# Uncertainty principle and digital camera

I recently got into a discussion in how far (miniaturized) digital cameras are affected by the uncertainty principle. Specifically the question was, whether the uncertainty principle is one of the reasons why smaller digital cameras (with smaller optics) tend to make blurrier images than larger ones. I couldn't image this would be the case and tried to prove that this is not the case, but I think I lack the knowledge to build a convincing reasoning.

What I came up with so far:

Leaving the optics aside, a "blurry image" would mean, that the sensor in the digital camera "localizes" the incoming photons in such way, that the uncertainty inequality $$\Delta p \Delta x \geq \hbar/2.$$ is violated.

The impulse of a photon is $$p = h/\lambda$$ with $h$ being a constant, so $$\Delta p \propto 1/\Delta\lambda$$

Assuming $\Delta\lambda$ to be the colour-accuracy of the sensor and $\Delta x$ the physical size of one sensor pixel and using common values for today's digital cameras, the inequality is satisfied and therefore no such effect exists.

• Is the uncertainty principle applicable at all in this concrete case and is this argumentation valid?
• How could the optics be also considered?

Or more generally:

• Does the uncertainty principle imply a theoretical limit on how far digital cameras can be scaled down?
• If so, can this limit be reached in practice and what would be the effect of going past that limit, blurred images?

I hope this is the right place to ask these questions and someone can shed some light on this issue.

• You should consider the uncertainty in the lateral component of the momentum of the photon. So, if geometric optics were 100% accurate, you would have perfectly sharp images. But this means that photons would have to travel on perfectly straight lines, the uncertainty in the momentum in the lateral direction would have to be zero. However the photons are entering the lens through a finite aperture, so the uncertainty in the position in the lateral direction is finite.... – Count Iblis May 5 '16 at 18:22
• All well designed high quality camera lenses are diffraction limited at intermediate apertures, for DSLRs that often happens around 8. A professional photographer will typically know at what aperture setting each of his lenses delivers the sharpest images that are the least impacted by geometric lens errors at large apertures and diffraction effects at small apertures. – CuriousOne May 5 '16 at 19:33
• @CountIblis I don't know if I understood correctly, but isn't the uncertainty in position smaller when I know that a photon has hit a certain sensor pixel compared to the uncertainty when I know it passed somewhere through the lens? – user116514 May 5 '16 at 20:57
• @dliw Yes, that smaller uncertainty gives you a far larger uncertainty in the angle that photon arrives at after it moves through the lens. When the photon coming from some direction moves through the lens the uncertainty in the position spans the entire lens. This has to be compatible with the uncertainty relation. If you choose the aperture too small, say F/16 then the smaller uncertainty in the lateral component of the momentum will lead to an uncertainty larger than one pixel, causing blurring. – Count Iblis May 5 '16 at 22:38

## 1 Answer

It is true that:

$$\Delta x \Delta p \ge \frac{h}{4\pi}$$

and,

$$p=\frac{h}{\lambda}$$.

However, the correct differential is:

$$\Delta p = -\frac{h \Delta\lambda}{\lambda^2}$$

So the uncertainty principle (ignoring the irrelevant minus sign) gives:

$$\Delta x \Delta\lambda \ge \frac{\lambda^2}{4\pi}$$

For visible light one may take $$\lambda\approx$$600 nm as a ballpark. The more difficult question is, what does one take for $$\Delta\lambda$$? This is not so straightforward, but one can conjecture that it is reasonably bounded by $$\lambda$$, in which case:

$$\Delta x \ge \frac{\lambda}{4\pi}\approx$$ 50 nm

This is far smaller than the pixel pitch on any digital sensor I am aware of (several microns is common).

We could also turn this around, and ask what $$\Delta\lambda$$ corresponds to the pixel pitch on a high resolution digital sensor, say $$\approx$$ 3 microns, from which I get:

3 microns $$=\Delta x \ge \frac{\lambda^2}{4\pi\Delta\lambda}\rightarrow \Delta\lambda\ge$$ 10 nm.