How does a laser work?

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Offline chris

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How does a laser work?
« on: 21/03/2009 10:42:04 »
What is the difference between "normal" light and laser light, and how is laser light produced?

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Offline RD

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« Reply #1 on: 21/03/2009 12:19:32 »
laser light is monochromatic: only contains light of one colour (of one wavelength).
"Normal" light, e.g. daylight contains light of different colours  (spectrum).

Laser light is coherent: all the light waves are in sync. Not true of "normal" light.

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Offline yor_on

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« Reply #2 on: 21/03/2009 23:05:11 »
Sure you want to know :)

'Ordinary' light contains a lot of different wavelengths (colours) in them, they are also called incoherent light meaning that they are not ordered in any way, and what light there is of the same wavelengths tends to be out of phase as well. In short that's sunlight.

Coherent light is ordered light and that is not a 'normal' state for light on Earth. They are light waves that are "in phase" with one another. It's created by for example having a gas with its atoms in a equilibrium that you 'energize' by injecting energy to those atoms. This is called 'pumping' and create together with the 'population Inversion', which is when there are more atoms of a high energy level than of a lower level, the possibility of 'lasing' which is the actual emission of a laser beam through stimulated emission. The concept of the beam is that all the photons will be in the same phase and wavelength, perfectly synchronized to each other.

"If an electron is already in an excited state (an upper energy level, in contrast to its lowest possible level or "ground state"), then an incoming photon for which the quantum energy is equal to the energy difference between its present level and a lower level can "stimulate" a transition to that lower level, producing a second photon of the same energy. When a sizable population of electrons resides in upper levels, this condition is called a "population inversion", and it sets the stage for stimulated emission of multiple photons. This is the precondition for the light amplification which occurs in a laser, and since the emitted photons have a definite time and phase relation to each other, the light has a high degree of coherence."

And the strangest thing about it is that those photons don't interact when being 'cajoled' into a coherent laserbeam, "the incident photon does not collide with the excited atom, apparently proximity alone triggers the stimulated emission." creating an exact copy of that photon in the process building up to that beam.

« Last Edit: 21/03/2009 23:55:05 by yor_on »
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Offline JP

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How does a laser work?
« Reply #3 on: 22/03/2009 04:11:13 »
You have to define what you mean by "normal" light, but I assume you mean light produced by an incandescent bulb, or by the sun, or by heating something up enough that it shines.  In those cases, you're adding energy to all the atoms present, which causes them to emit photons eventually.  However, all the atoms are acting somewhat independently of each other: an atom in one corner of the sun doesn't really care what an atom in the other corner of the sun is doing.  This means the light being emitted by the whole sun has a lot of disorder in it, which leads to a lot of the properties we take for granted in light sources some major points are that you can't form a really narrow beam of light from these sources and you don't see a lot of weird effects like interference and diffraction from this kind of light.  If you try to send this light through two pinholes, you won't form very visible interference fringes.

In laser light, all the atoms of the excited material are put in a cavity between two highly reflective mirrors.  The mirror at one end is slightly less reflective so that some light can escape from it.  Some light is sent into the cavity so it bounces back and forth until it finally leaves through the slightly-less-reflective mirror.  The interesting thing about the laser is that it exploits a property called "stimulated emission."  If the excited atoms see a photon pass by, they're likely to emit a photon that exactly matches it.  Now you can imagine as the light bounces back and forth in the cavity, gaining photons from the nearby excited atoms, you build up a lot of identical-looking photons.  Since these atoms aren't acting independently of each other, but are all tied to the photons they see passing by, there's a lot of order in this light.  This light is called "highly coherent."  Because of the high degree of order in the light, the light tends not to spread out quickly, so you can create very narrow beams that travel a long way without spreading out appreciably.  You can also easily notice diffraction and interference effects. 

Here's some things to try to see the difference between the incoherent thermal light, and the coherent laser light. 
-Try shining a laser and a light bulb onto a DVD or CD-ROM.  The laser will diffract and create a bunch of spots of light on the wall.  The light bulb won't.  Diffraction is a wave effect that is much more noticeable for coherent light, such as a laser.
-Try using a laser and a flashlight to put a spot of light on a distant object.  The laser will be able to create a spot at much further distances without spreading out, where the flashlight will spread out into a wide beam fairly quickly.  In this case, the more coherent laser light experiences less spreading-out than the relatively incoherent thermal light.

