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Position space - momentum space

  1. Feb 12, 2007 #1
    could somebody please explain to me why position and momentum space are related to one another by a fourier transform, meaning why do I get momenta when I do a fourier transform of an expression in position space?
  2. jcsd
  3. Feb 12, 2007 #2
    this is actually a wonderful way of looking at the HUP, it really takes the mystery out of it

    the HUP, and the relationship that you point out above, are a consequence of the fact that in quantum mechanics particles are (a) described by waves and (b) quantized in energy

    consider a gaussian wavepacket propogating through space..remember that the momentum and the wavelength are related by de Broglie's equation. The more accurately you can sample the positional "signal" of the wavepacket (i.e. make the digital "sampling interval" smaller and smaller) the worse you know the wavelength. Vice versa, in order to really know the wavelength well requires a greater sampling interval, ideally infinite, but this makes the positional interval large. The quantized energy of the waves that are being summed into the wavepacket place an upper bound on the standard error (i.e. <x^2> - <x>^2 and <p^2> - <p>^2) if you work it out exactly.

    if you don't believe me, then work out <q|X> and <p|X> knowing the <x|p> and <p|x> relations, and you will get your fourier factors. See Dirac, 4th edition pg.97 for details.

    these formulas can also be used to derive the Feynman path integral expression by inserting complete sets of states |q><q|, |p><p| into Tr{exp(-beta H)} and utilizing the Trotter identity.
  4. Feb 12, 2007 #3
    here's a concrete example.

    [tex]\hat{p} <x|p> = <x|p|p> = p<x|p>[/tex]

    but we also know that:

    [tex]\hat{p} <x|p> = \frac{\hbar}{i} \frac{\partial}{\partial x} <x|p>[/tex]

    consider the projection of the momentum into position space, <x|p>, as just some function f(x,p). Then we have the DE:

    [tex]\frac{ip}{\hbar} f(x,p) = \frac{\partial}{\partial x} f(x,p)[/tex]

    the solution of which (when normalized) is:

    [tex]<x|p> = \frac{1}{\sqrt{2\pi \hbar}} e^{\frac{ipx}{\hbar}}[/tex]

    and so it follows that

    [tex]<p|x> = \frac{1}{\sqrt{2\pi \hbar}} e^{-\frac{ipx}{\hbar}}[/tex]

    consider some arbitrary state projected into the momentum space:

    [tex]<p|\psi >[/tex]

    by the complete set of states, we have the following identity:

    [tex]\int_{-\infty}^{\infty} dx |x><x| = \hat{1}[/tex]

    and inserting this between the bra and ket:

    [tex]<p|\psi> = <p |\left( \int_{-\infty}^{\infty} dx |x><x| \right) |\psi> = \int_{-\infty}^{\infty} dx <p|x><x|\psi> = \frac{1}{\sqrt{2\pi \hbar}} \int_{-\infty}^{\infty} dx e^{-\frac{ipx}{\hbar}} <x|\psi>[/tex]

    and so the state projected into momentum space is given by the fourier transform of the state projected into position space.
    Last edited: Feb 12, 2007
  5. Feb 13, 2007 #4


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    Simple, the Fourier transform is the unitary application from the position coordinates Hilbert space [itex] L^{2}\left(R^{3},d^{3}r\right) [/itex] to the momentum coordinate Hilbert space [itex] L^{2}\left(R^{3},d^{3}p\right) [/itex]
  6. Feb 13, 2007 #5
    Thanks for all your replies. I am sorry for having phrased my question badly. I know the mathematics of how to transform from position space to momentum space and how to get the fourier factor. However what I really shoud have asked is: Is there a physically intuitve reason why we would expect to get to momentum space when we do a fourier transform? By a physically intuitive reason I mean something along the same lines as calculating commutation relations of poincare generators with the Pauli-Lubanski vector: We could either plug in definitions and do the mathematics(which is a similar method to the earlier replies) or we could say that the pauli lubanski vector is a vector and so we know how it will transform under poincare transformations(which is kind of physically intuitive).
  7. Feb 13, 2007 #6
    more intuitive? your joking, right? are you familiar with fourier analysis of wave phenomena?
  8. Feb 13, 2007 #7
    I am sorry, but I did not really understand your last reply. I am used to usual fourier analysis if that is what you mean, however I have never heard of a physical interpretation of such a transform and considering that it maps momentum space to position space and vice versa I was wondering whether such an interpretation exists. So if you know about any please let me know.
  9. Feb 13, 2007 #8
    i do. please see post #2.

    how do you typically transform from position space to wavelength space?
  10. Feb 13, 2007 #9
    Are you asking why a wave packet can be described by some equation

    [itex]\psi (x,t) = \frac{1}{\sqrt{2\pi \hbar}} \int^{\,\infty}_{-\infty} dp\, \phi(p) e^{i(px-Et)/ \hbar}[/itex]
    Last edited: Feb 13, 2007
  11. Feb 14, 2007 #10
    Thanks for the replies. I am sorry but I do not think that citing equations( for de Broglie wavelength in this case) has anything to do with physically intuitive reasoning. I agree that arguing along these lines the Heisenberg uncertainty principle becomes very nonintuitive(at least to me) and probably this is where my problem is as it is this principle which really gives rise to the fourier factors.
  12. Feb 12, 2009 #11
    Can anyone explain how to derive the schrodinger equation in momentum space?
  13. Feb 12, 2009 #12


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    No, no one can. Seriously. This is just the way it turns out that the world works. It wouldn't work that way if the world had turned out to be governed by classical mechanics (as was believed to be true for about 200 years).

