3
$\begingroup$

In QM a wave packet is modeled as an infinite, or almost infinite, Fourier series, and the Fourier transform provides a transformation between momentum space and position space.

To what extent is this superposition of plane waves physically realizable? Could it be that a wave pack is a physical object made up of a superposition of plane waves? Or is it only a mathematical model? Furthermore, how would one go about showing that there aren't plane waves that superimpose to zero everywhere except at the wave packet, if indeed that is the case?

$\endgroup$
  • 2
    $\begingroup$ "To what extent is this superposition of plane waves physically realizable?" To be sure, you're aware that wavefunctions don't live in physical space, correct? For example, an n-particle wavefunction lives in a 3N dimensional configuration space so this is an abstract space in the first place. In light of this, what do you mean by physically realizable? $\endgroup$ – Alfred Centauri Oct 30 '14 at 0:51
  • $\begingroup$ Do you understand the difference between a Fourier series and a Fourier transform? The former, having discrete frequency components, can only describe a repeating waveform; while the latter can describe the frequency components of any shape. After that it's just math... $\endgroup$ – Floris Oct 30 '14 at 2:35
3
$\begingroup$

A different angle on this that I DON'T believe is in conflict with Terry Bollinger's answer: whether you express a wavefunction in position co-ordinates, or, as its Fourier transform, i.e. in momentum co-ordindates, the two models are precisely the same. So neither the expression of a position co-ordinate wavefunction (such as you find from the solution of the chemist's wonted form of the Schrödinger or Dirac equation for the hydrogen atom) nor its Fourier transform are any less "real" or "physical" than the other.

More technically, most often in quantum mechanics we take the space of functions to be the separable (i.e. having countable basis) Hilbert space $\mathcal{L}^2(\mathbb{R}^N)$ of square Lebesgue integrable functions (well almost: see footnote). Such functions are a proper subset of the so-called tempered distributions, which are mathematical objects (broadened notions generalising functions) for which:

The distribution itself and its Fourier transform constitute precisely the same information

One can be reversibly transformed into the other by a one-to-one transformation (i.e. the Fourier transform) which is also unitary when we restrict to $\mathcal{L}^2(\mathbb{R}^N)$. If you need more information on tempered distributions, I give more discussion here in my answer to the question "What restrictions on time boundary conditions does it have to use Fourier transform to solve wave equation?" as well as here, in answer to the question "Are all scattering states un-normalizable?" and here.

What Terry's answer is about is that plane waves and delta functions, the "eigenfunctions" of the momentum and position observables, respectively, do not belong to the quantum state space $\mathcal{L}^2(\mathbb{R}^N)$ of physical wavefunctions: they are not Lebesgue integrable. So these observables actually have no eigenfunctions in this quantum space. It helps our intuition to extend the Hilbert space to the more advanced concept of a Rigged Hilbert space so that we can talk about these rigorously as vectors of the broadened notion of the observables in the advanced framework. But physical wavefunctions have finite extent and thus always have a nonzero spread in their Fourier transforms.

Footnote: More precisely: we take our quantum state space to be the Hilbert space of equivalence classes of Lebsesgue integrable function modulo the relationship equality almost everywhere (see Wiki page "Almost Everywhere"). So we're talking about $\mathcal{L}^2(\mathbb{R}^N)\,/\,\sim$, where $\sim$ is equality almost everywhere. Another way of saying the same thing, we can take it to be the span of a complex countable set of basis functions, such as quantum harmonic oscillator eigenfunctions)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This may be a little too purist, but as you generalized the domain of the Fourier transform to tempered distributions, I think you cannot say it is a unitary transformation anymore, since you are not dealing with a Hilbert space, although it is still a one-to-one map. $\endgroup$ – Mateus Sampaio Oct 30 '14 at 10:59
  • 1
    $\begingroup$ @MateusSampaio No, you are quite right to point that out, many thanks. I need to make that clearer, thanks. I just raised tempered distributions because these seem to be a convenient class of things for which the FT is bijective, thus "conserves information" and therefore the FT can, abstractly, be thought of as the "same information". I don't even know whether the tempered distributions are the biggest class for which this is true; I'm not up with the "state of the art" here: they're simply a fairly simply described class for which it is true and moveover seem a "reasonable" definition of ... $\endgroup$ – Selene Routley Oct 30 '14 at 11:11
  • $\begingroup$ ... something that is "anything you're likely to see in physics". $\endgroup$ – Selene Routley Oct 30 '14 at 11:12
  • 1
    $\begingroup$ @WetSavannaAnimalakaRodVance, wow, I have no disagreement with what you just said, especially on the fundamental equivalence of the real and momentum space descriptions. But to try to summarize your answer to Chris St Pierre, is this close? Perfect plane waves cannot exist for real particles. However, they can be derived as the mathematical limits of very consistent, experimentally proven behaviors that enable real particles to approximate such plane waves to any specified level of precision. They are a real-life, real-physics example of the epsilon-delta proof of calculus. $\endgroup$ – Terry Bollinger Oct 31 '14 at 22:09
  • 1
    $\begingroup$ @TerryBollinger That's a pretty good summary! $\endgroup$ – Selene Routley Nov 1 '14 at 4:30
1
$\begingroup$

A pure plane wave is not a physically realizable state for a real particle because it would require the particle's wave function to extend over infinite space. Even if the universe is open and infinite in size, it has only existed for a finite time, so no particle wave function would have had enough time to grow to infinity.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.