What exactly is a quantum of light?

I am currently trying to learn some basic quantum mechanics and I am a bit confused. Wikipedia defines a photon as a quantum of light, which it further explains as some kind of a wave-packet.

What exactly is a quantum of light?

More precisely, is a quantum of light meant to be just a certain number of wavelengths of light (something like "1 quantum = a single period of a sine wave" perhaps?), or is the concept completely unrelated to wavelengths? In other words, how much is a single quantum?

• No. It's intimately tied in with the concept of uncertainty, and how that's represented in waves. Check out this question: physics.stackexchange.com/q/18062/5223 Commented Dec 21, 2011 at 1:53
• Also, an excellent place to start wrapping your head around this is to study the double-slit experiment. Commented Dec 21, 2011 at 2:16
• @Dejan: ok, but the accepted answer gives a description of a photon as a wave-packet of E and B fields, which is grossly incorrect, and will mislead others. A single photon is purely quantum mechanical, and is described by a quantum superposition of E and B fields which has no definite classical analog in any circumstance. This is different from, say, a single pion, where the wavefunction can be formally described by a solution to the classical pion-field equation of motion, because the pion can be nonrelativistic. Commented Dec 29, 2011 at 16:31
• Possible duplicate of What exactly is a photon? Commented Aug 13, 2016 at 22:04
• @peterh Are you kidding me? You voted to leave open the newer question as a duplicate of this one and now you're voting to close this one as a duplicate of the newer one? Commented Aug 14, 2016 at 0:17

There are two meanings usually attached to the word "quantum" in quantum theory, one colloquial and one technical.

As you know, electromagnetic radiation behaves in ways characteristic of both waves and particles. For non-specialists, it's easy to think of a particle as being a "unit" of the wave, and since "quantum" means a unit of something, the word has gotten associated with "particle." But in reality, the idea of a particle isn't precisely defined. When people talk about a particle of light, the EM field associated with what they probably mean could be described as a wave packet, which you can think of as an electromagnetic wave that is localized to some small region in space. For example, something like this:

This is just an example, of course; wave packets can have all sorts of shapes.

The more precise, technical meaning of "quantum" has to do with Fourier decomposition. As you may know, any function can be decomposed into a sum of sine waves (or complex exponentials),

$$f(x) \propto \int e^{ikx}\tilde f(k)\mathrm{d}k$$

For any given momentum $k$, the amplitude $\tilde f(k)$ represents the contribution of the sine wave with that frequency to the overall wave. Now, classically the value of $[\tilde f(k)]^2$ at each $k$ represents a bona fide contribution to the energy of the light. But the assumption that makes quantum theory quantum is that $[\tilde f(k)]^2$ instead represents the probability that there is a contribution to the energy of the light coming from that frequency. The actual contribution that can come from any given frequency can only be one of a set of specific values, which are integer multiples of some unit $\hbar c/k$. "Quantum" is the word for that unit of energy.

• The question I feel puzzled is: how can a photon be represented by a wavelet when a wavelet can be decomposed into a fundamental and harmonics and still has to be a quantum. On the other hand: a gaussian could travel and not oscillate at all. link I didn't find a linear example, sorry
– Erna
Commented Aug 21, 2022 at 12:10
• @Erna Perhaps you could post that as a separate followup question. Commented Aug 21, 2022 at 20:21
• The point is: I'm asking stupid questions and so I'm downgraded.... See below where I showed how gaussians add up to perfect sinoids... This question physics.stackexchange.com/questions/721847/… was closed and I lost many reputations..
– Erna
Commented Aug 21, 2022 at 21:37
• Insight: atomic oscillators possess a "Q-factor," and the Q is roughly proportional to linewidth, and to the no. of cycles in these "packets." What's the Q of, say, an H atom? Something like 50,000! So, when hydrogen is fluorescing, each outgoing photon is associated with a wave-packet having about 50,000 cycles. That's how long it takes a single atom to "ring down." Unlike the little animation, actually each "wave packet" is very, very, VERY long. (And equivalently, it takes about 50,000 cycles before a hydrogen atom can become pumped up by an incoming EM wave, absorbing a photon.) Commented Aug 29, 2023 at 0:55

A quantum of light is a particle of light which can disappear, giving its energy to an atomic or particle system, or appear, taking energy away from a particle or atomic system. A quantum of light of wavelength $\lambda$ is the minimum amount of energy which can be stored in an electromagnetic wave at that wavelength, which is Planck's constant h times the frequency. The photon is not related to the wave in any concrete way, the classical wave is a superposition of a large number of photons which are coherent.

• ....not necesarily a large number of photons, but definitely an indeterminate number of photons, since field amplitude does not conmute with energy and/or mode occupation number Commented Dec 21, 2011 at 3:54
• @lurscher: No, a large number is the more precise statement. An indeterminate small number doesn't work to produce a definite field quantity, while a large definite number of photons can still produce a field whose local phase fluctuations are tiny, meaning that if you measure the phase at one point, the phase at a distant point collapses to a consistent wave. Commented Dec 21, 2011 at 9:39
• @lurscher: What does $[\hat{n},\hat{a}]$ have to do with Ron Miamon's reply? I'm not sure I understand your statement. Commented Dec 22, 2011 at 17:48
• @Antillar: the point is, when does a photon have a field description? He is saying that the limit needs not just a large number, but an indeterminate number of photons, just like the limit of "definite position" in a Harmonic Oscillator needs a "large indeterminate energy level". This is technically true, but I think it is better to just say "large number", because the relative phase can still be ok after a measurement, like after a position measurement of a large N HO, the particle oscillates. It's a minor issue, and the main point is unaltered. Commented Dec 22, 2011 at 18:58

Just a remark that might be helpful to understand what photon is: the "wavelengths of light" seems to be just a theoretical value calculated with the help of Planck model. What can really be measured in the experiment is the momentum/energy of photon, not the wavelength. For instance, the "colour" of the photon is fully determined by its momentum.

