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I have found closely related questions on StackExchange, but (surprisingly) not this exact question. Seems some answers say individual photons do not have amplitude, only when traveling with other photons, forming a wave. But even then, can't the amplitude be measured? If so, what are the range of amplitudes and how do they vary? Based on the frequencies and energies of the light? Does higher frequency correspond to lower amplitude?

Moderators: This question is not a duplicate to this question: Amplitude of an electromagnetic wave containing a single photon

That prior question is more about the formula for calculating electromagnetic energy amplitude of a single photon. My question is about an experimental way to MEASURE the positional (x) amplitude of the wave passing through a specific apparatus. Two very different things. Thanks.

I have updated my question to a more specific thought experiment:

If you had a vertical slit with a horizontal width that could be varied, and you passed horizontally polarized light through this slit, wouldn't the slit block any photons from passing through it if the width of the slit was smaller than the amplitude of the light waves?

Wouldn't we at least get a predicted or average amplitude, even with uncertainty principle?

Wouldn't the energy/frequency/wavelength of the wave also have some correlation to the amplitude?

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Brad Cooper - Purpose Nation
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  • amplitude is related to number of photons. frequency to the energy of each one of them. energy of light depends on both. no higher frequency does not correspond to lower amplitude. – Azad May 16 '15 at 14:34
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    Thanks, Azad. So a single photon has no amplitude? If not, it would seem this also means it has no polarity? – Brad Cooper - Purpose Nation May 16 '15 at 14:53
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    John, I did see that question, but it wasn't the same and none of the answers directly answered my question. Thanks – Brad Cooper - Purpose Nation May 16 '15 at 14:59
  • Photons aren't "objects" but the quanta of the electromagnetic field. What's a quantum? It's the measured change of state of a quantum object, in this case that of the quantum field that makes up the universe. It's much better to think of a photon as a set of numbers that are similar to the ones that describe the state of an atom than as a "particle". – CuriousOne May 16 '15 at 15:17
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    CuriousOne: I see frequency and velocity of EMF "waves" used all the time with no need to explain that EMF is both particle and wave. So, why then, when it comes to amplitude is there so much waffling and saying that we shouldn't think about photons as actual things, but rather a "set of numbers"? – Brad Cooper - Purpose Nation May 16 '15 at 15:22
  • @PurposeNation a single photon can only have circular clockwise or circular counter-clockwise polarity because it can have a spin of +1 or -1. all other polarities (linear, elliptical) are made by superposition of polarities of ensemble of photons. – Azad May 16 '15 at 15:34
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    Azad, ok, so can we measure the average or predicted amplitude of an "ensemble of photons" that are horizontally polarized passing through a vertical slit? If so, does this predicted/average amplitude go up or down based on the energy or frequency of the EM waves? – Brad Cooper - Purpose Nation May 16 '15 at 15:43
  • " Wouldn't the energy or frequency of the wave also have some correlation to the amplitude?" No. Not for light, not for sound, not for waves on a string. – dmckee --- ex-moderator kitten May 16 '15 at 15:54
  • @PurposeNation yes we can. it will still be horizontal it doesn't change polarity because EM waves have no physical extent that could get stuck in the slit. but there's a similar idea which do polarize https://en.wikipedia.org/wiki/Polarizer#Wire-grid_polarizer but by blocking the electric field – Azad May 16 '15 at 16:01
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    Isn't the very definition of AM radio the creation and receiving (ie measuring) varying degrees of EM amplitude? You are saying that waves with small amplitude and the same frequency have the same energy as waves with larger amplitude, same frequency? How can that be? http://en.wikipedia.org/wiki/Amplitude_modulation – Brad Cooper - Purpose Nation May 16 '15 at 16:07
  • @dmckee: Have read several places where the "amplitude of an electromagnetic wave that determines its energy density and intensity" and the specific equation is defined: http://www.quora.com/How-does-the-amplitude-of-an-electromagnetic-wave-affect-its-energy this article also articulated the quantum & classical differences: "A single photon has a well-defined frequency & energy but not a well-defined amplitude" Still, "not well-defined" doesn't mean cannot be defined. So, maybe it's a scatter with amplitudes in a range of values, varying with energy approximations based on these equations? – Brad Cooper - Purpose Nation May 16 '15 at 16:27
  • @Purpose I'm too busy today to field a string of questions, but I would strongly suggest not trying to match up the classical and photon descriptions of light until you have a thorough understanding of each on its own merits. That correspondence is hard enough that elderly theorists have told me that they have only partially mastered it. – dmckee --- ex-moderator kitten May 16 '15 at 17:43
  • My earlier comment was directed in particular at the notion of a relationship between frequency and amplitude, as you note there is a relationship between energy density and amplitude, but your question, but the coupling is not through frequency. But note the word density. Photons have energy, but classical waves have energy densities. – dmckee --- ex-moderator kitten May 16 '15 at 17:46

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Classically (since rob has done a thorough job on the quantum picture), the amplitude of a light wave is not related to any physical extent. It is not the size of the wave in space, it is the strength of the fields (electric and magnetic).

We often draw wavy lines, but if you look closely the transverse axes will be label differently for, say, waves on a string and electromagnetic waves. You should not take those lines to imply a displacement the way they do in ripple on a pond. They just mean differing values of the field.

