# 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?

Do note that my question has nothing to do with diffraction patterns or the process of interference, and is rather about the relationship between Heisenberg's Principle and diffraction.

• 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. – John Rennie May 24 '14 at 18:58
• 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. – John Rennie May 24 '14 at 19:14
• @CarlWitthoft Indeed the Uncertainty Principle has nothing to do with interference or diffraction patterns. I have no idea why you guys are discussing those processes. As for diffraction, it is fundamentally related to the Uncertainty Principle as I will show in an edit to my question – hb20007 May 25 '14 at 10:34
• you are so right. I don't know what their problem is. – Marty Green May 25 '14 at 20:53
• Similar/related question here. – knzhou Mar 27 '19 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.

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.

If you consider light to be a wave, then Huygens' principle explains the single slit experiment (which we call diffraction of light wave)

HUP can explain the same phenomenon as well, but in this case we consider the particle nature of light, i.e. it is composed of photons. So, obviously, idk why we call the phenomenon 'diffraction' if we consider using HUP because the term diffraction is solely reserved for the WAVE NATURE of anything.

All in all, HUP considers the particle nature of light while Huygens' principle considers the wave nature of light.