Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Arising of exp(-ipx) in QM in quantum field theory

  1. Apr 27, 2012 #1
    Wave functions and their exp(-ipx) are base of quantum mechanics and QFT. But how to derive them the most simply that we will understand why they are base of QM?
    It is interesting that Feynman in his book "QED: The Strange Theory of Light and Matter" very simplifies QED, but he did not give a reason how this exp(-ipx) (or sin(px)) arises.

    It is also interesting that his amplitudes succesively multiplies, but paralelly they suming. But one of the simplest form of multiplicaton function is exp(i..) function. Has this any deeper meaning?
    Last edited: Apr 27, 2012
  2. jcsd
  3. Apr 27, 2012 #2
    The fact that momentum eigenstates have the form psi(x) = exp(ipx) comes from the fundamental commutation relation [x, p] = i hbar. For instance you can verify from this commutation relation that if psi_p(x) is a momentum eigenstate with eigenvalue p, then
    [tex]\exp(i q \hat{x}) \psi_p(x)[/tex]
    is a momentum eigenstate with eigenvalue p + q. (Expand the exponential, then hit this expression from the left with the p operator, then commute p through each term and see what happens). Then if we agree that psi(x) = 1 is a momentum eigenstate with eigenvalue zero, we have that exp(ipx) is a momentum eigenstate with eigenvalue p.
  4. Apr 28, 2012 #3
    The Duck, you are right. I remembered that principle of uncertainty is a consequence of fourier transformation. p <--> x. If one (Gaussian) wave packet is short (p or x) in the other representation is long and oppositely.
    Of course, commutation relation is a cause for uncertainty relation. But I cannot follow you. There sould be something simpler which gives necessity for exp(ipx) in commutation relation?
  5. Apr 29, 2012 #4
    Hi Im not a physicist, just an amateur, but I had a doubt similar to yours before and I found a lot of help around here. I do not know If I can explain exp(-ipx) precisely but I think that I can organize your ideas. For simplicity, lets think in Non Relativistic QM knowing that QFT should converge to NRQM in c=inf limit:

    1) H is the logarithm of the evolution operator, and because of isotropy of space and similar things (no privilidged directions and such) it should be proportional to p^2 (see schweber page 11 of "An introduction to Relativistic Quantum Field Theory)

    2) there is a demostration in "Derivation of the Momentum Operator", (Chris Clark December 30, 2009) that says that, knowing 1), p should be proportional to i*dif(phy)/dif(x), asuming not many more than 1).

    3) Given 2), you see that if you are in a state of a given "p", then the state is described, in the position basis, by "exp(-ipx)" because of the eigenvalue - eigenstate equation of 2)

    I will be very happy if you find this helpful!
  6. Apr 29, 2012 #5
    Another reason is that, due to the formulation of QM in Linear Algebra, the "goal" of the Schrodinger Equation is, ultimately, to find a basis for any quantum system (not unlike the i,j,k basis of R^3). It happens that the time-independent S.E. is an eigenvalue problem which, after substitution of operators, and without a potential (the simplest case) one gets the equation for a free particle, which dictates that the wave function must satisfy [tex] \frac{-\hbar^2 d^2}{2mdx^2}\psi(x)=E\psi(x) [/tex] This has a solution proportional to [itex] e^{i\sqrt{\frac{2mE}{\hbar^2}}x} [/itex]. Using de Broglie's relations, you find that [itex] p=\hbar k [/itex] and that [itex] E=\frac{p^{2}}{2m} [/itex], we find that [itex] p=\sqrt{2mE} [/itex]. Plugging this into our free-particle wave, we get [tex] \psi(x)\propto e^{ipx/\hbar} [/tex] which turns into simply [itex] e^{ikx} [/itex] using the de Broglie relation (btw, k is the wavenumber =[itex] \frac{2\pi}{\lambda} [/itex]). Thus, the basis functions for a free particle are proportional to a linear combination of exp(ikx), and they will likewise arise in the behaviour of any system that can be approximated on a small scale to that of a free-particle. It also plays a curcial role in the fourier transform that is required to transfer from the position basis (x) to the momentum basis (p or k).
    Last edited: Apr 29, 2012
  7. May 1, 2012 #6
    We know
    [itex] \psi\propto e^{ipx/\hbar} [/itex]
    [itex] \psi\propto e^{iWt/\hbar} [/itex]
    but, what about maybe
    [itex] \psi\propto e^{il/\hbar} [/itex]
    where l is angular momentum or does exist any other options of
    [itex] e^{i.../\hbar} [/itex]
    in QM or QFT?
  8. May 1, 2012 #7
    In Feynman's Path Integrals, you see the appearance of [itex]e^{iS/\hbar}[/itex], and in problems with rotational invariance, you will find that (up to a degeneracy of dependence on r), the eigenbasis for a free potential is [tex]\psi(x)_n\propto e^{il\theta/\hbar};\ l=n\hbar[/tex]. It appears quite frequently in operators as well, since, for an infinitesimal generator of transformation, U, which corresponds to an observable, the finite transformation operator is something along the lines of [itex] e^{-iUq/\hbar}[/itex]. For a Hermitian U ([itex] U^\dagger=U[/itex]), which is transforming by conjugate variable amount q. Thus, given an operator U, and its conjugate variable, q, the finite transformation operator is of the form of a complex exponential.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook