Hilbert space of a free particle: Countable or Uncountable?

This is obviously a follow on question to the Phys.SE post Hilbert space of harmonic oscillator: Countable vs uncountable?

So I thought that the Hilbert space of a bound electron is countable, but the Hilbert space of a free electron is uncountable. But the arguments about smoothness and delta functions in the answers to the previous question convince me otherwise. Why is the Hilbert space of a free particle not also countable?

• A free particle and a harmonic oscillator both have the same Hilbert space $L^2(\mathbb{R}^n)$. – tparker Jan 27 '18 at 15:36

The Hilbert dimension of the Hilbert space of a free particle is countable. To see this, note that

1. The Hilbert space of a free particle in three dimensions is $L^2(\mathbb{R}^3)$.

2. An orthonormal basis of a Hilbert space $\mathcal H$ is any subset $B\subseteq \mathcal H$ whose span is dense in $\mathcal H$.

3. All orthornormal bases of a given non-empty Hilbert space have the same cardinality, and the cardinality of any such basis is called the Hilbert dimension of the space.

4. The Hilbert space $L^2(\mathbb R^3)$ is separable; it admits a countable, orthonormal basis. Therefore, by the definition of the Hilbert dimension of a Hilbert space, it has countable dimension.

• @joshphysics- How can free particle Hilbert space be $L^2(\mathbb{R}^3)$? The free particle states are momentum eigenstates and hence of the form $\sim e^{i\vec k\cdot \vec r}$, which are not square-integrable. Is there any countable basis in the free particle case? – SRS Feb 21 '14 at 7:06