What is wave of photon? oscillation of something in space?

[pastelchu]

I am curious what a photon looks like if we could observe it pass through space. It's also supposed to be oscillation of electromagnetic field. I don't understand what that means. Is it the field gets stronger and weaker and stronger and weaker, or spreads out and shrinks etc etc?

Or if it's supposed to be oscillation of a particle, wouldn't that mean that particle, already traveling at speed of light, is moving extra distances (amplitude of the wave) in space?

This is supposed to be a depiction of a photon wave packet. To me it seems like oscillation of a particle in space.

http://www.nist.gov/pml/div684/fcdc/quantum-physicstheory.cfm

This is how I understand frequency and wavelength equation. Higher frequency means higher energy, right? Because with more energy, you can cover more distance by going up and down more. So higher energy means higher frequency, and shorter wavelength. Am I wrong?

And how can we see it as a wave, if photon experiences length contraction. It should become infinitely small on its side length.
And if the photon also experiences time dilation, we should see it paused in motion because time is not moving at all. So how can we see it oscillating as a wave in space when it passes by?

 

[Drakkith]

A photon is the quantized interaction of an EM wave with matter. This means that an EM wave only interacts and transfers energy in small "packets" of energy. This energy packet should not be thought of as a physical object with a size. If we send out EM waves with just enough energy for one photon, this photon will have a chance of being found anywhere in the wave.

A photon is NOT a particle traveling in an oscillating motion through space. The diagrams you commonly see represent the oscillations of the electric and magnetic field vectors of the wave, not the motion of something through space. In other words, the diagrams you see tell us which direction the electric and magnetic forces will act in and how strong they are over time.

It's best to realize that quantum physics does NOT do away with an EM wave and replace it with a photon, but supplements the classical EM wave with quantum mechanical rules telling us how the wave interacts at the subatomic level.

 

[bhobba]

Just to elaborate a bit on what Drakkith said see our FAQ's:
https://www.physicsforums.com/showthread.php?t=511178

Thanks
Bill

 

[DParlevliet]

Or more basic: nobody knows yet what a photon is, so one cannot describe how it passes through space. We only know that if we measure it with classical instruments that it shows wave- and particle properties. So you can calculate with formulae, but must stick to that.

So you can calculate what the wave packet is (http://en.wikipedia.org/wiki/Wave_packet), but you cannot say that the photon looks like that.

 

[jostpuur]

Everytime I have attempted to discuss the wave functions of relativistic particles, the physicists change the topic to the creation and annihilation of particles and the particle collisions. So it seems that mainstream physicists do not yet understand the wave functions of relativistic particles. When a scientist doesn't know something, he (she) of course has the responsibility to know that he (she) doesn't know. The physicists of today are not doing very well with this responsibility.

 

[bhobba]

Or more basic: nobody knows yet what a photon is,

What exactly do you think QED doesn't tell us?

Thanks
Bill

 

[bhobba]

Everytime I have attempted to discuss the wave functions of relativistic particles, the physicists change the topic to the creation and annihilation of particles

That's because quantum fields turn out to be described by creation and annihilation operators - eg see section 3.3:
http://www.stfc.ac.uk/ppd/resources/pdf/quantumfieldtheory09.pdf

I have zero idea why you haven't gleaned this before - its pretty basic.

Thanks
Bill

 

[DParlevliet]

What exactly do you think QED doesn't tell us?

What a photon physically is, anyway according Copenhagen interpretation.

 

[jostpuur]

That's because quantum fields turn out to be described by creation and annihilation operators

I know that quantum fields can be described by creation and annihilation operators, but I also know that physicists of today don't understand the wave functions of relativistic particles. You see, I cannot be manipulated by the change of topic like many others.

 

[bhobba]

What a photon physically is, anyway according Copenhagen interpretation.

Its the excitation of the quantum field.

As per a previous post quantum fields are described by creation and annihilation operators. These in turn have number operators that give the number created and annihilated. Like any other operator in QM that is an observable they describe an observation - in this case the number of photons.

Scratching my heard what the issue is.

