# How big is a photon

1. Jul 15, 2007

### alvaros

When a photon is emitted it goes in all directions ? or just in one ? N, S, E..
Or there is a probability you can find it in any direction ?

2. Jul 15, 2007

### olgranpappy

emitted by what?

3. Jul 15, 2007

### alvaros

By my body, or by a dipole, or by an excited atom.
Is there any difference ?

4. Jul 15, 2007

### Schrodinger's Dog

A photon does not have a size in any real terms, it has no mass AFAIK or anyone else does, so if you're talking about how "big" is the wavelength or what is its direction that's not really indicative of size of a photon just it's path after emission, it's an unanswerable question. The question is phrased poorly I think...

Last edited: Jul 15, 2007
5. Jul 15, 2007

### meopemuk

This is true. In order to have an answerable question one needs to specify how photons are prepared, where are measuring devices, and what are they measuring. Then quantum mechanics will exactly predict the probabilities of measurement results. Still, QM won't be able to tell what each individual photon will be doing.

Eugene.

6. Jul 16, 2007

### alvaros

Schrodinger's Dog wrote:
"The question is phrased poorly I think...", shure it is.

The cuestion arises from:

A dipole ( an antenna= aerial ) is just made up by two wires that can be very thin. A dipole receives photons from a surface much bigger than their surface ( length * whidth of the wires ). So, I infer, the photons whose path is not exactly through the wires of the dipole can be "captured". This will give us a size of the photons.

But, if you put another wire in front of your dipole, it will "capture" more photons. "In front" means in the line between the emitter antenna and the receiver. So, whith this device ( a yagui antenna, used to receive TV signals ) you receive photons from a bigger surface.

Where is the limit ?

7. Jul 16, 2007

### meopemuk

I guess you are confusing the size of the wavelength of photon's wave function with the particle size. The wavelength can be made as large as you wish. Still the photon should be treated as a pointlike particle.

Eugene.

8. Jul 16, 2007

### alvaros

No, Im not confusing the size of the wavelength of photon's wave function with the particle size.

The photon cant be a "pointlike particle". Remember the experiments about interference: a single photon goes through two holes.

9. Jul 16, 2007

### Schrodinger's Dog

This is explained by wavelike behaviour: superposition of a wave, not by particle behaviour? I think you need to look at the two slit experiments carefully, they do not suggest a size of a photon at all or particle behaviour as going through both slits.

The photon itself only ever travels through one slit in terms of a single photon experiment and it does so in a random fashion and with a 50/50 certainty as shown by Feynman's two slit experiment.

http://www.upscale.utoronto.ca/GeneralInterest/Harrison/DoubleSlit/DoubleSlit.html

Last edited: Jul 16, 2007
10. Jul 17, 2007

### Gza

Why does this question come up every other post in this forum?

11. Jul 17, 2007

### Schrodinger's Dog

I don't know perhaps the sources people learn from are rubbish or it's hard to get a grip on the implications of the two slit experiment. Whatever the cause I think this could do with a FAQ, but then what would the question? Although the answers are invariably the same, the questions vary widely.

There's already a "does a photon have mass?" FAQ, but it doesn't cover it completely.

https://www.physicsforums.com/showpost.php?p=1285138&postcount=6 [Broken]

Last edited by a moderator: May 3, 2017
12. Jul 17, 2007

### lightarrow

Because it's related to the essence of QM weirdness and to dualism. If physicists talk about particles, and they never say these particles are in the detector, people will, rightly, think that these particles have to fly from source to detector. Then, the conclusion that they go in both slits is straightforward, as are the strange conclusions about their size.

I would like to tell the OP that, even if not original, his question is not a stupid question; it's the result of QM books/QM lesson's schizophrenic assumptions about particles.

Last edited: Jul 17, 2007
13. Jul 17, 2007

### alvaros

To Schrodinger's Dog: Thank you for the link "http://www.upscale.utoronto.ca/Gener...oubleSlit.html" [Broken] but Im not discussing about duality or Heisenberg Uncertainty Principle.

Define size of photon = surface ( or angle, I dont know ) where it can be detected.

