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Is the actual energy of a photon ever measured? How is it done?

I read that a photon is usually identified by diffraction, that means its wavelength is measured, is that right? In this way we determine that a red-light photon makes a full oscillation in roughly $700 nm$Then, by theoretical reasons, we deduce its energy is about $400 THz$. I wonder if, when this relation was established, ther was an instrument capable of precisely counting $4*10^{14}$ oscillations in an exact second.

What I am trying to understand is what exacly is a photon, when we can talk about the existence of a photon. I'll try to explain what is my problem:

we usually receive light (or radio waves) from a continuous source: the Sun, a flame, a bulb etc..., right? how/when can we isolate a single photon? Is a photon the whole set of oscillations during the span of a second? can we consider a photon of red light a single oscillation of the EM field lasting only $\frac{1}{4*10^{14}}$ second, even if no instrument would be able to detect it? Yet, that photo would propagate all the same at $c$ in vacuum for a second or more or forever, but, though its wavelength its still $700 nm$, its energy would probably not be in the region of THz anymore, or would it? To frame it differently, does energy change if the unit of time is halved or doubled?

I hope you can understand my questions even though my exposition is confused.

Edit

The energy of anything is a definite issue, it is the result of a meaurement, and does not depend on theoretical considerations, QM, classical or other models, or onthe fact whether it is a wave or not.

I'll try to clarify my main concern with a concrete example: consider something you can control and manipulate. You can produce low frequency EMR (short radio wave about 10m wavelength) making a charge oscillate up and down 30 million times a second, right?

Now, suppose you make that charge oscillate only for 1/1000 th of a second. all the same That wave will propagate at C and will oscillate 3*10^7 times a second and will be diffracted revealing a wavelength of 10m, is that right or will it oscillate only 30 000 times a second?

Furthermore, whenit hits something or you determine its frequency, will it still have the same energy of a wave that has been produced making a charge oscillate for one whole second or for ten seconds?

  • The speed of light was known, so was the wavelength of red light. Both together give you the frequency: $\lambda\nu=c$. – CuriousOne Feb 11 '16 at 08:48
  • To answer your question about the nature of a photon: a photon is a quantum of a quantum field. Detecting a photon is like detecting hydrogen in an s-state. – CuriousOne Feb 11 '16 at 09:02
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    @CuriousOne : OP wants to know how they measured such small wavelength of red light. – Anubhav Goel Feb 11 '16 at 09:09
  • @AnubhavGoel: With diffraction and interferometers. That's pretty easy, isn't it? Direct frequency counting has only been done recently.... sort of... Maxwell's paper mentions the frequency as "thousand million million cycles per second" if I am not mistaken. That's spot on. – CuriousOne Feb 11 '16 at 09:10
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    @CuriousOne : So, scientist never measured wavelength using distance traveled1 oscillation but just measured using tricks. Well, thanks for info. – Anubhav Goel Feb 11 '16 at 09:20
  • @AnubhavGoel: It's the other way round. In the 19th century they measured the wavelength, which is easy, but they couldn't measure the frequency. It took until the mid 20th century before we could measure frequencies well above a GHz directly and the THz range has just become accessible to "counting". There are no more tricks involved in this than in anything else... it's just straight forward physics. – CuriousOne Feb 11 '16 at 09:28
  • The fact that a photon is an oscillating wave is a misconception. This misconception has already been addressed in http://physics.stackexchange.com/questions/234861/why-doesnt-a-plane-wave-solution-represent-a-single-photon/234879#234879 and in http://physics.stackexchange.com/questions/234839/quantum-state-of-photon-question/234848#234848. An oscillating (electromagnetic) wave does NOT correspond to a photon. The frequency for which $E=\hbar \omega$ is the frequency of oscillation of the wavefunction of the photon, which does not correspond to the electric field produced by it. – Giorgio Comitini Feb 11 '16 at 12:13
  • This is what anna v was trying to deliver to you through her answer, though she wasn't as clear as she could be. In the limit of high wavefunction frequencies, it's easier and easier to build up oscillating classical fields of actual frequency $\omega$ from quantum superpositions of photons of wave-function frequency $\omega$. – Giorgio Comitini Feb 11 '16 at 12:20
  • And definitely: photons do not travel at some definite number of oscillations per second. In fact, they do not "travel" at all, no more than electrons or other quanta do, as by the Uncertainty Principle they don't have a definite speed and/or trajectory. On the other hand, they have definite energies and momenta. – Giorgio Comitini Feb 11 '16 at 12:30

5 Answers5

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The classical light beam, an electromagnetic wave, emerges from zillions of photons which travel with velocity $c$ and build it up.

