ELI5:Why can't electrons have a defined position?

There's absolutely no way to explain this like you're five. But, I'll try to explain it with the belief that you're capable of understanding it without a formal physics education if you're willing to struggle with the math. I'm going to try and answer your question by telling you a little bit about quantum mechanics and also by providing you with a system (and therefore some actual mathematics) you can play with. Hopefully the system at the very least illuminates what quantum mechanics is about. I'm only going to give you information I think is important, and even though you might not have studied most of the math in school, I'm going to assume your Wikipedia game is fire.

What is the theory of quantum mechanics, and what is the fate of classical position and momentum? In quantum mechanics, we entertain "states" with a somewhat mysterious role. The way we entertain them, for instance, is by solving the Schrödinger equation. The Schrödinger equation looks kind of like the heat equation, so you can expect some kind of 'spreading out' business at some point. Two cases, in general, arise when you solve the Schrödinger for a potential V(x). The energy levels you obtain are either discrete (hence quantum), or they're continuous. I'll tell you about the discrete case first because I believe it's simpler and I can tell you about the theory in general before moving onto the continuous case (the free particle specifically), hopefully answering your question at some point.

A specific discrete case you should take a look at is the infinite square well (or particle in a box, whatever). When you solve Schrödinger (2nd order partial differential equation, easier to solve in this case than it sounds) there is an infinite number of states, each with their own specific energy. The significance of these states is that they're probability density functions. This means you can find "expected values" of observable variables IF you know what's its "operator" is that can act on the wave function. Take a second now and look up what the operators for position and momentum are. This is important. An important question is, do these operators commute? If they don't, they have a "commutator". In quantum mechanics, if two operators don't commute, you cannot measure them both at once. The position and momentum operators DON'T commute. You can prove that any two operators that don't commute satisfy a general "uncertainty principle", that sets limits on how spread out their probability densities have to be. The uncertainty principle you're aware of is specifically about position and momentum, but the general principle is true for any two observable operators that don't commute with each other. Hopefully when I tell you about the free particle case, you'll begin to understand why this might be true in general.

For any discrete system, the states are "complete" and "normalizable". Being normalizable just means that the state can be scaled so that the area underneath it is 1, like it should be for any probability distribution. A set of functions being complete means you can add them together and build any function f(x) if you pick your coefficients properly. Experimentally, you'll have had your quantum system in some state f(x), and the squares of the coefficients represent the probability you'll end up observing the particle in that state, with that energy. Wait, didn't I just say the system was in state f(x)? Well f(x) is a superposition. It's dead AND alive, so to speak, but with an infinite number of other options potentially. When you observe it, the particle commits to a state. If you quickly observe it again, it'll be in the same state. If you wait awhile, the probability density kind of spreads out (like heat does, via the heat equation). You could know almost exactly where a particle is, but because of the uncertainty principle, the momentum probability density becomes super wide.

Okay, if you're still with me, we can do the continuous energy case. If we consider a free particle, the potential is zero. The Schrödinger equation is then super simple. The solutions are waves, each with a different "wave number" and therefore also momentum. But... wait... waves aren't normalizable. Therefore they're not physically real states, period. What we do do now? Well... we can build a state f(x) that is. Go on Wikipedia and look up "wave packet". If you allow for continuous energy and momentum values, we can put these waves together so they form a normalizable wave packet. We now have an interesting situation. You can create a wave packet that has a sharply defined position, so even though everyone is telling you that this isn't possible, it is. Particles can have defined position, just not perfectly defined. It must obey the uncertainty relation. If the wave packet is particle-like, the wave numbers (momentums) of the individual states that make up the wave packet spread over a large range of values. To ask what this particles "momentum" actually is would be a goofy question for reasons I hope are very clear to you now. We need a wide range of momentum values so we can build this particle-like normalizable wave packet in the first place. If we want a clearly defined momentum (as defined as is possible), then our wave packet is going to be spread out and will be more wave-like.

That's the mathematical fact of the matter for wave packets, and I hope the background information helped. If we're talking about different systems, (hydrogen for example), we won't be dealing with wave packets, we'll be dealing with the electronic states of hydrogen. However, to get the momentum wave functions that tells us what's going on with the momentum we do a Fourier transform, and the uncertainty principle always applies, so it's still "enforced". The mathematical facts of life will still enforce the uncertainty in momentum and position. If you want more quantum mechanics, and if your calculus, differential equations and linear algebra are decent, consult Griffiths. There are other good QM books I'm sure, but Griffiths is the only one I've read.

Source: Griffiths, and I'm in my 4th year of undergrad physics.

/r/explainlikeimfive Thread