From: rparson@spotNO.SPAMColorado.edu (Robert Parson)
Newsgroups: sci.chem
Subject: Re: atom and electron
Date: 21 May 1998 16:21:34 GMT

In article <356358FA.458E5585@ix.netcom.com>,
Eric Lucas  <ealucas@ix.netcom.com> wrote:

>By the way, I've never quite understood...how does quantum mechanics get
>around Maxwellian electrodynamics?  Granted the electrons aren't
>orbiting around the nucleus in any classical sense.  However, orbitals
>*do* have orbital angular momentum, and as best I can understand it,
>anything that has angular momentum *is* accelerating (when viewed from
>an inertial frame of reference), and thus photons should be emitted from
>the electrons.

 But remember that the ground state has _zero_ angular momentum. (Let's
 confine ourselves to Hydrogen for simplicity).
 States that do have angular momentum undergo spontaneous emission
 to other states - this is the quantum analog of classical radiation
 damping. Consider a hydrogen electron in a very highly excited state -
 classically it spirals into the nucleus, emitting an electromagnetic
 field; quantum mechanically it cascades down a ladder of states
 until it hits 1s, spitting out photons as it goes. (To improve the
 analogy start with a wavepacket rather than a stationary state for
 the initial quantum state; then the wavepacket will actually follow
 the classical path for awhile, although diffraction effects will
 build up as time goes by and confuse the picture.)

 Consider the frequency of the electromagnetic radiation that's
 emitted in this process. Classically the electron's instantaneous
 frequency of orbital motion increases as it spirals in, so the
 frequency of the emitted radiation will increase with time.
 Quantually, the electron makes bigger and bigger jumps as it
 follows the cascade to lower states, so the frequency of the
 emitted photons increases with time. Neat, huh?

 Hmm - what about multielectron atoms that don't have an S ground
 state - have to think about that, but my guess is that when you
 sort through everything you'll find that the quantum quantity
 that corresponds to an "acceleration" still vanishes.

 In fact, the equations of quantum electrodynamics can be written in
 a form that is very close to those of classical electrodynamics. You
 don't see this too much in standard textbooks because it's not the
 easiest way to calculate things, but IMO it dispells a lot of the
 mystery that surrounds QED. In this approach, a moving charge
 generates an electromagnetic field just as in the classical theory;
 however the position of the moving charge is an operator, not a
 classical variable. The resulting field is also an operator.
 These operators satisfy equations that are essentially the same
 as the classical equations, but at the end of the day you have
 to take expectation values to get an observable quantity and
 then the quantum effects reveal themselves.

 ------
 Robert

From: rparson@spot.colorado.edu (Robert Parson)
Newsgroups: sci.chem
Subject: Re: atom and electron
Date: 29 May 1998 18:12:27 GMT

In article <356EE0A1.4B2411E@ix.netcom.com>,
Eric Lucas  <ealucas@ix.netcom.com> wrote:

>What I was trying to say is that all classical objects that have angular
>momentum also have radial acceleration.  Yet we have a quantum object
>(an electron in an atom) that appears to have angular momentum (I'm
>specifically talking about orbital angular momentum, not spin angular
>momentum)

 Why do you keep saying this? An electron in a 1s (or any s) orbital
 has zero orbital angular momentum.

 Perhaps you are misled by the Bohr model, in which electrons are
 regarded as being in circular orbits even in the ground state.
 In fact, the correct classical analog to an s orbital is a straight
 line orbit. When you go to very high n, where the classical limit
 is appropriate, the high l states correspond to nearly circular orbits
 and the low l states to extremely eccentric elliptical orbits.
 Bohr-Sommerfeld, not Bohr, is the correct classical limit of
 quantum mechanics.

 By the way, this shows that angular momentum isn't the real answer to
 the paradox. The classical straight line orbit has zero angular momentum
 but obviously still involves acceleration. (My previous argument
 about the cascade down through l states as the quantum analog of
 the classical spiral orbit is still correct, it just isn't the
 whole answer.)