The energy of a photon is $E=h\times \nu$, where $h$ the Planck constant, and $\nu$ is the frequency which will appear in a classical wave built up by this energy photons. The way this happens is explained mathematically here, but is not simple to understand without quantum mechanics and field theory. The photon itself is not oscillating in (x,y,z,t). Only travelling with velocity $c$.

The energy of the classical wave is given by the average intensity, for example , for a plane wave it can be written

$$S=\frac{1}{c\mu_0}E^2\overline{\sin^2(kx-\omega t)}=\frac{1}{c\mu_0}\frac{E_m^2}{2} $$

where here $E$ is the electric field of the classical light wave.

The individual energy of $h\nu$ of photons will add up to the energy transferred by the collective electromagnetic wave.

The velocity of the photon is fixed and does not change unless there is an interaction, as in Compton scattering,. It is an elementary particle of the standard model .

The theoretical model, called quantum electrodynamics, is so well validated with experimental data that one can identify the energy of the photon with the frequency of the classical light beam, and use classical interference set ups. Individual photons from known frequency light beams have been observed through the double slit experiment, as dots on a screen. The existence of photons and their frequency with energy association is well validated.

When energies become large , as in X rays and gamma rays, different laboratory techniques can identify the energy of a single photon, as with the photoelectric effect, and with the electromagnetic calorimeters in particle experiments identifying single gamma rays of great energy.

For example this Higgs to gamma gamma event:

higgstogammagamm

The green lines display the energy deposited in the calorimeters by each gamma. It is known as a photon because it does not interact in the tracking chambers, and deposits the energy in the electromagnetic ones.

so to your questions

can we isolate a single photon?

Yes, as seen above.

Is a photon the whole set of oscillations during the span of a second? can we consider a photon of red light a single oscillation of the EM field lasting only 1 second/ 4*10^14 second

No. as said the photon is an elementary particle and the classical light beam rides on zillions of photons , each contributing in synergy a tiny part to the electric and magnetic fields of the electromagnetic field.

To frame it differently, does energy change if the unit of time is halved or doubled?

No the energy of the photon is always equal to $h\nu$, for the whole spectrum. $ν$ is just there for a photon only as a handle to inform what type of light beam a zillion of such energy photons will generate, see the table.

There is a mathematical quantum mechanical connection between the classical electrodynamic solutions and the quantum mechanical solution, because maxwell's equations are quantized and give rise to the wavefunction of the photon. In the wavefunction, which is complex, i.e not measurable, there are the same E nd B and $ν$ that will be built up by zillions of photons. This is the frequency appearing in the double slit experiment , one photon at a time. It is in the probability of detection of the photon that the frequency plays a role.

You ask in a comment:

if you oscillate a charge 3000 times in 1/10000 of a second you get a photon that discharges $3\times 10^7\ h$ of energy when it hits something

The units are not understandable, but no, an oscillating charge will not give one photon, it will depend on the boundary values and will give a spectrum of classical frequencies which will be composed of innumerable photons.

