# Understanding The ever elusive Photon Model

1. Jun 10, 2012

### toneboy1

Hi All!
PLEASE correct something if I say it like it's an axiom but it's not quite right, I'd like conformation of my research:

So as far as I can tell there is some contention about what an actual photon 'is'. So when an electron goes from high state to low state it releases a 'photon' in a packet of quanta 'particles'.
These travel according to the inverse square law AS A WAVE, together. Thus since this wave is made of particles moving straight outward, the wave DOES NOT HAVE AN AMPLITUDE but the AMOUNT OF PHOTONS present in a space (with superposition) affects the intensity. So the mechanical 'energy is proportional to the square of the amplitude' does not apply, only E = hf.

I have drawn a picture of a single particle photon as a black sphere, I'm not sure if this sphere particle actually exists or just the field lines that come out of it at different strengths as it moves. If someone could offer their thoughts that would be nice.

If this is all correct I have a question.

THANKS!!

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2. Jun 10, 2012

### Simon Bridge

It is not all correct. For example:

No it's not. I think you need to decide whether you want a wave model or a particle model of light or if you want to go into quantum mechanics with QED. So far you are mixing them up.

3. Jun 10, 2012

### toneboy1

I am sorry its not all correct, I didn't expect it all would be, though I fail to see how you think (if I was talking about a photon as a particle) it wouldn't be moving straight. It's not like it's doing a zig-zag and loop-de-loop.

I want to understand EMR in it's entirety, i.e. both.

4. Jun 10, 2012

### Simon Bridge

In that case you want to start with QED.
Light has wave and particle behavior. Where the photons end up are determined by wave equations but what you detect is a lump of energy that arrives in one go - the photon. The waves are not "made up of particles" in any way.

Look for Richard Feynman on Youtube
... this is part 1 of a lectures series covering your interest.
You need to know about reflection and refraction to a (junior) high-school level.

5. Jun 10, 2012

### toneboy1

I'll look that up immediately, thanks!

I thought by QED you meant 'quod erat demonstrandum' but I realise you meant 'Quantum electrodynamics' :P

6. Jun 10, 2012

### Simon Bridge

QED = "Quite Easily Done" ;)

7. Jul 4, 2012

### toneboy1

Thanks for telling me about Prof. Feynman, he's great. I watched that and an accompanying video, BUT...I'm still not 100% on if my original thread question is right or wrong.

From watching the video I would revise part of my question to: "the wave DOES NOT HAVE AN AMPLITUDE but the" probability of the "AMOUNT OF PHOTONS present in a space" is what we would percieve as the amplitude and is infact the intensity, which can be zero should two EM 'waves' be out of phase in that space.

But this would mean that the probability was zero of a photon being THERE at that time. So I don't understand how superposition works in that sense, and on any other conformation...

I'd appreciate any input. Thanks

8. Jul 4, 2012

### f95toli

If you are trying to get an "intutive" understanding of what a photon really then that won't work. We use "semi-classical" models of the photon all the time (particle, wave etc) but these are just models and have limitations; you can always think of situations where they don't work.
A fairly accurate definition of what a single photon "really" is would be that it it is a spatially and temporally localized excitatation of the vacuum. If you want to be more accurate you need math and the whole QED machinery

see e.g. M. Oxborrow, and A. Sinclair, "Single-photon sources," Contemporary Physics 46, 173-206 (2005).

Anyway, thinking about "where" a photon is when it is not detected is quie litterarly pointless, the formalism (as presened by Feynman) works exremely well but it can only tell us about what we will see when we actually measure something.

Btw, there is no controversy when it comes to what a photon is; the problem is how to EXPLAIN it.

9. Jul 4, 2012

### toneboy1

Valuable insite, thanks.

I read the abstract of 'M. Oxborrow, and A. Sinclair, "Single-photon sources," Contemporary Physics 46, 173-206 (2005).' As that is unfortunately all I am able to at present. Which is interesting but I don't think it would help me personally to further my understanding to a large degree.

Ok, sacrificing accuracy, in one semi-classicla model or another was what I said 'fair' to say?

Could I also postulate that from what you said, in one sense, QED or another that when phases 'cancel out' so that nothing is seen, that there is still what one MIGHT concider a photon or photons but there is no "temporally localized excitatation of the vacuum"?

Despite if I say something silly I do understand your point about the explanation being the problem and I endevour not to be pointless, but don't hesitate to tell me if I am being so.

