Do spin wavefunctions live on a Hilbert space?

Question

This question comes to mind while reading the quote in different books which is "Hilbert space is infinite dimensional". While the electron spin is 2 Dimensional problem. Secondly, we assume that spin wave function is square integrable which is a property of a function in Hilbert space. THen how do we know if spin wavefunctions belongs to Hilbert Space or not? Please help me clear this concept of Hilbert space.

Answer 1

A Hilbert space is a complex vector space with an inner product that is complete w.r.t. the norm induced by that inner product. It may be finite-dimensional, it may be infinite-dimensional - the dimensionality is not part of the definition. Neither is square-integrability. Either your sources are wrong - or you are misinterpreting them, not paying attention to a difference between the generic definition of a Hilbert space and the definition of a specific Hilbert space as the space of states of a physical system.

The space of states of position (or momentum) wavefunctions of objects without spin in quantum mechanics is the infinite-dimensional Hilbert space of square-integrable functions $L^2(\mathbb{R}^n)$.

The space of (spin) states for a particle with spin $s$ is the finite-dimensional Hilbert space $\mathbb{C}^{2s+1}$.

Since spaces of states are combined via the tensor product, the space of states of position (or momentum) wavefunctions for an object with spin $s$ is the Hilbert space $L^2(\mathbb{R}^n)\otimes \mathbb{C}^{2s+1}$, which is the space of $\mathbb{C}^{2s+1}$-valued functions that are square-integrable in each component.

Answer 2

A Hilbert space is a vector space $\mathcal H$ with an inner product which satisfies a particular metric condition with respect to the sense of distance induced by this inner product. (Specifically, the space must be complete, i.e., every Cauchy sequence in $\mathcal H$ converges to a limit in $\mathcal H$.)

Every finite-dimensional inner-product space is complete (because it is isomorphic to $\mathbb C^n$ and $\mathbb C^n$ is complete), so therefore every finite-dimensional inner-product space is a Hilbert space.

As such, anybody who claims that "Hilbert space [must be] infinite dimensional" is basically dead wrong about that. Hilbert spaces can be infinite-dimensional, but they don't need to be.