Is there a Hermitian operator on $L^2(\mathbb{R})$ which is outside the C*-algebra generated by $\hat{x}$ and $\hat{p}$?

Question

This question seems to have been popping up here in a variety of forms that I feel don't seem to really get at (and get criticized for being vague or less than perfectly defined) what I believe the underlying issue is, so I am posting another here to ask what I believe is the well-defined gist thereof, as I have also had the same question, and it is this:

Is the C*-algebra of operators generated by $\hat{x}$ and $\hat{p}$, defined as any pair of operators that meet $$[\hat{x}, \hat{p}] = i\hbar$$ with $\hbar \ne 0$, operatorially complete for the Hilbert space of positional wave functions $L^2(\mathbb{R})$, i.e. is every self-adjoint operator on that space contained therein, and thus an observable aspect of a position-momentum moving particle?

If so, how can we prove it? If not, what is a counterexample operator?

NOTE: It has been pointed out that this still doesn't quite nail it. The accepted answer, though, contains a correction that seems to make sense: if we consider the algebra of all observables that are bounded functions of $x$ and $p$, such as "clamping" $x$ or $p$ to each finite interval, and thus "covering" the infinite range of both.

Answer 1

The algebra includes all operators. If a bounded operator $A$ commutes with (the spectral measure of) $x$, then it commutes with the unitary group generated by it. If $A$ also commutes with (the spectral measure of) $p$, then it must eventually commute with the whole Weyl algebra on $L^2(R)$ which is known to be irreducible. Hence $A=cI$ for any complex $c$, in view of Schur lemma. The Von Neumann algebra generated by $x$ and $p$ is the commutant of the found operators by definition, i.e., the full algebra of bounded operators in the Hilbert space.

As a consequence, every bounded operator can be obtained as the limit in the strong operator topology of polynomials of bounded functions of $x$ and $p$ separately.

I guess that this last statement is the one you were looking for...

Answer 2

The notation $C^*$ used in the title of the questions refers to a type of topological algebra unsuited for coordinate and momentum in $L^2 (\mathbb R)$. Such an algebra is made up of bounded operators, while $x$ and $p_x$ aren’t. There are, however, two possibilities to make sense of this question:

1. Transform the operators $x$ and $p_x$ into bounded operators on a dense subset of $ L^2 (\mathbb R) $.
2. Replace the $C^*$ algebra with an $O^*$-algebra which is perfectly possible, because such an $O^*$-algebra on $L^2 (\mathbb R)$ can be defined because the Schwartz test function space $\mathcal S (\mathbb R) $ is both invariant for the two operators and dense everywhere in $L^2 (\mathbb R)$ (the operators are essentially self-adjoint as operators from $\mathcal S (\mathbb R) $ to $L^2 (\mathbb R)$ under the $L^2 (\mathbb R)$ topology).

Possibility #1 is the start of the so-called rigged Hilbert space formulation of Quantum Mechanics, in the sense in which the norm topology of $L^2 (\mathbb R)$ on such a dense subset is replaced by a stronger topology induced by a countable family of semi-norms. If the dense subset is chosen (again) to be the Schwartz test function space, then the RHS is $\mathcal S (\mathbb R) \subset L^2 (\mathbb R) \subset \mathcal S^\times (\mathbb R) $ and the seminorms on $\mathcal S $ are (up to equivalence):

$$ ||\phi||_p = \mbox{sup}_{k,q\leq p} \left|x^k \frac{\partial^q \phi(x)}{\partial x^q}\right|, ~ p = 0,1,2,... $$

So $x$ and $p_x$ (as initially an irreducbile set) become bounded self-adjoint operators on $\mathcal S (\mathbb R) $ and from here any other operators which can be written as (properly symmetrized) polynomial functions in the two variables $x$ and $p_x$ remain bounded everywhere self-adjoint operators on $\mathcal S (\mathbb R) $. Such an operator could be the Hamiltonian of a free particle in 1D or of a harmonic oscilator in 1D or any powers of it.

Possibility #2 is clearly now equivalent to Possibility #1, because the Schwartz space and a representation of the abstract Weyl-algebra on it (such as the Schrödinger representation) provide an example of an $O^*$-algebra. This $O^*$-algebra can be enlarged by adding any (again properly symmetrized) polynomial function of $x$ and $p_x$. So, if the question in the title is phrased: „Is there a self-adjoint operator on $L^2 (\mathbb R) $ outside of the $O^*$-algebra generated by $x$ and $p_x$?”, I would say: „No, but I would rather use $\mathcal S (\mathbb R)$ in phrasing of the question, because the full Hilbert space is too big for the $O^*$-algebra” (and ironically too small to solve the spectral equation of either of the two operators).