Talk:Photon/Archive 1

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I've just deleted this (hidden) comment from the article in order to discuss it here:

The problem with most of what follows here is that a photon is not a particle per se and doesn't move at the speed of light...photons are quanta, gauge bosons if you will...as they describe a single field mode they are nonlocal..

Of course, a photon travels with c. There is no contradiction here. Actually, the notion of velocity makes more sense in the particle picture than in the wave picture (where you have to sort out the amibuity of phase velocity and group velocity.

Also, a photon with energy uncertainty describes a not well-selected field mode.

The article overstates the wave-particle problem generally, as it in no way contradicts the photon picture. I wil think about a good formulation to make clear, that in quantum field theory, saying that the photon is a quantum does not imply that it is a particle in the old sense of being a localized corpuscular point thingy.

Sanders muc 21:03, 26 May 2004 (UTC)

I have added my tuppence, and tried to make the article a bit more "layman friendly". Someone with a better grasp of Foch states had better look at that part - I suspect that it could be rather interesting if explained a little further.

I have moved some text under a new "Properties" heading - lots of the text under "Symbol" was not about the symbol at all, and some text on physical propeties was buried in "Creation".

Comments welcome. -- ALoan 17:55, 27 May 2004 (UTC)

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Changed 'diffraction' back to 'refraction'. It really is refraction of light that causes rainbows. I'll explain the mechanism when I get around to doing rainbow.

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Is it really true that photons are slowed down when moving through a medium? I thought that they are constantly absorbed and reemitted, so it's not really one and the same photon moving through and being slowed down. --AxelBoldt

Not sure about in QM, but IIRC in classical oscillator theory, photons move at c (i.e. vacuum speed of light) between atoms. They're not really absorbed and re-emitted by the atoms, though. What happens is that the electric field of the photon drives charges into oscillation, and those oscillating charges radiate a field which is slightly out-of-phase with the photon. The superposition of the photon and the radiated field is slightly retarded w.r.t. the original field, and so the photon is 'delayed' a bit at each atom. On a large enough scale, this looks like the photon is slower. -- DrBob

what is a photon, really

This article and the vast majority of other such material about light, overlook completely what a photon actually is.

Firstly, a charged particle is surrounded by an electric field, which extends outwards in all directions, reaching a distance of at least 13.7 billion light years. When a charged particle is accelerated, due to conservation of energy and angular momentum, the change in movement must be conveyed to the electric field which surrounds it. This change propagates at the speed of light, and is in fact, what we call light. Any light we see is caused by a change in an EXISTING electric field. Thus, a photon probably should be called a virtual particle, since it does not really exist like a regular particle independent of its creator. It is as virtual as other such particles like phonons, solitons, or polaritons. If we observe a photon emitted long ago from a distant star, we are observing the change in the electric field of the particle which emitted the photon. There is no particle which is a photon. The quantization of light is solely an effect of the quantization of charge and the particulate nature of the electrons used to observe and emit light. There might, however, be a quantization of space-time, which would also lead to the observation of quantized light.

Furthermore, a photon is not a packet of energy with a simple frequency or polarization. A wave of light can find itself in any shape allowed by the propagation of change in an electric field. This means that a photon isn't necessarily a sine wave, or even periodic. It just so happens that we are designed to detect periodically changing electric field in a certain range of frequencies called visible light. And we've created devices which are designed to detect light of many other frequencies. But we shouldn't limit our thoughts to the special (albeit useful) case of the periodically changing electric field.

I use the term electric, rather than electromagnetic field, since it has been shown that the observation of a magnetic field is a special relativistic effect upon charged particles in motion with respect to an electric field.

In view of quantum theory, the electromagnetic field is made of truly virtual photons, and when a change is propagated through the field we observe "real" photons.

Would anyone else like to comment? Like why aren't these ideas common knowledge? And why does the idea of a photon as corpuscle still exist? (please spare commentary on the photoelectric effect, unless it can be shown that this is not the result of interaction with particles of definite charge and cross-section) --D. Estenson II 08:20, Mar 22, 2005 (UTC) (BTW,IANAP,BWBSS)

Yes, "photon" carries misconceptions, and even physicists aren't immune. Photons are quanta, not particles, since "particle" carries conceptual baggage (the word "particle" unfortunately implies something like a very small grain of sand.)

A good collection of recent articles by physicists is from the Optical Society of America : THE NATURE OF LIGHT: WHAT IS A PHOTON? [1]

The articles have some some suprising stuff: the Photoelectric Effect does not prove the existence of photons, Nobelist W. Lamb takes a very dim view on the whole photon concept, and even as late as 1966 some physicists had a running bet whether quantum phenomena (Lamb Shift) could be calculated without recourse to quantized EM fields. Other articles in the collection show where the "photon" concept is crucial in physics. --Wjbeaty 04:16, Mar 23, 2005 (UTC)

FYI, here is a link to the abstract for the original article from the Optical Society of America. The Nature of Light: What is a Photon?. There you can find a link to the article in pdf format. Thanks for pointing out these articles. --D. Estenson II 10:08, Mar 23, 2005 (UTC)

I support most of the said. Even if we don't know, what a electromagnetic field oder matter is, and even, if we don't know, what momentum and energy is, we believe: the amount of energy and momentum doesn't change, but can be moved from field to matter. This exchange of energy and momentum we call a photon. This will solve some problems with imagination. Field no longer consist of photons, but are excitated by "photonic events". Such an event is a change in action of size h. Action is, according to my knowledge, the only appearance, that is realy quantisized. Energy quantization depends on the properties of a "measuring instrument". Quantisation of action NOT. This is the key! ErNa 08:48, 14 November 2005 (UTC)

Truth is, photons are virtual particles, which means they really aren't anything at all. The concept of a photon is just a conceptual placeholder to describe what happens when matter interacts with an EM field and how that field interacts with matter. Dan 18:08, 9 August 2006 (UTC)

Who named the photon?

Wave-Particle duality is not unique to the photon and the EM field in QM. Every particle, for example the electron, behaves like a wawe and under the right conditions this can be (and was) detected experimentally.

Elad Tsur 13:30, 1 June 2004 (UTC)

Photon and Action

Is it true, that, whenever a photon is emitted or absorbed, the emitter/absorber state changes by h (Planck's quantum of action)?217.229.215.59 14:35, 17 May 2005 (UTC)

I threw the picture in

Hope you like it; the article needed some sprucing up.

I think a better picture given the caption would be two next to each other: a photograph of diffraction lines from the double-slit experiment with laser light, next to picture of a (maybe x-ray) micrograph showing single points of light (resulting from the photon collision with the detector). I'm no wikipedia expert, but such pictures may already be somewhere on here. The current picture, on the other hand, does nothing to enhance the understanding of photons, or how both their wave-like and particle-like features can be demonstrated. --D. Estenson II 11:09, Mar 23, 2005 (UTC)

Good point. I'll look around and see what I can find (and legally upload to Wiki). -- Zalasur 07:53, Mar 24, 2005 (UTC)

Here is a good illustration of diffraction of laser light, and it's already on wiki: (single-slit diffraction of laser light, with intensity graph) --D. Estenson II 12:32, Mar 24, 2005 (UTC)

• • Sorry, I think the picture should be removed. It makes people think that the wave behaviour of photons is spatial, and not electromagnetic. It's a common belief among students (and perhaps people in general), mainly because of pictures showing photons moving up and down as it propagates.

Yes, the photon wave is in a sence spatial, but the picture is still misleading. //Johan Falk, physics education researcher, Sweden

Mr Falk, please explain how the photon's wave is spatial in a sense. AFAIK, a photon follows a straight line through empty space (ignoring gravity and other complications). While it is in transit, the electric field oscillates (grows more and less intense) where the photon is along the path. This can be visualized using a sine wave with the x-axis representing space (or time, with appropriate conversion) and the y-axis representing electric field strength. But many laymen misinterpret such a graph as saying the photon wiggles up-and-down as it travels through space. This is the interpretation that should be avoided at all costs, and which the picture only reinforces. Am I mistaken? --D. Estenson II 19:20, Jun 20, 2005 (UTC)

To partially answer my own question, since individual photons are polarized, they have a spatial orientation to their electric oscillations. --D. Estenson II 20:11, Jun 20, 2005 (UTC)

Polarization

This article needs a section on polarization. I may add one within a few weeks if no one else does. --D. Estenson II 20:16, Jun 20, 2005 (UTC)

• Yes, and there's nothing about a photon wave as E × B; having electrical and magnetic components. — RJH 5 July 2005 22:45 (UTC)

Always an electron?

A photon is always created by an electron? Never by proton-proton bremsstrahlung, or by a positron? I'm just curious about that statement. Thanks. — RJH 5 July 2005 22:42 (UTC)

A photon can be created by the acceleration of any charged particle, not just an electron. --D. Estenson II July 8, 2005 08:48 (UTC)

History

This article could use some content on the history of the photon, or at least some links. See e.g. photoelectric effect. -- Beland 06:30, 14 August 2005 (UTC)

what photon is?

"Eenergy of a photon can equaly be shown in another way:

where T=1/f is a period of a photon. In this form it seems that Planck`s constant h might be interpreted as the energy of a ( "standard" ) photon with T=1s ( photon lasting 1 second ); and that ( single ) photon as ( single ) EM vawe, with period inversly proportional to its energy."

I am not physicist, it`s only my imagination, scientific method is somthing else. I expect comments.Sslavko 09:16, 12 September 2005 (UTC)

I'm not sure I'm following your objections(?). Please reiterate informally. Thanks. El_C 02:55, 15 September 2005 (UTC)

Understanding of the nature of photon and Planck`s constant is closely related with much more important question of understanding quantum processes and quantum world in the whole. Mathematical equations give correct results and predictions, but nobody knows what some of them describe in real, everyday world at microscopic level. Some most obvious examples are double-slit experiment, quantum entanglement, wave/particle duality and, quantum loops, strings and so on. It appears that quantum world is governed by rules quite different from our everyday expirience.

