# Do all photons have the same electric field

1. Jul 1, 2014

### tim9000

I deliberately wrote that heading not quite accurate, as a photon is more of a probability thing, and so might not have an electric field at all. But what I'm getting at is, we know a photon doesn't have a well defined amplitude, but it does have a frequency.
So what I wan't to know regarding the electric field of an EM wave, is E⃗=Eocos(ωt−kx)

and that Eo is dependant on the amount of photons in the wave, thus power is dependant on intensity rather than frequency of the individual photons. Does this mean that all photons are like a Plank Electric potential or something, regardless of frequency?

Thanks

Last edited: Jul 1, 2014
2. Jul 1, 2014

### Simon Bridge

The photon is the mediator particle for the electromagnetic interaction, so you sort-of have it backwards.

In terms of the wave model, the intensity is proportional to the photon flux and the energy per photon.
The photon energy is determined by the frequency.

But it is important not to mix up the models. It is not appropriate to equate waves and particles in this context.
Classical behavior like the wave is what you get on average in special situations.

Last edited: Jul 1, 2014
3. Jul 1, 2014

### tim9000

Hi Simon,
I don't think I'm inappropreatly mixing up models, I understand that classically Energy of a wave is only proportional to ampliude, and on a quantum scale a photon, that acts accordingly to probability to make up an EM wave-like-propagation, is only dependant on frequency.

I was intregued to hear that I had it "backwards" and I read about the phon flux you linked me to, however besides learning the definition of photon flux and learning that I was misusing the word 'intensity', it didnt' satisfy my curiosity. Infact it raises the point that I don't know what they meant by "intensity" because I looked up the definition for Radient Intensity and Irradiance and neither of those had the number of photons or frequency in their definition, only Electric field! (?)

It's what constitutes Eo that I'm plerplexed by, the whole 'dim gamma' vs 'insense RF' difference. Is Eo dependant only on photon flux and indepenant of frequency?

For context: The reason I asked is because I'm trying to get my head around how you can have two different sources of different frequencies and have the same power/m-2 (poynting vector) from them.

Thanks!

Last edited: Jul 1, 2014
4. Jul 1, 2014

### Simon Bridge

It is always inappropriate to mix up the models.
Quantum mechanics does have to include classical mechanics though, so it is fair to ask how the classical equations come from the QM ones. It is fair to talk about photons giving rise to an EM wave but the wave model does not apply to photons.

Classically the energy in an EM wave is proportional to the square of the amplitude.
The intensity is the time-average energy-density.
http://en.wikipedia.org/wiki/Intensity_(physics)#Mathematical_description
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/emwv.html

The energy per photon depends on the frequency. This is also the frequency of the classical light wave so... if $$E=E_0\sin(kx-\omega t)$$ then $E_\gamma = \hbar\omega$.

The intensity of light in a vacuum is, classically, given in terms of the mean power: $$\langle U\rangle = \frac{\epsilon_0c}{2}E_0^2 = NE_\gamma$$ ... where N is the expectation value of the number of photons per second.

If the flux is a constant, then it is fair to say: $$E_0=\sqrt{\frac{2N\hbar\omega}{\epsilon_0 c}}$$

... which means they have different photon fluxes.

i.e. A beam of blue light has fewer photons per second than a red one, at the same power.

5. Jul 1, 2014

### tim9000

What I meant was I didn't realise I was mixing up modles at all.

That was a tiptop answer, thanks a lot, I really appreciate it!!

I see what you meant by 'intensity' is Irradiance. I had no idea you could equate it to Nhf like that, brilliant. (It demonstraites how the E-field is depenant on intensity which is related to Number of photons and their freq)

Would it be fair to say:
Mean expected Number of photons * frequency*h = 1/2 * [epsilon*Eo^2 + mu*Ho^2] (which I think is <Energy>) ?

