, but with <x'|p'> as just an arbitrary function of x' and p'

    The basic idea is that, without getting too much into math behind it;

    |a'> is a vector in hilbert space, it denotes the state with quantun number a'

    a|a'> = a'|a>

    ok?

    these are called "kets"

    now the dual vector, called "bra":

    <a'|

    we can "think" of this as the ket's beeing column vectors and bra's as row vectors:

    <a''|a'> is then a number

    ok, this was assuming that a',a'' are discrete quantum numbers

    now, for x and p, which are continuous, we can use the same notation, but we can not imagine/represent them as discrete vectors as we do in introductory linear algebra.

    Please, also check out this recent thread:
    https://www.physicsforums.com/showthread.php?t=318899

    Good Luck, have fun
     
  6. Jun 11, 2009 #5
    r16 -> It's not that, say, "f(x) represent position" and its FT F(k) "represent momentum". What you do have is that you can express the same object, like the wave function, both in "position space" (i.e. as a function of x, psi(x)) or in "momentum space" (as a function of k, psi(k)). There is a 1-1 correspondence between a function and its FT and they contain the same information. But some problems are better described in position space and others in momentum space.
     
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