# A vector space with countable and uncountable basis at the same time

Let $V$ be a vector space over $\mathbb{C}$. Two self-adjoint, commutable linear operators $\xi$ and $\eta$ act on it. Both of their eigenvectors form a complete set of $V$, but $\xi$'s eigenvalues are discrete, while $\eta$'s are continuous. How is this possible? I mean how is it possible that the same vector space has both an uncountably infinite basis and a countably infinite basis? This is taken from Dirac's Principles of Quantum Mechanics, he uses these kinds of operators without considering this problem, which buggles me.

While thinking about this, I've realized, that the problem of Fourier-series is an example of the same vector space having two bases with different cardinality. A function which equals its Fourier-series can be viewed as a vector expanded in a continous basis in time domain and in discrete basis in frequency domain.

So this problem is definitely not a contradiction, there exist vector spaces with countable and uncountable basis at the same time. Still, if a vector space is countably finite dimensional there shouldn't be uncountably many independent vectors in it, and conversely, if it is uncountably infinite dimensional, then a countably infinite set of vectors shouldn't be enough to span it.

Last but not least we are considering a closed space equipped with a scalar product, so infinite sums of vectors are allowed.

UPDATE: I've asked the same question on physics.stackexchange.com and posted an answer for myself there with the help of the answers I got here and from one of my teachers (link). It contains the same info about the "denseness" Asaf Karagila suggested. However, there is an interesting part (at least for me) namely the Fourier series expansion of a function derived as a kind of basis transformation.

You should differentiate between a Schauder basis and a Hamel basis. The former only spans a dense subspace, whereas the latter spans the entire basis.

Assuming the axiom of choice we can show that every two Hamel bases have the same cardinality. Of course this is not the case with Schauder bases (because a linear basis is always a topological basis too).

Perhaps the crux of the difference here, is that a Hamel basis, and the notion of linear span, only permits finite sums. In the topological case this is not the case anymore, and because this is a normed space infinite sums suffice for talking about convergence.

• I'd be curious to see a concrete example of a function that can be represented in a Hamel basis but not a Schauder basis. (I suspect that one would consider such examples as possible but unphysical in quantum mechanics.) – Semiclassical Aug 5 '14 at 22:41
• I am not the person to ask that from. :-) I do recall a comment on this site, directed to me, that referenced a paper dealing with intangible objects in physics. It might interest you. – Asaf Karagila Aug 5 '14 at 22:43
• @Semiclassical - The spaces of Quantum Mechanics are Hilbert spaces that are separable, which guarantees a countable orthonormal basis $\{ e_{n} \}$. Finite linear combinations of the elements of $\{ e_{n}\}$ do not span the entire space, but such combinations are dense in the space, and that's good enough. A Hamel basis $\{ h_{\alpha} \}_{\alpha\in\Lambda}$ is one where every element can be written as a finite linear combination. That's not a suitable basis for dealing with Quantum because such a basis is not going to be countable for an infinite-dimensional Hilbert space. – Disintegrating By Parts Aug 6 '14 at 4:05
• I'm starting to get the point. I think the heart of the problem is that an uncountable set may have dense subsets with different cardinalities, which is exactly the case here. Just like in the set of real numbers the rationals form a countably infinite dense subset, while the irrationals form an uncountably infinite one. However I don't see mathematicians use uncountable bases, and express the elements of the vector space as integrals. Why is that? – user3237992 Aug 6 '14 at 7:54
• @user3237992: Yeah, that's the point. As for non-separable spaces, the issue is that with countable dense sets we can easily write them as a series like we're used to in calculus, and it makes things easier. Note by the way, that this is equivalent to being able to approximate every vector by a convergent sequence from the dense subset. So when you can't do it like that, you can't use sequences either, which makes everything a whole lot more complicated with nets and filters and whatnot. – Asaf Karagila Aug 6 '14 at 9:11

What you describe in the first paragraph is impossible, assuming you are talking about self-adjoint operators on Hilbert space. Eigenvectors for distinct eigenvalues of a self-adjoint operator are orthogonal. Hence the number is distinct eigenvalues is bounded above by the maximal size of an orthonormal set, a.k.a. the Hilbert space dimension. Any two complete orthonormal sets have the same cardinality, so you can't have one countable and the other uncountable.