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Offline techmind

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How does a laser work?
« Reply #4 on: 23/03/2009 22:17:12 »
There's a fairly reasonable WIkipedia page on lasers, although maybe it would benefit from a more gentle introduction...

http://en.wikipedia.org/wiki/Laser


Laser-light is monochromatic*, coherent, and usually well-collimated.
Everyday light sources are divergent, polychromatic (broad spectrum), and by definition broad-spectrum sources cannot be coherent.

*monochromatic: things can be monochromatic to varying degrees... a discharge lamp such as a low-pressure sodium streetlamp is substantially monochromatic for everyday purposes, but a laser is usually far more highly monochromatic.

Only a laser is coherent. This is due to the physics behind the stimulated emission and the laser cavity. It is the coherent nature which causes the laser speckle effect, and which is a requirement to make holograms. You need a known and constant phase relationship between the two light paths which combine on your photographic plate for the process to work.


The earlier type of lasers used a gas discharge in a glass tube, typically with a tube length of 250mm (8 inches) or more between the end-mirrors.
Modern solid-state semiconductor lasers ("laser diodes") have a laser cavity only a few mm long and consequently have less-well collimated beams and shorter coherence lengths than their larger counterparts.

The "coherence length" is a measure of the physical distance along the beam for which the light waves are "continuous" in phase. My understanding is that you can't use small solid-state lasers for making holograms - or maybe only of very small subjects (but I might be wrong).


Some applications of lasers, such as holograms, rely on the coherence properties.
Other applications, such as laser cutting, depend on the very high concentration of energy (which partly comes from it being a collimated beam) you can achieve with a laser beam.
A parallel (collimated) beam of light can easily be focussed down to an extremely small spot which you just cannot do for normal divergent light where the source is of finite size. The monochromatic nature of the light also means you don't have to worry about chromatic abberations in the lenses which is a great simplification. This is why laser sources are excellent for reading and writing CDs/DVDs/Blu-ray discs. You can focus the light down to a spot whose diameter is comparable to the wavelength of the light (eg 500nm).

In laser printers for example, the highly collimated light is focussed into a small spot ~1/600th inch (for 600dpi), and scanned across a photosensitive material to build up the image you want to transfer to the paper.
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lyner

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How does a laser work?
« Reply #5 on: 23/03/2009 23:35:14 »
yor_on
Quote
And the strangest thing about it is that those photons don't interact when being 'cajoled' into a coherent laserbeam, "the incident photon does not collide with the excited atom, apparently proximity alone triggers the stimulated emission." creating an exact copy of that photon in the process building up to that beam.

I keep batting on about how the photon shouldn't really be looked upon as a tiny particle at all. What you have written seems to  support my idea. The stimulated emission of a photon from an excited atom by the passage of a 'nearby' photon implies a very sizeable region for the photon to be affecting. Also, the 'avalanch' effect in the operation of a laser, also implies that one photon also affects more than one atom.
It would also appear that the implication of a sort of magic in the two slits experiment because of the 'one slit or the other' behaviour which 'particalists'  demand, becomes 'several atoms (/paths) at once' when a photon travels through a laser. The photon appears much more like a guided wave under these conditions.

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Offline lightarrow

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« Reply #6 on: 24/03/2009 12:45:19 »
If I remember well, it should depend on "coherence length": you can't in any way localize a photon better than that lenght. In a laser beam it can vary from millimeters to hundreds of meters...
« Last Edit: 24/03/2009 12:48:14 by lightarrow »

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lyner

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« Reply #7 on: 24/03/2009 14:17:00 »
It isn't just length, either. It involves width (csa), too.
The stimulated emission factor doesn't only apply in lasers, either. It is there all the time - just at a very low level. Is it the photon that changes, according to circumstances or is it just the circumstance that are different?
When you try to nail the particle down, as a model, it gets more and more slippery.
Is there, in fact, a phenomenon which necessitates the photon to be a particle? Classical em wave theory accounts for photon (wave) momentum, for instance.

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Offline Vern

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« Reply #8 on: 24/03/2009 15:58:43 »
I have always thought of a photon as one wave length of electromagnetic radiation. It consists of two points of saturated electric and magnetic amplitude always moving at the speed of light. The electric and magnetic fields exist in a spacial area around the saturated points and extend outward in space diminishing in amplitude with distance.