anna v
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  • I did not understand no. of oscillations part. When we can isolate a single photon, how many oscillations or part of wavelength it has? $\lambda , \lambda/2 , 2\lambda$ or something else. What is that tiny part? – Anubhav Goel Feb 11 '16 at 08:59
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    @AnubhavGoel The photon has a complex wavefunction that can build up the oscillating electromagnetic wave with other photons. (think of it like a stadium wave,one lifts once ones hand) It does not vary itself when single, It has energy hnu , the corresponding momentum and spin pointing either with the direction or against the direction of motion. That is all that can be measured by a single photon. – anna v Feb 11 '16 at 09:18
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    "..the energy of the photon is always h x nu, for the whole spectrum.", yes, but frequency depends indeed on the unit of time you choose. That was the gist of my post. If they had chosen a t ten times smaller or bigger, then the frequency of red light would have been 410^13 or 410^15 Hz. If you link energy to frequency, then... the energy of a single red light photon would vary accordingly, right? –  Feb 11 '16 at 10:05
  • Hi anna. I think that your answer struck the point: photons are no oscillating fields. However, I suggest you to elaborate a bit more on this concept, in order to make it explicit and more understandable to the questioner - the first paragraph is a bit unclear as what you're trying to say, in particular when you state that a photon has a frequency which emerges from a huge number of photons. – Giorgio Comitini Feb 11 '16 at 12:27
  • @GiorgioComitini, it doesn't really matter if a photon is an oscillating field or not, nor iss the fact that it has/n't a trajectory or speed. If you say it has energy , how do you measure it? that is a very concrete, down-to-earth issue. If you relate it to an arbitrary unit (time), there is a huge problem: how long does it take you to measure it, for how long do you have to measure it to get the exact value –  Feb 11 '16 at 13:27
  • It does matter, based on what you asked. "I read that a photon is usually identified by diffraction, that means its wavelength is measured, is that right?" No. Diffraction is a related to wave-like behavior and a single photon doesn't have one. Bunches of photon (more technically, superpositions of photonic quantum states), on the other hand, do. Any assertion/assumption you made thereafter about being able to measure the energy of a photon by measuring its wavelength is simply wrong, and so are the answers which are based on those assumptions. This is what anna, Whit3rd and Dimitri were sayng – Giorgio Comitini Feb 11 '16 at 13:44
  • As the photon has energy, you measure that energy by using the principle of conservation of energy. You wait until the photon strikes something and you measure the energy that it released in the impact from a knowledge of the physical properties of the object that it struck. This is done in colliders, as anna said, or through the photoelectric effect, as Whit3rd said. A single photon has a single, definite energy. If you were able to measure that energy, you would get one single value, independently of the time of the measure. – Giorgio Comitini Feb 11 '16 at 13:51
  • It should be clear that the units in which you measure time really don't make any difference. You always have a means to convert your units: the answer you get may formally depend on units, but units can be converted into one-another, so that the final result does not really depend on them. On the other hand, what you really have in experiments as an input is a superposition of photons, not a single one. Then you're simply not measuring the energy of a single photon. Then you'll be able to measure energies in a range which depends on the specific superposition. – Giorgio Comitini Feb 11 '16 at 13:55
  • In order to give to the electromagnetic field a wave-like appearance, i.e. in order to be able to speak of frequencies and wavelengths in a classical way, this superposition, as anna said, must contain not only an infinite number of photons, they must contain states with ANY number of photons from zero to infinity. So, again, unless you refer to the frequency in the internal, theoretical space of quantum mechanics (and you're not referring to that), when you talk about frequencies giving it a classical meaning you're not talking about photons. – Giorgio Comitini Feb 11 '16 at 14:05
  • The energy of a single photon is $\hbar \omega$ only as long as $\omega$ is the frequency in the internal theoretical space of QM, not the frequency of some classical wave. The energy of a classical wave of classical frequency $\omega$ is not $\hbar\omega$: it is infinite. Its (spatial, time-averaged) energy density isn't proportional to $\omega$. Again, if you want to measure the energy of a single photon, you'd better not talk about its wave-like behavior (as the photon doesn't have one). – Giorgio Comitini Feb 11 '16 at 14:10
  • @GiorgioComitini, it would be nice if you could write an answer and give your reply (especially to my edit) considering your expertise in QM. Does QM say that if you oscillate a charge 3000 times in 1/10000 of a second you get a photon that discharges 3*10^7 h of energy when it hits something? –  Feb 11 '16 at 15:09
  • If you're willing to accept a somewhat technical answers (as far as my knowledge goes: unfortunately I am not an expert, I only know the basics of QED and QM), I'd be willing to study your specific problem in the appropriate context (that of an oscillating charge producing an EM field) and present to you my results (I have no readily available reference to give). I have not studied the problem in full detail yet, but there is something I would like to stress from a general perspective. – Giorgio Comitini Feb 11 '16 at 16:53
  • In QM you don't have "charges". You either have a charge distribution, which is understood in a probabilistic way, i.e. not in the classical sense, or you have particles, which have definite momenta and as such do not have position. As they do not have position, you cannot oscillate them. Moreover, no exact solutions are known to the problem of matter interacting with radiation. This means that you can only make approximations (in a perturbative fashion, if you know what I mean), so I'd suggest not to analyze your problem by starting from the charges. – Giorgio Comitini Feb 11 '16 at 16:59
  • On the other hand, starting with the charges is not required as much as you don't require to know the charge distribution in classical mechanics to know that some specific field configuration exists in vacuum. You can always study the specific field configuration by itself. In this case, anna's answer (especially after her edit, some 2 minutes ago) addresses the question from the correct viewpoint (you may also have a look at the answers I linked you before). Your reasoning fails when you assume that a photon has a wavelength which you can derive from the frequency of the oscillating "charge". – Giorgio Comitini Feb 11 '16 at 17:09
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The easiest way to measure the energy of a photon, is to make a reaction using the photoelectric effect. The photon hits a surface, knocks out an electron, and the electron can be prevented from carrying charge away from the surface by putting a small attracting voltage onto that surface (this is called the 'stopping potential').
It is an experiment usually carried out with a vacuum phototube, and a LOT of photons, all of the same energy (color). You can adjust the stopping potential until the phototube doesn't generate any current when illuminated. The stopping potential (voltage) multiplied by the charge of an electron is the estimate of the photon energy (there's some corrections for the material of the phototube, and it doesn't work well on low-energy photons).

