Re: Quantum propagation from a Dirac initial point

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

Date: 13 Jun 2004 06:49:00 -0700

glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message news:<cagb8g$b30$1@hood.uits.indiana.edu>...

> In article <eca320d0.0406121604.7b65653a@posting.google.com>,
> Edward Green <spamspamspam3@netzero.com> wrote:
> >glhansen@steel.ucs.indiana.edu (Gregory L. Hansen) wrote in message
> >news:<caf0fs$uib$1@hood.uits.indiana.edu>...

<...>

> >> 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
>
> Yes, the second term is a normalization factor.

I realize now we have two meanings of normalization floating around.

<...>

> >Hmm... isn't the square of your expression exp(3*pi*i)? And isn't
> >that the same as exp(pi*i), or -1?
>
> That's the problem with ASCII math, it's hard to read. But I was using
> the convention where 3*pi*i/2 should be taken to have a 4 in the
> denominator.

:-)

> >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.

> >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?
>
> Well, you step (2) is a little vague.

Oh, we could certainly make it quite definite, if you force me to!

How's this:

 for all t>0, Int[+/-(oo)] |psi|^2 = 1 (ordinary normalization), and

 for all positive eps,delta there exist a tau, such that

   Int[+/-(delta)] |psi|^2 > 1 - eps; for all t < tau

In other words, pick any two small positive numbers eps,delta and I in
turn can pick a third small positive number tau, so that for all times
before tau the integral gets almost all of its juice (minus eps) from
integrating within delta of the origin. In plain English, as t goes
to zero from above, the integral gets indefinitely scrunched up on the
origin, but preserving its value.

I didn't want to get out those epsilons and deltas, but you left me no
choice. ;-/

> I'd think a delta function would
> qualify for that, because it's there only for t=0, but you didn't seem to
> like that.

As you know, Gregory, it is often crucially important in which order
you take the limits. I'm comfortable with my formulation being well
posed; I'm not so comfortable with "And let there be a delta
function". Of course, in your developement you didn't really do that,
but instead played some game with letting w -> 0, and only keeping
first order terms. That's a plausible sort of game, but I reserve
judgement.

However, I still put it to you: since now I've made my program
perfectly mathematically definite, what do you think is wrong with it?

<...>

> >[Or does <the momentum spectrum> persist <for all time>? 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".]
>
> It's a free particle, Edward. Newton's first law applies. The
> momentum-space wavefunction of the Gaussian is independent of time, and
> remains independent of time as w->0.

That was the imp on the left.

> Actually, I should check that. I just found the momentum wavefunction
> for psi(x,0). Presumably if you take the Fourier transform of psi(x,t),
> the t's will drop out. I expect it will since there's similar t-terms in
> the exponential and the normalization, but that should be checked.

That was the imp on the right.

<...>

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