The points of saturation appear to be particles; the fields appear to be waves.

« Last Edit: 24/03/2009 16:03:49 by Vern »

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lyner

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« Reply #9 on: 24/03/2009 19:39:06 »
If you try to make a single cycle of a sinusoid (in time), like you have drawn, it involves a huge range of frequencies (bandwidth). To send a pulse of energy with a well defined frequency, the pulse must build up very slowly and decay slowly, too. This gives the photon a length of many many wavelengths - if you want to think of it that way. I think we have to accept the standard ideas about time domain /  frequency domain relationships.
I, personally, think it's a non starter and that people who talk in terms of tiny tiny photons haven't really thought it through. Certainly, the model you have drawn says nothing about the effective 'width' of a photon.
I blame School Science and the use of the word 'particle' which is interpreted in the spatial sense (as in 'corpuscular theory of light' in the 16th / 17th Century).
Why should the the smallness of a photon refer to anything other than its small energy?
« Last Edit: 24/03/2009 19:40:39 by sophiecentaur »

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Offline lightarrow

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« Reply #10 on: 24/03/2009 19:44:05 »
If you try to make a single cycle of a sinusoid (in time), like you have drawn, it involves a huge range of frequencies (bandwidth). To send a pulse of energy with a well defined frequency, the pulse must build up very slowly and decay slowly, too. This gives the photon a length of many many wavelengths - if you want to think of it that way. I think we have to accept the standard ideas about time domain /  frequency domain relationships.
I, personally, think it's a non starter and that people who talk in terms of tiny tiny photons haven't really thought it through. Certainly, the model you have drawn says nothing about the effective 'width' of a photon.
I blame School Science and the use of the word 'particle' which is interpreted in the spatial sense (as in 'corpuscular theory of light' in the 16th / 17th Century).
Why should the the smallness of a photon refer to anything other than its small energy?
It refers also to the smallness of the "flash" in the detector screen.
For the rest, I agree with you.

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Offline lightarrow

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« Reply #11 on: 24/03/2009 19:47:02 »
I have always thought of a photon as one wave length of electromagnetic radiation. It consists of two points of saturated electric and magnetic amplitude always moving at the speed of light. The electric and magnetic fields exist in a spacial area around the saturated points and extend outward in space diminishing in amplitude with distance.

The points of saturation appear to be particles; the fields appear to be waves.
You can have a single photon in thousands of wave lengths of an EM wave packet. It can't be as you say.

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Offline Vern

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« Reply #12 on: 24/03/2009 20:05:11 »
Quote from: lightarrow
You can have a single photon in thousands of wave lengths of an EM wave packet. It can't be as you say.
So, do you see a single photon as composed of many wave lengths? Would this be a certain number of wave cycles?

Those types of questions lead me to suspect a single photon exists as a single wave cycle.

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Offline Vern

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« Reply #13 on: 24/03/2009 20:12:57 »
Quote from: sophiecentaur
Is there, in fact, a phenomenon which necessitates the photon to be a particle? Classical em wave theory accounts for photon (wave) momentum, for instance.
I suspect it is the maxima of the pointy photon wave that reacts with matter. The reaction of a single photon would be at a point because of the wave packet shape. Its exact reaction spot can't be predicted because the phase relation between the photon and its reaction partner affects the reaction location.

Edit: Here's a different model; the points should be sine functions but are a little distorted.

« Last Edit: 24/03/2009 20:19:42 by Vern »

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Offline lightarrow

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« Reply #14 on: 24/03/2009 20:41:21 »
Quote from: lightarrow
You can have a single photon in thousands of wave lengths of an EM wave packet. It can't be as you say.
So, do you see a single photon as composed of many wave lengths? Would this be a certain number of wave cycles?
I don't think so, the number of wave lengts depends on the kind of emission.

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Offline Vern

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« Reply #15 on: 24/03/2009 21:18:44 »
It is the amount of electric and magnetic change that gives the photon its action. For a single photon this has to be one quantum worth of energy-time. If the amount of wave cycles can vary, the amount of electric and magnetic change would vary right along with the number of cycles.

