Do Light Waves Have Amplitude?

[peter.ell]

Thank you all for the great, helpful answers.

You're right, I've never taken a class in quantum physics, but I'd love to.

In the mean time, let me just see if I understand all of this:

Light acts as both a particle and a wave, but a single light wave is not the same as a single photon. A single light wave can be quantized to be thought of as a group of photons. How many photons? That depends on the wave's amplitude which, in the classical wave model, is a measure of the peak intensity of the wave's electric and magnetic fields?

Is that correct according to our current level of understanding?

Based on this, I'm curious:
When light reflects from a transparent surface, some of the light passes through (most of it) and some is reflected back with less intensity. It is a matter of less individual light waves being reflected than those passing through, or is it a matter of for every individual light wave striking the surface, two are generated: one with a given amplitude passing through and one with less amplitude being reflected?

 

[Drakkith]

If photon has no PLACE to take, why such correspondence could occur? We have nothing to say about POSITION of photons, right? Or brightness and photons have no relation?

Regards.

Brightness of a light source is exactly dependant on the number of photons that are emitted per second. The more photons emitted the brighter the source.

Light acts as both a particle and a wave, but a single light wave is not the same as a single photon. A single light wave can be quantized to be thought of as a group of photons. How many photons? That depends on the wave's amplitude which, in the classical wave model, is a measure of the peak intensity of the wave's electric and magnetic fields?

Is that correct according to our current level of understanding?

I think the issue is that describing a "wave" of light is the classical way, while describing as a photon is the quantum way. Not sure really.

Based on this, I'm curious:
When light reflects from a transparent surface, some of the light passes through (most of it) and some is reflected back with less intensity. It is a matter of less individual light waves being reflected than those passing through, or is it a matter of for every individual light wave striking the surface, two are generated: one with a given amplitude passing through and one with less amplitude being reflected?

I personally wouldn't use the term "light wave" but I'm not sure how correct that may or may not be.

 

[Dale]

Drakkith, light waves clearly have amplitudes, otherwise they wouldn't be waves, as mentioned by the OP. There are a lot EM phenomena (both classical and quantum) that would not qualify as "light waves", but those that do certainly have an amplitude.

In classical EM it is pretty clear how the amplitude of a wave arises, but in QM it is a little less clear. In QM, the state that would best qualify as a light wave is called a coherent state: http://en.wikipedia.org/wiki/Coherent_states. A coherent state has an amplitude and a phase and all of the other properties you would expect from a wave.

 

[Drakkith]

Drakkith, light waves clearly have amplitudes, otherwise they wouldn't be waves, as mentioned by the OP. There are a lot EM phenomena (both classical and quantum) that would not qualify as "light waves", but those that do certainly have an amplitude.

Of course they do, did I somehow imply that I didn't think they did? If so I didn't mean to.

In classical EM it is pretty clear how the amplitude of a wave arises, but in QM it is a little less clear. In QM, the state that would best qualify as a light wave is called a coherent state: http://en.wikipedia.org/wiki/Coherent_states. A coherent state has an amplitude and a phase and all of the other properties you would expect from a wave.

Ah ok. I'll take a look at this as soon as I can. Thanks Dave.

 

[Philip Wood]

When light reflects from a transparent surface, some of the light passes through (most of it) and some is reflected back with less intensity. It is a matter of less individual light waves being reflected than those passing through, or is it a matter of for every individual light wave striking the surface, two are generated: one with a given amplitude passing through and one with less amplitude being reflected?

Your second alternative is much more like the wave picture of reflection. Your first alternative seems to regard waves as individual, countable, things, and I think you may have some misconception. [Cycles of waves are, of course, countable, but it would be quite wrong to think of some cycles going through, and others being reflected.]

 

[sweet springs]

Hi, DaleSpam

http://en.wikipedia.org/wiki/Coherent_states.

Thanks. From the article
---
In Figure 1: The average photon numbers of the three states from top to bottom are <n>=4.2, 25.2, 924.5
----
Amplitude of electric field in the graphs, 2.9, 7.1, 43 correspond 4.2/2.9^2 = 25.2/7.1^2 = 924.5/43^2 = 1/2
Growing width of wave lines by noise in lowering number <n> is imstructive on what's going on to me. Thanks.

Further may I expect amplitude of electric field for <n>=1, single photon, be sqrt of 2 = 1.41? Thus electric field of SINGLE PHOTON shows noise band of width about 3 with its center line is waving by amplitude 1.41.

Regards

 

[sophiecentaur]

I don't see how photons are not part of this discussion. The OP specifically used the example of an electron emitting a photon as it changes energy levels. Is there something different with using a photon than with a "wave"?

I just spotted this and have to take issue. A single photon is not 'enough' to describe a wave. In just the same way that a single electron can only arrive in one spot after passing through a diffracting structure, then a photon can also arrive at only one receiving atom. It is only when the statistics of large numbers are involved that the wave quantities become relevant.

There is no implication, whatsoever, that a single photon 'is' a wave. Its presence, along with all the other photons constitutes a wave, of course - but that is putting things totally the other way round.

None of these problems need arise unless one insists on finding contradictions between wave and quantum models. Both models are just models that happen to describe certain phenomena better or worse than the other model. Don't lose any sleep over the 'really' thing; it's pointless.

