# Why do we choose the operator or supremum norm while proving unboundedness of the momentum operator?

In most sources, I've noticed that while proving the unboundedness of the momentum operator $$\left(-i\hbar \frac{\partial}{\partial x}\right)$$ the operator norm (or supremum norm) $$\lVert\ .\rVert_\infty$$ is used. The proof is along these lines.

However, I'm not sure why we specifically choose the operator norm for this purpose. Is it somehow intrinsic to quantum mechanics? It doesn't seem that way because we generally only deal with Hilbert spaces with norm induced by the inner product.

• Interesting question. As far as I know, we specifically bake the operator norm into our definitions of (un)boundedness; the question of whether or not there are good, non-trivial alternatives to this fairly natural choice might receive better and/or more enthusiastic responses on MathSE. – J. Murray May 30 '19 at 3:42
• @J.Murray Is it really a fairly natural choice though? In what sense? I'm specifically interested in this, in the context of quantum mechanics, although yes, perhaps the math folks would be able to provide some insight too. – S.D. May 30 '19 at 3:44
• For a linear operator $A:V\rightarrow W$ with $V$ and $W$ normed vector spaces, the operator norm $\Vert \bullet \Vert_{op} : A \mapsto \sup_{\psi\in V} \frac{\Vert A\psi\Vert_W}{\Vert \psi \Vert_V}$ is natural in the sense that we are inheriting a norm on the space of operators from the norms on $V$ and $W$ without introducing any arbitrary additional structure. It's the only way to do this which is obvious to me, but I'm no mathematician - hence the suggestion to ask people who spend more time thinking about Banach spaces :) – J. Murray May 30 '19 at 3:51

Unboundedness is always relative to some norm. We don't set out to "prove unboundedness" and then choose a norm at random, we must have some reason for caring about the norm and the operator being unbounded in the first place. A mathematician would say that the operator norm is "natural" because it is canonically induced by the norms on the source and target space, but that you ask this question on physics.SE suggests you are not looking for the mathematician's reason for looking at this norm.

When we prove mathematical statements about the physical formalism, it is important to keep in mind how these mathematical facts are supposed to relate to a more "down to earth" view of physics. In this case, we should ask ourselves what it means for an operator to be (un)bounded in the operator norm. Here are two facts:

1. The operator norm is an upper bound for the eigenvalues of an operator, and in particular for self-adjoint operators it is simply the largest absolute value of the eigenvalues, if one exists. So one might physically care about an observable being unbounded because it means that the resulting values of measurements are unbounded.

2. Many "nice" statements about operators hold only for bounded linear operators, where bounded is w.r.t. the operator norm. This is a mathematical fact, and it often does us well to remember that position and momentum on $$\mathbb{R}$$ are unbounded operators to which not all of these nice statements apply.

The supremum norm of an operator is brought in by an almost coincidence of definitions.

An operator A on a pre-Banach $$\mathcal{B}$$ space is said to bounded, iff $$\exists C\in\mathbb{R}, \forall \psi\in D(A)\subseteq \mathcal{B}, \|A\psi\| \leq C\|\psi \|.$$ If no such C exists, then the operator is said to be unbounded. The supremum norm of a linear bounded operator A is the minimum value of all possible C's in the above definition.

This is just mathematics, no physics (quantum mechanics) involved. So the momentum operator in the position representation (thus a linear operator in $$L^2 (\mathbb R)$$) is unbounded by the very definition.

The nice comment#1 by ACuriousMind is the connection to physics for what I wrote above.