So, I am not convinced that single photons can exist as multiple cycles.


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Offline JP

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« Reply #16 on: 24/03/2009 23:08:26 »
I think trying to pick a photon out of a classical waveform is flawed, since coherent classical waves already assume a particular quantum mechanical state (a coherent state), in which there isn't a single well-defined number of photons.

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lyner

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« Reply #17 on: 24/03/2009 23:38:56 »
Quote from: sophiecentaur
Is there, in fact, a phenomenon which necessitates the photon to be a particle? Classical em wave theory accounts for photon (wave) momentum, for instance.
I suspect it is the maxima of the pointy photon wave that reacts with matter. The reaction of a single photon would be at a point because of the wave packet shape. Its exact reaction spot can't be predicted because the phase relation between the photon and its reaction partner affects the reaction location.

Edit: Here's a different model; the points should be sine functions but are a little distorted.


The problem with this model is the evidence.
The spectrum of the light from gases (i.e. well defined energy states) consists of very narrow lines. That implies a very well defined frequency.
A time function such as you describe (as the photon would be traveling, that implies the same shape of time function) would necessitate a very wide and undefined frequency.
The time function and frequency function are related by the Fourier Transform and I don't think you can ignore that.
So a narrow band frequency function makes it impossible for the time function of the photon to be as brief as one cycle.
For a 0.1% bandwidth, the time function would have to be in the order of  1000cycles long.
Also, there cannot be a 'specific time' at which the photon can be said to interact with the  atom. Because the energy change is so well defined, a high Q resonance is implied and this implies many many cycles of interaction between wave and charge system.

jpetruccelli
Each photon, in a normal beam of non coherent light would be 'guided' by its own wave. Superposition allows for this; it would simply involve as many wave functions as there are photons. In the case of laser light, the wave functions would be exactly (or very nearly) in phase, so the result would be a resultant set of em fields  which would be seen to behave just like a classical (coherent) wave.  There doesn't seem to be any clash with the quantum considerations -it doesn't involve "picking a photon from a waveform" as the waveform is not actually a classical one - it is a combination (fuzzy) of the effects of  many photons and that tends to the classical (time - windowed, remember) waveform.

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Offline Vern

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« Reply #18 on: 24/03/2009 23:47:08 »
Quote from: sophiecentaur
The time function and frequency function are related by the Fourier Transform and I don't think you can ignore that.
I don't think the Fourier Transform can be applied to single photons. It is a kind of statistical analysis requiring many photon cycles. The same problem exists with describing a monochromatic signal using statistical analysis. Maybe we have become so accustomed to statistical analysis and probability functions that we can't visualize nature's reality. [:)]

 
« Last Edit: 24/03/2009 23:50:54 by Vern »

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Offline techmind

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How does a laser work?
« Reply #19 on: 25/03/2009 00:52:25 »
Maybe I'm naive, but I do find the whole business of photons and this handwaving "wave-particle duality" quite unsatisfactory.

Has any significant progress been made on this problem in the past 50-60 years?

Given the classical two-slit results but also with more modern findings of stimulated emission and coherence length, is it about time we had some new improved models/expanations? Also, as sophiecentaur suggests, something which includes ideas of Fourier transform limitations of time and frequency functions.


Can we devise some new experiments to help crack this nut?
Or do I really need to do lots and lots of reading first? (Especially on quantum mechanics - not my favorite subject!)


Now I understand that coherence-length is a kind of statistical measure, but do we actually know for sure whether the laser light has abrupt changes of phase (at essentially perfectly constant frequency) or whether the effect is due to just a tiny amount of frequency instability so a continuous drifting of phase? Or, given the Fourier principles mentioned earlier, is this a silly question to ask? (an abrupt change of phase would require/imply a momentary broad frequency burst)

If you shine laser on a photodiode with MHz/GHz bandwidth (and amplifier) do you detect "noise" in a frequency range corresponding to the coherence length? If so, what is the spectrum/characteristic of the noise?


And how do we tie up ideas of polarization with "particulate" photons???