Thanks
Bill

 

[bhobba]

I know that quantum fields can be described by creation and annihilation operators, but I also know that physicists of today don't understand the wave functions of relativistic particles. You see, I cannot be manipulated by the change of topic like many others.

The point is the topic is not changed. A Quantum Field is exactly the same as creation and annihilation operators (it comes about by quantizing the Fourier transform of the field) - precisely what don't you get about the mathematics of the link I gave that proved it?

As it said:
'We thus find that the Hamiltonian and the momentum operator are nothing but a continuous sum of excitation energies/momenta of one-dimensional harmonic oscillators!'

So a mathematical analysis of the quantum field shows they 'are nothing but a continuous sum of excitation energies/momenta of one-dimensional harmonic oscillators!''

This is the so called Fock space you hear people talk about in QFT. It is this space that contains the the 'particles' of QFT.

Just as a general comment I really never understand people that because something alludes their understanding why the response is what they are having difficulty with is in error instead of the issue being within themselves.

I read a lot of physics books and I often don't understand stuff, but I understand it is VASTLY more likely that I have made an error than the people that wrote the textbook. It occasionally happens but it is VERY rare. So I leave it for a while and usually it resolves itself. If it doesn't I do a post to a forum like this and it gets sorted out. I leave it as the option of last resort that I am right and everyone else is wrong. In my entire time of doing this I can count on one hand when I am correct - and in those cases its always pointed out in more advanced texts that yes the more elementary treatments are in error - but that occasionally is the way of basic treatments - they gloss over some things.

Rest assured with QFT its the end of the line - its not fixed up by something more advanced. And also note that QFT is HARD - as the guy who I quote in my signature also says - one finds humility in field theory. Its very true.

Thanks
Bill

 

[jostpuur]

I am curious what a photon looks like if we could observe it pass through space. It's also supposed to be oscillation of electromagnetic field. I don't understand what that means. Is it the field gets stronger and weaker and stronger and weaker, or spreads out and shrinks etc etc?

I have wondered the same thing, and I'm capable of putting forward more precise questions conserning the same topic. Firstly I would like to ask that what kind of mathematical mapping is used to depict the photon's wave function. For example, is it $(t,x)\mapsto\psi(t,x)$, $\mathbb{R}^4\to\mathbb{C}$? I don't believe that that's right. A guess $\mathbb{R}^4\to\mathbb{C}^3$ seems more likely correct. Can somebody verify or correct on this? Does somebody here know how many components photon's wave function has? Once the number of components has been agreed on, I would like to know what pointwise transformation properties the wave function has under rotations and Lorentz boosts. Once the number of components and their transformation properties have been agreed on, then I would like to know what equation describes the time evolution of the wave function. Perhaps $i\partial_t\psi = \sqrt{-c^2\nabla_x^2}\psi$ with a pseudo-differential operator on right? I'm not sure, but this would be the best guess I have so far.

I know enough of relativity and quantum theory to know that these are extremely important questions. Unfortunately I have had difficutly finding clear answers during my studies. Those people, who have understood the creation and annihilation operators, can surely easily derive the aswers to my questions from the properties of the creation and annihilation operators themselves?

 

[WannabeNewton]

@jostpuur, why don't you spend more time researching instead of complaining about physicists' supposed "lack of ability" to construct meaningful notions of photon wavefunctions?

http://www.cft.edu.pl/~birula/publ/APPPwf.pdf
http://www.nist.gov/pml/div684/fcdc/upload/preprint.pdf

 

[bhobba]

I have wondered the same thing, and I'm capable of putting forward more precise questions conserning the same topic. Firstly I would like to ask that what kind of mathematical mapping is used to depict the photon's wave function.

Why do you think a photon in QFT has a wave function? It has one - but QFT is at a different level - like it is for all particles in QFT:
http://oco.uoregon.edu/sites/oco.uoregon.edu/files/TTRL5_V1.pdf

Could your issue be how does one arrive at that equation as an approximation to QFT. If so - good question which I will have to leave to someone more advanced in QFT than I am.