 Let's consider a slightly easier problem - a charged harmonic
 oscillator interacting with its own electromagnetic field. One
 dimension, so angular momentum doesn't enter at all. Couple
 the charges to their own field and you get an equation for
 the oscillator that has a damping term in it. The damping term
 is proportional to the time derivative of the acceleration, i.e.
 to the third derivative of the coordinate - one of the few physical
 equations I know of that has a 3rd derivative in it. (In physics,
 this derivative - the change in acceleration - is called 'jerk'.
 There is no standard terminology for the change of the jerk,
 although the word 'inauguration' has been suggested for this
 purpose [W. G. Harter]). You can solve this equation, and you get
 an oscillation that damps to zero.

 Now do it quantum-mechanically. You get the same _equation_, but
 now you interpret it differently: the coordinates, accelerations,
 and jerks are all operators, and at the end of the day you have to
 take expectation values to get observables. You solve the equations
 and calculate the observables, and now you find that the energy
 of the oscillator decays to a finite value as t->infinity (zero
 point vibrational energy), and the electric field generated by
 the system goes to zero (no more photon emission). The final
 state of the system is one in which the vibrational energy is
 nonzero but the system does not generate any net electric field.

 We can illustrate this with the motion of wave packets, but let's
 stop here for now.

 ------
 Robert

From: rparson@spot.colorado.edu (Robert Parson)
Newsgroups: sci.chem
Subject: Re: atom and electron
Date: 29 May 1998 22:13:47 GMT

In article <6kmtqb$5u9@peabody.colorado.edu>,
Robert Parson <rparson@spot.colorado.edu> wrote:

>
> We can illustrate this with the motion of wave packets, but let's
> stop here for now.

 OK, here comes part II: explaining everything with wave packets.

 Start with  harmonic oscillator all by itself, no electromagnetic
 field. Equivalently, a particle rolling in a parabolic well.
 Classically, solve an ordinary differential equation for the
 oscillator position, get cosines: the oscillator oscillates.
 Quantally, solve a partial differential equation for the
 time-dependent wave function, which contains within its
 vitals the answers to all the questions that you're allowed
 to ask about the system. Things are simplest if the initial
 wave function is a gaussian, then it stays gaussian at all
 later times. So you get a gaussian wave packet following the
 classical trajectory, back and forth. The effect of having
 a wave function rather than a single trajectory is that
 when you get down and ask what "the position" of the
 oscillator is, you get a distribution of answers, not
 a single answer. As you approach the classical limit, the
 wave packet gets very narrow and the distribution of
 possible answers at any time gets very tight around the
 average value.

 Now let the oscillator be charged, so that it creates a time-varying
 electromagnetic field as it oscillates. Classically the oscillator
 interacts with its own field, which produces a damping force, so
 the oscillation is damped as energy is radiated away. Quantum
 mechanically you have a wave packet again, which now follows
 this damped classical trajectory. (Unfortunately things are only this
 neat for harmonic oscillators; for other systems the wave packet
 does not exactly follow the classical path although as you approach
 the classical limit it does.) The center of the wave packet eventually
 settles down as t->infinity. But the wave packet is still there, so
 there is still a distribution of possible values for measurements
 of the oscillator position (or momentum). This distribution gives
 rise to zero point energy - as t->infinity, the wave packet becomes
 the ground state wave function of the harmonic oscillator. The zero
 point energy does not radiate away, because to do so would violate
 the Heisenberg uncertainty principle - the distribution of positions
 and momenta would be narrower than Heisenberg says it has to be.

 You can do the H atom the same way, though it's not quite as clean
 because of that stuff about wave packets not completely following
 classical trajectories in nonharmonic systems.

 ------
 Robert

From: rparson@spot.colorado.edu (Robert Parson)
Newsgroups: sci.chem
Subject: Re: atom and electron
Date: 29 May 1998 16:16:20 GMT

In article <6kler2$t8l@nnrp3.farm.idt.net>,
Joshua Halpern  <jbh@IDT.NET> wrote:
>
>:  Consider the frequency of the electromagnetic radiation that's
>:  emitted in this process. Classically the electron's instantaneous
>:  frequency of orbital motion increases as it spirals in, so the
>:  frequency of the emitted radiation will increase with time.
>:  Quantually, the electron makes bigger and bigger jumps as it
>:  follows the cascade to lower states, so the frequency of the
>:  emitted photons increases with time. Neat, huh?
>
>Yeah, but I think this argument is a lot like the Bohr theory
>of the atom, it works for hydrogen.  As a simple example
>consider Lithium.  The 3d-2p transition is 610.3 nm, but
>the 2p-2s transition is 670.8 nm, so the electron makes
>a big jump from 3d to 2p and then a small one from 2p to
>the ground state.  Moreover, in general s orbitals have
>larger mean radii than p and d orbitals so np-ns
>transitions will in general correspond to spiralling out.

