Do Light Waves Have Amplitude?

[peter.ell]

Waves in general have three properties: frequency (related to wavelength), amplitude, and speed. When referring to light as a wave, it's wavelength and speed are always referenced but never its amplitude, and I was wondering if light has a fixed amplitude for all wavelengths or if it changes.

Sound waves, for example, get louder in two cases: when there's more of them (multiple speakers) or when the amplitude of the sound waves is increased (more volume). Obviously having more light waves (multiple light sources) will result in increased brightness, but can the amplitude of a light wave change as well?

Since I've never heard the amplitude of a light wave referred to, plus the fact that higher frequency (shorter wavelength) light is referred to as more energetic, I was wondering if perhaps the intensity of light is only related to how much of it there is, and maybe the amplitude of a light wave is set. After all, how would a light wave of a given wavelength increase its amplitude? If an electron falls a greater distance, the light emitted won't be more intense, it will actually have a higher frequency, right?

Thanks for clearing this up for me!

 

[Drakkith]

I think I remember reading that the amplitude was planks constant or something related to it. Intensity of light refers to the number of photons and the energy of each is based on its frequency.

 

[sweet springs]

Hi,　peter.ell

Brightness of light corresponds to amplitude. Amplitude of light from two light bulbs are larger than that from one light bulb. Amplitude of light from each bulb is additive.

Regards

 

[Drakkith]

Hi,　peter.ell

Brightness of light corresponds to amplitude.

Regrds

Brightness does not correspond to the amplitude of each photon. Instead it corresponds to the number of photons.

 

[sweet springs]

Brightness does not correspond to the amplitude of each photon. Instead it corresponds to the number of photons.

...And number of photons also corresponds to amplitude of light, doesn't it?

Regards.

 

[Drakkith]

...And number of photons also corresponds to amplitude of light, doesn't it?

Regards.

If by amplitude you mean how bright something like a light bulb is sure. However when speaking of individual photons, which is what I think the OP is asking about, that is not correct.

 

[JaWiB]

I don't think you can define "amplitude" for a single photon. In the classical picture, light is an electromagnetic wave and the amplitude of the electric/magnetic fields is the property you talk about. In the quantum picture, light is made up of photons and the number of photons describe the amplitude. That's my understanding.

 

[Drakkith]

I don't think you can define "amplitude" for a single photon. In the classical picture, light is an electromagnetic wave and the amplitude of the electric/magnetic fields is the property you talk about. In the quantum picture, light is made up of photons and the number of photons describe the amplitude. That's my understanding.

Yes, I agree that we generally use amplitude to mean the number of photons. However in terms of a single photon I am unsure whether we can define amplitude or not. I swear I remember reading something about planks constant having something to do with the amplitude though.

 

[HallsofIvy]

Where did "photons" come into this? The original post was about the amplitude of light waves, not photons.

 

[Drakkith]

Where did "photons" come into this? The original post was about the amplitude of light waves, not photons.

After all, how would a light wave of a given wavelength increase its amplitude? If an electron falls a greater distance, the light emitted won't be more intense, it will actually have a higher frequency, right?

This plus the context of the OP's post suggests he is talking about a photon, not a large collection of photons. Unless I am missing some sort of difference between "light wave" and "photon".

 

[sweet springs]

Hi,　I write down energy density of monochoromatic light of frequency ν as

1/2 (ED + BH) = ε0 E^2 = N h ν

where E is electric field, D is density of electric flux, B is density of magnetic flux, H is magnetic field intensity, ε0 is electric permittivity in vacuum, h is Planck constant and N is number density of photon.
Amplitude of light, E, undertakes superposition or interference of lights from various sources.
Lights of frequency v1 and ν2 can have same amplitude E or same energy density when N1 h ν1 = N2 h ν2, higher the frequency of light, fewer the number density of photons. However, quality differs, for example, hundreds of green color photons entertain us but one X-ray photon of same energy may harm our cell.

Regards.

 

[Drakkith]

Of course, but can you define amplitude for a single wave or photon as the OP is asking?

