The question is a bit vague, but here's my attempt at two different interpretations.

In the context of quantum physics, "quantised vectors" are not themselves quantised, it is the (usually energy) eigenvalue which is constrained to be specified values.

Just because you can multiply the eigenvectors by any scalar, doesn't mean all those scalars are meaningful.

In a mathematical sense, you can define a set of vectors on an integer lattice (quantised in the sense that in a basis aligned with the lattice all coordinates are integers), but then this is a vector space over the integers, such that the axiom you are talking about is only valid if the scalar is an integer.

Technically speaking, such an integer lattice construction only gives you a free module over the integers, not a vector space, since the integers don't form a field.

Angular momentum for example is quantized, yes it might not be physically meaningful to multiply 1/2 and nh/2pi but mathematically it doesn’t satisfy the axiom which clearly states that the scalar is a real number not an integer

Integers are real numbers.

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Yes, but not all real numbers are integers.

technically that holds true for all real number representations of vectors we have. your computer computes stuff in "pretend" vector spaces and it just happens to approximately work out. Sometimes you have to care about the numerics, but most of the time you are fine.

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//edit Having said that, there are ways to make it mathematically work. A vector space is typically defined over a field, e.g., the real or complex numbers, and in this case the scalars are limited to that field. The same construction over a ring (like the integers) is called a module. So a quantized vector (for example the integers) limited to scalars being integers as well DOES work out. The only thing you have to deal with is that divison in a ring might not exist.

In what context are you seeing the term "quantised vector space"?

In any case, you may find it interesting to read about vector spaces over finite fields.

Thanks, your answer is really useful!

> multiply 1/2 and nh/2pi but mathematically it doesn’t satisfy the axiom which clearly states that the scalar is a real number not an integer

If you multiply a vector by an integer, that is still multiplication by a real number. It doesn’t matter than not all real numbers are integers.

Angular momentum in quantum mechanics is a vector in that it transforms under the vector representation of SO(3). The term vector unfortunately means two different things - vector spaces are one, and representation theory (tensors etc) are another.

Very useful! Thanks!