SECOND EDIT: actually isn't your intensity in (W/m^2) and wouldn't NE_\gamma$$be in Joules?? Last edited: Jul 1, 2014 6. Jul 2, 2014 ### Simon Bridge N is the expectation photon flux - so it has dimensions of number per unit area per unit time. hf has dimensions of energy, so Nhf has dimensions of... Note <U>=Nhf ... which is the expectation value of the power per unit area, i.e. not <E> At this level we would stop using "f" for frequency (we want to be able to use it for an arbitrary function) and use either omega or nu instead: $\omega = 2\pi\nu$ so we write: $E_\gamma = h\nu$. 7. Jul 2, 2014 ### tim9000 Ok, I see what you're saying, so expectation of photon flux, which is LIKE power density, It was the units confusing me, Nℏω is W/m2 and Photon Flux has the unit (per second square meter), because it's ('a number')/s*m2, but if that 'number' is timesed by energy (Joules) then it becomes J2/s*m2 Always good for me to be exposed to the nuances of effective phyics communication, thanks but I don't see why we would use 'nu'? (I've never seen that symbol infact) If what I've explained my understanding to be is correct, does that mean that I could say: N*hbar * omega= [epsilon_0*Eo^2 + μ*Ho^2]/2*m2 ?? (I did try using the LateX code but it didn't work, sorry) Thanks heaps for your replies, you have no idea how good it is to be finally putting the pieces together. Last edited: Jul 2, 2014 8. Jul 2, 2014 ### vanhees71 Photons are the most difficult piece of the whole Standard Model in the sense that they are massless quanta with spin 1. I just use the word "quantum" in this posting to avoid the word "particle" here, because people associate very wrong ideas when it's used without a lot of care; "elementary particles" are not just miniature "billard balls" as one might think, using the word "particle". The trouble with the Standard Model of elementary "particle" physics is that it is nearly impossible to explain entirely correctly in everyday language. It is based on relativistic local quantum field theory, which is a pretty abstract mathematical description of nature. It has been discovered by the physicists at about the same time when non-relativistic quantum mechanics was discovered and deals with situations, where the quanta you like to describe, scatter with each other at relativistic energies, which means that their energy is much larger than their "rest energy", $m c^2$ (note that I use mass always in the sense of "invariant mass" or "rest mass"; there's no place for the old-fashioned idea of a speed-dependent "relativistic mass" in modern physics anymore). Under such circumstances one has to use relativistic quantum theory, and it turns out that under quite general assumptions (local interactions, existence of a stable ground state, called "the vacuum") this implies the possibility to create and destroy quanta, as long as the fundamental conservation laws (conservation of energy, momentum, angular momentum, electric charge, and various other charge-like quantum numbers) are fulfilled. Particularly photons, the quanta of the quantized electromagnetic field, can be easily created (e.g., by hitting a heavy atomic nucleus with an electron, which is scattered and in the process emits a photon, socalled bremsstrahlung) and destroyed (absorbed). That's why you have to use the pretty abstract machinery of quantum field theory to describe them. This theory describes scattering processes like the example with the electron and the nucleus given above by something called the S-matrix (scattering matrix), invented by Heisenberg. It's modulus squared, with a grain of salt, is just the transition probability rate from an initial state (usually two quanta) to another state (usually two or more quanta) of asymptotically free quanta. Usually it is not even possible to define what's meant by the "number of quanta" as long as the quanta are in a process of interactions with each other, but you can define this observable for quanta, which are located so far from each other that you can neglect interactions with each other. For photons, even this is not the full truth, but that's usually neglected even in the quantum field theory lectures at university. The reason is that photons are massless. Now the electromagnetic field also describes the interaction among charged quanta, and the masslessness of the photons manfests itself in the long-range nature of the electromagnetic field. This is well known already in classical electrodynamics, which is an excellent approximation of quantum electrodynamics for macroscopic bodies: There you have the Coulomb field which describes a force between static charged bodies that only goes down with the 2nd power of the