The Heisenberg uncertainty principle arises from the description of matter as being represented as a wave packets, and hence its fourier decomposition having a range of multiple wave numbers in its creation, giving rise to a range of momentum associated with a specific wave packet.

Since as I understand it, the de Broglie relation can be "derived" from $E=pc$ and $E=hf$ (which are about photons) and then postulating they hold for matter as well (I'm sure this way of thinking is not the full picture), from this the imagination of a wave function was born.

So a natural question is does the Heisenberg uncertainty principle apply for photons?

More specifically, does the photon have a specific wave function, comprised of many different wave numbers(and or frequencies), in order to create a localised wave packet.

The reason this has interested me is consider the following:

$\sigma_{x}\sigma_{p} \geq \frac{\hbar}{2}$

$E=pc$ , $E=hf$


If there are some range of frequencies that comprised the photons wave function, then I'm pretty sure it follows that

$\sigma_{p} c = h \sigma_{f}$

Substituting into the uncertainty principle

$$\sigma_{x}\sigma_{f} \geq \frac{c}{4\pi}$$

This predicts that the colour of light from a single slit diffraction for example should vary as you go along the diffraction pattern(since the uncertainty principles explanation for diffraction varies the momentum of the photon), for an actual noticeable colour difference the slit should be on the order of *10^-7 m

However I cannot find this happening anywhere online so something in my logic is not correct, I suspect this being my assumption on the form of the photon wave function

  • $\begingroup$ The uncertainty principle has nothing to do with quantum mechanics. It applies to all linear systems that can be described with wave-like phenomena, e.g. surface waves on water or acoustic waves. A single photon does not have a wave function. Only the ensemble of photons has a wave function. A photon does not even have a frequency, either. E=hf is a relationship between photon energy and the frequency of a monochromatic classical wave as observed in the photoelectric effect. You are simply mixing two different pieces of physics. $\endgroup$ Apr 13, 2023 at 11:03
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    $\begingroup$ @FlatterMann "A single photon does not have a wave function". AFAIK this is not correct. see arxiv.org/abs/quant-ph/0604169 . Also this "A single photon does not have a wave function." is not correct. The photon is a quantum mechanical entity in the standard model of particle physics, of zero mass, spin one .en.wikipedia.org/wiki/Standard_Model . as such it must have a wavefunction for the QFT to work. $\endgroup$
    – anna v
    Apr 13, 2023 at 11:55
  • $\begingroup$ @annav I have built high energy physics detectors that have measured many trillions of photons. At most I got one energy, one momentum and one angular momentum value out of one of those measurements. What was the wave function of that photon? $\endgroup$ Apr 13, 2023 at 12:38
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    $\begingroup$ It's possible to define a single photon Fock state for any arbitrary normalizable solution of Maxwell's equations. This is covered by introductory QFT. This is a single photon wave function in the same sense that a single particle state of the electron field can be considered an electron wave function. $\endgroup$ Apr 13, 2023 at 14:21
  • $\begingroup$ @FlatterMann are you aware that the wave is a probability wave that the wave function discusses? Probability means that an individual event is not enough to check the measurements, but an accumulation is needed. see my answer here physics.stackexchange.com/questions/421181/… $\endgroup$
    – anna v
    Apr 13, 2023 at 17:05

2 Answers 2


The colour of light is determined by its frequency as this is what determines the response of the detecting compounds in our eyes or any other detector. The relation $f\lambda=c$ applies only to infinite plane waves. When a monochromatic beam passes through slits, its spatial variation is no longer a simple plane wave but its frequency remains unchanged.

Because the resulting wave must still satisfy the wave equation with an unaltered frequency the wave becomes a sum of infinitely many plane waves, all of which do have have the same wavelength, but with infinitely many different directions of propagation. The infinite sum gives the diffraction pattern.

Where the uncertainly principle comes into play is when you have a wave that wa created by a source that was active only for a finite time. In that case the wave cannot consist of only one one frequency. The finite-length wavetrain contains a range of frequencies whose bandwidth is inversely proportional to its length. That length is $c$ times the length of time that the source was switched on.


The comment statement by Flatterman

The uncertainty principle has nothing to do with quantum mechanics. It applies to all linear systems that can be described with wave-like phenomena, e.g. surface waves on water or acoustic waves.

is correct.

Photons are elementary point particles in the mainstream standard model,, and as such cannot be studied in interactions with matter using classical mechanics and optics, it is necessary to use quantum mechanics. The frequency that is used to define a photon's energy, $E=hν$ does not describe an individual photon. It appears only in the accumulation of photons

See here the experiment one photon at a time going through two slits.


Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames

Single photons scattering off the two slits are described mathematically by a Feynman diagram of a point photon interacting with the field (due to the atoms on the sides of the slits),and then act as free particles hitting the screen and leaving a point footprint. There is no "waving" of a single photon.Then as they are accumulated, an interference pattern appears. It is in this pattern that the Heisenberg constant $h$ is relevant and one can connect the interference pattern with the frequency $ν$ of the laser light used in the experiment.

Wave packets are a different story and are useful , see here a discussion .


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