a dummy question, but does a photon correspond to one wavelength of a wave...does wave mean one wave length?
Answers and Replies
i don't think so. i think a photon corresponds to a "wave packet". the number of photons per second depends on both the frequency and the intensity of radiation, but the number of wavelengths (cycles) per second depends only on the frequency.
RBJ is right. A photon is a finite wave train and does not have an exact frequency
but you can can make one with say 10 crests and troughs or 10,000 crests and troughs.
The one with only ten has a less well-defined energy and frequency than the
the one with 10,000 even though they may be centered on the same energy.
The shorter photon will have a broader spectrum and will have taken
1000 times less time to generate.
Photon corresponds to the particle nature of light and has no defined frequency which is a peculiar feature of a wave. A light possessing energy E will consist of number of photons each possessing energy 'E' , a photon is a packet of energy , and is a particle that moves with 'c' and doesnot exist when at rest.Its a particle that carries a momentum and a force is felt when it strikes a surface.
OK, I'll bite into this just to stirr things up.... :)
For those who say that a photon is simply a "wave packet", consider the following:
A "wave packet" is made up of a superposition of a number of waves of different wavelength. Not only that, the wave packet moves with the group velocity of the packet, while the phase velocity can be significantly higher.
So how do you reconcile the fact that a photon can carry only ONE wavelength, especially in a monochromatic source (even allowing for a finite spread due to resolution and thermal smearing - these spreadings do no make up the wavepacket)? I can also change the group velocity of the wave packet by mixing in different wavelengths. This means that different frequency mix travels at different speeds. We don't see this in photons with different wavelenths.
I found a very enlightening post in Marlon's journal (hope Marlon doesn't mind that I repost it here)
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.