If I was out of a job and could get a research grant, I'd be quite interested to have a go at devising experiments with the aim of better figuring out what a photon "is"!
« Last Edit: 25/03/2009 01:00:19 by techmind »
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lyner

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« Reply #20 on: 25/03/2009 08:56:06 »
Quote from: sophiecentaur
The time function and frequency function are related by the Fourier Transform and I don't think you can ignore that.
I don't think the Fourier Transform can be applied to single photons. It is a kind of statistical analysis requiring many photon cycles. The same problem exists with describing a monochromatic signal using statistical analysis. Maybe we have become so accustomed to statistical analysis and probability functions that we can't visualize nature's reality. [:)]

 
The Fourier transform is not statistical. It simply transforms from time to frequency domain. You drew a function in time and I told you something about the frequency domain version.
The whole particle thing about photons is a serious snare and delusion. It forces people to quite unnecessary conclusions- like your picture. Where is the evidence that they are localized in that way?
Coherence and polarization are just not a problem if you regard the photon as wavelike. What you see is just the result of many photons -identical ones.

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Offline Vern

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« Reply #21 on: 25/03/2009 12:37:26 »
Quote from: sophiecentaur
The Fourier transform is not statistical. It simply transforms from time to frequency domain. You drew a function in time and I told you something about the frequency domain version.
But as soon as you say frequency, you imply statistical analysis. You imply repeated cycles in a period of time. One cycle is not repeated cycles. One cycle can happen in a precise amount of time that can be as exact as you care to imagine. The time duration for that one cycle can be exact. But you can't know the frequency until you see some more cycles. [:)]

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Offline Bored chemist

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« Reply #22 on: 25/03/2009 20:10:34 »
"The time duration for that one cycle can be exact. But you can't know the frequency until you see some more cycles"
Nonsense, if I tell you that the time between the peaks on the electricity mains here is 0.02 seconds you can tell that the frequency is 50Hz.
Please disregard all previous signatures.

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Offline syhprum

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« Reply #23 on: 25/03/2009 20:35:46 »
You can only quote the frequency as 50HZ after measuring the time between two peaks if you make the assumption that time interval repeats indefinitely.
if you simply make a spot check the frequency could be varying wildly.

I think I have only repeated what Vern said!
« Last Edit: 25/03/2009 20:37:41 by syhprum »
syhprum

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Offline lightarrow

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« Reply #24 on: 25/03/2009 20:39:40 »
"The time duration for that one cycle can be exact. But you can't know the frequency until you see some more cycles"
Nonsense, if I tell you that the time between the peaks on the electricity mains here is 0.02 seconds you can tell that the frequency is 50Hz.

No. You can say the frequency is 50Hz *only if* you already know that the function is periodic. If you have a finite train of sinusoidal waves, then it's made of a superposition of an infinite number of monochromatic waves, so the frerquency is not perfectly defined. There is an indetermination relation between the length of the train and the frequency: the longer the train (and so, less defined the position of the train) the smaller the spread in frequency.
Heisenberg's principle can be proved mathematically from these considerations.
« Last Edit: 25/03/2009 20:42:00 by lightarrow »

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Offline lightarrow

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« Reply #25 on: 25/03/2009 20:45:03 »
You can only quote the frequency as 50HZ after measuring the time between two peaks if you make the assumption that time interval repeats indefinitely.
if you simply make a spot check the frequency could be varying wildly.

I think I have only repeated what Vern said!
That's correct syhprum.

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lyner

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« Reply #26 on: 25/03/2009 22:27:26 »
The Fourier transform is, as I said before, not a statistical thing.
It would be better for you to look it up on Wikkers  http://en.wikipedia.org/wiki/Fourier_transform, Vern.

What it says is that, for instance, a single pulse of finite duration can be looked upon as an infinite set of sinewaves of infinite duration and of different amplitudes and phase. They will interfere with each other at all other times to produce a zero value but, for the duration of the pulse, they add together to follow the pulse shape.  If you want to produce your particular shape of pulse then it must involve a vast number of frequency components which stretch over a very wide bandwidth. That is the pulse described in frequency domain. In the case of your single sinusiod, extending for one period, the FT would consist of a component with a maximum at the f (of hf) with a continuum of sidebands which would extend on either side of f.

This is very basic stuff and is not really open to question but it means that, to produce short pulses, you need a big bandwidth. If the only frequencies produced by a transition are in a very narrow band (a line spectrum), this means that the wavetrain must be extremely long.

