# How big is a photon?

1. Dec 2, 2005

### Revelationz

Does a photon have a 3-D envelope of a certain size? (Assume a photon travelling in free space, and by 'envelope', I mean some space outside of which the field strength of the photon's electromagnetic fields drops below some threshold, say 1% of maximum.) If so, how big is the envelope? Does the size vary with some other property of the photon, like wavelength?

Related question: Do multiple oscillations of the fields exist within one photon at any moment, spread over space? If so, how many within the envelope?

2. Dec 2, 2005

### ZapperZ

Staff Emeritus
If you look at the definitions of what a "photon" is, you'll notice the lack of "size" as being a part of it. It was never defined as a "particle" in the classical, ordinary sense. Einstein only refered to it (in translation) as the "corpuscular" of light.

However, there is a characteristic length that is associated with photons, and that is the wavelength of the light that we measure. But this should not be construed as being its physical size.

Zz.

3. Dec 2, 2005

### Revelationz

If I examine a bar magnet's magnetic field, I can define such an envelope. Are you saying it is not possible to define such an envelope for the magnetic field of a photon?

4. Dec 2, 2005

### ZapperZ

Staff Emeritus
An envelope? I'm not quite sure what you mean. If we use the simplest case of a dipole field, you'll see mathematically that the field extends over a large range in space. So where do you cut it off and call that as inside this "envelope"? Whereas you can visually see, and thus define, the edge of a ping pong ball, you can't do that with this case.

One can define the "envelope" for light (as in a gazillion of photons flowing through). That's how we get values such as wavelength - it defines the oscillation "length" of the light's electric field. However, when you ask for a physical size of a photon, and a photon was NEVER defined as to have a physical size, then the question ventures into the undefined territories. You might as well ask "what does blue smell like?"

Zz.

5. Dec 2, 2005

### marlon

Just to add. Keep in mind that a photon is defined as a quantum of energy. So it is designated by a number in an energy-coordinate space, NOT in a spatial coordinate space. So a photon is not defined as an entity with fixed spatial boundaries, designated by numbers in a spatial coordinate base.

marlon

6. Dec 2, 2005

### HallsofIvy

Really? HOW would you define such an envelope? The distance at which the field drops below some value seems very arbitrary. Of course, the field strength never becomes 0.

7. Dec 2, 2005

### Ratzinger

But what is it with cutting down intensity in the double slit experiment so that we will see many small spots on the phosphor plate?

8. Dec 2, 2005

### ZapperZ

Staff Emeritus
How does adding a filter have anything to do with physical size? It's not as if the aperture is changed.

Zz.

9. Dec 2, 2005

### DaTario

It is possible to add in the Fourier sense, different waves, with different wave vectors k, so as to obtain a localized wave packet (Mandel and Wolf - "Optical coherence and quantum optics" Cambridge).

I believe that it makes some sense to speak of coherent size of a photon, which could be defined as being constituted by the product of the coherence length and the coherence area.

Comments on my position here will be greatly appreciated.

Best Regards

DaTario

10. Dec 2, 2005

### Physics Monkey

It is true that one can construct a "localized" wavepacket out of single photon states. Because it is constructed of single photon states, it is an eigenstate of the number operator. However, it is not an eigenstate of the Hamiltonian so it doesn't represent a state of definite energy. Of course, the wavepacket can have a fairly well defined average momentum and average position consistent with Heisenberg. You might be tempted to associate the variance of your packet with some kind of "photon size", but its clear this size has nothing particular to do with any kind of inherent "photon size" and is most properly associated with how you prepare the system. In particular, such a wavepacket isn't exactly what people usually mean when they say photon. Such wavepackets are nevertheless useful for instance when one is talking about the continuous mode theory where it is difficult to give a precise definition of the photon.

11. Dec 2, 2005

### DaTario

No photon has really a precisely defined energy. It is a sort of idealization.

I would like to ask you another question related to this topic:

Suppose you have an isolated iron atom located at the center of an sphere made also of iron with radius of , say, 1 meter. The atom is standing still and it is at the time t = 0 in the excited state.