Expresion for the energy of the photon E=hf describes its energy, and is experimentaly confirmed, but tells nothing about what photon is. But, because f=1/T, the same can be writen as E=h/T, where T is the period of the vawe (in seconds). From the equation E=h/T it is seen that the single oscillation of electromagnetic wave lastig one second, contains energy of h , that lasting 0.1 sec contains h / 0,1=10 h etc. The greater available energy by craetion of photon, the shorter period of oscillation. On the other hand, Planck`s constant, in this case at least, is action (energy) of elecromagnetic wave (oscillation, photon) lasting one second ( „standard“ photon).

I wish to induce philosophical consideration with the intention to show that quantum and macroscopic events, although different in some aspects, are governed by the same laws and bz the same logic as human mind. I expect in the first place participation of laymen, because experts have their own view, but any contribution of experts is wellcome, of course.

Playing with numbers and equations is a pleasent activity. At least I think so... And sometimes it even produces interesting and ground-breakig results. De Broglie hypothesis is something of the kind. But notice that Wikipedia's purpose is not to do any kind of first hand investigation and I've never eard of "standard photon" nor I could find any reference in a fast web search (other than "standard photon-something").
«that quantum and macroscopic events [...] are governed by the same laws» is beyond the scope of this article, and is probaly covered somewhere else, namely at Hawking radiation. Nabla 21:24, 19 September 2005 (UTC)

This argument is well known. To me it seems to be just a "killer". The biggest nonsense, documented in an paper, that can be referenced, will have a mirror in Wikipedia, but, the most simple conclusion from very basis knowledge will be "first hand investigation".

It is very clear: in relation to a photon, we can not talk of "frequency". For: it is not possible to determine by measurement the slope of the electromagnetic field of an single photon event (Heisenberg). To talk of frequency needs at list two zero crossings. But who ever calculated the fourier transform of a wavelet know, that the result is another wavelet. So we always have a spread spectrum, we can not talk of a frequency.

Everything becomes very clear, if we understand, that any electromagnetic field can be quantisised. But, imagine:

I have a cubic box with black body radiation inside of a certain temperature. The dimension of the box defines the lowest possible energy & frequency of a photon. The temperature defines the highest possible energy of a photon: temperatur and volume of the box determine the energy content of the field, and all the energy can be focused (for an very short moment) into a single "Photon"

Now, in any moment, I freeze the field to determine the number of photons of different possible energies. I receive a sprectrum. But, whenever I determine the dimension of the box, there is a uncertainity and so all the energies will be lightly shifted. This will result in a different spektrum. This shows very clear: the electromagnetic field is not consisting of photons, but is created and by photonic processes. A photon is just the process of exchanging energy and puls between electric charge and field. And this process is quantisised in the manner, that the change in action is always equal h. Therefore, it takes more time to exchange less energy. I never found a reason, to see photons as particles. Who else did? ErNa 06:13, 17 October 2005 (UTC)

What am I missing?

If I'm reading this article right, light is made out of photons, and photons cannot decay on their own. So why is it that when I turn a flashlight off, the light goes away? What happens to all the photons that had been produced by the flashlight? -- Mwalcoff 00:00, 20 October 2005 (UTC)

Photons can be absorbed (and turned into kinetic energy) when they collide with normal matter. (Sometimes an electron is knocked free by this process, hence the photoelectric effect.) --Jasonuhl 07:39, 14 November 2005 (UTC)

Energy vs frequency

The energy of a photon is directly proportional to its frequency, correct? But this only applies to quanta, correct? A continuous sine wave of amplitude A at 100 MHz doesn't have "more energy" than a similar wave at 10 MHz, does it? The power level and amplitude would be the same. Perhaps the higher frequency wave has more energy per quanta, but fewer quanta? (And what would a radio quanta look like? Or is this one of those "don't go there" wave-particle duality kind of questions?)

I've heard this used in reference to sound (higher frequencies have "more energy"), and I don't believe it. Seems like a misunderstanding to me (though my own understanding is limited). — Omegatron 20:30, 22 December 2005 (UTC)

The power of a classical wave is proportional to its amplitude. This does mean that a higher frequency wave of equal power is made up of fewer photons. -- Xerxes 21:37, 22 December 2005 (UTC)

Right, energy E=hv applies to photons only. For example, a microwatt of gamma rays would have a few high-energy photons, while a microwatt of 60Hz would have way way more photons. On the other hand, with acoustic waves of the same energy but different frequencies, the energy is proportional to pressure, but I think the air moves far less for higher frequency waves. With a few watts at 30Hz you can see the loudspeaker cone vibrating perhaps 1mm back and forth, and the pressure variations in the waves are tiny, but if a 30KHz tweeter moved that much, the ultrasound wattage might heat your skin. Maybe they're referring to transducer motion, where a vibrating object with constant amplitude motion would put out higher pressure excursions and higher wattage at higher frequencies. --Wjbeaty 21:46, 22 December 2005 (UTC)

but I think the air moves far less for higher frequency waves

Hmmm... isn't sound pressure directly related to particle displacement, regardless of frequency?

Nope.

$\xi ={\frac {p}{Z\cdot \omega }}$

Where $\xi$ is displacement, p is pressure, ω is angular frequency, and Z is acoustic impedance. You learn something new every day. — Omegatron 22:45, 22 December 2005 (UTC)

Omegatron and Xerxes: the power of a classical wave is proportional to the square of its amplitude, and you are correct that if you compare two waves of different frequencies but the same power, the one with higher frequency is made up of fewer photons. A side point: the exact number of photons in the wave is not necessarily well-defined. Under certain pretty common circumstances, the number of photons present becomes uncertain.

Photons don't really "look like" anything. If you want to think about it quantum mechanically, there are neither waves nor particles, just states. Waves and particles are convenient abstractions that make things easier to understand, but don't always work well. In particular, quanta need not be small. It's perfectly possible to construct an experiment in which a "photon" is spread over a region a meter or more across.

Mechanical vibrations, including sound, can be treated quantum mechanically, and it's not uncommon in that context to talk about "phonons"—quanta of sound. This doesn't alter the fact that sound is a mechanical wave. Quantum mechanics does always give the correct answers for "classical mechanics" problems. We say that it "reduces" to classical mechanics in the "classical limit". Phonons are very commonly used in describing vibrations ("sound") in crystals.--Srleffler 21:55, 22 December 2005 (UTC)

Yeah, I've read our article on phonons many times and still don't understand it. I'll keep trying... — Omegatron 22:35, 22 December 2005 (UTC)

I would like to see some information about photon angular momentum, perhaps contrasted to spin. Spiral interfference pics and an explanation would be nice too. Also, I do not know if anyone is familiar with the red shifting out of existence 'paradox' (1. where does its spin go? 2. relativity says that a photons energy is not fixed, that it is relative to the observer, so can a photon be non existent to one observer and real to another? How is this different from the Unruh effect?)

Wow! I'm 100% surprised that Omegatron doesn't understand something. I'm so used to it being me.

"A at 100 MHz doesn't have "more energy" than a similar wave at 10 MHz, does it?"

What exactly do you mean by "simliar wave"? If their number of photons was "similar" then 100MHz would have more energy, if their power was similar, then 100MHz would have less photons. The same is true with sound, for the same power, the air would be pushed and pulled farther for lower frequencies. Given the same physical amplitude of air "push and pull", a higher fruency would have more power. I'm guessing i'm missing the entire point of your question. Fresheneesz 23:26, 7 April 2006 (UTC)

Perceived? Detected?

The caption says that photons can be perceived as particles or waves. "Perceived" implies with our senses. Should this be changed to detected? Experimentally observed? Bubba73 (talk), 05:43, 12 January 2006 (UTC)

No, perception reflects how the experimenter interprets what is detected. The photons are not "detected" as either particles or waves, but one may interpret or perceive the detected result to imply a wave-like or particle-like nature.--Srleffler 06:43, 12 January 2006 (UTC)

How about "observed" then? Or how about rephrasing it to say that they can behave as particles or as waves, depending on the experiment/situation? Bubba73 (talk), 16:03, 12 January 2006 (UTC)

The thing is that photons are neither particles nor waves, but those are abstractions that allow us to think about it more easily. In some situations the wave model is simpler, and in others, a particle model is. Fresheneesz 23:28, 7 April 2006 (UTC)

rephrase?

I think the sentence "Photons are always moving, and photons in a vacuum always move at the vacuum speed of light with respect to all observers, even when those observers are themselves moving." needs to be rephrased. There is too much information there for one sentence. I propose replacing it with something like:

• Photons area always in motion, and move at a constant speed in any medium (optics). The speed of a photon in a vacuum is known as the speed of light. This speed is the same with respect to all observers, even when those observers are themselves moving. Bubba73 (talk), 16:56, 12 January 2006 (UTC)

I'm fine with breaking up the sentence, but it would be better to avoid talking about what light does in a medium. Whatever you say, in this context, isn't going to be quite true. Light doesn't always move with a constant speed in any given medium. It's easy to come up with counterexamples, like birefringent crystals. There is also an issue about what exactly photons are doing in a medium. It's not clear that the photon that comes out is the same one that went in. Just restrict the statement only to vacuum, and it will be fine.--Srleffler 17:32, 12 January 2006 (UTC)

You can ommit some problems of understanding, if you just take for true, what you know: energy and momentum is transfered between matter and elmag field in certain quantities: this is what we call "photon". Whatever the field is, whatever matter is: it doesn't matter! If the field is exitated in place A at time T, there is a probability to exite a atom at place B after time (B-A)/c. That's all. It needs no "moving photons". ErNa 08:54, 13 January 2006 (UTC)

In case you missed it, the title of this article is "Photon". --Srleffler 13:44, 13 January 2006 (UTC)

What did I miss? In physics, the photon (from Greek φως "phos", meaning light) is the quantum of the electromagnetic field, and further quantum: a "light quantum", being a unit of light (that is, a photon). Can you tell me, what a "unit of light" is? Do you know more properties of a photon? ErNa 21:36, 13 January 2006 (UTC)

Actually, yes I do. It carries both momentum and angular momentum, for two examples. That's beside the point, though. I agree with you that one can describe light quantum mechanically without recourse to "moving photons". Moving photons are, however, a good simple conceptual model for explaining the behavior of light to people who aren't ready for the full quantum mechanical treatment yet, or who don't need it for what they are doing. Lots of physics is still done by thinking about moving photons. Since this page is Photon, it is an appropriate place to deal with light in terms of the "moving photons" model, rather than a more detailed quantum mechanical treatment. --Srleffler 02:26, 20 January 2006 (UTC)

Thank You, I intended to provoke an answer and it worked! Now, we have three properties. Any more? Do you think, it is possible do open an article on photon, that make the quantum mechanical treatment plausible to normal people, which will not be immediately deleted?

