From: carlip@physics.ucdavis.edu
On 9/14/17 4:32 AM, Gregor Scholten wrote:
> [I already tried to send this on Monday, September 11, this is the
> second attempt]
>
> Jos Bergervoet wrote:
>
>>>> In solving the hydrogen atom we assume a 1/r electric potential.
>>>> But since the electron wave function squared is a source for the
>>>> electric field, this should be a screened potential. But if we
>>>> do that, then it will be quite different from 1/r. At large
>>>> distances it will vanish, and at distances around the Bohr radius
>>>> the potential will already be significantly reduced.
>>>
>>> You mean, you interpret the wave function of the electron that way
>>> that there would be a classical charge cloud with charge density
>>>
>>> rho(x) = -e |psi(x)|^2 (1)
>>>
>>> This interpretation is wrong, though. According to quantum mechanics,
>>> one should rather imagine the electron as a point charge with an
>>> uncertain position.
>>
>> This seems to be a bit far-fetched. What quantum mechanics
>> says is merely this:
>> "This is the Lagrangian density, so now you can compute
>> the action. Have fun with it!"
>
> This might be true for Quantum Field Theory ("Second quantization"), but
> not for Quantum Mechanics ("First quantization"). Lagrangian densities
> are a property of field theories, not of mechanics. In Quantum
> Mechanics, there is a Lagrangian function, but no Lagrangian density.
>
> Below, you described how you think your "paradox" can be solve in
> Quantum Field Theory (or in QED, which is the application of Quantum
> Field Theory to electromagnetic interaction), but this "paradox" can
> already be solved in Quantum Mechanics, in the way I described it.
>
>
>> There is nothing said about how we should imagine things.
>
> You're wrong. Quantum Mechanics clearly tells us that imaging a
> classical charge clould with charge density -e |psi(x)|^2 (1) would be
> wrong.
There is a discussion of this issue in section 4 of Adler's paper
quant-ph/0610255. As he points out, imagining that an electron
interacts with its own "charge cloud" contradicts the Born rule.
Steve Carlip
>
>> Besides,
>> that wouldn't change the equations one bit, of course.
>
> You're wrong again. The fact that there is no classical charge cloud
> with charge density (1) makes clear that there is no shielding effect as
> you assumed it.
>
>
>>> Instead of a certain classical charge density, there
>>> is a charge density operator
>>>
>>> \op{rho}(x) = -e delta(x - \op{x0})
>>>
>>> with \op{x0} being the position operator of the electron. Calculating
>>> the average value = yields
>>>
>>> = -e |psi(x)|^2
>>>
>>> what is similar to (1), though, but this is only an average value,
>>> not a classical certain value.
>>
>> To resolve the paradox as presented this does not help. You
>> cannot refute the fact that hydrogen in fact *is* neutral and
>> has vanishing field at large distance
>
> The fact that the hydrogen atom is neutral means that if we assume a
> third charged particle besides the proton and the electron, e.g. a
> second electron, this third particle is not attracted or repelled by the
> hydrogen atom. This does not in any way contradict the fact that the
> electron that is part of the hydrogen atom does not shield the
> attractive force from the proton on that electron itself.
>
>
>> so also when using the
>> expectation value of an operator, you will have to get that
>> result (and you get it, presumably).
>
> No problem: when considering a large distance much larger than Bohr
> radius, the probability is very high that the distance between proton
> and electron is much lower, and so, the probability is very high that a
> third charged particles does not feel any attraction or repulsion.
>
>
>>> Therefore, there is no such shielding effect as you described it.
>>
>> Yes there is, Hydrogen is neutral.
>
> You're wrong. The neutrality of a hydrogen atom concerns the effect on a
> third particle, it does not imply a self-shielding of the electron.
>
>
>>> There
>>> is no portion of the electron charge inside the Bohr radius that
>>> would shield the charge of the proton and reduce the potential
>>> outside the Bohr radius. The electron always feels the full charge of
>>> the proton, independently from the distance.
>>
>> That is the other branch of the paradox! It must always feel
>> the full 1/r potential in the Schrodinger equation, in order
>> to get our well-known correct results. So why *does* it have
>> to feel the full potential?
>
> Because of the reasons I explained. There is no classical charge clould
> with charge density (1).
>
>
>>> You also can see this by considering the two-particles wave function
>>> Psi(xe, xp) of electron and proton. In the corresponding Schroedinger
>>> equation, a two-particles potential operator
>>>
>>> \op{W} = -e^2 / (4pi eps0 |\op{xe} - \op{xp}|) (2)
>>
>> Yes, I was wondering about that, but the difficulty is how
>> to choose the equation!
>
> This just follows from the classical situation by applying the rules of
> quantization.
>
>
>>> occurs, with \op{xe} and \op{xp} being the position operators of
>>> electron and positron. This is due to the rule for the transition
>>> from classical theory to quantized theory: if in the classical theory
>>> a classical two-particles potential W(x1,x2) occurs in the Hamilton
>>> function, in the quantized theory a corresponding two-particles
>>> potential operator like (2) occurs in the Hamilton operator.
>>
>> I think the whole point is how to find the potential in the
>> first place, even for the one-particle Schrodinger equation.
>>
>> In Weinberg (the QT of F's, part I,) we read in Sect. 13.6 ("The
>> External Field Approximation") that this potential is in fact
>> already summing a lot of multi-photon exchange diagrams.
>
> This is how you think that the "paradox" can be solved in QFT/QED.
> However, it can be already solved in Quantum Mechanics, in the way I
> described it. And as we will see below, both solutions turn out to be
> equivalent.
>
>
>> So the
>> potential in the Schrodinger equation is not meant as an expectation
>> value of a field, it is just a computational tool to represent a lot
>> of multi-particle interactions (obtained through the Bethe-Salpeter
>> equation). So this happens to give rise to an 1/r function in some
>> equation (the Schrodinger equation), whereas the expectation value
>> of the EM field in full QFT is falling off much faster than 1/r^n.
>
> BTW: this is equivalent to what I said. Let's consider some distance r0,
> e,g, one Bohr radius. The average value at that distance falls
> off faster than 1/r0^2, because of the shielding effect of the
> electron's charge: is determined by the average value of
> the charge in the range r < r0, where is given by the proton's
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