Those people, who have understood the creation and annihilation operators, can surely easily derive the aswers to my questions from the properties of the creation and annihilation operators themselves?

That's because the creation and annihilation operators that a quantum field is exactly the same as admits a different interpretation.

They are excitations of the field determined by the number operators.

Its mathematically exactly the same as the harmonic oscillator:
http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator

The particles are excitations of the field - the quantised energy levels of the harmonic oscillator. They are NOT the particles described by basic QM.

But as I said above QED should have the photons wave-function as some kind of approximation - but such is beyond my present limited knowledge.

Thanks
Bill

 

[jostpuur]

@jostpuur, why don't you spend more time researching instead of complaining about physicists' supposed "lack of ability" to construct meaningful notions of photon wavefunctions?

I know that physicists' explanations on these things are contradictory conceptual mess, and I know this precisely because I have spent time studying this.

http://www.cft.edu.pl/~birula/publ/APPPwf.pdf

From the introduction:

The very concept of the photon wave function is not new, but strangely enough it has never been systematically explored.

Well, I'm impressed! I didn't know that people can say this without being labelled as crackpots. This is the author: http://en.wikipedia.org/wiki/Andrzej_Białynicki-Birula Formally he appears to be a mathematician. The article is from 1993-1994.

Some textbooks on quantum mechanics start the introduction to quantum theory with a discussion of photon polarization measurements (cf., for example [1-4]). Dirac, in particular, writes a lot about the role of the wave function in the description of quantum interference phenomena for photons (in this context he uses famous phrase now: "Each photon interferes only with itself"), but in his exposition the photon wave function never takes on a specific mathematical form.

I like the Socratic spirit in this! Białynicki-Birula is not attempting to hide that which is not known. You cannot find similar scientific touch in the Feynman's explanations, for example.

Is this article confirming or debunking what I wrote above? I don't think its obvious. Of course it is true that I didn't know about the content of this article, and I still havent' read it. I can defend my position by pointing out that this is not a very well known article, and its contents are not covered in pedagogical books.

 

[bhobba]

Well, I'm impressed! I didn't know that people can say this without being labelled as crackpots.

So your concern is its only recently systematically explored and from this you draw the conclusion QFT is in deep do do. Amazing.

Is this article confirming or debunking what I wrote above?

Debunking what? Exactly what are you claiming other than physicists explanations are wrong.

Well detail how they are are wrong. Claiming they are side stepping the issue by talking about creation and annihilation operators when it is mathematically equivalent to quantizing a field is to be blunt - ridiculous.

Can you please read the link I gave previously where the issue of the wavefunction of a photon is examined:
http://oco.uoregon.edu/sites/oco.uoregon.edu/files/TTRL5_V1.pdf

But as the article correctly points out a photon is not a particle - its an excitation of a quantum field. But it answers the question if it was a particle what would its wave-function be.

In fact the complex electromagnetic field E + iB is the wavefunction of a photon - the detail is in the article.

Thanks
Bill

 

[Cthugha]

In fact the complex electromagnetic field E + iB is the wavefunction of a photon - the detail is in the article.

Well, one has to be a bit careful here as that is Mike Raymer's condensed opinion and I am quite sure he gave some more detail in the talk accompanying the slides. You cannot apply the traditional concept of a Schrödinger style probability wave function to photons as you run into the problem of localizing massless particles. Localizing them too well of course destroys the photon and localizing them just for a bit leads to strange phenomena like the energy density of the field and the probability density for detecting a photon looking very different and having maxima at different positions. So you do not get a position eigenstate representation.

However, that does not mean that one cannot introduce any meaningful concept of wavefunctions for photons. See John Sipe's PRA on the concept of a photon wavefunction (Phys. Rev. A 52, 1875 (1995)):
http://www.physics.utoronto.ca/~sipegroup/publications/PRA/1995_52_1875.pdf

The important quote is:
"In this paper we argue that a photon wave function can in fact be introduced, if one is willing to redefine, in what we feel is a physically meaningful way, exactly what one wishes to mean by such a wave function." Mike Raymer shows one possible path to do that in those slides.