Is it known how photons interact with particles? Since their postulated to have no charge (which seems odd to me), then they must interact in some other way. Fresheneesz 23:55, 7 April 2006 (UTC)

Quantum entanglement

isnt there somthing to do with seperatin photons, and then when one gets annilated, so does the other one, no matter how far apart they are, this recently came up while taking to a physicit i know.—The preceding unsigned comment was added by 172.202.236.88 (talk • contribs) .

Not exactly, but it sounds like you're talking about quantum entanglement. --Jasonuhl 22:37, 19 January 2006 (UTC)

The theory that two distant or adjacent photonic particles could share a bond would be possible and highly likely. If there was more than theoreticle proof of the existence of a photonic bond the subject could yeild results but since there is no proof it is simply discredited. —The preceding unsigned comment was added by 142.165.110.167 (talk • contribs) .

Not sure what you're talking about. Quantum entanglement is well established and has been demonstrated in many labs worldwide.--Srleffler 04:15, 14 February 2006 (UTC)

Article unnecessarily long and potentially confusing

Most of what's written in this article about "photons" is equally true about light in general and classical electrodynamics specifically. Radiation pressure is a classical effect. Calling an antenna or my microwave oven an emitter of photons seems to be an abuse of the language that brings to mind billiard balls. Similarly, talking about photons in the context of General Relativity, a nonquantum theory, is odd.

Why not merely devote this article to treating the photon as a fundamental particle (gauge boson/force carrier) and as the quantum of light?

Does Willis Lamb read Wikipedia? 09:30, 20 February 2006 (UTC)

Photons and color

I am unfortunately not a particle physicist, but I am trying to learn. What exactly is the relationship between an individual photon and its wavelength, e.g. color? I know wavelength times frequency equals the speed of light and all that. I also know that color perception (e.g. eyesight) is variable. But are there, for example, red photons, blue photons, or infrared photons? When light is separated by a prism, does that mean that "white" photons went in one side and individual photons for each color of the rainbow came out the other side? Someone please answer! oneismany 22:52, 23 February 2006 (UTC)

For the most part, you can think of each photon as having a specific frequency. The reality is a little more complicated than that, but that will do for most purposes. The relationship between the wavelength and the frequency depends on what medium the light is travelling through. When light enters a medium like water or glass, it slows down and its wavelength increases. The frequency stays the same. Don't try to imagine how a discrete "particle" of light can have a "wavelength". It won't help you understand this. What is helpful, though, is to understand that a photon's frequency is proportional to its energy. A photon with higher frequency has more energy than a lower-frequency photon.

Color is a perception created in the human brain by photons being detected by the retina in your eye. It is somewhat correlated with wavelength and frequency, but not entirely. For example, the well-known phenomenon that you can mix colors to obtain other colors (e.g. red and blue give purple) is entirely an artifact of the human visual system. Physically, if you mix red light with blue light, you have a mixture of red and blue light with no other colors present. The eye and brain perceive this mixture as purple.

So, there are sort of red photons and green photons and blue photons, but when you see green light (for example) you can't be sure there are any "green" photons there—you might be seeing a mixture of two other colors. There are no brown photons or pink photons, as those colors do not exist in the optical spectrum. Yes, there are infrared and ultraviolet photons. When light is separated by a prism, photons with a whole range of "colors" from red to violet (and beyond) go in, and they are separated in space by the prism. If photons with some range of frequencies are missing from the light that goes into the prism, the corresponding colors will be missing from the spectrum that comes out of the prism.

If you see green light it's because there are "green photons" because green is a primary color (same as red and blue). If, on the other hand, you see yellow light, you can't be sure whether it's due to "yellow photons" or a mix of red and green. There is an interesting experiment that uses this to highlight the variation in color perception between people. You have a source of "pure" yellow light, and a source of mixed red/green light. You then adjust the red/green ratio (by turning a knob or such) until you think that the two yellow lights look identical. You then show the result to someone else, who will almost certainly tell you that the two yellows are not identical! Similarly, when other people adjust the colors until they think they are identical, they will probably not look identical to you. Itub 16:22, 24 February 2006 (UTC)

Thank you, that is very helpful. Now, I wonder what is the mechanism for the color of shadows. Shadows are colored only among multiple sources of light with distinct colors. If you block red light, you get green shadows. If you block blue light, you get orange shadows. If you block purple light, you get yellow shadows. (And vice-versa.) What do you get when you block ultraviolet or infrared light? oneismany 16:32, 26 February 2006 (UTC)

Cancer-free and cold shadows? We humans can't perceive that stuff anyway... Melchoir 21:10, 26 February 2006 (UTC)

moving

The article says ...always move at a constant speed with respect to all observers, even when those observers are themselves moving. With respect to what are these observers moving? Should it say "with respect to each other"? Bubba73 (talk), 17:01, 24 February 2006 (UTC)

With respect to whatever you please. That's part of the point: there is no absolute reference for motion. You always have to specify with respect to what you are moving. Light is special, in that no matter what you pick as your reference, and no matter which direction you move in or how fast, the speed of light that you observe is always the same. You can be moving toward the source at 90% of the speed of light, and the light will be coming towards you at the speed of light relative to you. A friend can be moving away from the same source at the same time, and she will percieve the light overtaking her from behind, moving at the speed of light relative to her. This is extremely odd and unexpected behavior, but there is solid evidence that it's true, and relativity theory explains how it all works in a consistent way.--Srleffler 03:58, 25 February 2006 (UTC)

Yes, but the old version that you reverted to is still wrong. I'll take a crack at it. Melchoir 04:13, 25 February 2006 (UTC)

I like your version.--Srleffler 06:05, 25 February 2006 (UTC)

But my point is that if you have an observer and a photon and nothing else in the universe, you can't tell if the observer is moving. Bubba73 (talk), 04:25, 25 February 2006 (UTC)

Yes. I reverted your change, though, because it's irrelevant whether the observers are moving with respect to one another. You can pick any reference you please. Melchoir's new wording is better: instead of referring to the observers as moving, he merely says that the speed of light is constant regardless of their velocity. That covers all cases.--Srleffler 06:05, 25 February 2006 (UTC)

I like the new version better too. The old version said that they were moving, but not with respect to what. It could have specified "the light source', "other inertial reference frames", or something - but it idn't specify anything. It mentioned only a photon and observer. You and I known what the article intended to say (but didn't say), but many people viewing the article wouldn't know that. (i.e. "You can pick any reference you please." - the article didn't say that, but needed to say something to clarify that.) Bubba73 (talk), 01:00, 26 February 2006 (UTC)

Imaging in a camera does not require a wave description of photons

I don't think this a good example of wave behaviour. Imaging can be described with ray optics and thus is amenable to a particle description. This only breaks down when the diffraction-limit is reached but that is a secondary effect to the actual image formation. Why don't we use the standard example of interference (e.g. Young's double slit or diffraction at the edge of a shadow)?--J S Lundeen 22:15, 13 April 2006 (UTC)

A single photon

Can anybody confirm/refute this (somewhat paradoxical) interpretation of the photon? Consider a single photon emitted from some source. Is it not true that light is by definition EM radiation that ALWAYS propagates in every direction? (Every direction in a vacuum assuming no matter blocks it.) This means that the "particle" interpretation of a photon is quite misleading as the single photon is ALWAYS as sphere of EM disturbance whose radius increases with time (at the speed of light), which means the photon "exists" everywhere on the sphere at once. This is quite difficult to imagine if thinking of a photon as what most people consider a particle (a very small thing moving about in space with a definate position at any given time). Is this sort of like what is meant when, in explaining the wave/particle duality of matter/light, that the particle is "smeared" out over the wave front?

It is also interesting to consider the photon after a long time, when the radius of the EM sphere is very very large, meaning the photon "particle" exists simultaneously over an ever increasing spatial area. Because of this, I would almost expect that the energy of the photon decrease with time, as it has to "cover" more area. Is it not true though that in any given direction the energy of a photon does not decrease with time (or stated another way the magnitude of the EM disturbance does not decrease?) as the wave continues to propagate forever? It seems almost like getting something for nothing: while the energy at every point on the ever increasing sphere remains the same, the amount of space affected by this energy is ever increasing implying that the total energy of the photon in all of space is ever increasing with time. Perhaps this is just a property of PURE energy (which if I understand correct is what light is) as it relates to space and time that just does not sit well if one tries to give matter-like properties (a particle) to something that is just not matter. Is this a seriously flawed interpretation of quantum theory or an illumination of the paradoxical nature of it all? -Kyp4 19:51, 28 April 2006 (UTC)

Yes it is : ) . The photon model says that light does *not* "always propoage in every direction". However, for a point source, the probability that the photon will travel in any given direction is the same - thus for a large number of particles, statistics dictates that the light will form a virtual sphere of photons. Fresheneesz 06:15, 8 May 2006 (UTC)

Cleaning up the photon article

It seems to me that this article is not in very good shape. One of the first things that needs to be done to clean it up is to decide whether this is an article about photons or an article about light. (Obviously, it should be the former.) As an article about photons, we really ought to throw out all the bits dispersion and whatnot that are related to classical optics (except as it pertains specifically to the quantization of light). Also, the parts about the speed of light (in vacuum or not) are pertinent to light rays, but irrelevant to the photon. As a massless particle, the photon must always travel at c. The superposition of photons with some excited states of condensed matter are interesting, but need to be better phrased in the language of quantum mechanics, not of optics, to be appropriate for this article. Anybody agree or disagree? -- Xerxes 18:34, 1 May 2006 (UTC)

I completely agree. Melchoir 19:46, 1 May 2006 (UTC)

I agree too. However, I think we should have some explanation as to how the photon model agrees with the wave model. Fresheneesz 06:30, 8 May 2006 (UTC)

Origin of the term photon

The original Wikipedia article was correct. The term "photon" was coined by Gilbert Lewis in a 1926 letter to the editors of the journal Nature. The 1920 lecture by Planck, as posted by nobelprize.org, is a translation from the original German. In German, Planck uses the term "lichtquantum", meaning literally "light quantum". So this usage does not pre-date Lewis's 1926 coining. -- Xerxes 14:32, 18 May 2006 (UTC)

Yes. I've actually looked at the 1926 article. I don't have it right now, but I can vouch that it does say something like "I hereby propose the term 'photon'[...]" (paraphrased from memory). Itub 18:31, 18 May 2006 (UTC)

Luxons??

