Imagine we have a particle of mass M caught in a trap of depth M. The
trap
is centered at the origin and is very narrow spatially.
The form of the Klein-Gordon equation given is the free-particle
equation, i.e., constant potential, no barrier, no trap. I'll
consider this further as time permits, but this derivation strikes me
the same proving that 1/2 + 1/2 = 1 by arguing that 2/2 = 1. The
arithmetic is undisputable, but the proof has nothing to do with the
original problem, and won't work for 1/3 + 1/3.
I heard another story recently. A class of physics students were
given the following problem: A shepherd has 125 sheep and 5
sheepdogs. How old is the shepherd? A common answer was 25 years old
-- 130=125+5 is too old, 120=125-5 is also too old,but 125/5=25 is a
reasonable age for a shepherd.
Glenn
Subject: Re: virtual particles preferred frame?
Date: Mon, 8 Nov 1999 22:24:38 -0500
From: John Denker <jsd@MONMOUTH.COM>
Imagine that everywhere outside the trap, our particle obeys the
massive
scalar Klein-Gordon equation (aka the waveguide equation)
(d/dt)^2 - (d/dx)^2 + M^2 = 0 (0)
You can treat this as your basic barrier-penetration problem. The
particle
will "leak" out of the trap for a certain distance which we shall
calculate.
Notice that the escaped particle has exactly zero total energy -- it
has
its usual, normal rest-mass energy M, but it had to borrow one M to
climb
out of the trap. This is an artificial zeroing of the energy,
resulting
from my choice of the depth of the trap. But there was a reason for
choosing this depth, because it corresponds exactly to the creation of
a
virtual particle out of the previously-empty vacuum.
I assert that
psi = exp(-|x|M) (1)
is a solution to the equation of motion for all nonzero x. My
assertion of
zero total energy is confirmed by the lack of any time dependence.
You
could write this as
psi = exp(-|x|M) exp(i omega t) (2)
but we see we my solution has omega=0. Other solutions are possible,
but
this is the one that propagates the farthest, so it best answers your
original question.
(Shorthand 1) Among the cognoscenti it is a common shorthand to
pretend
that this nodeless decaying wave comes from the more-common nodeful
nondecaying wavefunction of the form
psi = exp(i k |x|) (3)
by means of an imaginary value of k. This is of course algebraically
the
identical thing.
(Shorthand 2) Furthermore, you have the option of interpreting this
imaginary k as an imaginary momentum hbar k, but now we are two layers
deep
in optional interpretation.
If imaginary momentum gives you indigestion, just go back to equation
(1). It solves the equation of motion and that's really all that need
be said.