Classically, you can not filter different amplitudes with slit width. You simple block more light and create more diffraction as the slit grows narrower.

dmckee --- ex-moderator kitten
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  • Thanks, but this doesn't seem to answer my updated question and doesn't explain polarized EM waves: http://en.wikipedia.org/wiki/Polarization_%28waves%29#/media/File:Electromagnetic_wave.png Are you saying there isn't really horizontal or vertical polarization of EM waves? – Brad Cooper - Purpose Nation May 16 '15 at 15:38
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    Recall that electric and magnetic fields are vectors (have direction)? That means that two waves can have the same amplitude but different electric field orientations which is what polarization is about. Conventionally we draw the field as an arrow, giving both direction and strength, but the longer arrow does not mean that a strong field is longer in space. So the answer to your question is that classically you cannot filter different amplitude with slit width. – dmckee --- ex-moderator kitten May 16 '15 at 15:45
  • what do you mean by "electric field orientations"? oriented with respect to what? The physical X/Y orientation of physical space? So you cannot have physical vertical slits that filter out horizontally polarized light? I thought that was the very definition of polarized light? http://www.olympusmicro.com/primer/lightandcolor/polarization.html – Brad Cooper - Purpose Nation May 16 '15 at 15:56
  • Nice answer, dmckee. I missed @PurposeNation 's confusion between the amplitude of the electric field and the physical size of the wave – rob May 16 '15 at 16:24
  • @Rob: What does the "physical size of the wave" mean? I didn't ask about an amplitude of a field, but rather the amplitude of a wave in that field. – Brad Cooper - Purpose Nation May 16 '15 at 16:31
  • Thanks, @dmckee. I think I get this now. However, now I am confused by what this means for interference patterns in double-slit experiments: http://physics.stackexchange.com/questions/188185/if-em-waves-are-not-physical-positional-waves-on-a-x-y-z-axis-why-does-inter – Brad Cooper - Purpose Nation Jun 07 '15 at 15:52
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If you twisted my arm and forced me to assign an amplitude to a single photon, I'd do it this way:

  • The energy density of a classical electromagnetic field is \begin{align} U &= \frac12 \left( \epsilon_0 E^2 + \frac1{\mu_0} B^2 \right) \\ &= \epsilon_0 E^2 &\text{(only for light in a vacuum)} \end{align} where $E,B$ are the amplitudes of the electric and magnetic fields

  • The total energy carried by the photon is $T=cp=\hbar c k$, where the wave vector $k=2\pi/\lambda$. (I shouldn't use $E$ for field amplitude and total energy.)

We can connect these if we come up with a reasonable estimate for the volume of a single photon. Which we probably shouldn't. But if you continued to twist my arm, I would say that the photon "fills" a long, skinny cylinder.

  • The length scale for a photon perpendicular to the propagation direction $\vec k$ is roughly the same as the wavelength, so the "area" of the cylinder will be roughly $\lambda^2$.

  • The length scale parallel to $\vec k$ is given by the uncertainty principle: \begin{align} \Delta x \Delta p = \hbar \Delta x \Delta k \approx \frac \hbar2 \\ \Delta x \Delta k \approx \frac12 \end{align} Relating the uncertainty in the wavevector to the uncertainty in the wavelength is a little tricky, because calculus, \begin{align} \Delta k = \Delta \left( \frac{2\pi}{\lambda} \right) = \Delta\lambda \frac{2\pi}{\lambda^2}, \end{align} which gives a position uncertainty $$ \Delta x = \frac1{2\Delta k} = \frac{\pi\lambda^2}{\Delta\lambda}. $$

So we have total energy $T = hc/\lambda$, smeared out over a volume like $V \approx \pi\lambda^4/\Delta \lambda $, to compare to a classical energy density $T/V \approx U = \epsilon_0 E^2$. That gives us $$ E^2 \approx \frac{hc \Delta\lambda}{\epsilon_0\pi\lambda^5} $$ It's probably less incorrect to use the uncertainty in the direction of $\vec k$ to find the "area" of the cylinder, but I'll stop here.

rob
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  • If you continued to twist my arm, I'd say that the whole concept is ill-founded – rob May 16 '15 at 15:25
  • Thanks, Rob. In your picture, with horizontally polarized light, it doesn't seem like it's really a cylinder, maybe a very vertically flattened cylinder? I'm wondering if you can get an estimate of the average or predicted horizontal diameter of that cylinder as the EMF waves pass through a slit? However, Does the horizontal diameter of that flat cylinder go up or go down as the energy of the light increases? Or are the wave energy and that diameter not correlated? – Brad Cooper - Purpose Nation May 16 '15 at 15:28
  • The "transverse size" of a photon is comparable to the wavelength; that's why you get diffraction from wavelength-sized slits, but not from macroscopic openings. – rob May 16 '15 at 15:33
  • So in my thought experiment, if the horizontal width of the slit was smaller than the wavelength, no light would pass through, then? If the horizontal width of the slit was larger than the wavelength, some light would pass through, somewhat dependent upon Schrodinger and also the quality of the polarity of the light (if it's not perfectly polarized). In this sense, seems amplitude = wavelength, then? – Brad Cooper - Purpose Nation May 16 '15 at 15:49
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    No, light can pass through sub-wavelength slits, but you can't think of it as a particle in that case. Don't push your luck with this analogy. – rob May 16 '15 at 16:19
  • ok, would you kindly, then, explain "The 'transverse size' of a photon is comparable to the wavelength" as it relates to my question? It doesn't seem to answer it. thank you. – Brad Cooper - Purpose Nation May 16 '15 at 16:33
  • Let us continue this discussion in chat. – rob May 16 '15 at 17:17