Good for you, Xerxes. What a needless word. Sounds like the particles that come out of the top of the big black pyramid casino in Las Vegas. Luxons. Yeah. And I may add to critics that while luxon may be a perfectly cromulent word, its use does not embiggen this article. Steve 00:28, 26 June 2006 (UTC)

I actually cannot see why luxon is not in this article...it gets its definition by photons and gluons alone. I also like how photons being gauge bosons gets like no play. --HantaVirus 20:33, 9 August 2006 (UTC)

In 100,000 physics papers I searched, there are 2 mentions of this word. So it's a word for which there is no pressing need (physicists say "massless" when they need to lump gluons and photons together), and which isn't much used. Let's not use Wikipedia to give it any airtime, or soon it will creap into the language like some horrid particulate neologism the writers of ST:TNG made up, especially for use by Ensign Crusher to save the day with. One of the papers which used it had a scheme for making tachyons too, which lets you know how odiferous the associations are. And let us not forget tardyons, which sound like something from Dr. Who. Or maybe they are the guage particle which explain school lateness, something like Higgs do for inertia. Surely a grand unified theory of this is needed [2]. And bradyons as the field particle of bad 70's TV sitcoms.

But opinions differ. If you think the word is cool, start your own website for it. As for myself, you remind me to tag the luxon wiki article for deletion, unless somebody can come up with a physics literature reference for it. I'm sure one exists (so it will be kept, probably), but having to FIND it may teach luxon-boosters out there a lesson. SBHarris 21:45, 9 August 2006 (UTC)

Photon Creation Section Missing Critical Information

X-ray photons are created whenever a charged particle accelerates, a well-known process called synchrotron radiation. When we even have an article at Wiki devoted to this very subject, how could x-ray photon production by this well-known process get overlooked? I've added a comment in the "creation" section about the production of x-ray photons by synchrotron radiation. Also, photons are also created in particle collisions, also called particle annihilation. This isn't mentioned either but I hesitate to add a comment for it because we have a section in this article specifically labeled "annihilation" and I feel it may be confusing to some readers without knowledge of particle physics. What are your thoughts on how to include photon-production from different kinds of particle annihilation (there are many)? Thanks. Cheers, Astrobayes 22:33, 26 June 2006 (UTC)

One final comment on my addition: while I know the article *did* mention the acceleration of charged particles, my desire and intention is to illuminate that this is an x-ray photon production process that is so widely familiar and used that it has a name - and this information is important for anyone who has either not taken a modern physics course or who is not familiar with advanced electromagnetism or particle physics. My addition was not to suggest that the article did not contain a cursory reference to acceleration of charged particles but rather that the link between this general process, and the popular method of producing x-rays was highlighted. Cheers, Astrobayes 22:38, 26 June 2006 (UTC)

Go for it. Be bold in edits-- they can only get reverted by others if wildly unpopular. And if good additions, they stay. I recall once being given a tour of the LLNL national synchrotron light source, which is basically this big electron synchrotron in UC Berkley, run to do nothing but produce X-rays, which are tapped off in beamlines for various materials science projects which need them. Whole thing was run by one master 486 computer at the time. :) These days I'm sure it takes a supercomputer, at least. But of course you're right that synchrotrons are very important sources of high intensity coherant X-rays. The mechanism is acceleration of charge, but the frequencies obtaineds are due to odd and specific relativistic effects, as you know. So they are worth mentioning for that alone.Steve 23:25, 26 June 2006 (UTC)

Polaritons

Are photons in matter really polaritons? A polariton is the composite particle that results from strong coupling between a dipole carrying particle (usually a quasi-particle) and a photon. The characteristics of the polariton are a combination of the composite particles. So in the case of matter, this other particle would in general be an atom (or molecule). But the dispersion of a photon in a matter is not a combination of the atomic dispersion and the vacuum photon dispersion. Moreover, I've never heard anyone refer to a photon travelling through matter as a polariton (other than very particular situations such as in EIT, or in a cavity). Could someone provide a reference? --J S Lundeen 17:48, 22 July 2006 (UTC)

After asking around and discussing with my colleagues, I see how a photon could be referred to as a polariton in a material. For photons in glass, it is definitely not usually discussed in this manner, nor is the optical dispersion typically calculated in this way. But it doesn't look like it is incorrect (in fact it looks like a very general explanation, applicable to a wide variety of materials).--J S Lundeen 15:46, 25 July 2006 (UTC)

Dubious "in vacuo" statement

Strait removed some of this paragraph, saying it was dubious and uncited:

Since photons move at the speed of light, by relativist time dilation they do not take any time to get from their source to where they are finally absorbed; that is, they have zero lifetime but can travel arbitrarily far. The emission and absorption events are at zero space-time interval. From this point of view, first articulated by Gilbert N. Lewis in 1926, the photon's energy never exists in the vacuum, but transfers from the source to the absorber without delay.

And shortened it to:

Since photons move at the speed of light, no proper time passes for them. The emission and absorption events form an interval of length zero in space-time.

It's not clear to me from the short version just what it that would be considered dubious. As to citation, I've placed copies of these two papers temporarily on my personal web site [3]. Gilbert N. Lewis, "The Nature of Light," Proc. National Academy of Science, Vol. 12, pp.22–29, 1926; and Gilbert N. Lewis, "Light Waves and Light Corpuscles," Nature, Vol 117, pp.236–238, Feb. 13, 1926. I haven't searched to see exactly what statement I was paraphrasing above, and perhaps someone else would like to have a go at it to make it less dubious. Dicklyon 20:22, 3 August 2006 (UTC)

The dubious things were:

• "they have zero lifetime". First, the word "lifetime" is ambiguous. If it means log_e(2)*half-life, then it's definitely wrong. If it means "time that it exists" (which I think is what was meant), then it is only true from the photon's perspective, which was already stated.

Good point. I perhaps over-reacted to the definition I saw somewhere that said photons have infinite lifetime (since they don't decay). I figured that by analogy with how partical lifetimes are measured, which is in their own frame of reference, if a photon is to be assigned a half-life or a lifetime, then the value must be zero, or at least shorter than the lifetime of any particle traveling at a finite speed. In some sense, though, I guess I put it there to provoke an explanation. The "lifetime" in this sense is less than a cycle of the frequency.

Well, I think it is meaningless to assign the photon a lifetime (meaning log(2)*half-life). But FWIW, the Particle Data Group says that it is "stable" [4]. I agree with them that this makes more sense than saying it is zero if you are forced to pick one or the other. --Strait 02:46, 4 August 2006 (UTC)

Stable? If that means it doesn't decay, OK. But on the other hand, there is no time so short that that you can't find an inertial frame in which the photon lasts longer than that. That is, it lasts shorter than any particle does, if you pick a reference frame moving from emitter to absorber at close enough to the speed of light. It's not at all like a particle that is long-lived in the sense of continuing to exist in its own frame. Dicklyon 04:09, 4 August 2006 (UTC)

• "the photon's energy never exists in the vacuum". I don't know what that could mean. I did a slow skim of the papers that you linked to and couldn't find that statement.

It means that there is no finite region of space of and time that excludes the emitter and absorber than can be said to have the energy of the photon in it. I guess I read that into what Lewis said, too. But if you look at the uncertainty principle, with photons have a definite frequency, the positional uncertainty is infinite, so includes the emitter and absorber; the energy is in their states, even during their quantum transition when they are in a mixture of states beating at the frequency of the photon.

Errrmmm... while perhaps there is some sense to this viewpoint, it is certainly not how most physicists think about it. They would say that the energy of the photon certainly exists between the emitter and the absorber. Perhaps a discussion of this in the article would be ok, but it needs to present the more usual viewpoint first and clearly contrast the two. --Strait 02:46, 4 August 2006 (UTC)

Good idea. I think I'll work on a section such as "Photons as transactions" to contrast this view with the more "Copenhagan" view. Dicklyon 04:09, 4 August 2006 (UTC)

• "but transfers from the source to the absorber without delay." See the first point.

In any case, I think the interesting point in Lewis's papers is not this, but the issue of whether or not a photon can be emmitted without having a predestined absorber. --Strait 23:06, 3 August 2006 (UTC)

Indeed, that is the interesting point, except that the concept of "predestined" is not time-symmetric. Following his logic, the reversibility of all known physical laws would make your question equivalent to the issue of whether or not a photon can be absorbed without having a predetermined emitter.

I may not be expressing it quite right, but this idea of Lewis is what later became Feynman-Wheeler theory and Cramer's Transactional interpretation and Carver Mead's "Collective Electrodynamics", none of which disagrees with standard QED in any way that I'm aware of, but is a difference of interpretation that can provide some useful insights.

