In my quantum mechanics class, my professor explained that the Hamiltonian along with position and momentum operators can be represented by matrices of countable dimension. This is especially usefull in harmonic oscillator problems. My professor explained that the eigenvalues of the Hamiltonian are (of course) the discrete allowed energies of the system, while the eigenvalues of the position operator are all possible positions, a continuum. How can a countable matrix have an uncountable number of eigenvalues? Why do the Hamiltonian and the position operator have the same dimension but different numbers of eigenvalues?
Answer
The countable and uncountable infinities are "different cardinals" according to set theory but in physics, the bases of that size produce equally large Hilbert spaces: the Hilbert-space is infinite-dimensional and all infinite-dimensional Hilbert spaces are isomorphic to each other (in other words, there is only "one unified kind of infinity" when it comes to the dimension of a Hilbert space). Quantum mechanics offers you infinitely many examples.
Perhaps the simplest example are Fourier expansions. Consider a particle in an infinite well so that the wave function $\psi(x)$ is only nonzero for $0\lt x \lt +\pi$. The operator $x$ has a continuous spectrum i.e. an uncountable number of eigenvalues and eigenstates (the basis of $x$-eigenstates is uncountable).
On the other hand, the operator $p^2 = -\hbar^2 \partial^2 / \partial x^2$ has a discrete spectrum and a countable set of eigenvalues and eigenstates. The eigenstates are standing waves $\sin (nx)$ for positive integer $n$ and the eigenvalues are $n^2$.
Nevertheless, every ("smooth enough" and/or $L^2$-normalizable etc.) function $\psi(x)$ that is nonzero in that interval – every combination of uncountably many wave functions $\psi(x) = \delta(x-x_0)$ – may also be written as a linear combination of the standing waves, $\sin(nx)$. This fact is what makes the Fourier series possible. (Normally, I would talk about periodic functions and complex exponentials but the sines in a well may be more beginner-friendly.)
There is really no contradiction with the different cardinality of the sets because the two sets, the uncountable basis of $x$ eigenstates and the countable basis of $p^2$ eigenstates, are not being identified via a one-to-one map. Instead, the map between one basis and the other is a general linear transformation that mixes them, and the different cardinality places no restrictions on such linear transformations of infinite-dimensional vector spaces.
Quite generally, cardinal numbers (the science about distinguishing many types) as well as most other related results in set theory (I mean especially Gödel's theorems) are completely inconsequential in physics. They're just some "recreational subtleties" in mathematical logic and physics doesn't find any of these operations relevant. So a physicist may do state-of-the-art string theory and interpret it in all corners of physics without even "knowing" that the real numbers are uncountable. The uncountability is unphysical. A physicist is generally agnostic about the existence of real numbers that cannot be constructed, about the validity of the axiom of choice, and other problems that cannot be operationally performed by an experiment. A physicist's typical reaction is that these questions are "philosophy" – they are empirically undecidable (we know that the axiom of choice is undecidable even in major axiomatic systems of set theory) so he doesn't care about the answers.
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