Is the computational basis for Hilbert space transfinite?

Question

In What is the Computational Basis? gIS states:

One also often speaks of "computational basis" for higher-dimensional states (qudits), in which case the same applies: a basis is called "computational" when it's the most "natural" in a given context.

The Wikipedia page for Hilbert space includes this snippet:

When that set of axes is countably infinite, the Hilbert space can also be usefully thought of in terms of the space of infinite sequences that are square-summable. The latter space is often in the older literature referred to as the Hilbert space.

If the Hilbert space has countably infinite axes, is the computational basis for it transfinite?

Additionally, would it be accurate to state that computations with finite bases correlate to classical computations, while computations with transfinite bases correlate quantum computations?

Answer 1

You have a finite basis for $(\mathbb{C}^2)^{\otimes n}$ if you have n qubits. Finite basis (with cardinality $2^n$)-still quantum.

$\mathcal{H}$ could be countably infinite dimensional. Like $L^2 (\mathbb{R}^3)$ for an orbital. Or an anti-symmetric combination of $N$ of those denoted $\wedge^N (L^2 (\mathbb{R}^3))$ for $N$ electrons in some molecule. Both countable dimensional.

The set of states is uncountable, but that is just like saying $\mathbb{C}$ or the set of phases $e^{i \theta}$ is uncountable as a set. True but not really relevant because you are using other structure besides just being a set.

Uncountable dimensional Hilbert spaces are hairier. They would not be separable so a bunch of theorems you need would disappear.

Edit: Was referring to orthogonal basis not algebraic basis. If you demanded everything written as a finite linear combinations of basis vectors then the sequence $e_i$ of vectors that have $1$ in position $i$ and 0's elsewhere would not be a basis for $\ell^2 (\mathbb{N})$. You would need infinitely many of them for something like $(1,\frac{1}{2},\frac{1}{4},\cdots)$. So to make an algebraic basis you would need more basis vectors, in fact uncountably more. Another way to see this is the set of $(1,t,t^2,\cdots)$ for all $0<t<1$ is uncountable and algebraically linearly independent so algebraically $\ell^2 (\mathbb{N})$ is at least uncountably infinite dimensional in the algebraic sense. In a Hilbert space usually using Hilbert space dimension not vector space dimension.