How Do Photons Relate to Light's Wave-Particle Duality?

[mace2]

Hi,

I was hoping you could help me in my understanding of light as both a wave and a particle. In particular, let's say we have light propagating at some frequency v. Does each oscillation of the electric field correspond to one photon? That is, if the wave has v cycles per second, does that mean we will observe v photons per second? Wait, I guess that doesn't make sense... if the light has a very long wavelength (say radio), does that mean we will receive multiple photons in one wavelength of the light?

My second question is related to the first. If a charged particle is emitting radiation via bremsstrahlung (decelerating in the process), does that mean the frequency of the emitted light is constantly changing? I picture the compression or expansion of the wave in a way akin to the Doppler shift. Is this a correct interpretation?

Sorry if these questions are not worded as best they could be. Hopefully you can see that I am a bit confused on how to relate the characteristics of the light to the number of photons observed per second.

Thank you.

 

[Galap]

The Photon essentially is a quantization of an amount of energy in an EM wave. The amount of energy for a given photon is dependent on its frequency. E=hv. energy is proportional to its frequency. To find number of photons per second, we must combine this fact with intensity of the EM radiation in question.

 

[mace2]

So does that mean the rate of photons is always constant? And it's just the energy of the photon that changes?

I thought the intensity was proportional to the square of |E|. But then how do I reconcile that with the frequency being the quantity that determines a photon's energy?

Thank you for your help. I'd really like to figure this out!

 

[Grep]

So does that mean the rate of photons is always constant? And it's just the energy of the photon that changes?

Ok, take, say, a red laser of a certain frequency v. You have it connected with a dial with which you can turn down the intensity (put less energy into it) with a fine level of control. And you shine it on a device called a photomultiplier.

I'd suggest you pull up a wiki on photomultipliers. It's worth the read. But in short, say we have a device that clicks when a photon hits the photomultiplier. And that click is louder or softer depending on the energy it absorbed.

Now suppose you turn that laser down so low that you only hear an occasional click. Those would be individual photons. You can tell that by turning it down, you get less clicks of the same loudness. Turning it up increases the number of clicks, but they are still the same loudness.

We'd have to conclude that, for a certain frequency, each individual photon will have a specific amount of energy. That is, red photons have a certain amount of energy each, as do blue, gamma rays, etc. As you turn up the intensity of the light, it just sends out that extra energy as more photons. Turns out that if you increase the frequency, the amount of energy per photon is increased and the clicks are louder (and vice-versa).

And as Gallap said, E=hv. So take the frequency and multiply by Planck's constant and you have the energy of photons of that frequency. Obviously higher frequencies mean more energy.

I'm not a physicist or anything, so someone correct me if I made a mistake, but this should be correct and hopefully understandable.

EDIT: Just wanted to add that I'm not the best one to explain the so called "wave nature" of light. It's actual meaning is more or less in probabilities (like how they can have "destructive interference" and you find a near zero probability of a photon being at a certain spot), but that's as I understand it. Perhaps someone else can chime in with a clear explanation. Sleep beckons me and I doubt I'll explain it well.

 

[mace2]

Hi Grep. Thanks very much for your explanation.

Okay, so now I understand the frequency simply determines the amount of energy contained within one photon (the "loudness" if you will), and only the intensity of the light affects the rate of photons.

My only outstanding question is about the intensity of light. I know intensity is the square of the electric field amplitude, and it is also defined as the energy per unit time per unit area. Is there some physical intuition linking the electric field amplitude (squared) and the photon rate? I don't have a gut feeling for how or why they are related.

Thanks!

 

[Phrak]

What photons? Where does this idea of little bullets of energy come from?

 

[Galap]

What photons? Where does this idea of little bullets of energy come from?

The concept was required to explain the photoelectiric effect.

 

[Phrak]

In the same Wikipedia article you will read, "It has been shown that it is not necessary for light to be quantized to explain the photoelectric effect."

 

[Grep]

In the same Wikipedia article you will read, "It has been shown that it is not necessary for light to be quantized to explain the photoelectric effect."

I should add to start that I, for one, would not characterize a photon as a "little bullet of energy". It's neither a particle or a wave, but rather something similar to both, and yet distinct from either concept. A photon can do things no bullet could do, to say the least. Bullets fired one at a time through two holes will likely not produce a diffraction pattern at the end.

Anyways, I see your point about the photoelectic effect not needing the light to be quantized. It does, of course, require the photoelectrons to be quantized. And since the light was probably emitted by quantized electrons... Alas, it's also such a simple way of explaining things, that my explanations may now need to be a little more complicated. Suggestions welcome.

From the paper that wiki article references, a relevant part at the end:

In conclusion, we understand the photoeffect as being
the result of a classical field falling on a quantized atomic
electron. The introduction of the photon concept is
neither logically implied by nor necessary for the
explanation of the photoelectric effect.

I see how that's logically sound. However, that's not exactly the same as saying that photons are not quantized. For example, another quote from that paper:

In fact we shall see that the photoelectric effect may be completely
explained without invoking the concept of "light quanta".
To be sure, certain aspects of nature require quantization of
the electromagnetic field for their explanation, for example:
1. Planck distribution law for black body radiation (1900),
2. Compton effect (1926),
3. Spontaneous emission (Dirac, 1927),
4. Electrodynamic level shifts (1947).
The photoelectric effect is definitely not included in the
foregoing list. It is an historical accident that the
photon concept should have acquired its strongest early support
from Einstein's considerations on the photoelectric effect.

In short, and correct me if I'm wrong here, but the conclusions stand, but the photoelectric effect isn't the best example as it can be explained without assuming discrete quanta of light. Really interesting stuff.

 

[Galap]

So ultimately the bottom line with the photoelectric effect is that while it did not require a corpuscular-like explanation, it is what led scientists to consider it.

 

[Phrak]

What is a quanta? It is something that takes on discrete values. What of electromagnetic force is quantized is only angular momentum. And even this, we could argue, is due to the discrete angular momentum of the electron field. However, this is not the usual way of looking at things, and for those studying Quantum Electrodynamics a photon is a discrete vertex on a Feynman diagram. The vertex is a world line of a photon. But even here, the vertex is not in any particular place but all smeared out over spacetime.

Of all the values that are quantized for fundamental particles, the one rarely mentioned explicitly in the collection, is total probability. This is the most interesting one, and also the one that makes the whole business of quantum mechanics so difficult to get a grip on.