It is not too surprising that the concept of a wavefunction must be interpreted differently for massive and massless particles. Applying the "standard" concept of a wavefunction to photons is about as pointless as asking for the physical description of a photon moving slower than the speed of light in vacuum.

 

[vanhees71]

I couldn't agree more. I have to read Sipe's paper first, before I can comment on this idea to introduce a "wave function" for photons, but I'm very sceptic about it. I've never seen a convincing argument what a single-photon wave function should be good for. If you stick to the observables, usual QED is enough, isn't it?

In my opinion, it is impossible to really popularize an answer to the question what a photon really is without making it partially wrong. Unfortunately, even many university introductory books on quantum theory start with the discussion of photons in an oldfashioned hand-waving way, perpetuating wrong ideas from the time before relativistic QFT has been established (and one should not forget that in a strict mathematical sense relativistic QFT has not been proven to actually exist beyond the perturbative treatment we use in the Standard Model and in Quantum Optics).

 

[bhobba]

If you stick to the observables, usual QED is enough, isn't it?

Of course it is.

That's the entire issue I have with Jostpuur.

For some reason he seems to doubt the creation and annihilation formalism which really has me beat.

Regarding the wave-function of a photon is something I have picked up from a number of places:
http://physics.stackexchange.com/qu...describes-the-wavefunction-of-a-single-photon

Thanks
Bill

 

[kith]

Everytime I have attempted to discuss the wave functions of relativistic particles, the physicists change the topic to the creation and annihilation of particles and the particle collisions. So it seems that mainstream physicists do not yet understand the wave functions of relativistic particles.

Do you know the definition of the wave function using the full formalism of QM? Do you realize that it depends on the existence of the position operator? Here are two links about why the concept of a position operator and hence a wave function for the photon is problematic:
http://arnold-neumaier.at/physfaq/topics/localization
http://arnold-neumaier.at/physfaq/topics/position.html

I agree with you that it is odd that this is rarely mentioned because it is such a natural question for someone who knows only non-relativistic QM. But I don't think it is something which is swept under the carpet because it is not understood. The main reason is probably that people don't need to think deeply about these issues in order to calculate things which can be compared with the experiment.

 

[jostpuur]

If you stick to the observables, usual QED is enough, isn't it?

QED is famous for not providing any information about the position operators. For example, when a photon hits your eye, the retina performs a measurement of the photon position. Measurement of position is an important process, and since QED doesn't describe it, the theory contains an incredible shortcoming. It is amazing that most physicists have chosen to ignore this, and prefer to pretend that photons would have already been reasonably understood.

 

[Cthugha]

QED is famous for not providing any information about the position operators. For example, when a photon hits your eye, the retina performs a measurement of the photon position. Measurement of position is an important process, and since QED doesn't describe it, the theory contains an incredible shortcoming. It is amazing that most physicists have chosen to ignore this, and prefer to pretend that photons would have already been reasonably understood.

You seem to have a fundamental misunderstanding about what QED does and does not describe. QED gives you a probability amplitude for detection of a photon at some position. It does not give you position eigenstates. If you have a look at your measurement at the retina, you will also find that you do not end up with the photon being in a position eigenstate after your measurement. It is simply gone. This is not an incredible shortcoming as there simply is no experimental situation where a strong measurement puts a photon into a position eigenstate. The probability for some photon to get detected at some position, however, IS described by QED.

 

[jostpuur]

QED gives you a probability amplitude for detection of a photon at some position. It does not give you position eigenstates.

An interesting claim, but I see no reason to believe it yet. Do you have proof for this? Which QFT or QED books explain something like this?

I know that most books don't explain anything like that, because most books deal with this question by changing the topic to the creation and annihilation of particles and the particle collisions.

Can you reveal some of the mathematical tools used to describe the spatial probabilities?

 

[Cthugha]

An interesting claim, but I see no reason to believe it yet. Do you have proof for this? Which QFT or QED books explain something like this?

Pretty much any book does that. See for example the Mandel/Wolf (Optical Coherence and Quantum Optics).