Cheers!

10. Jul 5, 2012

### Simon Bridge

f95toli answers this one (above). It's not "what it is" that's the problem, it's describing it to someone who does not have the prior knowledge. Usually, as students go through their courses, their understanding of a photon, or any particle, changes.

You'll have seen that Feynman has a special meaning for the word "amplitude". It is not the same as "amplitude" in classical wave mechanics... eg: a QED amplitude is often a complex number: the little phasors he draws are in the complex plane.

Also you'll see that he is very careful not to talk about what the photon is doing when it is not detected... QED does not tell us that. You'll remember his reflection example?

The probabiliities being worked out are the probability that a photon gets detected. The average number of photons that get detected is the total number of photons you've got multiplied by the probability ... it's all statistics.

If the total amplitude is zero at some place, then no photons will be detected there. There will be no photons there doing any interfering or anything. Remember the analogy to predicting eclipses by moving stones between different bowls? The bowls don't have anything to do with the eclipse and don't tell us how eclipses happen ... but they are very good at predicting them.

Similarly QED is a set of rules about putting marks on a bit of paper which is very good at predicting the results of experiments, yet don't say much about how those results came to be.

You'll have heard of some physicist or other saying that math is the language of the Universe? Basically, if you don't know the math, all anyone can do is give you poetic metaphors. If you want an intuitive understanding of fundamental particles, you will need to learn the math and have lots of experience with it and them.

11. Jul 5, 2012

### toneboy1

Nice Answer Mr. Bridge. I think I'm more or less satisfied not to understand so much so soon.

I gather that "single particle photon as a black sphere" I wasn't sure about existing; no one's really sure about either, but that modeling it like that works sometimes.

One thing I would like your opinion on is that E = hf, might that be more appropreate for individual photons than a packet? (in some instance)

Thanks

12. Jul 5, 2012

### Simon Bridge

Actually everyone is pretty sure it dos not exist as anything like that but the it is a useful model for things like Newtonian Ray optics.

E=hf is for a single photon - whose statistics may be described by a single photon wave-packet. Keep a clear separation between classical and QM descriptions of light.

13. Jul 5, 2012

### Simon Bridge

Thought about it - decided to elaborate on the original example.
Hmmmm ... OK, lets say we have a system set up so atomic decays are equally likely to release their energy in any direction. Detectors at different distances have different probabilities of detecting the resulting photon from a single decay. (It's the area of the detector's aperture divided by the surface area of a sphere with radius equal to the distance from the source.) This geometry automatically gives you an inverse-square law.

How the photons travel is unknown - these calculations are predictions of detection not travel. Once detected we can deduce that it must have travelled in a line from the source to the detector in the absence of any reason to believe it may not have (i.e. no mirrors).

The number of photons we'd expect at a detector over time is the number of decays in that time multiplied by the probability. The number actually detected has to wait for the experiment. How these photons are experienced depends on the detector. The flight-path has not been observed ... the act of observation would have changed the path.

The photons interact with the detector, transferring hf energy each ... as a result of many such interactions, the classical description of mechanical energy transferred in relation to an amplitude of a classical EM wave emerges from the quantum treatment.

We can conduct an experiment where the distinction between the classical and quantum models for light are important to the outcomes - one such is the photoelectric effect. The telling result that distinguishes the quantum and classical descriptions is when you get an ejected electron "too soon" for the classical model. Feynman points this out in his lectures.

14. Jul 8, 2012

### toneboy1

When you say 'atom...single decay' I'm assuming you mean like the closest thing to a single photon generator? Because photons are created in 'packets'.

I had to re-read your post manifold times, very interesting experiment. Did you mean the probability of the photon occuring on that detector was to do with the unpredictable nature of the atom decay, or the probabalistic nature of EM light?
Was your aim to point out that the light must be deduced to travel in straight lines?

Interesting! I was wondering, do you remember anything being said about an apeture that's diameter is smaller than the wavelength of 'a photon' or 'a wave', wichever is more appropreate, being said?
Because I heard with things like the faraday cage of the microwave door, that some small fraction of a percentage can still make it through, or there is a EM field bulging through the mesh still. Dispite the holes being smaller than the EM wavelength.

Enjoying talking with you!

15. Jul 8, 2012

### Simon Bridge

You know I once had this experience: a tourist asked me for directions to the waterfront. I told him to turn left and keep going. He thanked me, turned right, and walked off.