Suppose also that, after 10^-6 seconds, the atom decays ans the photon, generated by spontaneous decay, start travelling radially to the sphere.

OBS: I know these times are not so well defined experimentally and there are probabilities associated with them.

The question is the following:

The photon, which was gererated by an iron decay, has spectral features that permit excitation of any other iron atom, in particular those located on the sphere.

Thus, no matter the radius of this sphere, will that photon always have the power of excite an atom on the sphere ?

Best Regards

DaTario

12. Dec 2, 2005

### Revelationz

what I mean by an envelope

A bar magnet's field extends to infinity, but if we take an arbitrary cutoff value, like, say, 1% of the highest field strength inside the envelope, then a physical boundary can be defined. The boundary is crossed when the test point moves far enough away from the bar magnet that the field strength drops to less than 1% of the strength at the magnet's surface.

So, my question is, can such a "1% envelope" be defined for a photon's magnetic field, say when the magnetic field strength is at a maximum during an oscillation?

13. Dec 2, 2005

Perhaps off topic, but according to Ingrid Carey, "blue" smells like her breath--she has a condition called "synesthesia", which translates into "union of sensations" and she can in fact "smell colors". See this link:
http://www.livescience.com/humanbiology/050222_synesthesia.html
I wonder how the theory of quantum mechanics may help explain synesthesia" ?

14. Dec 2, 2005

### Tom D

I reckon so.
In fact, in principle it would be possible to find the exact dimensions of the 'envelope' which contains the spacial extent of a single photon thus:
consider a light source in an otherwise 'light-tight' box with a tiny hole which emits one photon at a time through the hole. Attached perpendicular to the box side with the hole and centred on the hole is attached a tube which is manufactured with the inside of the tube arbitrarily narrow.
A detector measures the arrival of any photon passing through the tube.
The tube is replaced with ever narrower tubes until no photon passes through. The last size tube successfully passed through by a photon is the size of the envelope.

15. Dec 2, 2005

### ZapperZ

Staff Emeritus
Maybe you'd like to explain how you would determine the magnetic field of a single photon?

Zz.

16. Dec 3, 2005

### McQueen

Maybe the photon itself has a field ???

17. Dec 3, 2005

### McQueen

Definitely.

18. Dec 3, 2005

### ZapperZ

Staff Emeritus
I know that EM radiation has fields. How do you think we accelerate particles in a particle accelerator?

But I want the description on how one would determine the "field" of a SINGLE photon to be able to determine its "envelope". So pay attention to the particular issue being asked here.

Zz.

19. Dec 3, 2005

### Physics Monkey

McQueen, you may be interested to learn that a single photon has an average magnetic field of zero. I would suggest you study a good text on Quantum Electrodynamics before assuring others that a photon "has a magnetic field" and that an "envelope" can be defined. One such good text is Rodney Loudon's "The Quantum Theory of Light".

20. Dec 3, 2005

### Revelationz

maybee a reeealy small probe made of superstrings?

While the average magnetic field of a photon should be zero, because it is oscillating + to - over time, at some point the field strength is at a maximum (say +) when the electrostatic field strength is a minimum (zero). So at that moment, all the photon's classical energy is contained in the magnetic field (does relativistic mass-energy have a part here? haven't a clue).

Perhaps some sharp-pencilled theoretical physicist could then say, well, infinite field strength doesn't make sense, so the magnetic field of known energy must have some kind of spatial extent at that point in time. Otherwise, if all the magnetic field is concentrated at a single point, the field strength is infinite at that point. Maybe inifinite field strength IS allowed, I don't know.

Since the magentic field strength varies with time, so would the envelope size, I suppose. Equally, so would the electrostatic field strength, although reaching a maximum out of phase with the magnetic field maximum. So it could have and envelope too, maybe different size than magnetic??? Dunno.

Perhaps the aperture experiment could be a way of physically looking at the extent of a photon, although you might just wind up with more of a probability envelope because of HUP, dunno

By the way, if a photon enters an aperture travelling along the X-axis, then hits the phosphor screen off the X axis as described by quantum theory, does the re-direction of its path cause it to lose energy? In other words, is the scattering of the photon at the aperture (to describe it in classical terms) perfectly elastic?