# The wavelength of a photon

1. Nov 19, 2013

### jaydnul

Does the wavelength of a photon actually mean wavelength? Because as I understand it, a force field, be it gravity, EM, ect..., has influence at any distance. Theoretically, the field never has an end, reaching out across the whole diameter of the universe, even though the effects would be small. Or is photon wavelength the mathematical value that represents wavelength when describing a photon as a wave. We know how much energy a photon has, therefore if we describe it as a wave, there is a number corresponding to that "wavelength".

Thanks

2. Nov 19, 2013

### sophiecentaur

The relevant thing about a Photon is its frequency - and that corresponds (E = hf) to the energy of the transition that it causes or is caused by. That is to say, the situation in which a Photon can be said to 'exist' in a particular place and at a particular time is only when it is interacting at either end of its journey. Whist the EM energy is on its way from source to receptor, the photon is a mythical beast as it can be anywhere or have any 'size'.

How could one even define the frequency of the photon whilst it is on its way, bearing in mind that it can have been produced by a different system thane the one that absorbs it? You have, in effect, a transmitter with some finite bandwidth and a receiver of finite bandwidth so all you know is that the energy was at some frequency in the 'overlap' between the two bandwidths with the same nominal frequency but with a finite difference. It cannot be looked upon as a little wiggly thing that zaps from place to place, any more than it can be looked at as a little bullet.
The problem you describe in reconciling the idea of waves and fields shows that the simple model of what goes on in between source and absorber is not good enough. There is a finite time involved with the atomic energy transitions and that corresponds to a spread in frequency (like the Q of a resonant circuit) the Q of the transmitting system could be different from the Q of the absorbing system yet the photon would have to be the same; it wouldn't know what sort of atom / system it was going to interact with it whilst it was on its journey.

3. Nov 19, 2013

### Khashishi

A photon has a wavelength given by its energy. You can't go out and measure the length of a single photon, because the properties of the photon only exist when it is emitted and when it is absorbed. So the wavelength of a photon refers to the wavelength of the electromagnetic field that the photon is a quantum lump of. The wavelength of the electromagnetic field is easy to verify using interferometric effects.

4. Nov 19, 2013

### sophiecentaur

Yes. And that wavelength can only really be assigned to the average effect of a large number of photons - large enough to produce a 'good' distribution (good enough to show a pattern). Each photon is, of course,interacting with a different detector / atom / molecule at different places on the projection screen (or whatever).

5. Nov 20, 2013

### f95toli

I am not sure I agree with this. There are certainly quite a number of examples of systems where we are dealing with single photons, and where the "wavelenght" matters.
A typical example would be cavity-QED experiments, where we can trap a single photon in a cavity with a lenght lambda/2.

Hence, although the concept of wavelenght becomes a bit fuzzy when dealing with single photons, it is not meaningless.

If you go through the math for a cavity-QED setup using a lambda/2 resonator you will also find that where the node would be for a classical EM wave is exactly where the coupling strength to the photon in the cavity is at maximum.

6. Nov 20, 2013

### sophiecentaur

Surely any cavity has a finite bandwidth (Q) - so the wavelength that you could assign to a photon in it would be, necessarily, approximate. Could one be certain that the wavelength, corresponding to the energy transition in the source, is exactly the same as the wavelength, corresponding to the energy transition in the detector?
My reason for bringing this up is to point out that the actual existence of a photon is only certain at either end of any process. Postulating its behaviour 'in between' as particular is only using an alternative model to describe the energy trapped in the cavity. (Is that being too controversial? )

7. Nov 20, 2013

### f95toli

Well, the cavity is in a number (Fock) state; meaning there is no fundamental reason for why the photon should not have an "exact" frequency. The Q of the cavity does not matter for this, it is only relevant for the decay time for the photon (T1): if you de-tune the cavity while there is photon still in it the photon will leak out faster than if it was exactly on resonance (of you de-tune quickly you get all sorts of interesting QM effects, e.g. the dynamical Casimir effect).
This can be used to create single photon sources.

Also, if a non-linear cavity/system is used there are ways of detecting photons non-destructively; meaning you can tell if there is a photon in the cavity without actually destroying it (a "simple" way of doing this involves using the non-linear Kerr effect to detect the number of photons in one mode of the cavity by monitoring another mode).

Haroche won his share of the Nobel prize last year for doing experiments in this field.

8. Nov 21, 2013

### sophiecentaur

There seems to be a difference between the bound state and the state of a 'travelling' photon - the same as for electrons. Heisenberg must apply(?) so, if the wavelength is known exactly, the position of the photon in the cavity is uncertain. The probability function would then be the standing wave function. Or am I harking back to QM too much?

9. Nov 21, 2013

### f95toli

No, there is no "Heisenberg" for photons. The photon does not have a position operator, meaning it never has a "position" in the usual sense of the word.
Also, the photon does not have wavefunction, so it does not have normal probability distribution (you can plot the Wigner function and other distributions; but they can be negative and so has no "classical" interpretation)

10. Nov 21, 2013

### sophiecentaur

Oh, right. Well, that relieves me in a way as the 'particle' nature is not stressed - rather the discrete nature of the photon - which has been my appreciation of the thing, for some while. You have cleared up the Heisenberg bit for me. That seems reasonable, so the whole of Physics can take comfort from my approval -haha.
I still have a bit of a problem with the concept of an 'exact' wavelength, though, in as far as the frequency of the source and the accuracy of the measurement must surely come into it. It's fair enough to have a mathematical model for what's going on in the cavity but it worries me that a popular view of this could not coincide with the actual meaning contained in the QED description. Any given photon would need to be able to interact with a range of atoms, in a spread of actual energy states or you couldn't be sure of ever getting an exact transition to occur. Do you get my point about this? It seems that a source and detector could only 'talk to each other' under exact conditions and this is never actually the case because there are always more Quantum Numbers involved than those which are talked about at a simple level. (Pauli forbids this, doesn't he?)
When you say that the photon does not have a wave function, is there something about the field pattern in a cavity with a single photon in a cavity that makes things different from what you get in a normal standing wave? Where does that Foch state come into it, if not to describe the pattern?