# Interesting relationship between diffraction and Heisenberg’s uncertainty principle?

I recently came across an interesting explanation of diffraction through an aperture which does not use Huygens’ construction but instead relies on Heisenberg’s uncertainty principle:

The uncertainty principle states that trying to pin a particle down to a definite position will create uncertainty in its momentum, and vice-versa. Therefore when you are confining a particle to go through a narrow aperture, one is very certain about its position (the aperture is very narrow and hence the particle must have been somewhere in that extremely small gap) So, by Heisenberg’s principle, the particle will now have a crazy momentum… so it can go in any direction.

However, it seems we cannot use the same principle to explain the case where diffraction occurs when light passes around an obstacle. We are not locking the photons in a small position in that case, are we? Keeping the uncertainty principle in mind, diffraction through an aperture makes sense; but in the case of a solid obstacle, my intuition says the wave should just be deflected. Is there a way to relate the uncertainty principle to diffraction around an obstacle?

Besides, given that the uncertainty principle is defined in terms of particles, can it be used to explain diffraction in mechanical waves such as sound which are not composed of photons?

• Diffraction through a slit is explained by the wave properties of particles, as is diffraction round obstacles. It's not obvious to me how you'd use the uncertainty principle to calculate a diffraction pattern. Commented May 24, 2014 at 18:58
• @JohnRennie I didn't mention using the Uncertainty Principle to perform any calculation and surely never mentioned diffraction patterns :/ Commented May 24, 2014 at 18:59
• @JohnRennie I agree with you however my question is about the process of diffraction itself - what causes it... as opposed to diffraction patterns created by interference Commented May 24, 2014 at 19:10
• I'm still not sure what you're asking, but the nearest I've seen to a fundamental explanation of why diffraction happens is the Huygen's construction. Commented May 24, 2014 at 19:14
• Similar/related question here. Commented Mar 27, 2019 at 16:31

Diffraction and the HUP are related because they have the same mathematical description.

The Fourier transform to the canonical commutation relationship and the Heisenberg uncertainty principle. The FT is the unitary (norm and inner product preserving, i.e. probability-preserving) transformation between position co-ordinates and momentum co-ordinates, and it can be shown that, given any pair of quantum observables $\hat{X}$ and $\hat{P}$ that fulfill the canonical commutation relationship $X\,P-P\,X=i\,\hbar\,\mathrm{id}$, the transformation between co-ordinates wherein $\hat{X}$ and $\hat{P}$ are simple multiplication operators is precisely the Fourier transform. I show how this must be true in this answer here. This leads to the Heisenberg inequality through the pure mathematical properties of the FT as I discuss in this answer here and here. A special case observation that summarises the behaviour intuitively is that a function and its FT cannot both have compact support (domain wherein they are nonzero): if you confine a wavefunction (i.e. quantum state) to a small range of positions, its Fourier transform is the same quantum state written in momentum co-ordinates, so the spread over momentums increases as you confine the positions more and more.

The analogy with diffraction is direct. Huygens's principle, or whatever method you want to use to explain diffraction is explained in detail in my answer here, this one here, this one here, or here. But a summary is this. A plane wave running orthogonal to a plane means that the phase on that plane is uniform. AS the wave tilts, its phase variation on the plane is of the form $\exp(i\,\vec{k}\,\cdot\,\vec{x})$, where $\vec{k}$ is the wavevector and $\vec{x}$ the transverse position on the plane. So, to find out what spread of directions you have in a light wave, you take its Fourier transform over the plane. The Fourier transform at point $k_x,\,k_y$ is simply the superposition weight of the plane wave component with direction defined by $k_x,\,k_y$. The more spread out in Fourier space a wave is, the wider the spread of propagation directions are important, and the more swiftly it will diffract. So a wavelength size pinhole in a screen means that the spread of directions will be wide, simply by dent of the Fourier transform uncertainty product. Indeed, for small diffraction angles, $\sqrt{k_x^2+k_y^2}/k \approx \theta$, where $\theta$ is the angle the plane wave component makes with the normal to the plane. Indeed the basic uncertainty product for FTs shows that $\Delta x\,\Delta k_x = \Delta x\,\Delta \theta\,k \geq \frac{1}{2}$ where $\Delta x$ is the slit width and $\Delta \theta$ the angular spread of diffracted light.

Strictly speking, the physics of diffraction cannot be explained as the HUP (i.e. as arising from the canonical commutation relationships) because there is no position observable $\hat{X}$ for the photon, so you can't think of $\Delta\,x\,\Delta p$. There are most certainly pairs of canonically commuting observables: for example the same components of the electric field and magnetic field observable for the second quantised electromagnetic field are conjugate observables. The reason HUP descriptions work is the mathematical analogy I have described above.

• "because there is no position observable X^ for the photon". Also, the explanation in terms of Fourier transforms works for the classical wave too, where there is no concept of probabilistic observables. Commented Dec 2, 2022 at 10:01

Let us look at this relationship for waves, particularly for electromagnetic waves:

where we see that for light v=c

A classical wave emerges from a large ensemble of photons, the quantized state of electromagnetism.

Assuming we have a single photon of frequency nu, if we multiply both sides by hbar and divide by c , we get a formula consistent with the Heisenberg uncertainty formula.

lamdahnu/c~h

delta(x)*delta(p)~h

where the delta symbol denotes that we have a quantum of the quantity.

The Heisenberg uncertainty principle introduces the larger > relation instead of equality, which is an assumption that does not exist within the classical electromagnetic description.

Thus there exists consistency between the classical framework, and the quantum mechanical one, but it is the classical that emerges from the quantum mechanical, and not the other way around.

So one could handwave the HUP to describe the diffraction from a slit because in this case one is essentially describing delta(x) and delta(p) which are consistent in both frameworks. The slit distances are chosen to be of order of magnitude of the wavelenth after all.

To compare with classical diffraction from an edge one would need the solution of the quantum mechanical problem with the edge boundary to explain how the probability distribution given by the photon wavefunctions agrees with the solutions of the maxwell equations.

P.S. photons are elementary particles and fall into the realm of the HUP

As like the photons/light waves emerge out from a very thin slit in single slit experiment, you are watching those photons/light waves which essentially pass through small region (of space) close to the edge of the obstacle. When you see the diffraction from obstacle, you are not seeing the light far away from the obstacle. So, the same explanation works.

Properties of light like diffraction and interference are explained using Huygen's construction, which takes into account the wave nature of light.

The Huygens–Fresnel principle states that each point on a wavefront is itself the source of a spherical wavelet, and the infinite secondary wavelets emanating from each of the infinite unique points on the wavefront mutually interfere. The sum of these spherical wavelets at any point in space, at any time, gives the phase at that point.

I can't see how HUP can explain such diffraction patterns of maxima and minima.

Heisenberg's principle is basically diffraction and it is clear from fourier optics. When a path of particle is describe, basically it is position measurement at various instances. A wave has not only wavelength in one direction but all directions, a spherical wavefront.

So any obstruction is kind of neasurement, if a thick block is inserted in between planar wavefront then maybe one can't get diffraction of shadow. Basically when a particle is subjected to probability, at atomic level to see or detect and track a single particle is almost impossible.

So in the case of probability, the many step to track, which is momentum by position, its final result is smeared. What a position measurement is doing that it re-arrange or re-distribute itself, diffraction is divergence so arranging as point source. So in diffraction grating, one gets fine result of interference pattern.

I think question is how interference by mach-zender or fabrey-perot interferrometer like devices doing. Aa they are not resdistributing its wave vector. Then surely they are redistributing energy. As quantum mechanics gains popularity from explanation of spectral pattern of radiation.