 

[Oudeis Eimi]

Hi, DaleSpam

Thanks. From the article
---
In Figure 1: The average photon numbers of the three states from top to bottom are <n>=4.2, 25.2, 924.5
----
Amplitude of electric field in the graphs, 2.9, 7.1, 43 correspond 4.2/2.9^2 = 25.2/7.1^2 = 924.5/43^2 = 1/2
Growing width of wave lines by noise in lowering number <n> is imstructive on what's going on to me. Thanks.

May I assume amplitude of electric field for <n>=1 Single Photon is sqrt of 2 = 1.41? This single photon graph shows almost flat noise of width about 3 but the center of noise vibrates subtly with amplitude 1.41.

Regards.

DaleSpam's reply was clearer and far more useful than my hurried non-explanation,
in which I tried to introduce the concept of a coherent state (quantum equivalent to
the classical wave) without actually explaining it (or giving a link). What I tried to
warn you about, and I'll do it again here, is the fact that a coherent state doesn't
(can't) contain a precise number of photons. So, a state with <n> = 1 is NOT the
same thing as a single photon. This is a common misunderstanding.

To illustrate the difference between a <n> = 1 coherent state and a single photon,
in the former the electric field is oscillating with a given frequency and has a very
small amplitude with a lot of noise; in the second, the electric field is all noise with
a not oscillating, constant mean value of zero. That is, the frequency of the associated (null!) electric field is not that of the photon.

 

[Drakkith]

There is no implication, whatsoever, that a single photon 'is' a wave. Its presence, along with all the other photons constitutes a wave, of course - but that is putting things totally the other way round.

I see. Well, that's interesting. What exactly causes multiple photons to be "wavelike" yet just one to not be?

 

[sophiecentaur]

I see. Well, that's interesting. What exactly causes multiple photons to be "wavelike" yet just one to not be?

The sheer quantity of them makes the wave more and more well defined. Think in terms of interference patterns for low photon counts. One photon arrives at the screen, telling you nothing about the wavelength (from the diffraction pattern). A few hundred will start to form a pattern that is better defined and tells you more about the pattern - wavelength. But it takes many more than that to establish, from the exact positions of nulls, what the actual wavelength of the 'wave' of which they are a part.

Now, before you come back with the following, I will deal with it. You will probably tell me that the wave equation λ = c/f should tell you the wavelength and that you could measure the energy / frequency of a single photon. But I say that is begging the question because, without a load of photons, you don't have an established wave for that equation to apply to. Remembering that the equation comes from the wave treatment and not the QM treatment. (I think that is a reasonable argument. How about you?)

 

[Dale]

Further may I expect amplitude of electric field for <n>=1, single photon, be sqrt of 2 = 1.41? Thus electric field of SINGLE PHOTON shows noise band of width about 3 with its center line is waving by amplitude 1.41.

Yes, this is correct except for the interpretation that <n>=1 is the coherent state of a single photon. A coherent state does not have a definite number of photons, in fact, the number of photons is Poisson distributed. So, the amplitude of a coherent state with <n>=1 is 1.41, as you suggested, but the actual number of photons in that state is uncertain (although 92% of the time it will be 2 or fewer).

In a coherent state the number and phase form an "uncertainty" relationship. The Poisson distribution has the property that its standard deviation is equal to its mean. So, the lower the mean of the Poisson distribution, the lower the uncertainty in the number of photons, and therefore the greater the uncertainty in the phase. So a coherent state with <n>=1 would have a very uncertain phase.

 

[Drakkith]

(I think that is a reasonable argument. How about you?)

I understand what you're saying, but I don't know enough to agree or disagree with it. For now I'll accept it as a reasonable example. Thanks!

 

[sweet springs]

Hi, Dalespam

Yes, this is correct except for the interpretation that <n>=1 is the coherent state of a single photon. A coherent state does not have a definite number of photons, in fact, the number of photons is Poisson distributed. So, the amplitude of a coherent state with <n>=1 is 1.41, as you suggested, but the actual number of photons in that state is uncertain (although 92% of the time it will be 2 or fewer).

In a coherent state the number and phase form an "uncertainty" relationship. The Poisson distribution has the property that its standard deviation is equal to its mean. So, the lower the mean of the Poisson distribution, the lower the uncertainty in the number of photons, and therefore the greater the uncertainty in the phase. So a coherent state with <n>=1 would have a very uncertain phase.

Thank you for your kind tutoring.

Now I think I understand that coherent state, which corresponds to classical em wave, is eigenstate of non-Hermite annihilation operator and one photon state we are looking for is eigenstate of Hermite number operator. Are eigenstates of the latter representation, Fock states, expressed by combination of coherent states and how?

Thank you in advance.

 

[Dale]

Are eigenstates of the latter representation, Fock states, expressed by combination of coherent states and how?

That is a good question. I think that the answer is yes, but I do not know how. I would recommend that you start a new thread in the QM section and ask this question. I would very much like to see the answer also.

 

[sweet springs]

Hi, DaleSpam.

That is a good question. I think that the answer is yes, but I do not know how. I would recommend that you start a new thread in the QM section and ask this question. I would very much like to see the answer also.

I posted new thread
https://www.physicsforums.com/showthread.php?p=3520757#post3520757

Regards.