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Reason for Fourier transform convention in QM?

  1. Jun 5, 2009 #1
    I always tended to think that we ought to use formulae which explicitly remind us that position and momentum are on equal footing in quantum theory (even though this may not be ultimately true) and write my transforms symmetrically

    [tex]f(x)=(2\pi)^{-3/2}\int{F(p)e^{ix\cdot p}d^3p[/tex]

    [tex]F(p)=(2\pi)^{-3/2}\int{f(x)e^{-ix\cdot p}d^3x[/tex]

    However, I frequently see the convention of putting the 2-pi's with the momentum volume element, that is,

    [tex]f(x)=\int{F(p)e^{ix\cdot p}\frac{d^3p}{(2\pi)^3}[/tex]

    [tex]F(p)=\int{f(x)e^{-ix\cdot p}d^3x[/tex]

    Here is an example. Folland here is acquainting mathematicians with physics conventions and explicitly states this convention.

    but is the there any reason behind this convention?
  2. jcsd
  3. Jun 5, 2009 #2


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    I'm pretty sure I've seen the first convention you mentioned in QFT books. I can't think of any other reason than personal taste for preferring one over the other.
  4. Jun 5, 2009 #3
    It's a notorious plot set out by the ancient theoretical physicists to confuse mathematicians.

    Either that, or it really is just a matter of taste (physicists tend to use the latter definition, since it looks prettier/easier to remember. Mathematicians use the former one, since the transformation is then unitary and more symmetric).
  5. Jun 5, 2009 #4
    It has to do with the fact that the number of states per unit volume in momentum space per unit of volume is 1/(2 pi)^3. In the case of infinite volume, you want to normalize things such that physical quantitiies are finite and such that it corresponds to the same quantity per unit volume in the finite volume case.

    If you put the system in a finite volume of V, then the free normalized momentum eigenfunctions will be:

    1/sqrt(V) exp(i k.x)

    And the number of momentum eigenfunctions in a volume V_k in momentum space will be:

    V V_k/(2 pi)^3

    This can be derived by taking the volume to be a cube and then imposing periodic boundary conditions. And then you argue that, if you have different kind of boundary conditions, that should not matter if the volume is large enough.

    Once the momentum eigenstates are fixed, the "Fourier transform" is fixed too:

    F(k) = <k|psi> = Int d^3 x <k|x><x|psi> =

    1/sqrt(V) Int d^3 x exp(-i k x) psi(x)

    psi(x) = <x|psi> = Sum_k <x|k><k|psi> =

    Int d^3 k V/(2 pi)^3 <x|k><k|psi> =

    sqrt(V) Int d^3 k/(2 pi)^3 exp(i k x) F(k)

    The momentum eigenstates are properly normalized, which is possible because we are working in a finite volume. We can now redefine the normalization of the momentum wavefunctions such that F(k) remains finite in the limit of V to infinity if psi(x) is properly normalized. So, e.g., if psi(x) is some properly Gaussian wavefunction normalized to 1 which falls off well before the boundary of the volume is reached and we make V larger and larger, F(k) as it is defined now, would tend to zero, simply because of the 1/sqrt(V) prefactor.

    If we redefine F(k) by multiplying F(k) as defined above by sqrt(V), we get:

    F(k) = Int d^3 x exp(-i k x) psi(x)

    psi(x) = Int d^3 k/(2 pi)^3 exp(i k x) F(k)

    The probability of a particular value of k has now become
    1/V |F(k)|^2. The probability that the k value will be in a volume element d^3 k is then obtained by multiplying this with the number of k values in this volume, which is V d^3k/(2 pi)^3. This is thus given by:

    |F(k)|^2 d^3k/(2 pi)^3

    If we integrate this over all of momentum space, then we'll get 1.

    Physically this convention makes sense, because all we've done is rescaling everything by a volume factor when passing to the infinite volume case. The measure d^3k/(2pi)^3 makes physical sense as it is the number of k states in the momentum space volume of d^3k per unit of ordinary volume. Distributing the 1/(2pi)^3 symmetrically by redefining F(k) again by dividing it by (2 pi)^(3/2) would only makes things more complicated.

    So, when doing compitations, all we have to do in the infinite volume case is to proceed exactly as in the finite volume case, except that you can omit volume factors and some quantities will now be per unit volume.
  6. Jun 6, 2009 #5
    Isn't this only true if we (arbitrarily?) hold to the convention which is the subject of this thread?
  7. Jun 6, 2009 #6
    This is not related to any conventions. The number of linearly independent quantum states below some energy in some volume is some given number. No changing of conventions can change that. Of course, you can change the units you use to measure the energy, or momentum and then say that the numerical values change. In this case that translates to changing the convention that hbar = 1, but this is not relevant to the question why the 1/(2 pi)^3 factor is not distributed symmetrically.
  8. Jun 9, 2009 #7
    Thanks, Count. Hopefully I can squeeze one more post out of you.

    Maybe I just don't understand what this means. Can you explain further? What is the number of states per unit volume in position-space?
  9. Jun 9, 2009 #8
    The number of states per unit volume in position space and per unit volume in momentum space is 1/(2 pi)^3 (times the number of internal degrees of freedom like spin etc.)
  10. Jun 9, 2009 #9
    Ok. I got it. Thanks.
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