I know that most books don't explain anything like that, because most books deal with this question by changing the topic to the creation and annihilation of particles and the particle collisions.

Well, of course you go to creation and annihilation of particles. If you detect a photon at some position, it gets annihilated and usually you create some photoelectron instead. Why would anyone want to describe a process in which a photon gets destroyed without using annihilation operators?

Can you reveal some of the mathematical tools used to describe the spatial probabilities?

I am not sure what you are after. It looks like you already know the formalism using creation and annihilation operators. You just define your initial state (maybe some excited atom which can emit a photon via spontaneous emission), your final state of interest (some detector at some position ending up in the excited state) and have a look at all possible ways to get from the initial to the final state and sum the probability amplitudes for each of them. Of course this involves annihilation operators. You would need a measurement that allows to detect the position of a photon without destroying it to avoid them.

 

[jostpuur]

Something like

$$\int\limits_{\Omega} \Big|\langle x| e^{-\frac{it}{\hbar}H}|\psi(0)\rangle\Big|^2 d^3x$$

with some set $\Omega\subset\mathbb{R}^3$?

An interesting claim. I remember that long time ago, early in my studies, I believed that it would work like this, but eventually I was "drowned" in the mainstream contradictory claims about these things, and I never figured out how to work out the details. I became convinced that the particle physicists don't know the computation details either.

 

[bhobba]

QED is famous for not providing any information about the position operators.

An interesting claim, but I see no reason to believe it yet. Do you have proof for this? Which QFT or QED books explain something like this?

You do realize because photons move at the speed of light there is no frame where it is at rest hence position is not an observable.

I know its easy to forget that, and I have fallen into that trap, but its pretty basic when you think about it, and I was a bit red-faced when it was pointed out to me - I should have picked it up.

That does not mean interactions can't happen at a point in space - which does confuse things a bit - but that does not mean the photons position is an observable.

That's one reason there is controversy about a wave-function of a single photon because it can't have the usual interpretation of being an expansion in position eigenkets - and why some posters in this thread have issues with the link I and others gave on such.

Kiths repose was pretty well spot on.

Thanks
Bill

 

[jostpuur]

You do realize because photons move at the speed of light there is no frame where it is at rest hence position is not an observable.

No I don't realize that. I think that photons move at the speed of light and I also think that position is an observable.

This conceptual mess also involves relativistic electrons, so I don't think that the photons' speed $c$ would be relevant.

 

[DParlevliet]

Its the excitation of the quantum field.

Elsewhere I read "the field is made by quanta" (Gravity does not exist, Vincent Icke). QFT is a theory, describing the behaviour of photons on a more detailed level then the simple wave or particle. It does not describe the physical nature of a photon, not does it explain yet "everything". The question of pastelchu was simple and basic: "what a photon looks like if we could observe it pass through space". The basic answer is that we don't know, that the question is meaningless because the path cannot be observer, that there is only other information which can be calculated by theories like QFT

 

[bhobba]

No I don't realize that. I think that photons move at the speed of light and I also think that position is an observable.

That really has me beat.

Since there is no frame where it is at rest it has no position.

Maybe others can explain it better:
https://www.physicsforums.com/showthread.php?t=418100

This conceptual mess also involves relativistic electrons, so I don't think that the photons' speed $c$ would be relevant.

Have you stopped to consider the reason you find it a conceptual mess is you are ignoring basic issues like the above.

Please see what I wrote previously regarding errors in well established theories.

Thanks
Bill

 

[jostpuur]

Since there is no frame where it is at rest it has no position.

How would existence of rest frame be related to the existence of position. We know that in classical context the photon does have a position and its time dependence comes from the formula $x(t)=x(0) + \frac{p}{\|p\|}ct$. The lack of rest frame has no effect on this.

Could it be that you first decided that the position operator must no exist, and then attempted to find justification for it?

 

[Nugatory]

How would existence of rest frame be related to the existence of position. We know that in classical context the photon does have a position and its time dependence comes from the formula $x(t)=x(0) + \frac{p}{\|p\|}ct$.

