> > Why would we expect superpositions of quantum states to be encoded as a vector sum of the individual state vectors?
> Because that's what the word superposition means.
No, that's not what superposition means a priori. I can think of many other ways to implement (mathematically) the idea of a system being in "two states at the same time", apart from vector addition. Yes, if you do Stern-Gerlach often enough, you might convince yourself that the vector space structure is a sensible choice but I take issue with OP's statement that
> there are logically no other possibilities
as if things had been obvious right from the get-go.
> No, that's not what superposition means a priori.
The word was originally used to describe the decomposition of waveforms into sums of sinusoids, which is as canonical an example of a linear system as you can get.
> the idea of a system being in "two states at the same time", apart from vector addition.
But that's not what's going on. A system is only ever in one state at a time: the ability to treat it as a sum (modulo the norm) of other states is linearity of all operators. This has immediate observable consequences: nonlinear operators can distinguish between different ensembles realizing the same mixed state.
> > No, that's not what superposition means a priori.
> The word was originally used to describe the decomposition of waveforms into sums of sinusoids, which is as canonical an example of a linear system as you can get.
You are right, I was being a bit too sloppy with my usage of the term "superposition". I guess once people realized that a QM system being in "two states at the same time" is just a linear sum like for waves, they started calling it a superposition. Anyway, my point (in my original comment) was a completely different one: You still have to assume all that linear structure to start talking about how canonical the tensor product is.
> But that's not what's going on. A system is only ever in one state at a time: the ability to treat it as a sum (modulo the norm) of other states is linearity of all operators
Again, you are right, that's why I put it in quotes. Nevertheless, if we start just from the observation that a system can occupy two states "simultaneously" (in the sense that sometimes we measure one, sometimes the other state) we might think of other ways to encode that beyond vector sums, e.g. Cartesian products without any linear structure.
Anyway, I don't think we disagree fundamentally, we're merely arguing about terminology.
A useful illustration: polarized light can be considered to be in a superposition if you use a basis rotated 45 degrees to the polarization axis. So whether or not polarized light is in a superposition depends entirely on how you choose to look at it. It's not a reflection of the underlying physical reality.
I think we're talking past each other. :) I fully agree with your comment. However, I was addressing your original comment[0] about the naturalness of the tensor product. My whole point was merely (and maybe I didn't phrase it particularly well) that you need to assume all the stuff we (now) know about quantum mechanics (vector space structure etc.) in order to conclude that the tensor product is pretty much the only option you have. Your comment didn't mention this and seemed to make a much broader claim. That was all.
(EDIT: I think your other comment[1] fully resolves any gripe I had with your original comment. :))
> Because that's what the word superposition means.
No, that's not what superposition means a priori. I can think of many other ways to implement (mathematically) the idea of a system being in "two states at the same time", apart from vector addition. Yes, if you do Stern-Gerlach often enough, you might convince yourself that the vector space structure is a sensible choice but I take issue with OP's statement that
> there are logically no other possibilities
as if things had been obvious right from the get-go.