I don’t know. I say this for anyone this has unintentionally clickbaited, or who’s looking at a search engine’s preview of the page.

I come to this question from a friend, though, and it’s got me wondering. I don’t have a good answer, either. But I’m putting the question out there in case someone reading this, sometime, *does* know. Even if it’s in the remote future, it’d be nice to know.

And before getting to the question I should admit that “why” questions are, to some extent, a mug’s game. Especially in mathematics. I can ask why the sum of two consecutive triangular numbers a square number. But the answer is … well, that’s what we chose to mean by ‘triangular number’, ‘square number’, ‘sum’, and ‘consecutive’. We can show why the arithmetic of the combination makes sense. But that doesn’t seem to answer “why” the way, like, why Neil Armstrong was the first person to walk on the moon. It’s more a “why” like, “why are there Seven Sisters [ in the Pleiades ]?” [*]

But looking for “why” can, at least, give us hints to why a surprising result is reasonable. Draw dots representing a square number, slice it along the space right below a diagonal. You see dots representing two successive triangular numbers. That’s the sort of question I’m asking here.

From here, we get to some technical stuff and I apologize to readers who don’t know or care much about this kind of mathematics. It’s about the wave-mechanics formulation of quantum mechanics. In this, everything that’s observable about a system is contained within a function named . You find by solving a differential equation. The differential equation represents problems. Like, a particle experiencing some force that depends on position. This is written as a potential energy, because that’s easier to work with. But it’s the kind of problem done.

Grant that you’ve solved , since that’s hard and I don’t want to deal with it. You still don’t know, like, where the particle is. You never know that, in quantum mechanics. What you do know is its distribution: where the particle is more likely to be, where it’s less likely to be. You get from to this distribution for, like, particles by applying an operator to . An operator is a function with a domain and a range that are spaces. Almost always these are spaces of functions.

Each thing that you can possibly observe, in a quantum-mechanics context, matches an operator. For example, there’s the x-coordinate operator, which tells you where along the x-axis your particle’s likely to be found. This operator is, conveniently, just x. So evaluate and that’s your x-coordinate distribution. (This is assuming that we know in Cartesian coordinates, ones with an x-axis. Please let me do that.) This looks just like multiplying your old function by x, which is nice and easy.

Or you might want to know momentum. The momentum in the x-direction has an operator, , which equals . The is partial derivatives. The is Planck’s constant, a number which in normal systems of measurement is amazingly tiny. And you know how . That – symbol is just the minus or the subtraction symbol. So to find the momentum distribution, evaluate . This means taking a derivative of the you already had. And multiplying it by some numbers.

I don’t mind this multiplication by . That’s just a number and it’s a quirk of our coordinate system that it isn’t 1. If we wanted, we could set up our measurements of length and duration and stuff so that it was 1 instead.

But. Why is there a in the momentum operator rather than the position operator? Why isn’t one and the other ? From a mathematical physics perspective, position and momentum are equally good variables. We tend to think of position as fundamental, but that’s surely a result of our happening to be very good at seeing where things are. If we were primarily good at spotting the momentum of things around us, we’d surely see that as the more important variable. When we get into Hamiltonian mechanics we start treating position and momentum as equally fundamental. Even the notation emphasizes how equal they are in importance, and treatment. We stop using ‘x’ or ‘r’ as the variable representing position. We use ‘q’ instead, a mirror to the ‘p’ that’s the standard for momentum. (‘p’ we’ve always used for momentum because … … … uhm. I guess ‘m’ was already committed, for ‘mass’. What I have seen is that it was taken as the first letter in ‘impetus’ with no other work to do. I don’t know that this is true. I’m passing on what I was told explains what looks like an arbitrary choice.)

So I’m *supposing* that this reflects how we normally set up as a function of position. That this is maybe why the position operator is so simple and bare. And then why the momentum operator has a minus, an imaginary number, and this partial derivative stuff. That if we started out with the wave function as a function of momentum, the momentum operator would be just the momentum variable. The position operator might be some mess with and derivatives or worse.

I don’t have a clear guess why one and not the other operator gets full possession of the though. I suppose that has to reflect convenience. If position and momentum are dual quantities then I’d *expect* we could put a mere constant like wherever we want. But this is, mostly, me writing out notes and scattered thoughts. I could be trying to explain something that might be as explainable as why the four interior angles of a rectangle are all right angles.

So I would appreciate someone pointing out the obvious reason these operators look like that. I may grumble privately at not having seen the obvious myself. But I’d like to know it anyway.

[*] Because there are not eight.