bodies' distance. So it's a long-range force, and it turns out that strictly speaking you can never neglect the interaction among the bodies, no matter how far they are appart. This is also the case in the quantized theory, and thus the "asymptotic states" are much more complicated than thought in the naive way we learn quantum field theory in the introductory lecture at the university. A charged quantum is always accompanied by its electromagnetic field, and this field cannot be neglected, when describing the scattering processes of charged quanta, no matter, how far apart they may be. Under many circumstances this makes no harm in the way we calculate the probability for a scattering process to happen (or the cross section, to put it in more experimental terms), using perturbation theory, which is nicely depicted by Feynman diagrams, that are ingeneous short-cuts to write down the S-matrix elements to evaluate the cross section of the process at interest. Sometimes, however, the sloppiness hits back, and one has to take into account that the asymptotic states are not the naive free-quanta states ("plane waves") usually used. Then one has to resum an infinite set of Feynman diagrams, the socalled "soft-photon ladders" or at least one has to take into account some higher-order diagrams. One example is the bremsstrahlung process mentioned above. The same methods show, what a "classical electromagnetic" field is in terms of QED: It's a socalled coherent state. This describes a state, which is given as a superposition of all states with arbitrary photon numbers! This implies that the number of photons is not determined for such a state. So it's dangerous to say, photons "mediate" the electromagnetic interaction in the sense of a (classical) field. This is also just sloppy language with a very detailed mathematical meaning when evaluating the Feynman diagrams of perturbation theory, where one often says, e.g., two electrons scatter at each other by "exchanging a virtual photon". What it in fact is, is just a symbolic internal line in the Feynman diagram, standing for a (free) propagator of the electromagnetic quantum field. It's not more than a clever way to write down the scattering amplitude for elastic electron-electron scattering. A "virtual photon" is not something you can observe, because what you can observe are only asymptotically free photons. As I said, it's really hard (if not impossible) to describe this without the sharp language of mathematics. 9. Jul 2, 2014 ### tim9000 Hi Vanhees71, There are still a few things I need to look up regarding what you just posted but I'd like your opinion on how I should think about things. Some background; last year I did 2nd year quantum mechanics and I just did Adv Electromagnetism, so I still don't clearly have these ideas fixed in my mind. I.e I understand the ramifications of Heisenberg's uncertainty principal and 'I know how much I don't know' about how a photon acts. (I just looked up an S-matrix then from your mentioning it, I had been unaware). I understand what you mean by needing the language of maths, I only started understanding things after deriving Schrodinger's equation, most of which I've forgotten now. Now when I was doing Maxwell's equations, it was kind of irritating me that it seemed the previous quantum stuff was less than relevant. The analogy I heard (on the internet) was like when a charged particle oscillates, imagine there is like an infinite amount of strings coming off it going off into space, and there is an impulse traveling along them outward from the charge accelaration. Is this of any use, or is it more harmful than helpful? Also, I always sort of had it in my head: there is rest mass, and there is relativistic mass; light only has relativistic mass (hence the momentum), is this not the right way to think of it? Also, do you have anything to weigh in on regarding my last post (i.e was what I said correct) Thanks Last edited: Jul 2, 2014 10. Jul 2, 2014 ### DrDu Hendrik, I am not quite convinced that the subtleties of the asymptotic states are really relevant here. After all, they are rather a problem of describing asymptotic states containing charged particles. Clearly you can study scattering processes which involve asymptotically only photons (light by light scattering). I agree with you on that for the description of electromagnetic fields, it is important to introduce coherent states. 