Detector = antenna. The photon is detected if it contributes to the received signal ( if its electromagnetic energy is converted to electric current ).

I think I didnt explain well the cuestion.

A dipole detects photons whose path does not cross the wires. It detects photons whose path is a little up or down the dipole . A parameter of an antenna is the Equivalent Surface.

But if you put another wire in front of the dipole the Equivalent Surface gets bigger.

If you put two wires ... bigger.

The real antennas have a lot of wires in front of the dipole as you can see.

So it seems that if you put enough wires you can detect the photons whose path is 10 m up or down the dipole.

How big is a photon ? ( Where a single photon can be detected ? )

Note that I started the thread from the very beginning: Have the photons a path ?

Last edited by a moderator: May 3, 2017
14. Jul 17, 2007

### Proof.Beh

It is obvious that your second question has "nowhere" answer. Since the calm mass of a photon is zero and therefor you dont detect a free photon in space (In fact it is a point because of that). Your question like that ask: where a single electron that has observed in space-time? Of course both of them (electron and photon) dont equal together really but indeed, we dont answer to these questions to when dont observe them in nature to form of single. But I think thus.

Thanks.
Mr Beh

15. Jul 17, 2007

### marlon

How does light propagate ? Indeed, isotropically...

The photon size is a concept that does not exist because size is defined in terms of spatial coordinates. A photon however is defined as a chunk of energy. Energy is a concept that is NOT defined in terms of spatial coordinates. Do you see the contradictio in terminis ?
Yes it can be because the particle (position and momentum) and wave (wavelength, frequency) nature are dual. They are like two different languages to say the same thing : ie a photon is a chunk of energy.

Besides, in an energy base (the coordinates are now values of energy), the photon is a nice point particle.

marlon

16. Jul 18, 2007

### Proof.Beh

There is no sense for your answer. If a light ray emits from a lighting source, we allocate to that length, diagonal of light ray etc and know that a light ray made of the number of bounded photons. Then the light has spatial coordinates and thus a photon also is this form.

Thanks.
Mr Beh

Last edited: Jul 18, 2007
17. Jul 18, 2007

### marlon

Yes but that does not imply the photon is a point particle in coordinate space, ie a particle with finite sized soatial boundaries. THAT is the point i was trying to make.

besides, what are "bounded" photons. I always thought that photons do NOT mutually interact (at least up the the first orders of EM interaction).

How on earth can you make this conclusion. Not only that, you are also talking about a photon's form. What is that ?

When you talk about form in this context, you are talking about a shape defined by finite spatial boundaries. A photon is NOT defined in this way. If you do not agree with me, i politely ask you to provide me with such a definition. Realise that you cannot use the wavelike photon concept to answer that question because of the reasons in gave in my previous post.

marlon

18. Jul 18, 2007

### Anonym

You repeat the same statements that you made about year ago. It reminds me the attitude of the average engineer. Energy is eigenvalue (diagonal matrix element) of the Hamiltonian operator when the QM state is the function of space-time coordinates. Moreover, it is the most fundamental conserved quantity. I suggest starting study of quantum theory.

To begin with it is useful to read posts in “size of photon particle” and M&W after that.

Regards, Dany.

Last edited: Jul 18, 2007
19. Jul 18, 2007

### Proof.Beh

"Bounded" imply to number of the photons in the length of a light ray that can reffer to finite. Like that in a bounded interval exist finite number of integer numbers. So if we consider the Photo-electric experiment, in a arbitrary time invertal, number of photons that osculate with the surface are bounded or finite. Thus there are concepts that emphasize if we wonna constrict the diagonal of location that photons are crossing from there, then the number of photons that will arrive to the surface reduse. It means that the accumulation of photons occupy a spacial region in coordinates of spase. Therefor can you say me that what is your supposal for relating this subject to energy of a photon? Of course I must go and I couldn't answer fully to your questions!

Thanks.
Mr Beh

20. Jul 18, 2007

### marlon

Sorry, but i don't get it :

1) What is the length of a light ray ?
2) What do you mean by "imply to the number of photons in the length of..." ?
3) How can you even be talking about the number of photons ? To what purpose ?

