Uncertainty principle and digital camera

Question

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.

Answer 1

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.