almightyJack t1_iscggi6 wrote on October 14, 2022 at 10:10 PM

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.

PersonUsingAComputer t1_isdi1tm wrote on October 15, 2022 at 3:14 AM

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.

1856NT t1_isdtxce wrote on October 15, 2022 at 5:17 AM

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

digitalsilicon t1_isdyzer wrote on October 15, 2022 at 6:21 AM

Integers are real numbers.

1856NT t1_iseb730 wrote on October 15, 2022 at 9:15 AM

Yes, but not all real numbers are integers.

digitalsilicon t1_isfh3gt wrote on October 15, 2022 at 4:04 PM

> 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.