• Not true. The wavelength can be measured directly with the aid of a diffraction grating. Commented Aug 4, 2022 at 12:56

A photon is a certain amount of energy carrying momentum. That is what we know for sure. We also know that Maxwells equations can model an electromagnetic field that can carry waves. Waves are characterized by wavelength (a spacial value) and frequency ( a time value). Planck's Gedankenexperiment imagined a hollow body where the walls carried oscillators that can be excited by exchanging energy with the electromagnetic field inside the body.

https://onlinelibrary.wiley.com/doi/epdf/10.1002/andp.19013090310

In §3 he breaks the energy Un into a finite number of parts.

He assumed the integral energy of the field to be distributed to different energy levels that are multiples of a certain base energy. His idea was to bring this base energy in the limit to zero to so get a continous spectrum. He then had to realize that this limit can not be reached and so discovered the quantum of action.

Later Einstein and Bose replaced the oscillators by the cavity itself as in any cavity there will be standing waves that are harmonics to a base frequency which is determined by the dimension of the cavity and the speed of light.

Planck himself remarked, that the base energy is arbitray, that is, the energy levels are only determined by the imaginated oscillators. So if you assume the base frequency is halved, the energy difference is halfed also, the number of spectral lines is doubled. But that doesn't change the outcome, the formula of the black body radiation.

So what we know for sure (energy, momentum, spin) doesn't necessarily imply a quantized field, but anyway interaction of matter with the field is.

Any try to imagine a photon as a travelling "wavelet" of a given wavelength leads to a paradox: if there is a wavelet, a fourier transform can determine the frequency spectrum of said wavelet, but a spectrum must consist of different frequencies, a.k. photons. What is a contradiction in its own.

But we all know by experience, that an exitation without a frequency can exist: any rope creates a pulse when lifted up and back again and this pulse looks like a Gaussian. A Gaussian indeed has no period to determine a frequency but as the FT created another Gaussian it can be thought to contain every frequency.

So taking what we know for sure there is an electromagnetic field thinkable that is a continuum but can be quantized by oscillator (matter) when interacting with the field.

One interesting fact is, that gravity also interacts with the field and exchanges energy continously.

To show how a close to perfect sin-wave is created by repeated emission of a pulse: https://www.wolframalpha.com/input?i=e%5E-x2%2Be%5E-%28x%2B3%29%C2%B2%2Be%5E-%28x%2B6%29%C2%B2%2Be%5E-%28x%2B9%29%C2%B2

If you care about the shift in Y you can alternate the sign: https://www.wolframalpha.com/input?i=e%5E-x2-e%5E-%28x%2B3%29%C2%B2%2Be%5E-%28x%2B6%29%C2%B2-e%5E-%28x%2B9%29%C2%B2

Now it's not such a nice sinoid, but this can be cured by changing the center of the time offset: https://www.wolframalpha.com/input?i=e%5E-x2-e%5E-%28x%2B2%29%C2%B2%2Be%5E-%28x%2B4%29%C2%B2-e%5E-%28x%2B6%29%C2%B2

As the function look like a cosine, you can substract cosine to see the residual: https://www.wolframalpha.com/input?i=-e%5E-%28x-6%29%C2%B2%2Be%5E-%28x-4%29%C2%B2-e%5E-%28x-2%29%C2%B2%2Be%5E-x2-e%5E-%28x%2B2%29%C2%B2%2Be%5E-%28x%2B4%29%C2%B2-e%5E-%28x%2B6%29%C2%B2-cos%28pi*x%2F2%29

Obviously there is now a negative cosine component so we adjust the amplitude: https://www.wolframalpha.com/input?i=-e%5E-%28x-6%29%C2%B2%2Be%5E-%28x-4%29%C2%B2-e%5E-%28x-2%29%C2%B2%2Be%5E-x2-e%5E-%28x%2B2%29%C2%B2%2Be%5E-%28x%2B4%29%C2%B2-e%5E-%28x%2B6%29%C2%B2-0.955*cos%28pi*x%2F2%29 what nearly cancels out the sinoidal in the center, but we have a third harmonic left. By recursion this harmonic can be eliminated to have the 9th as a residual..

The point with the Gaussians is: the FT of a Gaussian is a Gaussian of same standard deviation (normalized) but with a phase rotating proportionally to the shift. So if you define an interval to represent the Gaussian, this interval defines the base frequency and all the harmonics shift by a constantly increasing value so adding the two Gaussians eliminated all odd harmonics etc.

As this visualization of the generation of a wave by induced emission of gaussian peaks combines the continuum of a field with an quantized interaction to me this answers some questions I do no longer raise.

Reality needs three dimensions so there has to be another solution. I do not know how to handle vortices, but as vortex cannons can produce directed wavelets that should be a candidate.

Vortices are known from quantum fluids like superfluid helium. Being curious I just googled this: https://www.youtube.com/watch?v=TlEQbPSbYTQ and find it inspiring. So the math obviously exists.

Here are some things that might help:

Everything has wave-particle duality (even us). This 'effect' is not limited to the scale of single particles (microscopic/subatomic scale) like electrons. By the correspondence principle in quantum mechanics these quantum phenomena map onto the macroscopic scale (this can be loosely thought of as the scale of the world that we exist in).