 The theory would be much more complicated for multielectron atoms
 because you need to take electron-electron interactions into account.
 Very hard to do in QED. However, the correspondence principle
 must be obeyed: in the limit of large quantum numbers, quantum
 energy level spacings become classical frequencies, and the
 matrix elements of quantum operators between states go over
 to the Fourier components of the corresponding classical variables
 at frequencies corresponding to the difference between these states.
 See, for example, R. Parson and E. J. Heller, JCP _85_, 2569, 1986. :-)

>:  Hmm - what about multielectron atoms that don't have an S ground
>:  state - have to think about that, but my guess is that when you
>:  sort through everything you'll find that the quantum quantity
>:  that corresponds to an "acceleration" still vanishes.
>
>I would think this is true by definition.  Remember, the
>eigenstates are solutions to the time invariant Schroedinger
>equation.  If the energy of the states and all other measureable
>properties do not change with time, then there can be no net
>"acceleration"

 You are _completely_ missing the point. The atomic eigenstates are not
 stationary solutions to the QED equations that account for the electron
 interacting with its own electromagnetic field. Just as the classical
 orbits are not solutions to the classical equations of motion when radiation
 reaction is included. In the 2p state the <acceleration> is nonzero,
 in the 1s state it is zero. From this point of view the atomic
 eigenstates are just a convenient basis set, not an essential part of
 the theory. The basic problem is, given an electron, a proton, and
 the electromagnetic field, what are the solutions of the quantum
 equations of motion?

>In short...I hate this argument.

 Spontaneous emission is the direct quantum analog of classical
 radiation damping. Deal with it.

 ------
 Robert

From: rparson@spot.colorado.edu (Robert Parson)
Newsgroups: sci.chem
Subject: Re: atom and electron
Date: 29 May 1998 18:54:07 GMT

In article <356EE4B9.64D8F034@ix.netcom.com>,
Eric Lucas  <ealucas@ix.netcom.com> wrote:
>
>OK, fine, I'll ask about the lone 2p electron in an isolated boron atom
>(which I assume has the configuration 1s2 2s2 2p1.  Or perhaps
>specifically the lone 2px electron in a quartet nitrogen atom (1s2 2s2
>2px1 2py1 2pz1).

 This is tricky. We know that the answer has to involve the Pauli
 Principle (otherwise all these electrons would happily drop down
 to 1s). I think it goes like this: the electromagnetic field amplitudes
 generated by the various electrons must somehow interfere with each
 other destructively so that considered as a whole the multi-electron
 system generates no radiation. This destructive interference would
 arise from certain phase factors that would result when we require
 that the wave function of the multielectron system be antisymmetric.
 (i.e. obey the Pauli Principle.)

>> OK.  Take an electron and localize it in one dimension to
>> 10-15 cm.  Now use the uncertainty principal and figure
>> out how much kinetic energy the electron has.  Fast
>> little bugger eh
>>
>> Basically this is the same argument as to why an harmonic
>> oscillator has zero point energy.
>
>OK, that makes sense, but the fact is that electrons *do* fall into
>atomic nuclei.  Electron-capture mode of nuclear decay.  Your argument
>doesn't hold water in that case, why should it any other time?

 Electron capture involves a nuclear transformation: a proton becomes
 a neutron. The electron doesn't just fall into a hole, it reacts
 with the nucleus and is absorbed. (IIRC, the uncertainty principle
 argument bothered people back in the days before the discovery of
 the neutron, when it was thought that nuclei contained protons and
 electrons.)

> Or to put it a different way, why aren't electrons falling into nuclei
> all the time?

 That translates into "when is the corresponding nuclear reaction
 spontaneous?". Which we should let Rebecca Chamberlain answer. :-)
 Thinking of the simplest possible case, we know that neutrons
 spontaneously decay to protons, so the reverse reaction - electron
 falling into the H nucleus - clearly is not spontaneous. Presumably
 the details of nucleon-nucleon interactions are involved in heavy
 atoms where electron capture is spontaneous.

 ------
 Robert