Dicklyon 01:28, 4 August 2006 (UTC)

Uncertainly relations

I'm a bit unclear on the uncertainty relations section. It says: "...quantum mechanics forbids the simultaneous measurement of the position and momentum of a particle in the same direction. Similarly, it forbids simultaneous measurement of the number of photons in an electromagnetic wave and the phase of that wave." I'm familiar with the first part, but how are phase and number of photons conjugate variables? It also constrains the uncertainty of the product of energy and time. Is it fair to say that since the photon's momentum and energy are definite and known values (in some situations, at least), that the position and time are completely unknown? Is there a sensible way to temper this somewhat absurd-sounding statement, or does it have some deep meaning? Dicklyon 01:26, 9 August 2006 (UTC)

See Coherent state#Coherent states in quantum optics. "Specifically, coherent light is thought of as light that is emitted by many such sources that are in phase." "Contrary to the coherent state, which is the most wave-like quantum state, the Fock state (e.g. a single photon) is the most particle-like state." "Mathematically, the coherent state $|\alpha \rangle$ is defined to be the eigenstate of the annihilation operator $a\!$." In other words, measuring the phase assumes a coherent state which requires changing the number of photons so that that number cannot be measured simultaneously. JRSpriggs 05:36, 10 August 2006 (UTC)

OK, I see the point now. But the way it is written, I don't think "the number of photons in an electromagnetic wave" makes much sense as a way to describe those incoherent states. At the least, come clarification is needed. Dicklyon 06:49, 10 August 2006 (UTC)

Making quantum mechanics clear is beyond my abilities. However, let me add that instead of "electromagnetic wave", one could refer to "a vibrational mode of the electromagnetic field". Instead of "photon", one could say "excitation of the mode" or "energy level" (actually ${\frac {E}{h\nu }}-{\frac {1}{2}}$). Instead of "phase", one could speak of "time since the beginning of a cycle divided by the period of the cycle" or "time since the beginning of a cycle times the frequency". So you see that this is equivalent to the uncertainty relationship between energy and time in the context of quantum field theory. JRSpriggs 04:07, 11 August 2006 (UTC)

Photon geometry question

(What the -- a post not to do with a photon's supposed mass!?) I am trying to understand how to visualize a traveling photon. Is it more accurate to view, from a very casual standpoint, a photon as (1) a point-like "cart" on a sinusoidal imaginary "track", that progresses along the track, that does not leave a trail (save an EM disturbance wake ?). (2) A ribbon whose length is finite and shaped sinusoidally. (3) A ribbon whose length is infinite and shaped sinusoidal. (4) Something else? ...Does the same apply to the virtual photon that mediates the electromagnetic forces? (see fundamental forces) I'm putting my money on concept (1). Thanks to anyone who can help out. --HantaVirus 21:17, 10 August 2006 (UTC)

You're going down a slippery slope toward nonsense here. Photons don't propagate. Propagation of electromagnetic effects is exclusively a wave phenomenon. Photons are the quantized phenomena that are apparent only where photons are emitted or absorbed. That's my position, and I'm sticking to it ... for now. Dicklyon 21:30, 10 August 2006 (UTC)

The uncertainty principle comes from the Fourier analysis of waves (any waves, for EM waves are included). If you know the photon's wavelength perfectly, it's a very LONG (infinitely long) sine wave. As your knowledge of its wavelength (momentum) gets worse and worse, you get a spread in possible wavelength values, and your summed sin wave becomes a superposition of various sin waves of various wavelengths (momenta), and now you get a wavepacket which is shorter, and better localized in space. So momenta (wavelength) and localization trade off. The better you know the one, the worse off you are with the other. So when you ask: "How big is a photon?" the answer is partly "How well can you characterize it?" SBHarris 21:41, 10 August 2006 (UTC)

(Please note removal of wikimarkup from header; made edit summaries un-clickable.)

I think you're maybe not fully following the OP's question. I believe he's visualizing the photon actually moving transversally along his posited "sinusoidal path" as it travels. No, that's not the right picture. The sinus curve corresponds to electric and magnetic field strengths, not to the path of the photon. Think of the photon as going in a straight line. It's a slightly fuzzy straight line because of the uncertainty principle, but it's not a sine wave. --Trovatore 21:46, 10 August 2006 (UTC)

My point though is that the photon does not have a path. Certainly not a straight line. Consider the two-slit setup. Can you show a photon's path through it? Certainly not. But you're right in that the sinusoidal path is not sensible. Dicklyon 21:57, 10 August 2006 (UTC)

Well, this is general quantum weirdness, not about the photon path per se. The photon in the two-slit experiment is a quantum superposition of many photons (or "possible photons", if you like the many-worlds interpretation) that do have well-defined paths, and those paths are straight lines. --Trovatore 22:00, 10 August 2006 (UTC)

There's no point trying to reify the paths used for QED calculations into photons. These straight lines are not photon paths, just paths over which you can calculator phases of wave functions. If you want to view it as a superposition of photons, you can, but it's too weird for my taste. That's why I prefer approaches such as Wheeler-Feynman electrodynamics; less weirdness, more waves. —The preceding unsigned comment was added by Dicklyon (talk • contribs) 05:10, 11 August 2006 (UTC)

Don't actually know Wheeler-Feynmen ED, but there's no way around the weirdness in general, or at least no easy way. Look up the Bell inequality and the Aspect experiment. I think you might as well grasp the nettle on the photon two-slit experiment, because nothing's going to save you from non-local-realism in the long run. --Trovatore 05:29, 11 August 2006 (UTC)

The version I'm more familiar with is Carver Mead's "Collective Electrodynamics". Like Wheeler-Feynman, it uses advanced and retarded wave symmetrically. Weirdness of the Copenhagen entanglement sort is replaced by things whose arrow of time goes backwards, like the absorption of a photon influencing the emission, possibly light-years away and years earlier. It's less weird, but of course still can't be local and causal. Cramer's transactional interpretation is a similar kind of thing. Dicklyon 05:57, 11 August 2006 (UTC)

This to you is less weird than Copenhagen? Matter of taste, I guess. --Trovatore 06:07, 11 August 2006 (UTC)

All good points. That sin wave thing is really a GRAPH of how E and B vary along a 1-D path for a 1-D wave. The illustrators steal the other two spacial dimensions for this purposes, and in the process cause no end of confusion (as in the illustration that tops THIS article). The localization along that 1-D line is constrained by the uncertainty principle, though, in both transverse and longitudinal directions. As for other than 1-D paths, EM fields propagate in many forms and although some of the classical fields are vaguely visualizable as 2-D snapshots of fields moving in 3-D (like the plane waves around an antenna), none of this really works very well for photons. A photons takes all possible 1-D paths straight from HERE to THERE, and you can always only pick only ONE of them (or at least just a few) to illustrate. And then you're stuck with using your extra dimensions in the usual way, unless you just want to show a ray-trace. SBHarris 22:09, 10 August 2006 (UTC)

Reference for application of relativistic mass to photon

So I'd been thinking about WillowW's challenge, and my eye happened to wander across a Scientific American compilation on "Particles and Fields" in my bookcase, and I wondered idly if I could find it there. Wasn't very hard:

The photon always travels at the speed of light (denoted by the letter "c"); it can never be at rest. Because of its motion it possesses energy. It therefore also possesses mass, according to the famous relation E=mc². Gell-Mann, Murray (1957). "Elementary Particles". Scientific American. {{cite journal}}: Unknown parameter |coauthors= ignored (|author= suggested) (help); Unknown parameter |month= ignored (help)

Now, strictly speaking, this is a "popularization"; no one is going to claim that Scientific American is a physics research journal. Still, when I hear "popularization", I think Asimov in F&SF, or maybe even Hofstadter, but not Gell-Mann in SciAm. I think this is good enough evidence for going back to my earlier version of the footnote, the one that spoke of relativistic mass for the photon as an "older notion ... now little used". --Trovatore 04:26, 21 August 2006 (UTC)

infinite half-life?

How can you tell, given that the photon experiences no duration? Isn't this a case of dividing 0 by 0? --Trovatore 22:14, 21 August 2006 (UTC)

Hi, Trovatore! I have to get back to my tomato canning, but I have time for a quick explanation. It's true, in the photon's own rest frame (if that can even be defined, which seems uncertain), the photon would experience zero proper time between its emission and absorption events. But in our oh-so-improper world, the photon does not decay spontaneously, to the best of our experimental knowledge. Hence, it is indeed a stable particle with an infinite half-life. Here's a reference:

• Perkins DH. (1987) Introduction to High Energy Physics, 3rd ed., Addison-Wesley.

As for the Gell-Mann reference, thanks! I'll definitely look it up; Gell-Mann is always a good read. But I'm still hoping for a reference in the scientific literature, like an article from Phys. Rev. It's too early to give up looking, don't you agree? Willow 22:35, 21 August 2006 (UTC)

Well, as to the first point, surely it would be natural to define half-life in the substance's rest frame. It seems a bit strange to say a photon has infinite half life just because it never has time to decay. It might also be a bit misleading -- I have the idea that a photon bigger than two electron masses can "decay", when it's traveling through a non-vacuum medium. No doubt there are other interpretations of that, but one kind of obvious one is that the photon slows down, experiences proper time, and therefore has a chance to split into electron and positron. --Trovatore 22:53, 21 August 2006 (UTC)

I changed the article before I read these comments, but I have to agree. It's fine to say it doesn't decay, but to say that it is stable or has an infinite half-life is unsupportable in the normal way such things are quantified. I know it's been said by respectable authors, but that doesn't make it so. We might as well just state it in the way that is more accepted, which is what I tried to do by saying it has no half-life and does not decay. Dicklyon 22:58, 21 August 2006 (UTC)

Consensus on what should be covered, inter alia

Hi all,

Somehow I had gotten the impression that we had agreed to cover only topics that pertain to photons per se, and not to light or electromagnetic radiation. If we're indeed agreed, I would recommend that we not cover the gravitational bending of light here, and leave that for the general relativity and light.

I moved the uncertainty relations into the section on the wave-particle duality, which I meant to do yesterday — sorry!

You're right about the helicity correction and the absence of editorial comments in the section titles; I merely wanted to alert readers that that section was "under construction" and not really coherent.

