In my understanding, the energy of a photon, in wave terms, just translates to the frequency of the wave. If I make a photon with more energy, it will use the extra energy to just oscillate faster.

Where does the amplitude come in here?

In contrast, if I am creating sound waves by beating on a drum, if I consider the oscillation of the drum fabric as simple harmonic, if I beat on it harder, the drum will just produce a louder sound, i.e. its amplitude would increase. Since the motion of the drum fabric is simple harmonic, beating the drum harder doesn't change frequency, so energy only goes into increasing the amplitude. Tapping on a wooden table harder increases both the amplitude(loudness) of the wave and the frequency (pitch) of the wave, since I guess it's not simple harmonic motion.

Coming back to light waves, the only similar concept I can find is wave intensity. If more number of waves come in (constructively interfering at the observation point) then the brightness will increase, while keeping the color (based on frequency) constant. This constructive interference seems to be represented by increasing the amplitude in the wave diagram.

*Is amplitude just a way of representing light intensity in the wave diagram, possessing no physical differences from intensity?

In other words, can a light wave with more amplitude only be created by constructively interfering light from coherent sources?

If so, what is the amplitude of just a single photon? Does it even mean anything?

  • $\begingroup$ You won’t find much support for the Particle aspect of photons around here. But if you did you could imagine a photon Oscillating between positive and negative amplitudes at a very high frequency as it travels along at the speed of light. $\endgroup$ Commented Jun 19, 2021 at 16:29
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    $\begingroup$ @BillAlsept Why do you saw the photon would be oscillating at very high frequency? That would only be true for a very high energy photon, right? $\endgroup$
    – Andrew
    Commented Jun 19, 2021 at 17:47
  • $\begingroup$ @Andrew you are correct. I said high frequency to help visualize. If we’re talking visible light, even the slowest frequencies for red light are about 430 THz per second. That’s 430,000,000,000,000 Oscillations per second. $\endgroup$ Commented Jun 19, 2021 at 18:15

3 Answers 3


The irradiance of light (Watt per square meter) is directly proportional to the square of the amplitude of the electromagnetic wave. In contrast, the energy of a single photon (the wave consists of) only depends on the frequency. The link is the number of photons, which is frequency-dependent. Meaning an electromagnetic wave of given amplitude (energy) consists of less photons if it's frequency is large. If the frequency is smaller, the number of photons is larger (under the condition that the amplitude of the electromagnetic wave is the same).

  • $\begingroup$ I feel like the wording of this answer could be misleading, but I think it touches on the key point - the number of photons in the EM wave. $\endgroup$ Commented Jun 19, 2021 at 19:56
  • $\begingroup$ Thanks, I feel like this made it clear. So, if we keep the number of photons constant, and increase the frequency of the photons falling upon a surface, the amplitude will increase, since the irradiance increases. So I guess that amplitude isn't a fundamental characteristic for an individual photon. When it falls on a surface, it's frequency can be used to calculate irradiance and hence amplitude. Is this right? $\endgroup$
    – VVidyan
    Commented Jun 20, 2021 at 5:43
  • $\begingroup$ Basically yes, the number of photons is proportional to the squared amplitude of the electromagnetic wave, and the proportionality factor depends on the frequency. $\endgroup$ Commented Jun 20, 2021 at 7:52

Photons are not really particles as we imagine them (with our classical intuition), but quanta of excitations of electromagnetic field. A mode of EM field with frequency $\omega$ can be excited multiple times, e.g., $n$ times, in which case the energy of this mode is $n\hbar\omega$, and we say that it contains $n$ photons. There are of course also modes of electromagnetic field of different energies, including a mode with frequency $\omega_1=n\omega$, but this is not the same mode. This is to say, that energy given to the EM field may not necessarily go into exciting a higher energy mode/photon, but into exciting multiple lower energy mdoes, or even several modes of different energies.

Furthermore, electric field intensity can be expressed in terms of photon creation/annihilation operators, as, e.g., described here. These operators do not commute with the photon number operators, $n_{\mathbf{k},\mu}=a_{\mathbf{k},\mu}^\dagger a_{\mathbf{k},\mu}$, but rather have coherent states as their eigenfunctions. Thus, the amplitude and number of photons describe really different situations (e.g., coherent laser light vs. thermal radiation) with different physical properties.

Intensity, in classical terms, is the square of the field amplitude - because this is what is measured by most detectors. Obviosuly, even if the amplitude has zero average (e.g., in a sign wave), the intensity will not.

Finally, the distinction between the light waves and the sound waves discussed in the OP is really the distinction between quantized and classical waves. Elastic waves however can be quantized in the same way as the electromagnetic radiation, and this is routinely done in case of cristal vibrations, called phonons. These quantized waves have the properties very similar to quantized EM radiation, in particular in what concerns their amplitude and the excitation numbers.


Amplitude shows the strength of the electric field. Intensity is the square of the amplitude (with a coefficient).


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