# I Should the position basis be quantized?

1. Sep 17, 2016

### nashed

In most situations in QM we would get a quantized energy basis, that is a countably infinite basis ( I think it's called having a cardinality of aleph 0), In the meanwhile we take the position basis to be continuous ( cardinality of aleph 1?) and I'm pretty sure that there is a theorem stating that the cardinality of the Hilbert space doesn't change under change of basis, so how does this sit with QM?

2. Sep 17, 2016

### Staff: Mentor

The position basis is not a proper basis - in particular, a particle cannot be at a single point in space. Only a subset of the vectors you can get with the position basis are in the Hilbert space of physical wavefunctions.

3. Sep 17, 2016

### andrewkirk

The cardinality of the set made up of all points in the space is not the same thing as the cardinality of the basis. Consider Euclidean 3-space. The cardinality of a basis is finite (3) whereas the space is uncountably infinite.

Perhaps the theorem that says all bases have the same cardinality only applies to finite-dimensional vector spaces.

Addendum: It turns out they do, provided the space is complete and that the bases we are talking about are Hamel bases. See this paper. Since a Hilbert Space is by definition complete, it follows that all bases should have the same cardinality.

Last edited: Sep 17, 2016
4. Sep 17, 2016

### PeroK

... and, in fact, the position basis vectors themselves are not even in the Hilbert space!

5. Sep 17, 2016

### vanhees71

You are right. The Hilbert space of non-relativistic QT of a fixed number of particles is a separable Hilbert space, i.e., it has a countable complete orthonormal basis. You can take the harmonic-oscillator energy eigenbasis as such a basis (it's of course not a Hamel basis but you allow for the usual "generalized Fourier series", i.e., an infinite series converging in the sense of the Hilbert-space metric induced by the scalar product).

In physics we need another concept, namely the decomposition of a vector in terms of "generalized bases" which live in the dual of the domain of the self-adjoint operators making up the observable algebra of the theory (in the most simple case the Heisenberg algebra; in a more general sense an algebra realizing the Galilei group in terms of a ray representation). The handwaving way physicists treat this subtle issue can be made rigorous in an elegant way using the socalled "rigged Hilbert space" (G'elfand) construction.

What one should, however emphasize more in the introductory QM1 lecture is that these "generalized basis vectors" are not representing pure states. Then a lot of confusion is avoided, e.g., about how to square the energy conserving $\delta$ distribution in the S-matrix elements of potential-scattering theory (or the energy-momentum conserving $\delta$ distribution in particle-scattering theory).

6. Sep 17, 2016

### secur

It's true that the cardinality of a Hilbert space (and any vector space) is equal to the cardinality of any basis (they're all equal). BTW the continuous real numbers are aleph 1 only if we assume the continuum hypothesis, as is normal.

It's also true, of course, that the "naive" position vectors (and, momentum vectors) are not actually in the space because they're not square-integrable. That's why for a rigorous treatment the rigged Hilbert space is necessary. This certainly seems relevant to the OP. But in fact, I think, it isn't. The issue here is much simpler than that.

Consider free particle vs. particle in a box. A free particle can have any energy - any momentum. Physically it must actually be a wave packet but we can ignore those subtleties (I think). However in a box only a discrete set of energy levels is allowed, as shown in any introductory course. The possible energy levels are reduced from a continuous spectrum (aleph 1, ignoring the "rigged" issue) to a discrete set (aleph naught).

For the particle in a box (or harmonic oscillator, or any case where this happens) the energy eigenvectors no longer span the entire, original Hilbert space. They span only a very small (effectively, measure 0) subspace of it, consisting of linear combinations of those eigenvectors, whose eigenvalues are the allowed energy levels. Of course that's exactly what we want - it's the solution to the problem.

We loosely say that the eigenvectors form a basis of "the" Hilbert space - but which Hilbert space? The basis spans only the reduced subspace of physical interest to the particular problem - which is itself, of course, a Hilbert space. And that subspace has a different cardinality than the original - again, ignoring the rigged-space subtleties, which I think are not relevant to OP's interesting question.

