Conceptual question on eigenvectors in quantum mechanics Page number 1 on Quantum Many-Particle Systems by John W. Negele and Henri Orland says the following about quantum mechanical position eigenvector $|r\rangle$ & momentum eigenvector $|p\rangle$ in a Hilbert space:

Although these vectors do not belong to the Hilbert space because their norm is not finite, they span the whole Hilbert space (as reflected through their closure relations).

I'm familiar with closure relations but can't understand what the above statement means - How can a vector not be part of a vector space but still span it? It's like saying $\hat i$ is not part of the rectangular Cartesian vector space.
Any explanation would be welcome.
 A: There is already a good answer from J. Murray. Here we merely want to answer OP's last question.

How can vectors not be a part of a vector space but still span it?

Answer: E.g. a basis $(\vec{e}_1,\ldots,\vec{e}_n)$ of a finite-dimensional vector space $V$ spans by definition $V$ and therefore a subspace $U\subset V$ even if none of the basis vectors belong to $U$.
A: The word "span" is meant in a slightly different way than in elementary linear algebra, where (as you say) it wouldn't make sense for a set of vectors to span a space without actually being in it. As mentioned in the comment by Charlie, this requires the notion of rigged Hilbert spaces to actually formalize.
If you want the end-user result, it turns out that you can define the symbol $|x\rangle$ to have the following properties:

*

*$\hat X|x\rangle = x|x\rangle$, so $|x\rangle$ acts like an eigenvector of the position operator with eigenvalue $x$, and

*$\int_{-\infty}^\infty dx\ |x\rangle\langle x| = \mathbb 1$ is the identity operator

If you do this, then any state $|\psi\rangle$ can be expressed as
$$|\psi\rangle = \mathbb 1 |\psi\rangle = \int_{-\infty}^\infty dx \langle x|\psi\rangle |x\rangle \equiv \int_{-\infty}^\infty dx \ \psi(x) |x\rangle$$
where $\psi(x)$ acts as the position-space probability amplitude for the state $|\psi\rangle$.   Other properties also follow, such as the fact that $\langle x|y\rangle = \delta(x-y)$.
If you restrict yourself to following these rules, then all will be well.  Note that we have not made any claims as to what $|x\rangle$ actually is - we haven't said it's an element of the Hilbert space (it isn't) - we are treating it as a purely formal mathematical symbol.

Justifying the existence of such an object more rigorously takes some work.  What we first do is identify a special, extremely well-behaved subspace $S\subset \mathcal H$ of our Hilbert space.  The defining characteristic of $S$ is that any vector $\phi\in S$ can be acted on by $\hat X$ and $\hat P$ as many times as you want without leaving $\mathcal H$.  This is a very special property, which is not shared by most vectors in $\mathcal H$; it does turn out, however, that $S$ is dense in $\mathcal H$.
We then consider the set of all linear maps from $S$ to $\mathbb C$.  This is called the (algebraic) dual space $S^*$.  For example, let $\psi\in L^2(\mathbb R)$.  The following  are ways we could linearly map $\psi$ to $\mathbb C$:

*

*$\psi \mapsto \psi(x)$ (we could evaluate it at a point)

*$\psi \mapsto -i\psi'(x)$ (we could differentiate it, evaluate it at a point, and then multiply it by $-i$)

*$\psi \mapsto \langle \phi|\psi\rangle$ (given any $\phi\in L^2(\mathbb R)$, we can take the inner product of it with $\psi$)

Convince yourself that these are all linear maps from $S$ to $\mathbb C$.  The last example demonstrates that every element of the Hilbert space can be thought of as an element of $S^*$ (more properly, every element of $\mathcal H$ corresponds to an element of $S^*$, but nevermind that).  However, the first two actions cannot be expressed as the inner product of some element of $\mathcal H$ with $\psi$, so it seems that $S^*$ is even larger.  That's why you often see $S\subset \mathcal H \subset S^*$.
If you look at example (1) above, that looks an awful lot like $\langle x |\psi\rangle = \psi(x)$, and indeed this is the case.  $\langle x |$ is an element of $S^*$, the algebraic dual space to $S$.  More generally, $S^*$ corresponds to the space of "bra vectors", whereas the kets are elements of the nearly-identically-constructed antidual space $S^\times$, consisting of the conjugate linear maps from $S$ to $\mathbb C$.
That's a (hopefully gentle) introduction to the idea of a rigged Hilbert space. The answer linked in Charlie's comment gives a more thorough description, and for a real understanding you can read this wonderful paper by Rafael de la Madrid.