It means that, if one actually insists on the photon's length being important, then, depending upon the system which produced it (a discrete energy state transition or a transition within a band energy structure) the photon would have a different length. Furthermore, for a photon to  be absorbed by a system with a particular energy structure, it would also have to be the right length. So laser light couldn't be absorbed by an atom in a solid, for instance.

The only thing one can say about a photon is that it has a particular energy associated with it, which corresponds to a particular frequency. There is nothing about its 'length' in its specification. SO, I say that assigning a length to a photon must be meaningless.

I think that it is very optimistic to assume that the photon can be characterised by something so concrete and familiar as a short 'squiggle', bearing in mind what a sophisticated concept it entails.

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lyner

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« Reply #27 on: 25/03/2009 22:40:38 »
techmind
You're talking my kinda language.
At least one person seems to be appreciating something of what I am saying.

The problem goes back to the Corpuscular theory of light, which was a very tangible idea. After the idea of light as a wave and then the reconciliation between the two ideas when duality was introduced, people went back to the corpuscle idea because it was / is so attractive and concrete.
Einstein's photoelectric effect experiments confirmed that em waves occur in energy dollops - nothing more. Teaching in School has prolonged what I think is a myth about the little bullet model and the statements by the great and good in the Science world have also been interpreted as confirming it. Whether or not they really have confirmed it, explicitly, is another matter.

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Offline techmind

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« Reply #28 on: 26/03/2009 02:00:28 »
The stimulated emission concept is quite interesting.

Now we know that inherently the energy "gap" in the excited-state atom is perfectly matched to the rest of the laser-radiation in the cavity.
We also know that the excited state has to be relatively "long lived" (metastable or whatever) to as to "hang around" long enough to be taken to the ground state predominantly by stimulated emission rather than random emission (sorry I can't think of the proper word).
The fact that it "hangs around" could be interpreted/considered as some kind of energy potential-barrier.
So if the excited atom momentarily "borrows" some energy from a passing "photon" then it can return to the ground-state.

Buuuut... I said before the energy gap is "perfectly matched", which might imply a concept a bit like a high-Q resonant system has some relevance? ... something that would couple to an EM wave of the precisely correct frequency...? And if this was the case you might expect the stimulated "photon" to be phase-matched and polarisation-aligned to the stimulating "photon"?


A bit handwavy I'm afraid - tis late at night, plus a decade or more since I studies undergrad photonics.  [;)]

What do you think? sophiecentaur?

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Offline swansont

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« Reply #29 on: 26/03/2009 17:48:14 »
The stimulated emission concept is quite interesting.

Now we know that inherently the energy "gap" in the excited-state atom is perfectly matched to the rest of the laser-radiation in the cavity.
We also know that the excited state has to be relatively "long lived" (metastable or whatever) to as to "hang around" long enough to be taken to the ground state predominantly by stimulated emission rather than random emission (sorry I can't think of the proper word).
The fact that it "hangs around" could be interpreted/considered as some kind of energy potential-barrier.
So if the excited atom momentarily "borrows" some energy from a passing "photon" then it can return to the ground-state.

Buuuut... I said before the energy gap is "perfectly matched", which might imply a concept a bit like a high-Q resonant system has some relevance? ... something that would couple to an EM wave of the precisely correct frequency...? And if this was the case you might expect the stimulated "photon" to be phase-matched and polarisation-aligned to the stimulating "photon"?


A bit handwavy I'm afraid - tis late at night, plus a decade or more since I studies undergrad photonics.  [;)]

What do you think? sophiecentaur?



You've got it right.  The concept of cavity Q and phase matching is very relevant to laser operation.  Though because of the high frequencies involved, and therefore really large Q's, the "free spectral range" is sometimes introduced
http://en.wikipedia.org/wiki/Free_spectral_range

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Offline Bored chemist

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« Reply #30 on: 26/03/2009 19:52:59 »
You can only quote the frequency as 50HZ after measuring the time between two peaks if you make the assumption that time interval repeats indefinitely.
if you simply make a spot check the frequency could be varying wildly.

I think I have only repeated what Vern said!
That's correct syhprum.
No it's not.
Th FT is a mathematical operation like reciprocal.
You can apply it to  a data set and get an answer. That answer is correct, no matter what happened at some other tme.
If the peaks were 20 msec apart then the frequency at that time was 50Hz.
Please disregard all previous signatures.

