Re: Quantum propagation from a Dirac initial point
From: Gregory L. Hansen (glhansen_at_steel.ucs.indiana.edu)
Date: 06/13/04
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Date: Sun, 13 Jun 2004 01:39:28 +0000 (UTC)
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>...
>
>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)
Gaussians are done to death, I want to throw a square or a triangle at
you. But the propagator is a Gaussian that would be integrated over x,
and how many people have a ready understanding of erf(x)? I thought the
triangle would let me integrate by simple substitution, but there are
other terms, too.
>>
>> 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+ .
And try to get some intermediate step between a delta function at t=0, and
the wavepacket blowing up for 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
Yes, the second term is a normalization factor.
>really need a blackboard ... but an idea is beginning to germinate!
>~8^}
I do see t's in the denominators. If w<<t, the denominator will be
dominated by whatever t near zero is doing. Is that what you had in mind?
I think you might want to factor out some w^2's and let (t/w^2)->inf. Or
factor out some t's and let (w^2/t)-0. Or something like that.
>
>> 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?
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.
>
>> 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?
"...And then a miracle happened..."
>
>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?
Well, you step (2) is a little vague. 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. Or we could go back to the Gaussian and let (w^2/t) be a
constant (which we can later take to zero), and see what happens when we
fiddle with t.
>
>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?
But we do have a normalization with the Gaussian procedure. The magnitude
goes to zero as the width goes to zero, but we can slide the width down
and watch what happens as we do so. That's why I took w small enough that
w2<<w. The bell flattens out faster than the function goes to zero.
>
>[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".]
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.
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.
And if something can be found that can be solved analytically, it would be
a nice demonstration (and homework problem!) to show that if the particle
has momentum p1 with probability P(p1) at time t=0, it will have momentum
p1 with probability P(p1) for all t unless you switch on the external
force.
>
>[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.]
--
"What are the possibilities of small but movable machines? They may or
may not be useful, but they surely would be fun to make."
-- Richard P. Feynman, 1959
- Next message: Jim Greenfield: "Re: Syncronized atomic clocks"
- Previous message: Dennis: "Re: Sundial Time"
- In reply to: Edward Green: "Re: Quantum propagation from a Dirac initial point"
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