# Is Heisenberg's Uncertainty Principle applicable to light?

Can we apply the Uncertainty Principle to light? If so wouldn't it violate it because we just need to find the position of light, since we would already determine it's momentum(from the wavelength of light used)?

• Monochromatic light doesn't exist. Dec 4 '17 at 17:55
• @ Wavelength tells you the length but not direction of the momentum vector, whose components each still have uncertainty.
– J.G.
Dec 4 '17 at 19:52

No, it is not at least in the usual form (3D momentum-position inequality), because differently from massive particles there is no position operator for photons as is known from the theory of representations of Poincare' group and imprimitivity theory. See for instance Barut-Racza's textbook on representation theory. Extending the formalism some interesting results exist like this one. I do not know if this proposed re-formulation of HUP corresponds to experimental facts.

I do not think that Fourier analysis is enough, "without introducing $\hbar$", i.e., without quantum phenomenology, to say that HUP takes place for classical systems like waves of sound.

Yes, one could argue that if the Fourier transform of a wavepacket of sound has extension $\Delta k_1 \Delta k_2 \Delta k_3$, then the spatial extension satisfies $\Delta X_i \geq \frac{1}{4\pi \Delta k_i}$.

The point is that this spatial extension has not the meaning of the statistical uncertainty of the position of a "particle of sound" because no localization phenomena exist and also because momentum is not related to wave number vector through quantum relation $P= \hbar k$.

Instead, if we are dealing with a Schroedinger wave of a massive particle and we perform a sufficiently precise (3D) position measurement, the particle localizes inside the support of the packet. In other words, a relatively small spot of space (also very very small with respect to the extension of the wave) is abruptly and discontinuously chosen by the system.

However, repeating the position measurement for identical waves, the position where the particle localizes fluctuates with a dispersion satisfying $\Delta X_i \geq \frac{1}{4\pi \Delta k_i}$, i.e., $\Delta X_i \Delta P_i\geq \frac{\hbar}{2}$.

These phenomenological facts are completely absent for classical waves, because they do not describe quantum particles.

Regarding photons the situation is quite complicated. As far as I know they localize on screens their wavefunction encounters during its evolution. But this is not a three-dimensional localization and it is difficult to handle this phenomenon with the standard Fourier analysis machinery (though we know that the three components of the momentum are related with the wave vector components according to the standard quantum relation $P_i = \hbar k_i$). The mathematical counterpart of this fact is the absence of the standard position operators along the three spatial direction. So it is difficult to state a precise form of HUP for photons.

• This seems misleading to me. The HUP is really just a fact about the Fourier transform, and that fact applies to things like sound waves. There are wavenumber-position and frequency-time uncertainty principles that have no h in them, and they can be applied to Maxwell's equations, so they apply to light, and the answer to the OP's question is basically yes. Whether we call these the "Heisenberg uncertainty principle" is a matter of taste, but it's not as though we somehow evade this fundamental limitation because there is no position operator for photons. (So "not at least in the usual form?")
– user4552
Dec 4 '17 at 23:30
• Please write down a precise statement and we can discuss on it. Dec 5 '17 at 5:16
• However I disagree on extending HUP to sound waves: there is no particle localization in that case while there is for quantum particles. Fourier analysus is not enough to say that something like HUP takes place. Dec 5 '17 at 5:22
• For photons spatial localisation exists just on a (two dimensional) screen encountered by the wave. I do not know if it is enough to produce a sort of 2D HUP referring to components of momentum tangent to the screen itself. Dec 5 '17 at 5:30
• A gamma ray in an electromagnetic calorimeter has a three dimensional Δ(volume), the momentum can be defined from the vertex to this x,y,z and the frequency from the energy summed up in the calorimeter, but the experimental errors are such that the HUP is fulfilled with no problem. Dec 5 '17 at 6:59

Using wavelength you can find only the magnitude of momentum $\sqrt{P_x^2 + P_y^2 + P_z^2}$. For finding its direction you will need its wavevector $\vec{k}$. Heisenberg's Uncertainty principle applies to each component of momentum and the corresponding component of position, for example, you can't find simultaneously $P_x$ and $x$. Heisenberg's Uncertainty Principle in vector form: $\Delta\vec{P}.\Delta\vec{r} =3\frac{h}{4\pi}$

There is a nice demonstration of this in an experiment of diffraction of light through single slit in this video by Veritasium: https://www.youtube.com/watch?v=a8FTr2qMutA&t=4s

In the experiment, when the light is allowed to pass through a thin slit, it's position is being confined to the slit. So as the slit becomes smaller and smaller it's position gets confined to a smaller region of space. Now, Heisenberg's principle has to work, so its momentum spreads out in different directions (which is uncertainty), giving us the pattern of diffraction.

The Heisenberg Uncertainty principle, HUP, is a basic quantum mechanical observation, modeled by the commutation relations of the appropriate operators.

Light is the name of classical electromagnetic radiation, described by Maxwell's equations. It can be mathematically shown that light, (classical electromagnetic waves,) is emergent from an enormous number of photons of energy h*nu where h is the Planck constant and nu the frequency of the classical EM wave.

The HUP is only relevant for elementary particles and their composites, like atoms and molecules. Photons as individual elementary particles obey the Heisenberg uncertainty principle when their momentum and position is detected . If the interaction point of a photon in space is detected with great accuracy, its momentum will have an uncertainty bounded by the HUP.

I searched for experiments checking the HUP for single photon momentum and position, but I suspect that the experimental errors in position and frequency measurement are so large that the HUP is always obeyed.

The linked video given by Yaman Sanghavi does show the correlation collectively for photons building up light, showing that for a fixed momentum/frequency the beam has to spread out going through a narrower and narrower slit, but does not test the bounds of HUP.

Thys the question in the title :

Is Heisenberg's Uncertainty Principle applicable to light?

is answered with "yes" for for the photons uilding up light and other sets of non commuting variables.

This study checks a HUP relationship of photons for non commuting polarizations states and validates the principle.

• It seems to me that the quoted work deals only with different types of linear polarizations whose observables does not commute (like the spin 1/2 components) and thus they give rise to a generalized Heisenberg principle...However all that has nothing to do with the usual interpretation in terms of position an momentum... Dec 4 '17 at 19:21
• @ValterMoretti I found the link while checking whether a position/momentum experiment has been done using single photons and this came up. I expect no experiment exists because the experimental errors in the photon position detectors and the errors for determining the frequency of the incoming photon are so large that the HUP is always obeyed ( as with tracks in track detectors) so no check could be made on those two non commuting variables. The xp ones are just one of the sets possible. Dec 5 '17 at 5:12
• this may be useful for teachers researchgate.net/publication/… Dec 5 '17 at 7:28