 

[sweet springs]

How about N = 1, say single photon in unit volume, in the formula 1/2 (ED + BH) = ε0 E^2 = N h ν?

 

[Drakkith]

How about N = 1, say single photon in unit volume, in the formula 1/2 (ED + BH) = ε0 E^2 = N h ν?

But what is the amplitude referring to in this equation? The amount of energy found in N photons?

 

[sweet springs]

But what is the amplitude referring to in this equation? The amount of energy found in N photons?

E in this formula is amplitude of light,　say square root of h ν / ε0　for single photon in unit volume.

Regards

 

[Drakkith]

E in this formula is amplitude of light,　say square root of h ν / ε0　for single photon in unit volume.

Regards

Amplitude of light meaning that it is the energy of N photons?

 

[Naty1]

Photons exhibit wave particle duality, right?? What doesn't?? ..so they DO have amplitude and wavelength...and experience wave inteference and polarization, for example.

Einstein's photon model accounted for the frequency dependence of light's energy, E = hv.
I n a sound wave, extra energy is louder, in light, it's a higher frequency...say X-ray instead of visible.

I was wondering if perhaps the intensity of light is only related to how much of it there is, and maybe the amplitude of a light wave is set.

See here for an illustration of an electromagnetic wave: note the amplitude is given by the amplitude of the electric (E) and magnetic (B) field vectors...amplitudes vary.

"intensity" is one thing to the human eye, another to objective instruments. Maybe lumens is most closely related to "brightness" or "intensity"...I'm not sure that's the BEST term.

 

[Philip Wood]

When we have large numbers of photons, the photon stream behaves as an e-m wave, in which energy is spread out over broad wavefronts. A natural meaning to give to amplitude is the maximum value, E_max of the electric field strength, as the wave passes a point. Alternatively, it could be the maximum value, B_max of the magnetic field strength. One would know which one from the units. In any case, they are related very simply:
E_max = cB_max.

The wave intensity, i.e. the wave energy per unit area per second passing through an area normal to the waves' direction of travel is $\frac{1}{2}c\epsilon_{0} E_{max}^{2}$. So, assuming monochromatic waves of frequency f, for which the photon energy is hf, the photon flux density (i.e. the number of photons per unit area per second passing through an area normal to the waves' direction of travel) is $\frac{1}{2}\frac{c\epsilon_{0}}{hf} E_{max}^{2}$. So the photon flux density is proportional to E_max², i.e. to amplitude².

 

[Redbelly98]

A natural meaning to give to amplitude is the maximum value, E_max of the electric field strength, as the wave passes a point. Alternatively, it could be the maximum value, B_max of the magnetic field strength. One would know which one from the units. In any case, they are related very simply:
E_max = cB_max.

Yes, exactly. The amplitude refers to the peak electric or magnetic field strength.

It is not necessary to bring photons into the discussion, the original question was about the classical description of light as a wave.

 

[klimatos]

Waves in general have three properties: frequency (related to wavelength), amplitude, and speed. When referring to light as a wave, it's wavelength and speed are always referenced but never its amplitude, and I was wondering if light has a fixed amplitude for all wavelengths or if it changes.

It has always been my understanding that the frequency, wavelength, and energy content (amplitude of a sort) can be rigorously related by two equations:

The first equation relates frequency to wavelength to the speed of light; that is, the speed of light in meters per second is equal to the frequency in cycles per second times the wavelength in meters.

The second equation relates the energy content of the photon to both frequency and wavelength. This energy content is equal to Planck's Constant in Joule seconds times the frequency and is also equal to Planck's Constant times the speed of light, the product divided by the wavelength.

I believe that these two equations will work if you consider a single wave to be the equivalent of a photon.

 

[ZapperZ]

Waves in general have three properties: frequency (related to wavelength), amplitude, and speed. When referring to light as a wave, it's wavelength and speed are always referenced but never its amplitude, and I was wondering if light has a fixed amplitude for all wavelengths or if it changes.

Sound waves, for example, get louder in two cases: when there's more of them (multiple speakers) or when the amplitude of the sound waves is increased (more volume). Obviously having more light waves (multiple light sources) will result in increased brightness, but can the amplitude of a light wave change as well?

