Non-wave solution to wave equation and virtual particles

[Fefetltl]

TL;DR

Solutions of "vanishing particles" may arise from the very basic 1D wave equation, what about their physical meaning ? Could we link them to vacuum fluctuation?

Hello everyone. The 1D wave equation is written:

$$ \left( \partial_t^2/c^2 - \partial_x^2 \right) \Psi = 0$$

An electromagnetic wave or matter wave, like free electron (unnormalized here), can be written with the following wave function $\Psi_m$ of energy $\hbar k c$:

$$ \Psi_m \propto e^{-i (kx + c t)}$$

is a solution of wave equation. Now if we take $\Psi_v$ ($x > 0$):

$$ \Psi_v \propto e^{-(kx+ c t)} $$

is also a solution of wave equation but it is not a wave (it does not oscillate): it is something that appears (or is already here, depends on initial condition) and disappears then. I heard about Feynman using of Klein-Gordon equation to describe vacuum fluctuations, with some unclear statements (but we are talking about QED...) like the mass problem, the energy-momentum which is not the same that for matter (negative kinetic energy and son on)...

So you guys have an idea if such a solution could exist in physics (we can extend it to 3D also) ? As an appearing pairs of particles ? Or it is something completely wrong and not relevant ?

 

[pasmith]

The general solution of the 1D wavew equation is f(x + ct) + g(x - ct) for arbitrary functions f and g. If \hat f and \hat g are the fourier transforms of these functions, then f(x + ct) + g(x - ct) = \int_{-\infty}^\infty \hat f(k)e^{ik(x + ct)}\,dk + \int_{-\infty}^\infty \hat g(k)e^{ik(x - ct)}\,dk is a superposition of waves.

 

[Fefetltl]

Thanks Pasmith, I Know. Any function $f(x \pm c t)$ fits the wave equation ;). But do you have any idea if this kind of solution $e^{-k(x+ c t)}$ or even (why not) $e^{- k^2 (x \pm c t)^2}$ can be found in physics (even for mechanical perturbations)? Like for virtual particle ? I am interested in this subject for the moment, I have computed some stuff using Dirac-like matrices and I found some evanescent "apparition" wave function like these.

 

[renormalize]

But do you have any idea if this kind of solution $e^{-k(x+ c t)}$ or even (why not) $e^{- k^2 (x \pm c t)^2}$ can be found in physics (even for mechanical perturbations)? Like for virtual particle ?

1. As @pasmith states, these solutions are just particular infinite superpositions of plane waves $e^{i \kappa(x\pm c t)}$. You can think of them as wave-packet solutions that arise from very specific initial conditions imposed on the wave equation.
2. As such, they have nothing to do with virtual particles, which by definition do not satisfy the wave equation, i.e., virtual particles must be "off-shell".

 

[Fefetltl]

1. As @pasmith states, these solutions are just particular infinite superpositions of plane waves $e^{i \kappa(x\pm c t)}$. You can think of them as wave-packet solutions that arise from very specific initial conditions imposed on the wave equation.
2. As such, they have nothing to do with virtual particles, which by definition do not satisfy the wave equation, i.e., virtual particles must be "off-shell".

Thanks Renormalize ;) (it is a good name to speak about QED).

1) Good hint with the specific condition, but very very specific (since we have to use an oscillatory Green function to solve it... but ok with that).
2) Thank you. I read the wiki article on off shell particles, so as I understood, these are particles that do not obey to energy-momentum relation, i.e $E_{on}^2 = p^2 c^2 + m^2 c^4$.
I am interested in this topic, if you may I would like to ask you a question: do you have example of off-shell energy-momentum relation? Like I don't know $E_{off}^2 = - p^2 c^2 + m^2 c^4$ ?

 

[PeterDonis]

do you have example of off-shell energy-momentum relation?

There is no such thing. "Off shell" does not mean that virtual particles obey some different energy-momentum relation from the standard one. It means they don't have a well-defined energy-momentum relation at all.

 

[Fefetltl]

There is no such thing. "Off shell" does not mean that virtual particles obey some different energy-momentum relation from the standard one. It means they don't have a well-defined energy-momentum relation at all.

Ok, thank you.
In fact they do not know anything in this field, except dividing by infinite quantities (^^ just kidding), thanks guys for your replies

 

[Demystifier]

So you guys have an idea if such a solution could exist in physics (we can extend it to 3D also) ? As an appearing pairs of particles ? Or it is something completely wrong and not relevant ?

The exponential solutions diverge at (minus) infinity, so they are unphysical.

 

[Fefetltl]

The exponential solutions diverge at (minus) infinity, so they are unphysical.

Consider $x > 0$. A plane wave is not physical neither by the way .

Moderators can close the topic if they want.

 

[PeterDonis]

Moderators can close the topic if they want.

Fair enough. Done.