Whit3rd
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    Of course, this means you trust the electrons ejection energy (at 0V stopping potential) to be nearly zero. But if one assumes we have good understanding of electron orbital potentials, etc., this works. – Carl Witthoft Feb 11 '16 at 14:25
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It is convenient to use wavelength as a unit of measurement when it comes to photons but really it starts with the frequency that the photon is oscillating at. We know the speed of light and if you diffract it from one surface to another we can measure the distances using Pythagorean's Theorem to determine the frequency or wavelength. A photon oscillating at 400THZ and traveling along at the speed of light will complete one oscillations every 700 nm. The energy (frequency) of a photon does not change if the unit of time is halved or doubled? Wavelength is not a wave, its just the distance a photon travels in one oscillation. The speed stays the same and if the photon has less energy (lower frequency) then the so called wavelength will be longer.

Bill Alsept
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I think you get confused because you try to picture a photon in a classical framework. There is (unfortunately) no accurate way of explaining what a photon is without the use of quantum mechanics. A photon is much more complicated than just "a set of oscillations during the span of a second" : it is in essence a quantum of excitation of the electromagnetic field in the vacuum. I know it might sound a bit technical, but you need the framework of quantum field theory (more precisely, quantum electrodynamics) to fully understand the nature of the photon.

There are number of classical experiments that were reconducted in a "quantum way" that might give an awnser to some of your (very legitimate) doubts on the topic. For instance, many versions of the famous Young's double-slit experiment have been studied with single photons, or in other complicated situations that reveal the quantum nature of light.

Dimitri
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I worked on the firmware of the femtosecond laser Maitai. Each pulse of light last approximatively 20 fs. The infrared light at 800 nm has time to complete about 100 oscillations.

The repetition rate is 80 MHz. The duty cycle is 1/1000 so given a mean power of 2.5 watts, the intensity of the light pulse is equivalent to 2.5 mega watt.

My understanding is that a single photon is actually one sinusoidal oscillation starting at zero, going up to reach positive peak, going down to negative peak and returning to zero.

Fourier analysis tells us that such wave shape would need a large (infinite) number of harmonics to generate using the sum of sinusoidal waves which extend in the past and future indefinitely. That is where quantum physics intervene, allowing one electron to absorb a single oscillation and instantaneously jump to one or more energy level.

The equivalent with sound wave would be, suppose a bunch of guitar laying of tables and I produce a sound wave at 441 Hz. If the first guitar is well tuned, it would only capture a wave at 440 hz. The next guitar happen to have one string imperfectly tuned so that it's resonant frequency happen to be 441 Hz. That string would capture all the energy of the sound wave and start vibrating while all the other string of that same guitar would detect nothing. And all the other guitar around would not perceive any air molecule vibration because the entire wave energy has been entirely absorbed by the string on the second guitar.

The energy of a photon correspond to the frequency which can be measured with prism, an optical equivalent of a FFT. The photoelectric effect is another way to measure the energy using a blocking voltage as described in another answer. The wavelength, of course, change with propagation speed.

Christian Gingras
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