We may have some disagreement on virtual photons. My limited understanding suggests that the photon propagator is generally taken to be "on shell" (i.e., the invariant mass of virtual photons in vacuo is always zero) but that one can have two unphysical polarization states. Indeed, I seem to recall a theorem by Feynman that the unphysical polarizations are needed to preserve unitarity? I'll try to find the reference when I get a free moment; maybe it depends on the gauge? I also don't understand your comment about "negative kinetic energy". Do we agree that the four-momentum is conserved at every vertex of a QED Feynman diagram in momentum space? Willow 18:01, 22 August 2006 (UTC)

I too am confused. It's not clear to me how a virtual photon relates to an actual photon, or how/whether an actual photon can be said to travel at a speed other than c. From the photon's point of view, there is no medium, so no alternative to vacuum. Propagation of "waves" on the other hand can be said to vary their speed by interacting with matter, but in terms of photons, that's absorption and re-emission, is it not? Dicklyon 18:25, 22 August 2006 (UTC)

On one point: I'm not sure whom Willow is talking to, but I gather that by "invariant mass" you mean the magnitude of a momentum 4-vector?

Actually, virtual particles (including photons) are usually off-shell. I mean "usually" in two ways. First, if a virtual photon in a Feynman diagram has an undetermined momentum, then you integrate the propagator over off-shell momenta as well as on-shell. Second, when a virtual photon's momentum is determined by conserving momentum at one of its vertices, then the "mass" of the photon is not likely to turn out zero. For example, if an electron pair annihilates to a virtual photon, which splits to a muon pair, the photon's "mass" is not zero but the CM energy of the system.

Given the history of this talk page, I should point out that in my experience, no one uses the word "mass", invariant or otherwise, to refer to a dynamical property of virtual particles, so don't nobody get any weird ideas! Melchoir 18:36, 22 August 2006 (UTC)

You're definitely giving me weird ideas. Why are you talking in terms of mass if those aren't the right terms? And is it the case that photons as used in Feynman diagrams are the same as the photons that this article is mostly about, or is that a made-up (virtual) construct for working problems in particle interaction? Dicklyon 18:45, 22 August 2006 (UTC)

Because Willow did, and it seemed appropriate to answer the question in the same language. This article is not (mostly) about virtual photons. Melchoir 18:55, 22 August 2006 (UTC)

Thanks muchly, Melchoir! I tried to amend the article accordingly. I never got much further than the $\phi ^{4}$ theory, and even that was kind of vague and hazy for me. The $e^{+}e^{-}\rightarrow \gamma \rightarrow \mu ^{+}\mu ^{-}$ diagram is a wonderful counterexample; thanks for clearing up my confusion! Hoping to see more of your contributions, Willow 20:58, 22 August 2006 (UTC)

P.S. Sorry, I should've clarified that I was speaking mainly to JRSpriggs, who made a lot of edits last night. But I also think we should keep up a sense of community and of working towards a common goal (improving the Photon article), so I was also inviting everyone into the conversation.

Ah, shucks. Melchoir 21:11, 22 August 2006 (UTC)

Multiple instances of footnote

Willow, why do you object to the footnote appearing in a couple of places? There's this gorgeous mechanism and it seems a shame to waste it. It's also much less distracting than multiple wikilinks to things like physics and light that you seem to be quite fond of. --Trovatore 22:34, 22 August 2006 (UTC)

Of course it's not really just about having a hammer and looking for a nail. The reader who skips the intro and starts in the middle may still need to see that there's an asterisk on the claim that the photon is massless; that may not have the intuitive meaning the reader has in mind. --Trovatore 22:41, 22 August 2006 (UTC)

... and perhaps more to the point (given that this is a reference work rather than a teaching tool) conflicts with an attested, though older, usage. --Trovatore 22:51, 22 August 2006 (UTC)

Hi Trovatore,

I appreciate your concern for readers who might have a false impression of the sense in which photons have zero mass, and who skip the introduction completely. However, please consider these two points:

• such readers seem likely to be a relatively small fraction of our readership; and

• the disadvantages of repeatedly calling attention to the concept of relativistic mass might outweigh its advantages.

I hope we agree already on the first point, so it doesn't bear much discussion. Point two is more critical, so let's focus on that.

My concern is that repeatedly calling attention to the relativistic mass of the photon gives that concept more "air time" (and, hence, more legitimacy) than it deserves. That concept is undeniably useful in popularizations for explaining why massive particles cannot reach the speed of light. However, I hope that we agree that it has been roundly denounced by working physicists from as early as 1921 to the present (see the FAQs and references cited above) and has never been a mainstream concept in physics research since 1906.

Can we quantify how much "air time" the concept deserves? Fermi problem! Let's assume that the published research articles in Physical Review and Phys. Rev. Lett. are a representative sampling of the field of physics. The concept of the photon is roughly 100 years old, and let's also assume that, over the century, roughly 10,000 pages of scientific articles have been devoted to it per year (1000 articles per year, 10 pages per article). That amounts to one million pages of scientific literature. In that statistical sample, we have not yet identified even one page that applies the concept of "relativistic mass" by name to the photon, whereas the competing concept of invariant mass appears in a very high fraction of them. Treating it as a Poisson process, the probability of relativistic mass appearing on the Wikipedia photon page should be less than 10^-6.

Thus, to my mind, one reference to relativistic mass suffices to fulfil our obligations for "encyclopedic coverage". To cite it more often seems risky, since it seems likely to misrepresent the consensus view of working physicists.

However, this is only my perspective. I'm open to other ideas and we should reach consensus as a community, in a friendly way. What do other people think? Willow 15:26, 23 August 2006 (UTC)

I think your perception that it has "never been a mainstream concept in physics research" is probably not true. If that were so, Gell-Mann would not have used it, even in a SciAm article. I frankly feel that your hostility to the usage is undue.

As for your analysis of the literature: Permit me to doubt that you would necessarily have found such a reference if it were there. It's not an easy task to sort through old papers looking for something that would not likely have been called out by specific keywords; physicists writing in research journals would assume that their readers would understand them, whichever convention they used.

The traditional definition of mass is "quantity of matter", which is quite obviously not a good description of invariant mass, for the simple reason that the latter is not additive. For the "amount of stuff" to depend on the observer's reference frame is counter-intuitive, but in a way we've had to get used to in relativity. For the amount of stuff in two objects taken together to be different from the sum of the amounts in the objects separately; well, that's just out of bounds. Now, granted, there are measures that are both additive and frame-independent (say, baryon number), but they do not correspond to the classical notion of mass in the classical limit. There is only one additive measure that does, and it's relativistic mass, and you have to count photons or it's not additive anymore. --Trovatore 15:46, 23 August 2006 (UTC)

Wow, that's a little surprising. Perhaps my letter was interpreted more strongly than I intended. I can happily say that my personal feelings (hostility or otherwise) are not engaged on the question of whether we footnote relativistic mass once or twice in this article. We're all just trying to make a good article.

I think we should discuss it maturely as a group and appeal to concrete references. I've offered some data that suggest that the application of relativistic mass to massless particles has never been a mainstream concept in physics research. Whether this is true is a well-posed question with a definite answer that will influence our presentation. Some key points that support the idea that it was never mainstream:

• Einstein and Planck didn't use the concept, Pauli, Sommerfeld and Dirac didn't use it, and Feynman and Dyson didn't use it, to the best of my knowledge. Apparently, there has never been an era when the concept was used in physics research.

• It appears in no textbooks intended for training physicists that I surveyed.

• I have personally never seen it in the scientific literature, to the best of my memory.

• Even for massive particles, the concept is deprecated explicitly; "Ouch!" is the expression in the French and Taylor textbook. Why should we help propagate such concepts to Wikipedia's readership, if they're considered so misguided by working physicists?

However, I am all too aware of my limited training in physics and I would be perfectly happy to be informed better by someone who can cite scientific literature. I encourage everyone here to scour the literature for such references. However, Scientific American is not scientific literature, as I think we agree, since working physicists do not publish novel research there. Looking forward to others' comments and reaching consensus, Willow 16:32, 23 August 2006 (UTC)

JRS Edits of 23 August 2006

Dear JRSpriggs,

Thanks very much for your tireless work in improving equation formatting and catching oversights. On my own screen, the equations render fine even without the "\!", but I can well imagine that it might be different for other readers. You are completely correct, too, about my having overlooked "polarization" as a parameter; I had been thinking only of "continuous parameter", and missed an important contribution.

I agree with your change of symbol for the invariant mass from $m_{0}$ to $m$; I had written that myself originally, but added the zero subscript to avoid more confusion about mass in special relativity.

I hope that you'll agree with my reversion back to the original (correct) equation for the four-momentum $p^{\mu }=(E,\mathbf {p} )$. The contravariant form $p^{\mu }$ is more common than the covariant form $p_{\mu }$, and has the advantage of not needing to explain the minus sign. I also clarified that virtual photons may have up to four polarization states, which is my understanding; in the customary transverse gauge, the two unphysical polarizations are longitudinal (along the direction of photon propagation) and time-like, respectively. These four polarization states correspond to the four components of the potential $A^{\mu }$. But I would be happy if an expert would improve my understanding!

Please consider joining our discussion here. It would be nice to have your thoughts on the article. Willow 09:50, 23 August 2006 (UTC)

No. Actually your equation for momentum is NOT correct. First, p^0 = E/c^2, not E. Second, you should not be using the contravariant form. The right-hand side of the equation is covariant (being the gradient of the phase), so the left-hand side should also be covariant.

No again. Photons, like all elementary particles, are irreducible representations of the universal covering group of SO(3), i.e. they are spinors (not tensors). A spinor of rank k has k+1 components. Since spin 1 particles, like the photon, have rank 2, they must have 2+1 = 3 components. It is similar to the m quantum number of an atomic orbital when the l quantum number is one. Another way to see that it is three rather than four is to realize that the four components of the vector potential are reduced by one because of the need to choose an arbitrary gauge (which fixes one of the components in terms of the others). When moving at c, the number of components is further reduced by one because of the degeneracy of the longitudinal direction of space with time.

The mass of the photon is zero always, even for virtual photons. What changes is that the equation (m·c²)² = E² - (p·c)² is no longer exact. The more the two side differ from each other, the faster the probability amplitude decreases over distance or time.

