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

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