Since I've never heard the amplitude of a light wave referred to, plus the fact that higher frequency (shorter wavelength) light is referred to as more energetic, I was wondering if perhaps the intensity of light is only related to how much of it there is, and maybe the amplitude of a light wave is set. After all, how would a light wave of a given wavelength increase its amplitude? If an electron falls a greater distance, the light emitted won't be more intense, it will actually have a higher frequency, right?

Thanks for clearing this up for me!

When you solve Maxwell's equations, you "wave" equation for light (or EM radiation in general) has amplitudes defined by the E and B fields. So yes, light does have amplitudes.

With analog radio signals, you can see that the amplitude of a signal is larger the closer you get to the transmission source, all without requiring changing the frequency.

Note that when you start solving classical E&M for fields from antennas, you'll notice that you can simply have electrons to oscillate with a larger amplitude, at the same frequency, to increase the amplitude of the EM signal.

Zz.

 

[Drakkith]

Yes, exactly. The amplitude refers to the peak electric or magnetic field strength.

It is not necessary to bring photons into the discussion, the original question was about the classical description of light as a wave.

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"?

 

[nonequilibrium]

Drakkith, you're assuming the poster knew that photons are essential when talking about an electron changing orbit, but I can well imagine someone having heard of the wave formulation of light (being aware that accelerating particles radiate), but not having had a class on quantum mechanics, and I think this is the case.

 

[klimatos]

I believe that these two equations will work if you consider a single wave to be the equivalent of a photon.

Strike that statement. That has to be one of the stupider statements I have made recently.

 

[Drakkith]

Drakkith, you're assuming the poster knew that photons are essential when talking about an electron changing orbit, but I can well imagine someone having heard of the wave formulation of light (being aware that accelerating particles radiate), but not having had a class on quantum mechanics, and I think this is the case.

I'm not assuming anything, I actually don't know where you would use photon and where you wouldn't. In any case I think the classical answer has been given, and I'm curious about the other end. What would amplitude represent for a single photon? Something to do with the strength of the electric and magnetic fields?

 

[Redbelly98]

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"?

Well, you're right that the OP apparently referred to photons -- not explicitly, but his comment "if an electron falls a greater distance, the light emitted..." refers to photons.

But I'll maintain that a discussion of light as a wave does not require talking about photons. People talked about waves for a long time before Planck came up with the idea of quanta or photons.

 

[sweet springs]

Hi, all
Let me do a little correction to my previous post below.

Hi,　I write down energy density of monochoromatic light of frequency ν as
1/2 (ED + BH) = ε0 E^2 = N h ν
where E is electric field, D is density of electric flux, B is density of magnetic flux, H is magnetic field intensity, ε0 is electric permittivity in vacuum, h is Planck constant and N is number density of photon.
Amplitude of light, E, undertakes superposition or interference of lights from various sources.

Uncertainty of position of photon is greater than wavelength λ=c/ν according to relativistic quantum mechanics. For this reason one cannot invent probability amplitude of photon as we do for electrons or other particles with rest mass in Shrodinger equation. Energy of sinusoidal electromagnetic wave in this rough spatial resolution should be considered as spatial average, so I have to multiply factor 1/2 to 1/2 (ED + BH) = ε0 E^2 in the formula. Let me state again that:
There is a box of x-length a, y-depth b, z-height d. In the box there are n photons moving z direction of wavelength λ << d. Answer to the question :

What would amplitude represent for a single photon? Something to do with the strength of the electric and magnetic fields?

Electric field and magnetic flux density in the box corresponds to photons.
E = sqrt { 2nhν / (ε0 abd)}, B and E/c
For example
For four photons in the box, E = sqrt { 8hν / (ε0 abd)} and B = E/c
For three photons in the box, E = sqrt { 6hν / (ε0 abd)} and B = E/c
For two photons in the box, E = sqrt { 4hν / (ε0 abd)} and B = E/c
For SINGLE PHOTON in the box, E = sqrt { 2hν / (ε0 abd)} and B = E/c
For zero photon in the box, E = B = 0

E and B depend on not only wavelength of light and number of photons in the box but also volume of the box.

