Can the Heisenberg Uncertainty Principle be explained intuitively?

The best intuitive analogy I've heard is with classical sound waves. Consider a musical instrument playing a pure sine wave of frequency $\nu$ and amplitude $A$, and no other harmonic frequencies at all. Graphing this in frequency-amplitude space ($x$-axis=frequency, $y$=amplitude) gives you a $\delta$-function-like point function with value $y=A$ at $x=\nu$, and zero everywhere else. That represents your exact knowledge of the note's frequency.

But at what time was the note played? A pure sine wave extends from $-\infty<t<\infty$. Any attempt to play a shorter note necessarily introduces additional components/harmonics in its Fourier decomposition. And the shorter the interval $t_0<t<t_1$ you want, the broader your frequency spectrum has to become. Indeed, imagine an instantaneous sound. Neither your ear, nor any apparatus, can say anything about its frequency at all -- you'd have to sense some finite portion of the waveform to analyse its shape/components, but "instantaneous" precludes that.

So, you can't simultaneously know both a note's frequency and the time it's played, due to the Fourier conjugate nature of frequency/time. The better you know one, the worse you know the other. And, as @annav mentioned, that's analogous to the nature of conjugate quantum observables.

Edit:

to address @sanchises remark about some "crude MSPaint drawings"...

For simplicity (i.e., my own simplicity generating the following "crude drawings"), I'm illustrating an almost-square wave below, rather than a sine wave. Suppose you wanted to produce a sound wave with a one-cycle duration, looking something like,

So the "tails" are zero in both directions, indicating the sound's finite duration. But if we try generating that with just two fourier components, we can't get those zero-tails. Instead, it looks like,

As you see, we can't "localize" the sound's duration with just two frequencies. To get a better approximation, four components looks like,

And that still fails to accomplish much by way of "localization". Next, eight components looks like,

And that's beginning to exhibit the behavior we're looking for. Sixteen looks like,

And I could go on. The initial illustration above was generated with 99 components, and looks pretty much like the intended square wave.

Comment:

you guys coincidentally stepped into one of my little programs when mentioning drawings. See http://www.forkosh.com/onedwaveeq.html for a discussion, although not about uncertainty. To get the above illustrations, I used the following parameters in that "Solver Box" at top,

nrows=100&ncols=256&ncoefs=99&fgblue=135&f=0,0,0,0,0,0,1,1,1,1,1,-1,-1,-1,-1,-1,0,0,0,0,0,0,0&gtimestep=1&bigf=1

Just change the ncoefs=99 to generate the corresponding drawings above.

The explanation you have heard, extended, goes as follows: suppose I want to find the position of a particle in a box. To do so, I shine light on it, and, in a very similar way to what happens in the macroscopic world, by how the light bounces off I understand where the object is. However, the particle is so small that the momentum of a photon can push it and change its momentum. So: if I use a low-energy, large-wavelength photon, it won't change the particle's momentum much (because of low energy) but also won't tell me its position with high precision (because of large wavelength). If I want a higher precision in the position you need a short wavelength photon, which is unfortunately a high-energy photon and will change the momentum of the particle in an unpredictable way. See Compton Scattering for the physical details.

This, however, is only an example of a consequence of the uncertainty principle. Heisenberg's uncertainty relation is actually far more general and holds in principle, in the same sense as the conservation of energy is not "proven" by explaining why a certain kind of endless energy source cannot work.

A more general statement would be any kind of measurement changes the state of a system. I can only explain this in an axiomatic way, I'm personally unable to convince you based on physical arguments. But there is a sound reason for this. No physical argument based on our intuition of physics can explain quantum uncertainty, because it is fundamentally different from our intuition of physics.

To a person ready to accept this change of paradigm you can explain that the concept of state is different in qm. As someone writes in a comment, position and momentum don't simultaneously exist in qm (as, by the way, the angular momenta along different axes). Some states can have definite position, some can have definite momentum, but not both.

As you are a mathematician, I can explain to you axiomatically why this happens. In the standard theory of QM, it is usually taken as true that:

• States are vectors in a complex Hilbert space
• Observable quantities such as position and momentum correspond to operators in this Hilbert space. Their explicit form depends on the basis you choose, but the important thing is they don't commute in the case of $x$ and $p$.
• States with a definite value for an observable are eigenvectors of the corresponding operator. The definite value is the corresponding eigenvalue.

If two operators don't commute, maths shows they cannot have simultaneous eigenvector bases, and therefore the two physical quantities are never well defined simultaneously.

Another way of putting this mathematically is by showing that the wavefunction (whose modulus squared is the probability of finding the particle at a certain location) and the "wavefunction" in the momentum space are Fourier transforms of each other. You can easily show that if you choose a low-variance distribution on one side, the variance increases on the other and vice versa.

I think: Yes there's an intuitive explanation for the Uncertainty principle. The explanation is the following:

The most important thing to convince the non-scientist listener is that particles in Quantum Mechanics ARE NOT OBJECTS! This is observed in interference experiments and is a fact that we are very certain about. So they are waves. Once they wrap their heads around this idea, things become much easier to explain.

Show them this picture or similar:

And tell them. Electrons look like the wave you see in this picture on top. Can you tell me what's the position of this electron? The listener will fail, and will start to understand that instrumentation errors has nothing to do with it. It's all about what these subatomic particle are. Then explain to him that scientists have a way to say where the electron may act as a particle most likely (which we call the probability of finding the position of the electron). This is defined by where the wave has a higher amplitude (or even naively, where it's further from the x-axis). Now if we would like to map that kind of position and make a position for the electron, the lower picture is how it will look like.

So from this, the listener learned:

• Electrons are waves

• The problem is mapping waves to particles.

• Mapping waves to particles gives uncertainty, which is given by the Heisenberg uncertainty principle.

Good luck!