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.

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    $\begingroup$ Relevant question. $\endgroup$
    – Charlie
    Commented Jan 2, 2021 at 15:30
  • 1
    $\begingroup$ The question I linked doesn't necessarily answer your question, but what you're asking about is why the position and momentum eigenstates are being singled out here and the reason is that they are not elements of the Hilbert space we use in quantum mechanics because they are not square integrable. This is usually something that can be ignored in QM unless your intention is to do QM in full rigour, in which case you will need a "rigged Hilbert space". $\endgroup$
    – Charlie
    Commented Jan 2, 2021 at 15:45
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    $\begingroup$ It actually answers, $S^*$ is the wanted extension of the Hilbert space... $\endgroup$ Commented Jan 2, 2021 at 16:00
  • $\begingroup$ See also this Answer. $\endgroup$ Commented Jan 2, 2021 at 23:20

2 Answers 2


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.

  • $\begingroup$ thank you for the answer, I think I get it now. Question - Is this method of simply defining a vector as a formal object (with some desired properties) not necessarily being part of the concerned vector space more general, i.e., can (naively speaking) something like this also be done in any other vector space (eg. Banache, etc.)? $\endgroup$
    – user263315
    Commented Jan 3, 2021 at 13:47
  • $\begingroup$ @user263315 I don’t know the answer to that, but my gut reaction is no. My suspicion is that an inner product structure is necessary in order to make sense of the idea that these “generalized vectors” should actually span the space. It’s possible that this idea can be generalized to Banach spaces in some way, but I’ve never seen it done nor have I put much thought into the idea. $\endgroup$
    – J. Murray
    Commented Jan 3, 2021 at 14:46
  • $\begingroup$ @user263315 More generally though, defining a formal object with a certain desired set of properties is a recurring theme in mathematics which ultimately gives rise to new structures. For example, things like the covariant derivative in GR can be defined by saying “I want an object which acts like a derivative on tensor fields” and then imposing some set of properties to try to define it uniquely and consistently. One often finds that new data needs to be supplied-the Christoffel symbols in this example-and together this new data and your defined symbol constitute a new mathematical structure. $\endgroup$
    – J. Murray
    Commented Jan 3, 2021 at 14:55
  • $\begingroup$ that's a really helpful example. Thanks. $\endgroup$
    – user263315
    Commented Jan 3, 2021 at 15:29

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 $U$.


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