11. Jul 2, 2014 ### vanhees71 The question of how to define asymptotic states is crucial to the understanding, what the term "particle" or even "vacuum" means. Have a look at the following, and you know what I mean ;-)): http://dx.doi.org/10.1016/j.aop.2013.05.021 [Broken] http://arxiv.org/abs/1305.1425 Last edited by a moderator: May 6, 2017 12. Jul 2, 2014 ### tim9000 ook, or disregard my response, that's cool too. 13. Jul 2, 2014 ### Simon Bridge What is the education level you are trying to understand this at? Here let me help: LaTeX: N \hbar \omega = \frac{ \epsilon_0 E_0^2 + \mu H_0^2 }{2m^2} gets you:$$ N \hbar \omega = \frac{ \epsilon_0 E_0^2 + \mu H_0^2 }{2m^2}$$... show derivation and reasoning. What does "m" stand for in that equation? Have you tried checking it by dimensional analysis? Wikipedia on frequency and notation ... online discussion of the notation - elsewhere. You'll find it all through college notes online and in scientific papers. 14. Jul 2, 2014 ### DrDu Of course, but the asymptotic state containing a single photon is not very complicated, in contrast to asymptotic states containing charged particles. 15. Jul 2, 2014 ### tim9000 Thanks again Simon, Engineering degree, so my physics is not quite as up to scratch as someone who is doing optics or something. Hence it's purely curiosity rather than necessity. I fully understand that Equation might be wrong, infact I think I left off the seconds unit on the RHS, so it'd be:$$ N \hbar \omega = \frac{ \epsilon_0 E_0^2 + \mu H_0^2 }{2*s*m^2}$$where$${( \epsilon_0 E_0^2 + \mu H_0^2 )}{/2}$$is the energy (from equipartitioning of EM energy) s is seconds and m is meters, would this equation be accurate? the units were being a bit of a mine field for me with this thread. P.S I will read up on that wiki notation. 16. Jul 3, 2014 ### Simon Bridge It is not suitable to include units explicitly in an equation. i.e. What if you don't want to use S units? (Very uncommon in QM ferinstance.) Do you mean to divide some energy by some time and some area? Which time? Which area? Does the equation for "the energy from equipartition of EM energy" (whatever that means) actually have dimensions of energy? Does it make sense to adjust the magnetic field for non-vacuum when you did not do this for the electric field? I don't know if the equation is "accurate" but it does not make any sense to me. Bear in mind that a relation may be accurate but also wrong. I'm trying to get you used to this sort of troubleshooting. Do you now know how the energy carried in an EM wave is calculated? 17. Jul 3, 2014 ### tim9000 Fair point, I shouldn't have done that. yes, by (per) 1 meter2 (per) second. Two things here, in trying to answer you I found out that it was actually energy per volume: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/engfie.html and I assumed energy was in Joules, is that incorrect? Also according to that link I'm assuming they mean per cubic meter? Sorry that was meant to be μo Well I definitely have a new appreciating for double checking units. [/QUOTE] Well like: B=Bocos(kx-wt) for the electric field too, then poynting vector for power/m2 And as the link above states for the total energy of the wave per cubic meter? Given what I just said came to my attention, I can see what I previously posted was incorrect in atleast that it was equating: W/m2 to what I now think is: J/m3 is this statement accurate and is there anyway to modify the RHS to make it also W/m2 ?? Thanks for your help and bearing with me, hopefully it'll clear up these misconceptions. 18. Jul 3, 2014 ### Simon Bridge That is still just including dimensions ... I mean physically which area? The area of your desk? The area of a window? The surface area of a sphere centered on the source of the radiation? See what I mean? What is the 1m^2 area to the subject of the investigation - you don't get to use arbitrary areas and times: they have to come from somewhere? i.e. why pick one second for time? What if we don't want to do physics in SI units? It's common to use length in angstroms and energy in electron-volts for example. Your equation does not specify units for E and H so why do you do that for area and time? That link contradicts your equation. Read it carefully - the expressions on that page are for computing energy in a device, right at the bottom it redirects you to the energy in an electromagnetic wave via the Poynting vector. The equation for that is very different from yours. Never "assume". Do the actual dimensional analysis - do you not know how to do that. In physics there are three "dimensions" for units. length L, mass M, and time T. i.e. [Energy]=ML2T-2 Here the square brackets are read "dimensions of" Energy may be electron-volts or horse-power-minutes or Plancks or anything you're energy meter is marked out in. ... it's very powerful but you should do it by dimensions not units. The link talks about the total energy density for the wave. When you average something over a volume