# 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 $L^2(\mathbb{R}^n)$. Commented Jan 27, 2018 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.

There is another notion of basis that is usually not being referred to when one discusses Hilbert spaces, namely a Hamel basis (aka algebraic basis). There is a corresponding theorem called the dimension theorem which says that all Hamel bases of a vector space have the same cardinality, and the dimension of the vector space is then defined as the cardinality of any Hamel basis.

One can show that every Hamel basis of an infinite-dimensional Hilbert space is uncountable.

As a result, the dimension (in the sense of Hamel bases) of the free particle Hilbert space is uncountable, but again, this is not usually the sense in which one is using the term dimension in this context, especially in physics.

• @Jim If you're using the rigged Hilbert space formalism, your state space is actually a subspace of a countable Hilbert space. See here: physics.stackexchange.com/questions/43515/… Commented May 22, 2013 at 11:39
• @user1504 : ok, but then it's not a Hilbert space. Commented May 24, 2013 at 18:02
• @jjcale Jim asked a second question (in his comment) about rigged Hilbert spaces. In this formalism, the state space is not actually a Hilbert space. Instead, it is a vector subspace of some Hilbert space. This is sufficient to answer Jim's question. For more details about the formalism, go see the answer to the question I linked to above. Commented May 24, 2013 at 19:19
• @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
Commented Feb 21, 2014 at 7:06
• @joshphysics- Okay. But I don't understand why do we throw away this set $\{e^{i\vec k\cdot\vec r}\}$ out of the Hilbert space? Since these are not square integrable it implies that they does not belong to $L^2$. That is fine. But we could have extended the Hilbert space from $L^2$, to something that includes these functions as well. Isn't it? Does it imply this solutions are not physically acceptable? But during scattering problems and other cases we do use such functions? What is the Hilbert space there? Do we always use $L^2$ in quantum mechanics? To save Born's probability interpretation?
– SRS
Commented Feb 22, 2014 at 7:14