7. Sep 17, 2016

### vanhees71

No, $\mathrm{L}^2([a,b])$ is a separable Hilbert space as is $\mathrm{L}^2(\mathbb{R})$. Both have a countable complete orthonormal basis and both spaces are thus equivalent. In fact there is only one separable Hilbert space.

8. Sep 17, 2016

### secur

If you want to understand it, re-read my post, more carefully. Any questions, just ask! I'll be glad to help.

9. Sep 17, 2016

### vanhees71

??? You should reconsider your answer. There's (up to equivalence) one and only one separable Hilbert space. That's a mathematical theorem.

10. Sep 17, 2016

### secur

My point was, that has nothing to do with my post, or OP's question. But - as long as you mention it - being equivalent, or isomorphic, doesn't make them the same space. For instance - and this is relevant to my post - a proper separable subspace of a separable Hilbert space is a separable Hilbert space, but obviously not the same one. Proof: the original space is not a subspace of its own proper subspace, although the converse is true. So they're distinct entities.

11. Sep 17, 2016

### PeroK

When you say:

A finite dimensional vector space has a finite basis. The vector space has the cardinality of the underlying field ($\mathbb{C}$ in this case), but the basis has finite cardinality.

Similarly, a compelx vector space with a countable basis will itself be uncountable.

So, what do you mean here?

What was the "original" Hilbert space? What is the space spanned by the energy eigenvectors? And why does this subspace have "measure 0".

12. Sep 17, 2016

### secur

You're right. The dimension of the Hilbert space is equal to the cardinality of the basis. I always try to follow OP's terminology, but shouldn't, of course, when it's incorrect.

The whole Hilbert space, as opposed to the subspace spanned by the particle-in-a-box energy eigenvectors. The space spanned by all energy eigenvectors, for the free-particle case. It's a slightly tricky point, but I don't see how to express it better. It's necessary to answer OP, who wants to know how the dimension of the "original" Hilbert space seems to become reduced when considering quantized energy levels.

Google it, the first hit is a paper titled "A note on function spaces spanned by complex energy eigenfunctions". I mean exactly what that paper means. Just basic linear algebra. (BTW eigenfunction = eigenvector = eigenstate.)

Ok, that's more complicated to explain and probably should have been left out. Still, the explanation helps communicate my main point. First consider this recent post from Nugatory:

He's referring to the same issue involved here. The fact that position eigenstates (also, momentum) aren't physically realizable necessitates rigged Hilbert space with its subtle complications. In his case Nugatory judged that digressing into those details wouldn't help OP. So he mentions the existence of the issue, for completeness, and leaves it at that. Well, that's exactly what I want to do. OP's question, at first glance, seems to involve rigged space, but actually it doesn't. It's much simpler. Unfortunately I had to deal with the issue a bit because others had already brought it up, incorrectly, in their answers.

Following Nugatory, I mentioned, or indicated - four times, just to make sure -, that the issue does exist but I don't want to get into it:

"for a rigorous treatment the rigged Hilbert space is necessary"
"The possible energy levels are reduced from a continuous spectrum (aleph 1, ignoring the "rigged" issue) to a discrete set (aleph naught)."
"(effectively, measure 0)"
"ignoring the rigged-space subtleties"

IF we pretend that momentum eigenvectors are actually physically realizable, then the large set of energy eigenvalues for the free particle is continuous: all the positive real numbers, cardinality aleph 1. But the smaller space of energy eigenvalues for the particle in a box is countable, aleph 0. That contrast provides the answer to OP. Rigged space is irrelevant. You may not agree with this pedagogical approach, matter of opinion.

Anyway, those particle-in-a-box eigenvalues, as a set of discrete points embedded in R, (the continuum), has Lebesgue measure zero. Probably shouldn't mention this small point, even as a quick parenthetical aside - just causes confusion.

The key point is: this is the right answer to OP's question. The detail of rigged space should have been mentioned for completeness, then ignored, from the start. No doubt the answer can be expressed better; give it a try!