you get a density. Think "energy per unit area" I think I have already provided you with the correct relation much earlier. What is wrong with that one? The main trouble here I think is a bit of a bad habit picked up in secondary school. Probably worth kicking it in the head: namely - think about the equation in terms of what they have to describe. When you have a distance - what distance is it? The distance between New York and Chicago or the distance between a source and a receiver (may be the same thing!) or what? Think about what things are rather than how they are represented. It's not "Joules" it's "energy" etc. 19. Jul 4, 2014 ### tim9000 OK, I'll conceed that what I was wondering has no context to energy of an EM wave or relevant connection to Nh(bar)omega. I need to work on understanding the dimensional differences. That's a good question actually, I never concidered it, I suppose I was interpreting it as being an area perpendicular to the direction of propergation, i.e the plane wich both B and E oscillate. Umm, I wasn't aware where they came from was important, say, a dipole? I chose a second to meet the same units on the LHS, as Nhbarω as is in Watts per square meter, which is per second. Ok, I didn't realise that was only for energy in a device, I sort of thought mine was related to the components that made up the cross product of it. Until you mentioned doing a dimensional analysis, I'd never heard of it, but I'll have another look at the wiki page on it, I thought it was just a matter of writing all the units in context and making sure they matched up but now I realise there is a distinction between units and dimensions. There was nothing wrong with the explanation you provided earlier, I was just trying to understand how it could relate to what I thought was the total energy of an EM wave, that apparently isn't related. Yes you're right, it is an habbit worth kicking, but when you say "not joules, energy" do you mean 'well it is Joules but that has a context of energy in some form, think about what the form represents'? Last edited: Jul 4, 2014 20. Jul 4, 2014 ### Simon Bridge You seem to be thinking that the energy in the wave is related to the vector sum of the amplitudes of the E and B components, but you have refused to be drawn as to your reasoning so I cannot be sure. If so, then you have been trying to relate a correct answer to an incorrect one. Not a useful exercise. All that stuff about dimensional analysis is to get you used to checking your own work, so you are less reliant on asking others. Please read: Energy carried in an EM wave. http://hyperphysics.phy-astr.gsu.edu/hbase/waves/emwv.html ... compare with how you've been thinking. The Poynting vector, for instance, is $\vec S = \vec E\times\vec H$ i.e. you picked those particular dimensions without thinking about what they are for. You seem to be having trouble thinking of a physical property without the units. To get the correct dimensions - you would take an energy, then divide by area and time. Thus any equation of form:$$I=\frac{U}{At}$$will work. But to make sense we have to say that energy U has accumulated in time t over an irradiated area A. [*] In SI units, that would be W.m-2 - but I can use other units if I like. Notice how there is no need to specify a particular area and a particular time? However, in real life, the energy U need not have arrived at a constant rate, and need not be distributed evenly over the area A ... so the calculation is actually for an average. This is why the standard derivations talk the way they do. No - "Joules" is a man-made construct that we use to mark out the dial on our energy-meters. "Energy" is what Nature does - it is the real thing, the thing that we are measuring. We do not measure Joules, we measure energy. Just like we do not measure meters, we measure distance; we do not measure kilograms, we measure mass; we do not measure seconds, we measure time; etc. -------------------------- [*] ... the definitions of the variable has to be relevant. For instance, the power dissipated by a resistor is $V^2/R$, so$$I=\frac{V^2}{AR} has the correct dimensions ... but what area should we use for A? The area of the circuit board? The area of the ceiling of the lab? Maybe the surface area of the resistor would make sense right?

Having decided that you will take the voltage across the resistor, square it, divide by the product of the resistance and the surface area of the resistor, then you have just calculated an intensity ... but what is it the intensity of?

This is what I mean about where something comes from being important.

This sort of thing is absolutely paramount for an engineer btw!

Last edited: Jul 4, 2014