# Conceptual question on eigenvectors in quantum mechanics

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:

1. $$\hat X|x\rangle = x|x\rangle$$, so $$|x\rangle$$ acts like an eigenvector of the position operator with eigenvalue $$x$$, and
2. $$\int_{-\infty}^\infty dx\ |x\rangle\langle x| = \mathbb 1$$ 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 $$\mathbb C$$:

1. $$\psi \mapsto \psi(x)$$ (we could evaluate it at a point)
2. $$\psi \mapsto -i\psi'(x)$$ (we could differentiate it, evaluate it at a point, and then multiply it by $$-i$$)
3. $$\psi \mapsto \langle \phi|\psi\rangle$$ (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.

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 $$U$$.