Re: Quantum propagation from a Dirac initial point

From: Edward Green (spamspamspam3_at_netzero.com)
Date: 06/13/04 

Date: 12 Jun 2004 17:04:11 -0700

glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message news:<caf0fs$uib$1@hood.uits.indiana.edu>...

Greg! As always, a real pleasure to hear from you.

> In article <eca320d0.0406101921.230beb4@posting.google.com>,
> Edward Green <spamspamspam3@netzero.com> wrote:

> >I am 95% sure zigoteau is wrong in applying the delta function to the
> >wave function itself, but only 65% certain your BC is in fact
> >compatible with a square integrable solution -- his argument about the
> >delta function implying completely indefinite momentum and hence
> >vanishing out from under us _sounds_ like a plausible sort of argument
> >in this case, but it's a "physicist's argument", not a solid
> >mathematical insight into what is going on.

I'd like to clarify what I meant by that remark in a bit.

> Would you feel better if we constructed a delta function from scratch?

Sure. Construct away! I will defer discusion of my "feelings". ;-)

> Suppose
>
> psi(x,0) = exp(-x^2 / 2w^2) / (pi*w^2)^(1/4)
>
> Note that
>
> |psi|^2 = exp(-x^2/w^2) / sqrt(pi*w^2)
>
> We localize the particle at x=0 by letting w->0. Then the peak height
> 1/sqrt(pi*w^2) -> inf.

Ok. I'd interject here that by taking the limit _at_ t=0, we've
already deviated from my idea for a program for the problem: I would
look for a solution whose square went to a delta function _as_ t->0+ .
 
> The time-dependent wavefunction, according to Shankar and rearranged a
> little by Greg, is then
>
> psi(x,t) = exp(-x^2 / (2 w^2 + 2 i hbar t / m))
>
> * sqrt(m w / sqrt(pi)(m w^2 + i hbar t)

Ah... ascii math. ;=|

I take that the second line is outside of the exponent, at least. We
really need a blackboard ... but an idea is beginning to germinate!
~8^}
 
> Clearly, if we let w->0 it will vanish. But let's just make w small
> enough that w^2 vanishes, and note that sqrt(-i)=exp(3*pi*i/2).

Hmm... isn't the square of your expression exp(3*pi*i)? And isn't
that the same as exp(pi*i), or -1?

> Then
>
> psi(x,t) = exp(-i (m x^2 / 2 hbar t) - 3*pi/2)
>
> * sqrt(m w / sqrt(pi) hbar t)
>
> We no longer have a Gaussian, exp(-x^2/w). We have exp(-i*x^2/w).
> Something funny is going on with the phase there, it's not quite a plane
> wave, but the square amplitude is
>
> |psi| = m w / sqrt(pi) hbar t
>
> I.e. same amplitude in all of space, but an amplitude decreasing as t
> increases.
>
> If you Fourier transform the original wave packet,
>
> psi(k) = exp(-k^2 w^2 / 2) * sqrt(w / sqrt(pi))
>
> |psi(k)|^2 = exp(-k^2 w^2) * w / sqrt(pi)
>
> The momenta form a Gaussian wvepacket centered on k=0. This also vanishes
> when w->0, but if we again let w get small enough that just w^2 vanishes,
> the momentum-space wavefunction goes to
>
> psi(k) = sqrt(w / sqrt(pi))
>
> and all k become equally represented as w->0.

That's cool, but if you think I wanted to deny that, I deny it. :-)

Look ... humor me a space. What we have here, and I'm sure you
recognize the phenomenon, is a "No, don't do the problem _that_ way,
do it _this_ way" -ism. As you know, the inquiring mind is never at
rest until it both sees what is right _and_ what was wrong: otherwise
we entertain cognitive dissonance to infinity. Putting your this way
on the back burner for a moment, please look at my proposal in outline
and tell me what you think is wrong:

Suppose we discover a psi(x,t) with the following properties:

(1) exists and is square integrable for all t > 0.

(2) as t->0+, (pours self shot of Wild Turkey as thinking aid),
epsilon-deltas itself into something which only has support for the
integral at x=0.

If we _could_ demonstrate such a solution, and if it _did_ offer an
adquate representation of the boundary condition, then it would
contradict the assertion that the boundary condition implied a
solution which failed to be square integral for all time, because
(insert just-so story here), would it not?

Your ball:

Do you think (a) my program fails to adequately capture the boundary
condition, even if the set of psi's meeting (1) and (2) is not empty,
or (b) that capturing the boundary condition or not, the solution set
is in fact empty?

Note that even if because of some hidden flaw in my thinking my
program fails to capture the boundary condition, or else has no
solution, it doesn't imply that your program _does_ capture the BC,
though you do demonstrate a solution, which is always supportive.

Oh yeah: physicist's argument. My stigmitization of the argument "all
momenta are equally represented at t=0, therefore the particle escapes
to infinity and we have no normalization" is based in the following
thoughts: (alpha) that the identification of k with momentum is
something outside the immediate mathematical structure of the problem,
hence dubious though suggestive, and (beta) anyway, on the same
hand-waving level, I could claim that the flat spread of momentum only
persisted for zero time. So now what?

[Or does it persist for all time in your solution? Imp on left
shoulder says "k-spectrum persists for all time for a free particle,
and does not evolve", but imp on right shoulder says "don't listen to
him! they were playing some fast and loose games with differentials to
get this so-called result".]

[Social administrative addendum: I don't want to come off claiming I
have nothing to learn here. I've a great deal to learn here. But
that doesn't imply I knew nothing relevant to start with, or that any
adept who tries to snow me knows all, present company excluded.]