PDA

View Full Version : Amplitude of Light



mugaliens
2007-Aug-25, 04:59 AM
I was reviewing Wiki's entry on light and came across a disturbing comment on the "The three basic properties of light, intensity, or alternatively amplitude, which is related to the perception of brightness of the light."

I was under the impression that light has a fixed amplitude and that it's "intensity" is simply a matter of how many photons are passing through space.

JohnD
2007-Aug-25, 07:38 AM
Save us searching, Mugs and give us the link.

On the quote you give, it seems to me that by using "alternatively" the Wiki contributor is hedging their bets in the particle/wave conundrum, or even just playing semantics.

A particle cannot have an amplitude, but a wave can. If we consider light as particles, then it's the number of photons; if as a wave, then it's amplitude.

John
John

01101001
2007-Aug-25, 12:52 PM
Wikipedia: Light (http://en.wikipedia.org/wiki/Light)


Light is electromagnetic radiation with a wavelength that is visible to the eye (visible light) or, in a technical or scientific context, the word is sometimes used to mean electromagnetic radiation of all wavelengths. The elementary particle that defines light is the photon. The three basic properties of light (i.e., all electromagnetic radiation) are:


Intensity, or alternatively amplitude, which is related to the perception of brightness of the light,
Frequency, or alternatively wavelength, perceived by humans as the color of the light, and
Polarization (angle of vibration), which is only weakly perceptible by humans under ordinary circumstances.


Due to its wave–particle duality, light can exhibit properties of both waves and particles. The study of light, known as optics, is an important research area in modern physics.

NEOWatcher
2007-Aug-27, 12:59 PM
I was reviewing Wiki's entry on light and came across a disturbing comment...
And thanks to 01101001 providing the link, I can clearly see from the context that they are not talking about a single photon by itself.

As far as the shape of a photon, I always imagined it coming in shaped as a football, and the longer the wavelength, the more elongated the shape. Thus the ball is thinner as it gets longer. Would I be wrong in imagining this as amplitude? (of course it would be directly proportional to the wavelength). Or is it just a concept that can't be described because the photon has no mass?

Ken G
2007-Aug-27, 02:06 PM
I was under the impression that light has a fixed amplitude and that it's "intensity" is simply a matter of how many photons are passing through space.There is really no sense to which light has a "fixed amplitude"-- even the wave function of a single photon does not have a fixed amplitude at each point in space. For me, the problem with the Wiki statement is that amplitude is a very different concept from intensity, and is not an "alternate" concept the way frequency and wavelength are. The concept of amplitude is really what separates a wave from a beam of particles, as you say, and is where the concept of interference comes from. If that was in some way equivalent to intensity, interference would be impossible. In short, you can get the intensity by knowing the amplitude, but you cannot get the amplitude by knowing the intensity.

mugaliens
2007-Aug-28, 05:12 PM
You mean that this light diagram (http://en.wikipedia.org/wiki/Image:Light-wave.svg)on Wiki is wrong?

mugaliens
2007-Aug-28, 05:23 PM
I'm still not convinced that light is a particle, but rather simply the E-M fluxing of space itself that exists first here, then there, and the distance between the here and there is Plank distance and the time between the two points is Plank time.

As for light's frequency, that's a vibration of space-time that occurs over much larger times.

Tobin Dax
2007-Aug-28, 06:18 PM
You mean that this light diagram (http://en.wikipedia.org/wiki/Image:Light-wave.svg)on Wiki is wrong?

No, it's correct. It also doesn't imply that all electromagnetic waves have a fixed amplitude, no matter what. A single wave will usually have a consistent amplitude, but that doesn't mean that all waves have only a single amplitude.

mugaliens
2007-Aug-28, 07:39 PM
But light's "amplitude" of a single wave (single photon) is related to it's frequency, is it not?

Ken G
2007-Aug-29, 01:05 AM
But light's "amplitude" of a single wave (single photon) is related to it's frequency, is it not?

Partially, if you mean the amplitude of the associated elecric field, but it is also related to the volume in which the particle is contained. That isn't fixed, it depends on how the photon is made. It also depends on whether by "photon" one means the particle emitted by a quantum transition, or a quantum of electromagnetic excitation (they're different meanings).

tusenfem
2007-Aug-29, 01:08 PM
It is all due to the wave-particle duality of light. You can see light in two ways:

1. Light it a "particle," a collection of so-called photons. The intensity of light will then be the number of photons hitting your detector. Note that a photon (as mentioned by mugaliens is NOT a single wave, it is a wave package. Also the "amplitude" of the photon (if you decide to speak of an amplitude, but it is not defined) is not related to it's frequency. The energy of a photon is related to the frequency through the equation E = h f, where h is Planck's constant and f is the frequency of the light.