I would appreciate someone point out mistakes here for my progress.

Regards

 

[jtbell]

What would amplitude represent for a single photon? Something to do with the strength of the electric and magnetic fields?

When you are dealing with a single photon, it's no longer possible to talk meaningfully about the classical electric and magnetic fields. There is a "photon field" $A_\mu$ which gets quantized in QED, and it's tempting to associate it with a probablity density similar to the way we do it in ordinary Schrödinger-type QM: $P(x,t) = \Psi^*(x,t) \Psi(x,t)$. But I don't remember doing this sort of thing when studying QED many years ago; it was all about calculating scattering amplitudes and cross-sections.

Electric field and magnetic flux density in the box corresponds to photons.
E = sqrt { 2nhν / (ε0 abd)}, B and E/c

This is valid for large numbers of photons, i.e. in the classical limit, but not (as far as I know) for small numbers such as these:

For four photons in the box, E = sqrt { 8hν / (ε0 abd)} and B = E/c
For three photons in the box, E = sqrt { 6hν / (ε0 abd)} and B = E/c
For two photons in the box, E = sqrt { 4hν / (ε0 abd)} and B = E/c
For SINGLE PHOTON in the box, E = sqrt { 2hν / (ε0 abd)} and B = E/c

 

[Oudeis Eimi]

Hi, all
Let me do a little correction to my previous post below.

Uncertainty of position of photon is greater than wavelength λ=c/ν according to relativistic quantum mechanics. For this reason one cannot invent probability amplitude of photon as we do for electrons or other particles with rest mass in Shrodinger equation.

This is true, but the situation is even more serious than that. Photons seem to
be truly unlocalisable objects. They don't have a position operator, period. That
makes them fundamentally different to, say, electrons.

Energy of sinusoidal electromagnetic wave in this rough spatial resolution should be considered as spatial average, so I have to multiply factor 1/2 to 1/2 (ED + BH) = ε0 E^2 in the formula. Let me state again that:
There is a box of x-length a, y-depth b, z-height d. In the box there are n photons moving z direction of wavelength λ << d. Answer to the question :


Electric field and magnetic flux density in the box corresponds to photons.
E = sqrt { 2nhν / (ε0 abd)}, B and E/c
For example
For four photons in the box, E = sqrt { 8hν / (ε0 abd)} and B = E/c
For three photons in the box, E = sqrt { 6hν / (ε0 abd)} and B = E/c
For two photons in the box, E = sqrt { 4hν / (ε0 abd)} and B = E/c
For SINGLE PHOTON in the box, E = sqrt { 2hν / (ε0 abd)} and B = E/c
For zero photon in the box, E = B = 0

E and B depend on not only wavelength of light and number of photons in the box but also volume of the box.

I would appreciate someone point out mistakes here for my progress.

Regards

More an observation than a correction: the above isn't strictly correct, because
a single-photon state (or in fact, whenever the e-m field contains a well defined
number of photons) the electric and magnetic fields are *zero*! The lesson here
is that single photon states are VERY non classical. They really are rather different
from the classical monochromatic em wave.

So, for four, three, two, one photon, E is zero (but its indeterminacy isn't).

This doesn't mean you can't include photons in a discussion of em waves, though.
If you have a classical, monochromatic wave E = E0 cos(w t), its amplitude
E squared is proportional to the expected value of the photon number operator.
That is, E squared is proportional to the 'mean' number of photons. For very
low intensity waves, this can be of the order of one photon. Just note that you
still don't have a single photon - your state is pretty different to a single photon
state. It's a coherent state with a mean photon number of one, a significant
difference.

 

[sweet springs]

Hi. Thank you for nice teachings. Let me know it better by clarification.

This is true, but the situation is even more serious than that. Photons seem to
be truly unlocalisable objects. They don't have a position operator, period. That
makes them fundamentally different to, say, electrons.

When I switch on light bulb in this room, it makes here bright. There are photons in this room.
When my brother switches on light bulb in that room, it makes there bright. There are photons in that room.
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.

 

[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.]