In classical context there are no photons, just electromagnetic waves described by solutions of Maxwell's equations. The waves don't have a clearly defined position.

When we move from classical to quantum physics, we add this notion of photons to describe the experimentally observed phenomenon that light, regardless of its intensity, transfers energy in discrete chunks. Why should we expect classical mechanics to apply to something that was developed to explain a quantum phenomenon that has no classical counterpart?

There is one heuristic that we use often and with great success: in the limit where quantum mechanical effects become insignificant, QM must make the same predictions as classical mechanics. But this heuristic argues against assigning definite positions to photons, because in the classical limit light is a wave and a wave has no definite position.

 

[bhobba]

How would existence of rest frame be related to the existence of position. We know that in classical context the photon does have a position and its time dependence comes from the formula $x(t)=x(0) + \frac{p}{\|p\|}ct$. The lack of rest frame has no effect on this.

Does it? Operationally how do you measure velocity classically? And exactly how would you apply that to a photon? I think you will find that any observation with a photon that happens at a position destroys it making that determination rather difficult - but I am all ears.

Could it be that you first decided that the position operator must no exist, and then attempted to find justification for it?

For me it was the other way around.

I originally thought it was an observable but recently was corrected (see my post 13 where I finally saw the light - pun intended ):
https://www.physicsforums.com/showthread.php?t=760945

Its like that around here - you often get erroneous views corrected - I have been through that frequently in my time here - and expect to continue in the future.

If you read the link I gave in my previous post some think the idea of having no rest frame is even not quite on the mark - that's another thing you find around here - sometimes things are a bit deeper than you may think.

Thanks
Bill

 

[Geometry_dude]

Well, I think jostpuur is criticizing the lack of realism in QM and QFT, something which is, IMHO justified, but one gets very quickly into interpretations of quantum theory when doing this and we all know how deep that hole is. Add to that the naive attempts of some textbook authors to reintroduce this realism by talking about single photons and electrons and you got your mess.

On an elementary level, one can say that QFT is a statistical theory that is experimentally well-established and does not talk about single particles like ordinary QM does in terms of wave functions. The reason for this goes back to the beginning of QFT with the relativistic analogue of the Schrödinger equation, i.e. the Klein-Gordon equation. The precise reason is that the Klein-Gordon "wave function" does not admit an interpretation in terms of a probability density and "thus" (beware: ad-hoc step) has to be reinterpreted in terms of a field that needs to be second-quantized to get back to the probability amplitude formalism.

jostpuur, if you expect quantum physics to be as nice and well-defined as "Classical physics", you will have a hard time as it evolved out of playing around with the math in an ad-hoc manner and cross-checking with experiment.

I wish I could give you a more satisfying answer, but I don't have one. Personally, I have aquainted myself with the idea that QM and QFT are phenomenological theories that are not well-understood on a conceptual level.
Of course, that doesn't mean that one shouldn't try to understand them. :)

I agree with Bhobba when he says that localizing photons is inherently problematic in the special relativistic picture due to the fact that they are assumed to "move" at c.

 

[bhobba]

The precise reason is that the Klein-Gordon "wave function" does not admit an interpretation in terms of a probability density and "thus" (beware: ad-hoc step) has to be reinterpreted in terms of a field that needs to be second-quantized to get back to the probability amplitude formalism.

That's true - but it runs deeper than that IMHO.

In ordinary QM time is a parameter and position an operator. But that conflicts with relativity which treats time and space on equal footing.

QFT's solution is for both position and time to be a parameter ie everything is a field.

Thanks
Bill

 

[bhobba]

Personally, I have aquainted myself with the idea that QM and QFT are phenomenological theories that are not well-understood on a conceptual level.

On that front you may find the following interesting:
http://arxiv.org/pdf/quantph/0101012.pdf

And also the modern version of Gleason's beautiful theorem (see post 137):
https://www.physicsforums.com/showthread.php?t=763139&page=8

QM basically from one axiom? Thought provoking isn't it?

Thanks
Bill