2. You can see the light as an electromagnetic wave which is described by Amplitude * exp(i ω t). This is a wave like a water wave. It's intensity is given by the Amplitude2. So, now we do not have discrete particles but sinusoidal wave trains coming into the detector.

Naturally, you can put things together (call this a homework assignment) and calculate how many photons would agree with an electromagnetic wave of a certain Amplitude.

Ken G
2007-Aug-29, 02:36 PM
Note that a photon (as mentioned by mugaliens is NOT a single wave, it is a wave package. Also the "amplitude" of the photon (if you decide to speak of an amplitude, but it is not defined) is not related to it's frequency. The energy of a photon is related to the frequency through the equation E = h f, where h is Planck's constant and f is the frequency of the light. But note this last sentence contradicts the first. It is those "two meanings" of "photon" I was talking about, that confusion comes up quite a lot!

Naturally, you can put things together (call this a homework assignment) and calculate how many photons would agree with an electromagnetic wave of a certain Amplitude.You can do this for a given detecting surface and get the rate of photon arrival, but to really know "how many photons" you had everywhere, you will also need to know the total volume that these photons are confined to, as I mentioned above. This is related to the quantum mechanical meaning of the term "amplitude", which is a little different and has to do with the wave function of a single photon and will depend on the volume that the photon is confined to. Unfortunately the words get a bit confusing when we start mixing classical and quantum mechanical descriptions!

tusenfem
2007-Aug-29, 02:43 PM
But note this last sentence contradicts the first. It is those "two meanings" of "photon" I was talking about, that confusion comes up quite a lot!


Yep, I fell into the trap here!

Ken G
2007-Aug-29, 02:44 PM
I still don't even know the "correct" use of the word "photon"-- seems to me it gets used both different ways all the time with no apologies!

mugaliens
2007-Aug-30, 12:55 PM
I still don't even know the "correct" use of the word "photon"-- seems to me it gets used both different ways all the time with no apologies!

Well, if you're confused, I certainly hope no one invents the field of gravitostrong-weak electromagnetics, for if that were to happen, I'd stub my toe (http://en.wikipedia.org/wiki/Theory_of_everything)on it so hard it would hurt like 10^19 GeV!

mugaliens
2007-Aug-30, 01:04 PM
Note that a photon (as mentioned by mugaliens is NOT a single wave, it is a wave package. Also the "amplitude" of the photon (if you decide to speak of an amplitude, but it is not defined) is not related to it's frequency. The energy of a photon is related to the frequency through the equation E = h f, where h is Planck's constant and f is the frequency of the light.

With respect to the E-M field, the integral of a single wavelength with an amplitude of a single photon, by definition, must be it's energy.

That's how he solved for his equation, having obtained the constant from his experiments with black-body radiation.


2. You can see the light as an electromagnetic wave which is described by Amplitude * exp(i ω t). This is a wave like a water wave.

Actually, it's more like a sound wave, with a variance in pressure (amplitude) as it passes.


It's intensity is given by the Amplitude2. So, now we do not have discrete particles but sinusoidal wave trains coming into the detector.

Naturally, you can put things together (call this a homework assignment) and calculate how many photons would agree with an electromagnetic wave of a certain Amplitude.

I'd be willing to bet that if could isolate a single photon with a sensitive and fast enough detector, it would detect a sinusoidal rise, fall, reversal in both the E and M fields of a single photon.

Ken G
2007-Aug-30, 06:55 PM
I don't think so. It's a fairly central idea in quantum mechanics that you cannot observe an amplitude. Observations correspond to operators that act on amplitudes, and the amplitude is used to predict, statistically, those observations, but the amplitude itself is not an observable.

mugaliens
2007-Aug-30, 07:25 PM
I don't think so. It's a fairly central idea in quantum mechanics that you cannot observe an amplitude. Observations correspond to operators that act on amplitudes, and the amplitude is used to predict, statistically, those observations, but the amplitude itself is not an observable.

And, as a photon behaves in a quantum manner...

So, it would simply register at the sensor. Am I correct?

