don@allegra.UUCP (D. Mitchell) (03/28/84)
Our best understanding of the nature of photons is through quantum field theory. To give the flavor of this view of reality, let me describe how a laser works in terms of very fundamental physical concepts. First, consider an atom in an excited state. At any moment, there is a probability of seeing the atom in a lower energy and a photon coming out, or one may just see the exited atom with no photon. In a sense, the true state of the system (when no one is looking!) is a SUPERPOSITION of these two states. ~~~~~> (( o )) o The probability of emission is derived from a low level axiom, namely the coupling between the photon field, and the electron field. Excitations and de-excitations of these fields correspond to what we mere mortals understand as the CREATION AND DESTRUCTION OF PARTICLES (photons or electrons). So in a crude sense, the exited atom is destroyed, and a photon and a lower energy atom are created at some instant, and a probability for that happening is some fundamental constant. That is what quantum field theory (sort of) says. Now, suppose we have these two possible states: ~~~~~> ~~~~~> ~~~~~> ~~~~~> ~~~~~> ~~~~~> ~~~~~> ~~~~~> ~~~~~> (( o )) o All particles are Bosons or Fermions, and photons are Bosons. Bosons like to be in the same state, Fermions loath being in the same state. Why? Well, if you have a bunch of IDENTICAL PARTICLES, quantum mechanics says the state of the system is really the superposition of all possible permutations of these particles. With bosons, these states INTERFERE CONSTRUCTIVELY, so the amplitude of the state is much higher than classical physics predicts. With fermions, half the permutations cancel the other half, so the amplitude is zero unless there is just one particle. I know that sounds confusing. I just say it to show how strange things are. So, in the second picture, there is much higher probability of the second state (emission of a photon) than there was in the first picture. We call this STIMULATED EMISSION, but that is somewhat misleading. The emission is more probable because it results in a big group of bosons in the same state. Inversely, if you have a big herd of bosons traveling along, it is very hard to make one of them break off and do something else. At low temperatures, electricity is carried by bosons called "Cooper Pairs", and that is why one sees SUPERCONDUCTIVITY. Once the current starts, individual cooper pairs don't want to scatter off the metal atoms, and so there is no resistance. The herd just keeps moving along.
dub@pur-phy.UUCP (Dwight U. Bartholomew) (03/28/84)
Your article on the nature of light was very nice. It was very understandable (in a general sense.) Just one tiny point. I have heard people mention your explanation of how a laser works, but I have also heard people who swear that the fact that photons are bosons has nothing to do with it! You can think of a laser as a tuned optical cavity. Such a cavity will only allow light wavelengths exactly fit inside of the cavity (which consists of two parallel mirrors) (well, one on the mirrors is only 99% reflecting. you want SOME light to get out.) The idea is that an atom gets excited by some external means and then reradiates some light as it "drops" down into its unexcited state. The problem is to get MOST of the atoms excited it any one time so that not all of the reradiated light is reabsorbed by the atoms. One way to do this is to cleverly choose a transition (an excited state) that stays excited for a long time. That way you can "pump" alot of atoms into their excited states before any atoms reradiate and can thus reabsorb. I haven't mentioned anything about bosons. I am not trying to knock your arguement down for it sure is an elagent agruement. So, which is it: Boson or Cavity? D. Bartholomew
mam@charm.UUCP (Matthew Marcus) (03/29/84)
The article this follows up on interprets laser action in terms of cavity
resonance and ends "which is it? Cavity or boson?". Answer: Both!
You need a cavity with a high Q to define a single mode. You need the radiation
to be made of bosons so that lots of them can be in this mode at one time.
Imagine the following electron gun (you'll have to *imagine* this one!):
Between electron-reflecting walls (E-fields?) put a LOT of free
muons. Muons are metastable, and decay into electrons (+ pair of
neutrinos, but let's not worry about those). Usual arguments suggest that
a beam of electrons, with momentum such that an integer number of wavelengths
fit inside the cavity, will exit if one wall has a hole in it. However,
this WON'T work! Why? The reason I have in mind is that electrons are
fermions, so you can't get more than two in one mode. There can't be such
a thing as a coherent electron beam for this reason.
