What does the frequency of a photon actually mean?

Main Question or Discussion Point

Say you have one photon travelling through free space.

It has its energy level with its corresponding wavelength and frequency.

What does the frequency mean? What is it doing x amount of times per second?

Also, the same goes for the wavelength.

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Mk
marlon said:
WHAT IS THE DIMENSION OF A PHOTON ???

You all know that QM provides us with 1 (and not two) way of describing physical processes : particle wave duality. We apply our classical ideas of what "wave" is, and what a "particle" is. A particle, like a grain of sand, has a definite boundary in space, i.e. a grain of sand doesn't appear spread out that it's exact shape and boundary are vague. Thus, it has what we classically define as a particle. A wave, on the other hand, can spread out over space.

Now, here is the clue : A photon description in QM is NOT defined as having an exact shape and boundary in space, thus a photon is NO classical particle. It is defined as clumps of energy. So in energy coordinates, it has definite "points", but it has no definite "size" in real space! So when talking about 'size' of a photon you must realize that we work in energy space (more formally we work with momentum-eigenstates)

Having said that, the most common explanation for the "wave-particle duality" is that light behaves as waves in experiments such as the double slit, and behaves as particles when we do things like the photoelectric effect

A photon has a perfectly well-defined wavelength only when it's in a momentum eigenstate, i.e. when it has a perfectly well-defined momentum (and energy). This never happens. A photon is always in a superposition of momentum eigenstates:

The only quantity that we might want to call the "size" of the photon is the width of the Fourier transform of the momentum-space wave function, f, i.e. the uncertainty in the photon's position. This uncertainty could be anything between zero and infinity. (I'm ignoring Planck-scale effects here). Since it can be arbitrarily close to zero, it makes sense to call the photon a "point particle".

However, if we assume that the uncertainty in momentum is proportional to the magnitude of the momentum (which is the only thing we can assume if we know nothing about the state), the uncertainty in position is proportional to Planck's constant divided by p (the magnitude of the momentum). Since p is inversely proportional to the wavelength, the uncertainty in position is proportional to the wavelength.

So it makes sense to think of the wavelength as the "size" of the photon (or at least as something proportional to it). This may seem strange, but it is at least consistent with e.g. the fact that microwaves (with wavelengths of order 1 cm) won't go through a metal net with millimeter-sized holes (like the net that covers the window of your microwave oven), but they will go through a net with much larger holes.
Electromagnetic waves moving through a vacuum, when v = c, can be described as velocity=speed of light/wavelength.

When light travel from one medium to another, frequency remains more or less the same, but wavelength and/or speed changes.

Awesome, that was sort of what I was thinking. Thanks for the clarification.

Claude Bile