Ken G
2007-Aug-31, 08:45 PM
Yes, I think so. In quantum mechanics, the way you will describe the system is intimately dependent on the experiment you are performing. If your experiment is to measure a field at one point, you will be making a classical measurement and will need a lot of photons. If you only have one photon, then you can't really measure field, because field is not a quantum measurement. Or at least, I don't know how you would do it. Maybe there's a way to get statistical kinds of measurements, but they would only be meaningful if you averaged them over many runs of the experiment, which effectively turns the quantum measurement into a classical one anyway.

mugaliens
2007-Sep-01, 11:08 AM
Ok, so when the one photon enters the photomultiplier tube, it passes the photocathode, on which one atom captures the photon, releasing one electron. That one electron passes through the focusing electrode where it begins an ever-increasing series of cascades on the dynodes, and the cascade is what's registered on the anode?

In other words, it either kicks off the electron or it doesn't. Quantum. And you're right - when you get lots of measurements, whether or not it's all at the same time or over time, you wind up with the classical experiment, much like the wave-slit phenomenon where it works over time as well as it does all at once.

mugaliens
2007-Sep-01, 11:09 AM
accidental duplicate - please delete.

Ken G
2007-Sep-01, 02:59 PM
Ok, so when the one photon enters the photomultiplier tube, it passes the photocathode, on which one atom captures the photon, releasing one electron. That one electron passes through the focusing electrode where it begins an ever-increasing series of cascades on the dynodes, and the cascade is what's registered on the anode?
Right-- that's an expressly quantum measurement.

In other words, it either kicks off the electron or it doesn't. Quantum. Exactly.
And you're right - when you get lots of measurements, whether or not it's all at the same time or over time, you wind up with the classical experiment, much like the wave-slit phenomenon where it works over time as well as it does all at once.
Yes, you can do one quantum experiment and repeat it, but if you want a classical observable you'll only get that by taking a big average.

mugaliens
2007-Sep-02, 01:20 PM
The question is, why do successive, independant quantum measurements produce the wave interference pattern?

I believe I know the answer: It has nothing to do with interference with other photons, per say, but due to the state of the photon's wave function as it entered either slit.

Have they polarized the incoming photons parallel to the slit and repeated the experiment? How about perpendicular?

Ken G
2007-Sep-02, 02:59 PM
Yes, interference is a property of the photon wave function. So the appearance of the pattern is because the wave function is the correct way to get the right prediction. You may as well ask why the wave function is the right way to predict photon behavior-- we don't get to know why the mechanics of waves rule the behavior we observe, but the closest I could come to a reason is to say that the fundamental thing waves do is called "Huygens' Principle", wherein a wave loses track of its source and instead acts as if every part of the wave was a new source. In other words, propagation is a form of self-actualization, in concert with a continuity property in space and time. Reality appears to act by making similar but evolved copies of itself as time progresses. The copy must do both propagate and cancel the previous reality, which is the source of constructive and destructive interference. That is wave mechanics in a nutshell.

mugaliens
2007-Sep-05, 08:49 PM
I think it's because a photon is constantly oscillating, and whatever it's state when it hits the slits will determine how much it's refracted.

Throw enough of them at the slits, regardless of the source, but all of the same wavelength, and you observe what appears to be standard wave interference, but what is in reality merely a mapping of the varying (sinusoidal) wave states that could exist when any photon of that frequency hits the slits.

Ken G
2007-Sep-05, 09:17 PM
I think it's because a photon is constantly oscillating, and whatever it's state when it hits the slits will determine how much it's refracted.But that's precisely the "hidden variables" approach to quantum mechanics. It was pursued vigorously early on, and pretty much discredited as a possible approach, though there may be some hangers-on and maybe it's not completely dead. Stranger things have happened, but it certainly has been looked at very carefully. For example, I can put a filter on the slits that only admit photons at a given phase-- yet I still get the full interference pattern.



Throw enough of them at the slits, regardless of the source, but all of the same wavelength, and you observe what appears to be standard wave interference, but what is in reality merely a mapping of the varying (sinusoidal) wave states that could exist when any photon of that frequency hits the slits.
That sounds a lot like standard wave interference to me in the classical limit, it's not telling me where the individual photons will go. And you are aware, are you not, that electrons do exactly the same thing? So the answer cannot have anything expressly to do with photons-- all particles obey the mechanics of waves.