    But classical mechanics turns out to be just a limit, approximately applicable in certain circumstances, to quantum mechanics, in which position and momentum are postulated to be related by Fourier transformation. (Note that this relation requires the introduction of a constant with the dimensions of momentum times length; this is Planck's constant.) This postulate turns out to be necessary to explain tons of experimental data, so as good scientists, we provisionally believe that it is likely to be true.
  14. Feb 12, 2009 #13
    Ok i see... Incidently, how is the schrodinger equation in momentum space written?
  15. Feb 12, 2009 #14
    I think it's related to the fact that momentum generates translations.
  16. Feb 12, 2009 #15


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    Because the relevant dynamical equation ( Schrodinger's in your case) is LINEAR PDE and admits PLANE WAVE solutions. Physics (classical or quantum) has nothing to do with it.
    Take, for example, the equation

    [tex]\frac{\partial^{2}}{\partial t^{2}} u(r,t) = c^{2} \nabla^{2}u(r,t)[/tex]

    Does u represent a massless boson or a sound wave? Who cares, the fact remains that the plane wave

    [tex]\exp [i(\vec{k}.\vec{r} - c|k|t)][/tex]

    solve our equation. Mathematics then tells you that

    u(r,t) = \int d^{3}k \ U(k) \exp [i(\vec{k}.\vec{r} - c|k|t)] = \int d^{3}k \ U(k,t) e^{i \vec{k}.\vec{r}}\ \ \ (F)

    is a general solution. Now, why is it that coordinate and wave number spaces related by the above Fourier transformation? well, the one and only answer is; because eq(F) represents the most general SUPERPOSITION of PLANE WAVES.


  17. Feb 12, 2009 #16
    Doesn't the interpretation of k as a wave-number depend on a dispersion relation between k_0 and k? If you instead put the time dependence into the Fourier coefficient, i.e., a(k,t)e^(ikx), then this is not a wave, and also a(k,k_0)e^(ikx)e^(ik_0t) is also not a wave - these are just general Fourier transforms. The differential equation will give you a relation between k_0 and k, but unless it's particularly simple, can it be called a wave?

    And does a taut string undergoing only transverse motions have momentum in the longitudinal direction, and is it related to the wave-vector k?
  18. Feb 13, 2009 #17
    Here my take.

    In quantum mechanics, as in classical mechanics, the physical system at hand obeys a ten dimensional symmetry group, the Galilean symmetry group. Three translations, three rotations, three reflections (?), time translation.

    In quantum mechanics, as opposed to classical mechanics, we have vectors representing physical systems. When these vectors are infinite dimensional, for example a wavefunction in position space, the symmetry groups acting on these infinite dimensional vectors need to be in an infinite dimensional representation.

    Say when you want to translate a wavefunction in space, you see that the generator of this group is id/dx (i so that the transformation is unitary). Or when you want to rotate, then id/dL. From that you can conclude, that they correspond to momentum respectively to angular momentum.
  19. Feb 13, 2009 #18
    I dont think anyone would be thinking of a deeper (intuitive) meaning if it had been some 'no-name', unique transform.
    Fourier transform fully preserves the information and so states can be described completely in such a transformed space.

    As to why the fourier variable is interpreted as momentum...
    It is because the eigenvector of the momentum operator is of the form [tex]e^{ipx}[/tex] and has the momentum eigenvalue, p, in place of the fourier variable.
  20. Feb 13, 2009 #19
    In classical mechanics momentum generates translations. So an eigenstate of translation T is an eigenstate of momentum P. Call this eigenstate q(x). Then Tq(x)=q(x+a)=kq(x). The function q that obeys this is q(x)=c^(dx). The fact that translations ought to be unitary forces d to be the imaginary number called 'ip' where "p" is the real. 'c' somehow turns out to be e^(1/h), where h is Planck's constant. So the momentum/translation eigenstate q(x) is of the form e^(ipx/h).

    So given an arbitrary function f(x), the fourier transform allows you to express it as a sum of the basis functions e^(ipx/h) which are eigenstates of momentum.

    This is just a hand-waiving argument I came up with, but even in classical mechanics translations generate momentum. The difference here is that the eigenvalues of operators represent the actual value of whatever it is your measuring (I think).
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