I am sorry but according to me, there are no photons "osculating" with the surface. I

Ok, i get your point here.

First of all, the double slit experiment learns us that we CANNOT make any claim onto the wherabouts of any atomic particle in the region in between the emittor and detector !. If you want to know its position, you need to measure and this causes the wavefunction collapse, remember ? But i guess you know this stuff so i will not get into that.

Secondly, in wave-lingo, lets indeed assume that the EM intensisty is lowered when going from source to detector. This indeed means that, if we now switch to the particle-lingo, some photons have left the bundle (i.e. that have scattered off atoms that constitute the medium through which the EM ray passes). But, that is about all you CAN say. Nothing more. You cannot talk about their position (x,y,z) coordinates because you did not measure. But suppose you WOULD measure, you would indeed observe that there is a photon present through its interaction with the detector (emitted EM radiation for example : the light signal from the detector). The point is however that you are not observing the actual photon, you are not measuring the photon's (x,y,z) coordinates !!!

You observed a packet of energy (as it is defined) through its interaction with the detector.

The clue is to understand what people mean by "measuring a photon" !

Suppose that a photon indeed has finite spatial boundaries, how would you measure them ?

greets
marlon

21. Jul 18, 2007

### Anonym

For example, I take the rigid ball with the internal surface being mirror and small inlet. I hope that you know that pi is irrational number. Therefore, I will take rigid rod and will measure it diameter.

W.Heisenberg, “The Physical Content of Quantum Kinematics and Mechanics”, Zeitschrift fur Physik, 43, 172 (1927):

“We turn now to the concept of “path of the electron” By path we understand a series of points in space (in a given reference system) which the electron takes as “positions” one after the other. As we already know what is to be understood by “position at a definite time”, no new difficulties occur here. Nevertheless, it is easy to recognize that, for example, the often used expression, the “1s orbit of the electron in the hydrogen atom”, from our point of view has no sense.

There is no difference between the electron and photon with this respect. In addition I desperately tried to explain to you in the past that the position (x,y,z) and the size (delta(x),delta(y),delta(z)) are different self-adjoint operators.

Regards, Dany.

P.S. In addition, the detector absorbs the photons and does not emit them.

Last edited: Jul 18, 2007
22. Jul 18, 2007

### Shahin

I think we can answer the question by antoher method. Has the photon a wavefunction? it it has, we cab use |psi|^2 as a magnitude that can give us an idea of te size of the phton, considering, for example, that the photon size is "the volume" where, again for example, |psi|^2>1/10maximun(|psi|^2). Ii is like whe you want to know the size of a nucleo, you can use several magnitudes to determine this, the mass, the charge... and you say: when de density is less than.... i consider the nucleo has ended.

23. Jul 18, 2007

### Anonym

It is E.Schrödinger conjecture. I am close to prove that. |psi(x)|^2 is what we observe as a photon. However, the photon size is a parameter (number) that defines it (among others). You don’t need guess; you may calculate it and compare with the experimental data.

Regards, Dany.

24. Jul 18, 2007

### Proof.Beh

Why you connect the structure of a photon (such as spatial coordinates) to the "measuring" concept? You said that "they are the packets of energy only that we are observing them", so have these packets the spatial coordinates or not? You did not answer yet my essential question. But first of all, I answer your elementary questions about the PERCEPTION of some parts of my last posting:

1- length of a light ray means that if we consider a laser ray that emmited from its source, then we can measure its duration of arriving for example to point B ("A" is location of sourse), then by name its duration "T" and use the formula c.T=L ("c" is speed of light) we shall define L or length of light ray.

2- imply sth means PREDICATE to sth.

3- Because I meant that show if we guess in the double slit experiment will constrict the diagonal of each slit, then the your energy packets because of their spatial coordinates and volume will cross for the first situation less.

Thanks.
Mr Beh

25. Jul 18, 2007

### alvaros

I think it would be helpful if you speak about operational definitions instead of "chunk" "size" "bounded"...

If a photon has no size nor path, how can you get a photo ?

If a photon makes a spot on the photografic film we can infer that it collided in that point. How big can be this point ?