quantum mechanics - Infinite dimensional vector spaces vs. the dual space

I just happened across this over on Math Overflow. It references the following theorem from linear algebra:

A vector space has the same dimension as its dual if and only if it is finite dimensional.

I would like to ask a physical question using the infinite square well (ISW) in quantum mechanics as motivation. For the ISW we obtain $$\psi_n=A_n\sin(\frac{n\pi x}{a})$$ as the eigenfunctions of the Hamiltonian. Here $n=1,2,3,4...$ enumerates the states. If I understand correctly this is an infinite dimensional vector space, because the $\psi_n$'s form an infinitely large basis (ie there is no largest value of $n$). If the dual space is the set of functions $\psi_n^*$ (which I think it is) how can the vector space and the dual space have different dimensions?

Answer

There are two concepts of duality for vector spaces.

One is the algebraic dual that is the set of all linear maps. Precisely, given a vector space $V$ over a field $\mathbb{K}$, the algebraic dual $V_{alg}^*$ is the set of all linear functions $\phi:V\to \mathbb{K}$. This is a subset of $\mathbb{K}^V$, the set of all functions from $V$ to $\mathbb{K}$. The proof you can see on math overflow uses, roughly speaking, the fact that the cardinality of $\mathbb{K}^V$ is strictly larger than the cardinality of $\mathbb{K}$ if $V$ is infinite dimensional and has at least the same cardinality as $\mathbb{K}$.

So for algebraic duals, the dual of any infinite vector space has bigger dimension than the original space.

The other concept is the topological dual, that can be defined only on topological vector spaces (because a notion of continuity is needed). Given a topological vector space $T$, the topological dual $T_{top}^*$ is the set of all continuous linear functionals (continuous w.r.t. the topology of $T$). It is a proper subset of the algebraic dual, i.e. $T_{top}^*\subset T_{alg}^*$.

For topological duals, the restriction to continuous functionals makes the previous statement false (i.e. there exist infinite dimensional topological vector spaces whose topological dual has the same dimension of the original space).

The usual example are Hilbert spaces, where the Riesz representation theorem holds (see my comment above): any object of the topological dual $H^*_{top}$ of a Hilbert space $H$ can be identified via isomorphism with an element of $H$. So an Hilbert space and its dual are the "same".

Note however that the topological dual is always thought to be "bigger (or maybe equal)" than the original space. I am very non-precise here, but I think the following example clarifies. Think to the distributions $\mathscr{S}'(\mathbb{R})$. This is the topological dual of the functions of rapid decrease $\mathscr{S}(\mathbb{R})$. Any $f\in \mathscr{S}$ is isomorphic to a distribution in $\mathscr{S}'$, but the converse is obviously not true: there are distributions that are not functions (the Dirac's delta), and in general any $L^p$-space is thought as a subset of $\mathscr{S}'$ (so $\mathscr{S}'$ is quite "big").