<Philosophy=On ,or post -philosophical , to taste>
All physics is one subject. The more physics you know, the more
connections you see between apparently disparate areas, methods, or limits.
Case in point: You can derive laser action semiclassically by assuming the
active medium to be one with a negative absorption coefficient. Many of
the statistical properties of light can be derived without invoking the
notion of 'photons' at all. The QED description of light is built around
Maxwell's equations, so both 'field' and 'boson' interpretations are
perfectly correct.
Photons - You can make light of them (From a 'Mooney's Module'
cartoon in Isaac's SF Mag.)
{BTL}!charm!mamb12958@ihuxp.UUCP (Pete S) (03/29/84)
This is in response to the ongoing discussion of whether or not a laser
depends on the boson property of photons or on the cavity construction.
Since astronomers have detected maser action in the atmosphere of Mars
and in interstellar space associated with dust clouds, the answer is clear.
A cavity is not necessay for a maser or a laser (which, I believe, can be
thought of a as a short wavelength maser). For us on Earth, working in a
lab, a cavity is an easy method to take advantage of the limited path
length available and to increase it to the point that there is enough
amplification.
P Schultz
AT&T Bell Labs, Naperville, ILmam@charm.UUCP (Matthew Marcus) (03/29/84)
ihuxp!b12958 (Pete S.) states that a cavity is not necessary for laser action
and points to the natural 'lasers' discovered in the Martian atmosphere as
example. He goes on to say that on Earth we use cavities because you get a
bigger path length that way.
Actually, if you want to get technical, a cavityless 'laser' is really
a 'superradiant device'. The distinction is that a superradiant device
has the photons only going through once, and thus cannot be single-mode.
It's coherence, divergence, and monochromaticity are much reduced compared
to those of a true laser. It has been proposed to build a signalling system
to reach the stars by putting huge mirrors in Mars orbit to create a true
laser. The output power wouldn't be all that impressive, but think of the
narrow beam you'd get from km-diameter mirrors!
As another example of superradiance vs. lasing, I built a nitrogen
'laser' for an experiment that never materialized. Nitrogen has a huge gain,
but only sits in its upper laser level for several ns. Because of this gain,
you get superradiance with a path length of only 6". With no mirrors, the
output beam at 1m looked like:
.....
.........................
.............................
........................
.....
as near as I can recall. With a rear mirror, but none in front, I got:
***
***
showing the advantage of even 1 more pass through the active medium. Because
the active medium is active only for a few ns, a front mirror is not very effective
at increasing output - the canonical thing here is a 20% reflector, giving
30% (?) more output. If anybody's interested, I'll send you some references
on how to build this thing.
Anyway, I hope I've cleared up the role of the cavity in lasing action.
{BT}!charm!mam "I whip out my laser pistol and blast 'em!"stekas@hou2g.UUCP (J.STEKAS) (03/30/84)
Doesn't LASER mean "Light Amplification by Stimulated Emission of Radiation"?
It seems that no restrictions are placed on overall gain, monochromicity, ...,
so why does one need a cavity to have a LASER? You might need a cavity to
make a useful LASER , but "useful" isn't part of the definition.
JimShinbrot.WBST@Xerox.ARPA (04/16/84)
D. Bartholomew, re. lasers requiring bosons, There are 3 types or statistics governing particles and their behavior. Bose-Einstein: These particles 'like' to occupy the same state. They do more, therefore, than resonate when dumped into a cavity together - they tend to re-enforce one another when appropriately "pumped." Maxwell-Boltzmann: These particles are indifferent to other particles' states. Air is a common example of this. They can be made to resonate in a cavity, but cannot lase. That is, although pre-existing particles may move a wave back and forth within a cavity, they do so by banging into each other and by and large moving each other in the appropriate direction by virtue of the fact that the particles moving in inappropriate directions die out. Fermi-Dirac: These particles (degeneracy aside, please) will not occupy the same state. They cannot even be made to resonate, therefore, much less re-enforce each other's wave. In short the former category encompasses the particles which predispose themselves to 'enforced resonance'. Since they (photons) are emitted by atoms hanging about in the cavity (or matrix - remember ruby lasers?) at frequency (recall that they 'like' to be in the same state as their mates) that the cavity happens to be designed around, they are doubly appropriate as laser constituents. -Troy Shinbrot