# Can observables with discrete and continous eigenvalues be commuting?

In Dirac's The Principles of Quantum Mechanics he often uses the notion of "a complete set of commuting observables". This means a set of observables $\{\xi_1, \xi_2, \ldots, \xi_v, \xi_{v+1}, \ldots, \xi_u\}$ where $\{\xi_1, \xi_2, \ldots, \xi_v\}$ have discerete eigenvalues, and $\{\xi_{v+1}, \ldots, \xi_u\}$ have continous eigenvalues and each pair of observables commute, and to each set of eigenvalues only one simultanious eigenvector belongs.

The definiton of observables by Dirac is that each $\xi_i$ is a real linear operator, and the set of eigenvectors of the operator is a complete set. Now my question is: how is it even possible that obsevables with discrete and continous igenvalues are commuting? I mean if an opeator with discrete eigenvalues is an observable, that means that the set of basis vectors for the vector space is a countably infinite set. However if an operator with continous eigenvalues is also an observable (acting on the same vector space), that means that the set of basis vectors for the same vector space is an uncountably infinite set. For me, these two properties contradict with eachother.

Not to mention that Dirac also gives a proof of two commuting observables having a complete set of simultanious eigenvalues, the "side result" of his proof being that if two observables commute, their sets of eigenvectors mutually contain eachother, which means that the two sets are equal. To me this is also a contradiction.

Also can even two observables with discerete and continous eigenvlaues act on the same space?

Can somebody give me an explanation?

• A standard example of commuting operators with discrete and continuous spectrum are the angular momentum operator $L$ and the kinetic energy operator $p^2$ on ${L^2}(R^3)$. Operators with continuous spectrum are somewhat tricky since they do not have square integrable eigenstates so the formalism needs refining. See en.wikipedia.org/wiki/Complete_set_of_commuting_observables for more details. – Urgje Jul 25 '14 at 20:37
• @Urgje But the way I see it these operators are both continous or both discrete at the same time, not one discrete and one continous. Or am I mistaken? – user3237992 Jul 26 '14 at 19:29
• No, the angular momentum operator has discrete spectrum. $L^2$, for instance, has the eigenvalues 0,1,2,... etc, whereas the kinetic energy operator $p^2$ has continuous spectrum $R^+$. Note that commuting operators can have the same eigenfunctions but in general different eigenvalues. In my example, in spherical coordinates eigenfunctions are spherical harmonics times spherical Bessel functions. The angular momentum operator acts on the spherical harmonics only. See further: en.wikipedia.org/wiki/…. – Urgje Jul 27 '14 at 10:41
• @Urgje Okay, I have to think about this. One more thing: by $L^2(R^3)$ do you mean square integrable functions whose domain is $R^3$? – user3237992 Jul 27 '14 at 12:13
• Yes, that is the standard notation. – Urgje Jul 27 '14 at 17:43

## 1 Answer

After some inquiry on math.stackexchange.com and writing a letter to one of my teachers, I've found the key: in infinite dimensional vector spaces the basis of the space is a dense subset of the vector space, and two dense subsets may have different cardinalities.

A very primitive example of this is the space of real numbers (over the field of real numbers). In this space the rational and irrational numbers both form dense subsets (that is any real number may be "synthesized" as the limit of a sequence of rational numbers, and the same is true for irrationals). This way both the rational and irrational numbers form a (Scahuder) basis of the vector space.

A more advanced example of the same space having bases of different cardinalities is the space of square integrable functions which equal their Fourier series. In bra-ket notation we may write that any such function $f(t)$ has an abstract $|f\rangle$ vector representing it, and there are two bases on the vector space, the time basis: $|t'\rangle, t' \in (-\infty, +\infty)$; and the angular frequency basis: $|k'*\omega_0\rangle, k' \in \mathbb{Z}$.

Since in the continuous basis: $|f\rangle=\int_{-\infty}^{+\infty}dt'|t'\rangle\langle t'|f\rangle=\int_{-\infty}^{+\infty}dt'|t'\rangle f(t')$, the function in the time domain may be looked upon as the coordinates of the vector $|f\rangle$ in the "time" basis.

At the same time $|f\rangle=\sum_{k'=-\infty}^{+\infty}|k'*\omega_0\rangle\langle k'*\omega_0|f\rangle=\sum_{k'=-\infty}^{+\infty}|k'*\omega_0\rangle c_{k'}$. Here the $c_{k'}$-s are the Fourier coefficients of the function, and may be seen as the coordinates of the vector $|f\rangle$ in the "frequency" basis.

From these one may calculate the transformation functions between the two bases.

$f(t')=\langle t'|f\rangle=\langle t'|\sum_{k'=-\infty}^{+\infty}|k'*\omega_0\rangle\langle k'*\omega_0|f\rangle=\sum_{k'=-\infty}^{+\infty}\langle t'|k'*\omega_0\rangle\langle k'*\omega_0|f\rangle=\sum_{k'=-\infty}^{+\infty}\sqrt{2\pi/\omega_0}e^{jk'\omega_0t}c_{k'}$.

Also $c_{k'}=\langle k'*\omega_0|f\rangle = \langle k'*\omega_0|\int_{-\infty}^{+\infty}dt'|t'\rangle\langle t'|f\rangle=\int_{-\infty}^{+\infty}dt'\langle k'*\omega_0|t'\rangle\langle t'|f\rangle=\int_{-\infty}^{+\infty}dt'\sqrt{2\pi/\omega_0}e^{-jk'\omega_0t}f(t')$

Here $\langle t'|k'*\omega_0\rangle=\sqrt{2\pi/\omega_0}e^{jk'\omega_0t}$ and $\langle k'*\omega_0|t'\rangle=\sqrt{2\pi/\omega_0}e^{-jk'\omega_0t}$ are the representatives of the frequency vectors in the time basis and of the time vectors in the frequency basis, respectively. They are also the transformation functions between the two bases.

Here is the link to the same question I've asked on math.stackexchange.com.

• "in infinite dimensional vector spaces the basis of the space is a dense subset of the vector space" That's completely wrong! "having bases of different cardinalities" that's make no sense, all Hilbert bases have the same cardinality. – Valter Moretti Aug 13 '14 at 13:15
• You are using a notion $|t>$ which is not a basis in any sense. You are confounding, some practical physical idea as bra-ket Dirac formalism with properly defined mathematical objects. – Valter Moretti Aug 13 '14 at 13:17
• It is not clear the status of the question and of the answer: if it is mathematics it should be handled by means of the appropriate formalism. Otherwise it makes things even more confused. – Valter Moretti Aug 13 '14 at 13:22
• @ValterMoretti It depends on the way you define "basis". In the case of Hamel bases what you say is true, but in the case of Schauder basis it isn't. Correct me if I am wrong, and please give a correct answer if you can. – user3237992 Aug 13 '14 at 13:23
• It is also true in the case of Hilbert bases. Those relevant in QM. – Valter Moretti Aug 13 '14 at 13:24