Now, will you have some more respect for the work of other contributers instead of acting like you own this article? JRSpriggs 08:31, 24 August 2006 (UTC)

I'm really, really sorry if I've come across as high-handed to you and, presumably, other people. I'm the first to admit my imperfect physics training, and I really have striven to keep the atmosphere here more friendly and cooperative, more consensus-y. I do respect the contributions of other editors. Admittedly, I've changed the article significantly over the past week or so, but I've mainly been re-arranging or re-wording earlier contributions to improve the flow of the writing. For example, the "radiation pressure=photon momentum" idea was previously in two places in the article, neither of which was under a really appropriate section. It's still not perfectly integrated, but I hope that people here feel that the article is better for my work. The Photon article is of "Top" importance, and I would like the article to be good enough so that even a Physics professor at M. I. T. wouldn't cringe at it. That means good coverage, good writing, good references and accurate formulae, criteria that I'm sure we all agree on.

Turning to the other issues, you're right about p⁰; I was taking $c=1$ to keep the formula simple, but that's inconsistent with the earlier formulae. However, I think it's $p^{0}\equiv {\frac {E}{c}}$, not ${\frac {E}{c^{2}}}$, as you can see by dimensional analysis. Regarding the covariant-contravariant question, I understand your point of view, although the other is equally valid; a contravariant 4-gradient is possible, just by taking the derivative w.r.t. a covariant vector. It's really not important, though, so let's just leave it as is.

The question of the virtual photons having invariant mass seems to be just semantics. The essential idea is, as you say, that the four-momentum magnitude (which, by the usual definition, is proportional to invariant mass) is no longer zero. That was the essence of User:Melchoir's excellent example. I'm not picky about how we phrase it; my wording seemed shorter, but also more prone to misunderstandings.

Finally, I'm aware of the spinor representation of particles and I understand that fixing the gauge reduces the number of free components in $A^{\mu }$ to three. However, I have consulted several modern books on quantum electrodynamics, and they seem to include all four polarizations in the summation: the two physical polarizations, as well as the unphysical longitudinal and time-like ones. Perhaps an expert can clarify this for us? Willow 10:44, 24 August 2006 (UTC)

Well, perhaps we are beginning to converge on an article which we can both accept. I only had to fix one thing this time.

Your dimensional analysis of p⁰ and p₀ is overly simplistic. The components of a tensor will not, in general, have the same units. That would only happen if we gave all the four coordinates of space-time in the same units, e.g. either give time in meters as well as space or give space in seconds as well as time or give both of them in arbitrary (pure number) units. If we follow the International System of Units as I think we should and give time in seconds and space in meters, then the units of a component of a tensor are obtained by: (1) beginning with some units characteristic of the physical nature of the tensor as a whole; (2) multiply by a factor of "second" for each contravariant time index; (3) multiply by a factor of "meter" for each contravariant spatial index; (4) divide by "second" for each covariant time index; and (5) divide by "meter" for each covariant spatial index. The covariant momentum vector has physical character of action, i.e. kg meter² second^-1. In the spatial directions, we divide by meters to get units of momentum, i.e. kg meter second^-1. In the time direction, we divide by seconds to get units of energy, i.e. kg meter² second^-2.

A contravariant gradient is only possible if you are taking the gradient with respect to a covariant space, e.g. in momentum space rather than normal space. But the frequency and wave vector are obtained by differentiating phase by normal time or space. So it is covariant. And it is important to get these things right.

Mass is NOT defined as the length of the momentum 4-vector. It is an intrinsic and constant property of the KIND OF ENTITY in question. The reason that it is usually very nearly equal to the length of the momentum 4-vector is precisely for the reason that I mentioned above. The amplitude for the particle to exist decreases rapidly to zero as the two sides of the equation get further apart. This is not just semantics. If you were right, the photon or any other kind of particle could willy-nilly change its mass to whatever. But that does not happen.

I cannot comment on what your textbooks say about polarization without seeing them.

Sorry, I thought "wave number" meant what you call "wave vector", rather than just its magnitude.

In one of your earlier comments, you said that we should focus on photons per se, rather than light generally. I took this to mean that you want this article to emphasis the particle nature of light and not delve into its wave nature except where they meet. Since the momentum vector and energy are particle attributes and the wave vector and angular frequency are wave attributes, I thought we should emphasis the former in the introduction rather than the latter. JRSpriggs 08:56, 25 August 2006 (UTC)

Please calm down. WillowW has been incredibly calm and constructive (and respectful), yet people insist on being hostile about this article. There is no need for this.

Mass is defined through the norm of the momentum 4-vector. That is why they call it invariant, because all frames agree on the value (not because it is just defined as some constant and is therefore invariant). How do you wish to define mass? Consider for instance the mass of the top quark, where the lifetime is short enough that the distributions in mass are quite evident in the experimental data. To say they all had the same mass would be to ignore the quantum nature of it. Willow is quite diplomatic and called your complaint a semantic issue. If you'd wish to propose an alternative definition, maybe we can just leave this as semantics.

As for the four vectors, I have always seen the components listed as dimensionally consistent, such as (ct,x,y,z) or (E/c,p_x,p_y,p_z). I guess you could do it another way, but it is non-standard and would require the components of the metric to have units (which I feel misses the point). Many books just call c=1 for simplicity, but the ones I've seen that do carry it around (Griffiths for example), they have the components of a four vector dimensionally consistent (and even Wikipedia follows this standard convension, for example look at the Electromagnetic field tensor).

If we're having issues on definitions or conventions, just say so and we can cite sources and sort it out. But in the mean time, if there are any misunderstandings, please assume good faith and work together. -- Gregory9 10:09, 25 August 2006 (UTC)

I appreciate the supportive letter very much but, really, the article is bigger than any of us and there shouldn't be any reason for personal feelings to become involved. Let's just consider what the evidence is, and what would be most intelligible for our readers.

I was surprised by JRSpriggs' idea that four vectors need not be dimensionally consistent. That would seem to suggest that the flat space-time metric tensor $\eta _{\alpha \beta }$ should be (-c^{2}, 1, 1, 1) instead of (-1, 1, 1, 1) or its opposite (1, -1, -1, -1), which are what I remember. For example, the invariant of the position four-vector is always defined as

$ds^{2}=x^{\alpha }x_{\alpha }=x^{\alpha }\eta _{\alpha \beta }x^{\beta }=-c^{2}t^{2}+x^{2}+y^{2}+z^{2}$

or its opposite. If we choose $x^{\alpha }=(t,x,y,z)$ instead of the dimensionally correct $x^{\alpha }=(ct,x,y,z)$, we have to modify $\eta _{\alpha \beta }$ to include the c in the metric. That's certainly possible, but that is not done in any of the references I checked (listed chronologically):

• Bogoliubov NN and Shirkov DV (1959) Introduction to the Theory of Quantized Fields, Interscience, p. 8.

• Weinberg S (1972) Gravitation and Cosmology, Wiley, p. 26.

• Landau LD and Lifshitz EM (1975) The Classical Theory of Fields, 4th ed., Pergamon, pp. 27-28.

• Jackson JD (1975) Classical Electrodynamics, 2nd. ed., Wiley, p. 575.

• Goldstein H (1980) Classical Mechanics, 2nd ed., Addison-Wesley, pp.288-293, 309-311.

• Ryder LH (1985) Quantum field theory, Cambridge University Press, pp. 27-29.

I won't revert immediately, though, since JRSpriggs may have some references that escaped my attention in which the spacetime metric is defined differently.

The two QFT references above (B&S and Ryder) both refer to longitudinal and time-like photons. I'll wait a little while longer on this as well, before reverting to four virtual photon polarizations. As an aside, the Ryder reference (a standard QFT textbook for physics graduate students) also clearly shows the contravariant form of the 4-momentum (Equation 2.11 on page 29) — with no apologies. ;)

Let's stay light-hearted about ourselves and serious about the article, Willow 19:03, 25 August 2006 (UTC)

I am calm. I am trying to focus on improving the article rather than personalities. I am sorry, if I gave a different impression and offended anyone.

Would you agree that the momentum and energy of a particle are not the cause of its mass? The mass appears as a constant in the relativistic energy-momentum equation, not as something to be determined by it. If anything were to be determined by the equation, it would be the energy. Thus it is inappropriate to call that equation a definition of mass. However, you are certainly correct that mass is an invariant. But I do not think that we need that equation to tell us that. I cannot define the mass of a type of particle in terms of anything else any more than I could define $\hbar$ or c in terms of something else. The particle masses are just physical constants which we have to obtain experimentally.

See "E.J.Post, Formal Structure of Electomagnetics: General Covariance and Electromagnetics, Dover Publications Inc. Mineola NY, 1962 reprinted 1997" for an explanation of how and why to calculate the units of the components of a tensor. One reason that many people use "natural units" is to avoid doing this, perhaps because they do not understand how to do it.

Yes, Willow. In SI units the metric of special relativity is:

$\eta _{\alpha \beta }={\begin{pmatrix}-c^{2}&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}}.$

For support of the idea that spin 1 vector bosons (such as photons) have three spin components, see the articles Spin (physics) and Vector boson. The article Pion in a section on isospin (which is not spin, but is mathematically similar to it hence the name) says:

$SU(2)\otimes {\overline {SU(2)}}\approx SO(3)\oplus U(1)$

The point is that if one combines two spin 1/2 particles each with two components together into one composite particle, the result can be either a spin one particle with three components (if the spins are parallel) or a spin zero particle with one component (if the spins are anti-parallel). So the fourth component would be a distinct particle, not a photon. Just as when you combine a quark and an anti-quark to get either a rho meson spin 1 or a pion spin 0. JRSpriggs 06:45, 26 August 2006 (UTC)

So this isn't an issue of semantics. You actually believe invariant mass is invariant because it is just a defined constant and not because it is the norm of a four vector. Please answer the following questions to clarify your suggestions for this article:

1] Do you at least acknowledge that this is a non-standard claim?

2] In GR, the norm of the momentum four vector can be related to the gravitational mass. If you want to define the "mass" as some constant (instead of related to the norm of the momentum four vector), your definition will violate the equivalence principle. Your definition seems to be inconsistent. What simplifications / conceptual advantage if any does your definition provide over the standard one?

As for the metric:

1] Do you at least acknowledge that your claims here are also non-standard (considering all the sources cited by Willow and I, as well as the current consensus across Wikipedia articles/editors)?

