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.]

<On the correct boundary condition for a particle obeying the
time-dependent Schroedinger equation, and completely localized at t=0>

You said in one post of this thread that it is not psi(x,0) that
should be equal to delta(x), but mod(psi(x,0))^2. That is
unfortunately *>meaningless<*. The delta function was introduced by Dirac
to fill a perceived need, but it is not a nice function. It is defined
as a limit, and in quite a few cases expressions involving it converge
when you take the limit in the prescribed way. However not all
expressions involving the delta function converge correctly. They are
not meaningful. Whenever you deal with the delta function, you must
keep these problems in mind.

<emphasis added>

Fine words, but I think destined to impress nobody who's opinion is
worth a tinker's damn.

Let's look at this in detail:

We have two functions connected in some way with the problem. For
neutrality, lets call them g and h: "g" is understood to be psi, "h"
|psi|^2. _You_ assert the correct interpretation of the boundary
condition is to let g be a Dirac delta function a t=0,

I suggest that
we want to let h have this property. You respond not by saying my
suggestion is wrong for such and such a reason, or will not work, but
that it is _meaningless_. But you produce no evidence for this
assertion other than a general pandemonium about the subtle issues
surrounding the blessed delta functions, which I am supposed to be on
the wrong side of, because ... well, because you say so.