Sometime later, he came back and complained.
I told him he'd done the opposite of what I told him, I pointed down the street and told him to "go that way". A look of realization crossed his face and he thanked me. Turned right, and walked off.

Sometime later he came back and complained.
Each time I explained he said he understood and each time started to walk off the opposite way but I was ready for him and physically stopped him.

In the end, I grabbed him by the shoulders, forcefully rotated his body to face the right way, and pushed him down the street for a ways. Then he got it ... and found the waterfront.

I still think there was something subtle about my communication that kept mis-cueing him. I had failed to account for something in his expectations.

I'm afraid that talking to you has been like that. I suspect someone actually has to sit down with you and draw pictures and get you to draw pictures until you get it right. I and others keep pointing you in the right direction and each time you seem to head off in the opposite direction. Let's try one more time and I'll try to get someone more experienced than me.

One energy level transition produces one lump of energy = one photon - what are you talking about "packets"? We sometimes describe photons in terms of "wave packets" - the wave packet is still only one photon.

I suspect you still imagine the QM waves being made out of particles like water waves are made out of water molecules. Did you watch the lectures? Did you hear Feynman talk about photons as wave packets? Try not to mix your models up.

This was explained in the lecture series I pointed you to. The probability refers to the statistical nature of photons.
No - I am saying that the idea that light travels in a particular way is an assumption - it has not been observed. What has been observed is the presence or absence of a photon at a particular place. Watch the lecture about reflection again - compare what Feynman describes with the classical law of reflection.

That would be part of a classical description - you wanted to explore the quantum mechanics. Lets not get side-tracked. You are having enough trouble coping with the revelations in the Feynman lectures.

Go watch them all again - one of the last ones he answers a question specifically about wave-particle duality, iirc "isn't your theory really a wave theory?". Summarize the answer to me.

16. Jul 8, 2012

### toneboy1

I do intend to watch them again, along with everything I haven't already.

I'm glad these things are being dispelled while I'm young. The things I learnt in HS physics were very misleading....well if you misunderstood the vague explanation you weren't corrected.

Thanks, all I needed was clarification that it was statistical nature of photons and the motion was unobservable, I thought as much.

H'mm. I see, I did concider that, though I thought it must have been like a singular photon creation which cascades into a packet.
Interesting, this packet is obviously divisable, yet even when there is just one left it must exert the same characteristics as if it was still with the rest of the packet.

I might start a new thread asking about how a smaller appature than the EM 'wavelength' 'stops' it from going through. I tried to find data on if it was just the majority or all but I couldn't.

Thanks for your guidance

17. Jul 8, 2012

### Simon Bridge

http://en.wikipedia.org/wiki/Wave_packet
The wave packet is just another statistical description.

Wave-packets are always divisible - but the particle remains in one bit. Remember - the wave just describes the way the "probability of detecting a particle" behaves ... not the particle itself.

Einstein did consider that EM waves could only exist in packets - but that's a historical use - in error. Photons are well understood via QED - hence the videos.

18. Jul 9, 2012

### Hernik

Hi! Maybe I can help. As a layman I have been asking directions for the waterfront for some years now, and I'm still wandering about asking anybody I think can help :-)

Your problem is, that you want to explain what a photon IS. Physics describes accurately (very impressing I think) how it acts, how it behaves. But it is not absolutely clear what light IS.

Apart maybe from the fact that it is an amount of energy that leaves one entity and arrives at another entity at the speed of light. An "entity" being an antenna, an atom, a molecule, the annihilation of an electron and a positron.. and many other possibilities, I'm sure.

You read things about light like "It's a particle" or "It behaves like a wave". But it cannot be waves.

If it was a wave emitted from a point it would spread evenly in three dimensions. Like the rings on the surface of water in two dimensions. But clearly light is not like that. It has direction. And it arrives at a single point. No wave in the ocean lands on a single grain of sand on the beach.

Neither can it be particles. If light was particles how could a single photon ever go through two slits at the same time?

Sometimes measurements show light has wave-LIKE properties. Sometimes experiments document light has particle-LIKE properties.

So physicist have a fine collection of descriptions of light and how it interacts with matter which each work in different situations. Just not one description which fits all situations.

They don't know where that waterfront is themselves ;-)

Hope this helps, (and hope it is correct!)

Best regards, Henrik