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« Reply #31 on: 26/03/2009 23:36:02 »
You can only quote the frequency as 50HZ after measuring the time between two peaks if you make the assumption that time interval repeats indefinitely.
if you simply make a spot check the frequency could be varying wildly.

I think I have only repeated what Vern said!
That's correct syhprum.
No it's not.
Th FT is a mathematical operation like reciprocal.
You can apply it to  a data set and get an answer. That answer is correct, no matter what happened at some other tme.
If the peaks were 20 msec apart then the frequency at that time was 50Hz.
Can you please find the FT of this function:
(hint: 1<x<15 is not a choice of visualization of the x, it's *the domain* of the function).
« Last Edit: 26/03/2009 23:47:09 by lightarrow »

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« Reply #32 on: 27/03/2009 00:24:55 »
BC
What you say about the FT is not correct. The FT involves integrals over all time / frequency (depending on which way you are going. You can frigg things to get a sort of answer by 'windowing' to reduce the integration span.
Only a continuous infinite sinewave has a single 'frequency'- . Any other function has an Frequency domain function which is not single valued but extends around the central frequency peak. Just looking at a row of maxes, mins or zero crossings does not define a frequency fully.
« Last Edit: 27/03/2009 00:32:50 by sophiecentaur »

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« Reply #33 on: 27/03/2009 00:30:09 »
lightarrow
As far as I remember, the FT of your function (I assume that it has zero value above and below the x values given) will be that of a single, top hat function modulated sinewave. That will be, basically a sinx/x  spectrum extending on either side of the 'carrier' frequency - the sinewave you have drawn. The sidebands cover an infinite range.

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« Reply #34 on: 27/03/2009 12:43:19 »
It struck me that people may be confused between the FT and the DFT (discrete FT) which assumes a repeated waveform.

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« Reply #35 on: 27/03/2009 14:00:30 »
As far as I remember, the FT of your function (I assume that it has zero value above and below the x values given) will be that of a single, top hat function modulated sinewave. That will be, basically a sinx/x  spectrum extending on either side of the 'carrier' frequency - the sinewave you have drawn. The sidebands cover an infinite range.
More precisely (if I computed correctly):
f(ω) = [1/(ω2-1)]*[e-i15ω(sin15+ωcos15) - e-iω(sin1+ωcos1)].

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« Reply #36 on: 27/03/2009 14:46:53 »
What does that look like when you  plot it?
How are your graphing skills?

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« Reply #37 on: 27/03/2009 16:34:38 »
What does that look like when you  plot it?
How are your graphing skills?
This is the real part:

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« Reply #38 on: 27/03/2009 17:58:36 »
OH!
Well, the left hand bit looks like a sinx/x function shape to me but the right hand bit is very strange.
Does the picture differ a lot if you choose a different actual length for the 'burst' of cycles? Does the the x axis zero  correspond to zero frequency?
btw, your use of degrees rather than radians in the sums could be the source of a 'scale' problem. Is it ok?

If you had a very wide window, with many cycles in it, I think you'd expect to see two very sharp peaks (+/- the basic frequency of the squiggle). Is it easy to confirm that / apply the test of your maths?
« Last Edit: 27/03/2009 18:00:40 by sophiecentaur »

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« Reply #39 on: 28/03/2009 00:06:30 »
Ok, you have decided to make me work this evening  [:)]

It's simpler (for me, at least) to study the function f(x) = cos(ω0x) in the interval (-a,a). Its Fourier transform is real and it is:

f~(ω) = [1/(ω0-ω)]*sin[a(ω0-ω)] + [1/(ω0+ω)]*sin[a(ω0+ω)]

Choosing ω0 = 1 and a = 10 (first case), a = 100 (second case), the resulting plots are, respectively:
« Last Edit: 28/03/2009 00:09:23 by lightarrow »

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« Reply #40 on: 28/03/2009 00:15:35 »
To precise: yes, the x axis zero means zero frequency; you notice that when the "window" goes from 20 to 200 you have two much sharper peaks (as you say).
« Last Edit: 28/03/2009 00:18:30 by lightarrow »

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« Reply #41 on: 28/03/2009 15:29:33 »
Your second graph looks like a pretty good sinx/x curve which is what one (I) would expect.
The difference between a long burst and a short burst is very significant, when you're talking about the form of spectra from a substance emitting light.

