I saw some videos where a person points a laser through a slit. As they reduce the width of the slit, the diffracted image spreads out, like this:
Can this pattern be viewed as a consequence of Heisenberg's uncertainty principle, applied to photons?
Quantum Mechanics – How the Single Slit Experiment Demonstrates Heisenberg’s Uncertainty Principle
diffractionheisenberg-uncertainty-principlequantum mechanicswave-particle-duality
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The precise, mathematical statement of the uncertainty principle is $\sigma^2_x \sigma^2_k \geq 1/4$. The use of deltas is just an informal way of talking about it. Nevertheless, it's pretty common to say, for instance, that the width of a peak is either the standard deviation or some quantity proportional to it--see, for example, full width at half maximum, which ends up being about $2.35\sigma$.
I'm not really sure what a slit would look like in 1 dimension. It's easier for me to consider a particle in a 1d infinite square well. Note that the infinite well absolutely forbids any leakage of the particle into the forbidden region, just like the classical case. In this case, the variances depend on the energy of the particle. For a particle in one of the $n$th energy eigenstate of an infinite well with width $L$, the variances are (per wikipedia)
$$\sigma^2_x = \frac{L^2}{12} \Bigg ( 1 - \frac{6}{n^2 \pi^2} \Bigg), \quad \sigma_k^2 = \frac{n^2 \pi^2}{L^2}$$
The product of the variances is then $\sigma_x^2 \sigma_k^2 = (n^2 \pi^2/12 - 1/2)$. For $n=1$, this is about $.322 \geq .25$, as required.
You can't really see what the uncertainties will be by inspection, by the geometry of the problem. These are the uncertainties for energy eigenstates, and there's no reason to expect that a particle will be in an eigenstate (which would then make the computation more complicated).
Really, one just calculates the variances of the wavefunction with respect to $x$ and $k$. You might be able to get a rough idea from the quantities in the problem (for instance, the standard deviation with respect to $x$ is indeed proportional to $L$, but only proportional, not exactly $L$), but that's all.
You ask if a particle has nonzero probability of existing everywhere. To be pedantic, a particle has zero probability of existing at any specific point, but it typically has a nonzero probability of existing in a region of any finite size. This infinite square well is an exception, as the infinite potential around the box absolutely forbids particles.
Uncertainty really is just a loose, loose word to use. It almost always really means standard deviation of the wavefunction.
The uncertainty principle is a simple consequence of the idea that quantum mechanical operators do not necessarily commute.
In quantum mechanics, you find that the state which describes a state of definite value of an observable $A$ is not the state which describes a state of definite value for an observable $B$ if the commutator of both observables $[A,B]$ is not zero. (Formally, the two operators are not simultaneously diagonalizable.)
You just write down the definition of the standard deviation of the operator $A$ on a state $\psi$, $$ \sigma_A(\psi) = \sqrt{\langle A^2\rangle_\psi - \langle A\rangle^2_\psi}$$ where $\langle\dot{}\rangle_\psi$ is the expectation value in the state $\psi$ and with a bit of algebraic manipulation (done e.g. on Wikipedia) we find that $$ \sigma_A(\psi)\sigma_B(\psi)\ge \frac{1}{2}\lvert \langle[A,B]\rangle_\psi\lvert$$
Now, the standard deviation (or "uncertainty") of an observable on a state tells you how much the state "fluctuates" between different values of the observable. The standard deviation is, for instance, zero for eigenstates of the observable, since you always just measure the one eigenvalue that state has.
Plugging in the canonical commutation relation $$ [x,p] = \mathrm{i}\hbar$$ yields the "famous" version of the uncertainty relation, namely $$ \sigma_x\sigma_p\ge \frac{\hbar}{2}$$ but there is nothing special about position and momentum in this respect - every other operator pair likewise fulfills such an uncertainty relation.
It is, in my opinion, crucial to note that the uncertainty principle does not rely on any conception of "particles" or "waves". In particular, it also holds in finite-dimensional quantum systems (like a particle with spin that is somehow confined to a point) for observables like spin or angular momentum which have nothing to do with anything one might call "wavenature". The principle is just a consequence of the basic assumption of quantum mechanics that observables are well-modeled by operators on a Hilbert space.
The reason how "waves" enter is that the uncertainty relation for $x$ and $p$ is precisely that of the "widths" of functions in Fourier conjugate variables, and the Fourier relationship we are most familiar with is that between position and momentum space. That the canonical commutation relations are equivalent to such a description by Fourier conjugate variables is the content of the Stone-von Neumann theorem.
However, it is the description by commutation relations and not that by Fourier conjugacy that generalizes to all quantum states and all operators. Therefore, it is the commutation relation between the operators that should be seen as the origin of their quantum mechanical uncertainty relations.
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Yes, light diffraction may be viewed both as a classical phenomenon and as a quantum mechanical consequence of Heisenberg's uncertainty principle. However, since both explanations work equally well, it doesn't provide any direct evidence for quantum mechanics.
Let me explain why the two explanations are equivalent. I'll do the classical uncertainty bound first.
You may have noticed that in far-field diffraction, the product of the width of the aperture and the size of the pattern on the screen is a constant: a slit half as wide makes a pattern twice as big. In fact, plugging in the formulas for any kind of diffraction whatsoever will give something like $$\sigma_{\text{slit}} \sigma_{\text{screen}} \gtrsim D \lambda$$ where $D$ is the distance to the screen. This is the "uncertainty principle" for classical diffraction.
Now let's view this on the quantum level. The position uncertainty is simply $$\sigma_x = \sigma_{\text{slit}}.$$ The momentum is given by the de Broglie relation $$p = \frac{h}{\lambda}$$ but we want the uncertainty in the $x$-component of the momentum, $p_x = p \sin \theta \approx p\theta$. Now, the uncertainty in the angle $\theta$ is just $\sigma_{\text{screen}}/D$ by trigonometry, so we have $$\sigma_{p_x} = p \sigma_{\theta} = \frac{h}{\lambda} \frac{\sigma_{\text{screen}}}{D}.$$ Putting this all together, $$\sigma_x \sigma_{p_x} = \frac{h}{\lambda D} \sigma_{\text{slit}} \sigma_{\text{screen}} \gtrsim h.$$ Up to a constant, this is the usual Heisenberg uncertainty principle; the two pictures are equivalent.
From a mathematical perspective, both uncertainty principles listed above special cases of a more general fact: the product of the width of a function $f$ and the width of its Fourier transform is bounded.
In the quantum case, the Fourier pair is position and momentum. In the classical far-field diffraction case, the pair is the screen and source, as proven here. This result also has applications in signal processing.