2] Do you understand our complaint regarding the components of the metric having units?

3] Are you suggesting wikipedia articles need to be changed to match your convention, or only that comments should be added at appropriate places to note that other conventions are possible?

As for the polarizations:

I spoke to a theorist today, and he agreed with Willow that all polarizations (even unphysical ones) are integrated over for a vertex with a virtual photon.

1] Does everyone agree on this at least?

If we agree on that, then this seems only a question of how many total polarizations there are for a photon (including unphysical ones).

2] Do we agree this is the issue?

I believe spin zero is possible for a virtual photon. Consider the coulomb interaction mediated by virtual photons. No spin need be exchanged between the charges, so the virtual photon can have spin zero. I admit I do not feel very confident on the intricacies of this last subject, however Willow cited a source which seems to confirm her statements (and my rough understanding).

Hopefully your answers to these questions can put us on solid understanding of everyone's position, and how our conflicting conventions should affect (if at all) the formulas and phrasing in the article. -- Gregory9 08:49, 26 August 2006 (UTC)

(Edit conflict, so I inserted this out of order.)

Mass 1: No. The physics texts I have read implicitly use m as a constant. They do not define it in terms of something else. In "Relativistic Quantum Mechanics" by Bjorken and Drell, they begin on page 4 talking about the Hamiltonian as being equal to p^2 / (2 m) in classical physics and then move onto defining it as sqrt(p^2 c^2 + m^2 c^4) in relativistic physics. See they are defining energy in terms of mass and momentum, not the other way around.

Mass 2: "Gravitational mass" is an over simplification. Actually, the source of the gravitational field is the Stress-energy tensor. See Einstein field equations. I do not see how you figure that I am violating the equivalence principle. The advantage is not giving a false and misleading "definition". On page 95 of Bjorken and Drell, to get the propagator, they integrate in 4-momentum space over an expression which has p^2 - m^2 + i&epsilon (with c=1;h=2pi); in the denominator where by p^2 they mean the square of what you are calling mass (the squared "length" of the 4-vector).

Metric 1: In general relativity you are free to use any curvilinear coordinate system. The usual practice is to just ignore units altogether. But the components of the metric tensor vary all over the map. They have signature -,+,+,+ and are thus non-singular and the tensor is symmetric and smooth (infinitely differentiable). If one is going to use units (as one does in special relativity), why not use SI units?

Metric 2: No. Why do you find it objectionable for the metric components to have units?

Metric 3: I am saying that we should use SI units. That implies that the time component of covariant momentum is -E. Do you not think that it is confusing to be using different systems of units in different articles? Or even in different places in the same article? Even different places in the same equation -- since x^0 and t are being used with different units even though they are both time?

Polarization 1: Yes. We integrate over all polarizations.

Polarization 2: Yes. The issue is how many polarizations are possible for a vector boson such as the photon. It is two when going at c (helicity = -1 or +1), and three otherwise (-1, 0, +1).

I hope that this reply will bring us closer to resolving these questions. JRSpriggs 10:24, 26 August 2006 (UTC)

Rereading your post, I think I understand your mass comment better now (and it appears it is just a semantics issue). What you refer to as the invariant mass, I would call the on shell mass (the mass parameter that appears in the propagator). Whereas what I call the invariant mass, you refer to simply as the norm of the momentum four vector. The essence of the difference in our views is whether we state that E² - p² = m² holds true (and m varies) or that the equation can be violated (and m is constant). I found your view briefly mentioned in one of my text books, and it finally dawned on me what you meant.

So it is a semantics issue. Which is the standard interpretation? I only have my undergrad text book (Griffiths), which isn't a large enough sampling to really let me state there is "a" standard interpretation for that. You two seem to have a larger library on this subject, so I'd be interested in hearing how your texts present this issue. -- Gregory9 10:13, 26 August 2006 (UTC)

Hi, I have family visiting for the next few days, and I still have to clean up the house, so this will have to be short and I might not be able to work on the article for the next few days. Sorry!

I think we can leave the mass issue. The present wording is fine by me; I kind of like the fact that it doesn't mention mass, since that can be confusing — as we have all seen!

The chief physics question remaining seems to be the number of polarization states of virtual photons. We all agree that real particles of spin 1 have three possible components.

Speaking for myself, I prefer the (-1,1, 1, 1) metric, since that is most common in physics textbooks and here at Wikipedia. Look at the definition of four-momentum, for example. I think dimensional consistency is less confusing that sticking with conventional units like seconds for $x^{0}$. Perhaps, though, we should ask at WikiProject Physics for a consensus?

Where to go from here? I think we should all think about the "Photons and Matter" section at the end. Maybe mention Brillouin and Raman scattering and multi-photon processes? It could be a cool conclusion! :)

Talk to you all soon, Willow 12:01, 26 August 2006 (UTC)

Experimental proof?

I notice the article never address experimental proof of quantized light (photons). To my knowledge, the first true proof was: P. Grangier, G. Roger, and A. Aspect (1986). "Experimental Evidence for a Photon Anticorrelation Effect on a Beam Splitter: A New Light on Single-Photon Interferences". Europhysics Letters 1: 501-504. Should this be added somewhere? Waxigloo 08:58, 24 August 2006 (UTC)

The photoelectric effect and Compton scattering aren't experimental proof? Melchoir 09:11, 24 August 2006 (UTC)

The photoelectric effect is definitely not proof as it can be explained with a semi-classical model where atoms are quantized and the light is not. I am fairly certain Compton scattering is derived without quantized light. Any experimental evidence of a photon should rest upon a characteristic of light that requires the second quantization (since that is what the photon is...quantized light). I am pretty sure the reference I sight is the first evidence that can't be explained semi-classically; the only other thing as evidence of quantized light is the Lamb shift. Waxigloo 20:09, 24 August 2006 (UTC)

The Lamb shift brought about the birth of modern QED. I guess if you want to call QED the first truly quantum theory of photons, there's an argument to be made. But the distinction between light being quantized in its interactions or by itself is, as far as I know, an anachronism at best. And if you want to focus on second quantization, what's wrong with the Planck spectrum?

Anyway, what do other physicists write about the experiment you mention? Melchoir 20:38, 24 August 2006 (UTC)

If we accept that the electron has no internal structure, then the Compton experiment cannot be explained semi-classically, isn't that right? A perfectly free electron should radiate light of the same frequency.

A simple detector for individual photons might be a photomultiplier tube, or maybe a Geiger counter/scintillation detector for gamma rays. I don't know when those were invented, though — maybe the 1940's?

If I remember correctly, the first experiment to show that individual photons can exhibit interference was done in 1908 by Geoffrey Ingram Taylor. Basically, he did the classic two-slit experiment at ever lower intensities and saw no change even when only one photon at a time was passing through the apparatus. I'd need to look it up, though, to find the details and check whether he was really looking for photons.

Hope that this helps, Willow 21:15, 24 August 2006 (UTC)

I don't think any of that helps a whole lot, as it's all proof within the current paradigm. The inadequacies of these proofs won't be evident under we got to the next paradigm. Personally, I think it will happen, but not soon. The immune reaction of the physics establishment to such things as the transactional interpretation is too strong. Dicklyon 21:29, 24 August 2006 (UTC)

I think you are right about Compton scattering. But photomultipliers and Geiger counters don't demonstrate that photons exist: they click even when classical light is incident upon them. Same with a two-slit experiment: simply attenuating classical light does not prepare light in a state requiring the quantization of light. These two items give the same results as classical EM. To prove the existance of quantized light you need a state that acts differently than a classical state. Indeed, there is a big difference between the coherent state $\ket{\alpha}$ and the Fock state $\ket{1}$. To my knowledge, the reference I gave is the first experiement showing both the anti-correlation effect of a single photon at a beam splitter as well as the first single photon interference with states of light that were actually prepared as a single photon state and not attenuated classical light. Waxigloo 02:57, 25 August 2006 (UTC)

I think it would useful and interesting to have articles (or sections) on those experiments so that we understand them without finding and reading the papers. Dicklyon 03:16, 25 August 2006 (UTC)

Please explain what you mean by "classical light". How can you know (experimentally) what it would do, given that we cannot turn off quantum mechanics to see what would happen "classically"? JRSpriggs 07:46, 25 August 2006 (UTC)

Classical light is any state of light who's quasi-probability distribution in the Glauber-Sudarshan P-Representation behaves like a probability distribution, i.e. always non-negative and no more singular than a delta function. Any light that is classical in this sense can be explained by classical EM. Waxigloo 10:38, 25 August 2006 (UTC)

To help explain my point of view, I refer to this quite accessible paper on the quantum nature of single photons. The introduction gives a good summary of what I am trying to express. Waxigloo 11:39, 25 August 2006 (UTC)

Waxigloo is correct. The first clear evidence for the quantization of the EM field (as opposed to a quantization of matter-field interaction or a semi-classical model) was the Lamb-shift. The photoelectric effect, compton scattering and blackbody radiation were all evidence that a photon model was useful but they didn't strictly require it. The first clear evidence for photons as particles was the Kimble experiment in 1977 which showed antibunching in a Hanbury-Brown Twiss style measurement. Other similar experiments by Aspect and Clauser should also be cited as they were the first to actually create a beam of light containing a single photon (within some experimental caveats). This stuff should definitely be in the article. Here is a reference: Phys. Rev. Lett. 39, 691–695 (1977) Photon Antibunching in Resonance Fluorescence, H. J. Kimble, M. Dagenais, and L. Mandel. --J S Lundeen 14:38, 25 August 2006 (UTC)

And here is another: Phys. Rev. D 9, 853–860 (1974), Experimental distinction between the quantum and classical field-theoretic predictions for the photoelectric effect, John F. Clauser. I'm not sure why this experiment isn't usually given as the first experimental evidence given that it was published before the Kimble paper. --J S Lundeen 14:46, 25 August 2006 (UTC)

Oh and 'Classical light' is any state of light whose properties can be perfectly modelled as combination of oscillating electromagnetic waves. A laser beam can be modelled in this way. A beam containing a single photon can not.--J S Lundeen 14:52, 25 August 2006 (UTC)