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« Reply #42 on: 28/03/2009 20:24:57 »
Your second graph looks like a pretty good sinx/x curve which is what one (I) would expect.
The difference between a long burst and a short burst is very significant, when you're talking about the form of spectra from a substance emitting light.

Sinx/x actually is the FT of the rectangular impulse of unit height:

f(x) =  1  if  -1< x <1

      =  0  if    |x|>1

f~(ω) = 2sin(ω)/ω

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« Reply #43 on: 28/03/2009 21:52:21 »
That's right. And when you modulate a sinusoid with it, you get sinx/x sidebands (convolving one with the other, i seem to remember - which is the equivalent of modulation only in the f domain). We ex broadcast engineers used to do that stuff for breakfast (but sometimes the theory may be a bit rusty).

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« Reply #44 on: 29/03/2009 11:51:48 »
If I record that sound of next door's cat and FT the amplitude vs time trace I get an amplitude vs frequecy trace. If I look at the trace and find the lowest frequency peak in it (apart from any DC component) it's reasonable to call that the frequency of the cat's howl.
I think that if I take a 50Hz sine wave like the mains and crop a 20msec part of it then FT it, I will get the convolution of the sinx/x from the cropping functiion and a spike at 50Hz for the sinewave.
The lowest frequency peak in that will be at 50Hz. I can therefore conclude that the mains frequency is 50Hz.
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« Reply #45 on: 29/03/2009 13:01:01 »
BC
In practical terms you could often say just that.
BUT, if you were half way across the Atlantic and you looked at the 'mains hum' on your amplifier, you would still get a low frequency wave which went up and down. It would also be 'beating' at 10Hz. If you froze the signal and looked at zero crossings, what would your simple analysis tell you?
1. If you took a very long sample and did a DFT on it, then you would get two peaks (50 and 60 Hz).
2. If you took a short burst, you might find that there was  a predominance of 50Hz or 60Hz, depending on when you actually took your sampled waveform and how long it actually was, before you made an endless loop of it. The DFT would have lost some information about the signal.

In the context of the photon, you mustn't do a DFT on it (which is what your treatment is implying) because it really is a 'one off'.

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« Reply #46 on: 29/03/2009 14:04:10 »
Techmind, how do you think when you write "So if the excited atom momentarily "borrows" some energy from a passing "photon" then it can return to the ground-state."? Can one see a photon as something you can 'borrow' energy from? And still be a 'photon'. Also what are we talking about here, how does it lend that energy, by what 'mechanism' do you see it. The only way I see that happen is by HUP, and then one still have the problem of defining what 'energy' has been 'loaned' as it then will have the same 'energy' after the 'loan' as it had before? Wouldn't this be a violation of the idea of 'conservation' laws?
---

Maybe you are looking at as SC wrote? "a single pulse of finite duration can be looked upon as an infinite set of sinewaves of infinite duration and of different amplitudes and phase. They will interfere with each other at all other times to produce a zero value but, for the duration of the pulse, they add together to follow the pulse shape"?

But this still seem to be a result of HUP?
« Last Edit: 29/03/2009 14:10:33 by yor_on »
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« Reply #47 on: 31/03/2009 00:11:24 »
I think HUP is consistent with this. The precision of the energy measurement (frequency / momentum /spectral width ) and the precision with which the pulse can be said to be passing (length of pulse) are mutually limiting. I imagine 'h' is in there somewhere.

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« Reply #48 on: 31/03/2009 10:50:20 »
Techmind, how do you think when you write "So if the excited atom momentarily "borrows" some energy from a passing "photon" then it can return to the ground-state."? Can one see a photon as something you can 'borrow' energy from? And still be a 'photon'. Also what are we talking about here, how does it lend that energy, by what 'mechanism' do you see it. The only way I see that happen is by HUP, and then one still have the problem of defining what 'energy' has been 'loaned' as it then will have the same 'energy' after the 'loan' as it had before? Wouldn't this be a violation of the idea of 'conservation' laws?

But this still seem to be a result of HUP?

Indeed I was thinking HUP (in a vague sort of way). There's no violation of conservation of energy as you're just overcoming a potential barrier (or tunneling through it).  You still come down the other side, so no net energy transfer.
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