However, if at least one of the operators need not be self-adjoint, the situation you describe can easily happen. E.g., consider the Hilbert space $\ell^2$ of absolute-square-summable sequences of complex numbers. Let $(a_n)$ be a bounded sequence of real numbers, and define the operator $A:\ell^2\to\ell^2$ by $A(x_0,x_1,x_2,\ldots)=(a_0x_0,a_1x_1,a_2x_2,\ldots)$. Then $A$ is a self adjoint operator with $\{a_n\}$ as a set of eigenvalues, and it has a (countable) orthonormal basis of eigenvectors.

On the same space, consider the operator $S$ defined by $S(x_0,x_1,x_2,\ldots)=(x_1,x_2,x_3,\ldots)$. This operator has all complex numbers with absolute value less than $1$ as eigenvalues, and correspondingly, an uncountable linearly independent set of eigenvectors. Namely, for $|\lambda|<1$, $\lambda$ is an eigenvalue for $A$ with eigenvector $(1,\lambda,\lambda^2,\lambda^3,\ldots)$. There is no contradiction here, because these vectors are not orthogonal, so the size is not constrained by the (countable) Hilbert space dimension. This shows that the vector space dimension (size of a maximal linearly independent set) can be greater than the Hilbert space dimension (size of a maximal orthonormal set).

• The "observable" in QM has the property that it's self-adjoint and it's eigenvectors form a complete set. For observables with discrete eigenvalues the normalisation condition is $\langle\xi^r|\xi^l\rangle=\delta_{rl}$. For observables with continuous eigenvalues the normalisation is $\langle\xi'|\xi''\rangle=\delta(\xi'-\xi'').$ A set of observables is a complete commuting set (CCS), if the observables in it commute, and to each set of eigenvalues only one simultaneous eigenvector belongs. – user3237992 Aug 7 '14 at 9:30
• Dirac often uses observables forming a CCS among which there are some with discrete and some with continuous eigenvalues. To me it means that it is possible that a vector space may have a continuous and discrete basis at the same time. Am I wrong? Btw, in QM we use continuous basis in the sense that any vector $|f\rangle$ may be written as an integral: $|f\rangle=\int dt'|t'\rangle\langle t'|f\rangle=\int dt' f(t')|t'\rangle$. Discrete basis is used in the same as usually (infinite, convergent sum). – user3237992 Aug 7 '14 at 9:34
• @user3237992: Ah, that is interesting, so "eigenvectors" and "eigenvalues" are not to be taken literally in the continuous case, in the context you describe! Thank you. There was a miscommunication, I guess because the standard terminology used in physics and mathematics don't quite align. I have to study up on these continuous CCS. – Jonas Meyer Aug 7 '14 at 19:32
• @user3237992 Sorry for replying to an old question. You may know this in the meanwhile, but the "eigenvectors" for the continuous part of the spectrum are not really eigenvectors because they do not belong to the Hilbert space. Example: Physicists would say that the (unbounded) momentum operator $\hat p$ (on $L^2(\mathbf R)$, say) has as eigenvectors the plane-wave solutions $\psi_p(x)=\exp(ip x)$, but these are not in $L^2(\mathbf R)$. I believe it is possible to adopt this view if you consider a rigged Hilbert space (or Gelfand triple): a triple $(V,H,V^*)$ with $V$ a (see next comment) – a... Jan 5 '15 at 1:06
• topological vector space which is densely and continuously included in $H$. In applications of this to quantum mechanics we would typically have $V=\mathscr S(\mathbf R^d)$ the Schwartz space, $H=L^2(\mathbf R^d)$ and $V^*$ the space of tempered distributions and I think there is a spectral theory of self-adjoint operators on rigged Hilbert spaces which allows you to speak of eigenvectors like a physicist. – a... Jan 5 '15 at 1:12