rarily large elements, and this contradicts compactness.
This result is not needed for what follows, but it does imply that it is not possible to describe the position and momentum operators as elements of a Banach algebra. Equivalently, it is not possible to represent position and momentum as bounded operators on a Banach space. In their usual representation on , both position and momentum are unbounded (discontinuous) and densely defined (can only be applied to certain dense subspaces of ).
Fortunately, there is a way out: instead of dealing with position and momentum directly, we can deal with the strongly continuous one-parameter unitary groups they generate via Stone’s theorem (see also the Stone-von Neumann theorem). Indeed, these unitary groups are defined on by
for position and (up to normalization)
for momentum, both of which are defined everywhere on .
The Gelfand representation
Recall that for any commutative ring we can define a homomorphism
where the product runs over all maximal ideals of and is the image of in . This represents as a “ring of functions” on , although ideally all of the residue fields should be the same for this interpretation to really make sense, and the kernel of this homomorphism is the Jacobson radical .
We now know that if is a Banach algebra, we always have , so the above homomorphism lands in the space of functions . It’s natural to want to equip , the spectrum, Gelfand spectrum, or Gelfand space of , with a topology such that the corresponding functions are all continuous. This is usually done by identifying an element of with the corresponding map , which is in particular a continuous linear functional, hence naturally sits as a subspace of . The initial topology on making all evaluations against elements of continuous is the weak-* topology, and the topology we want on is the subspace topology with respect to this topology.
(An equivalent way to define the weak-* topology, which makes sense for replaced by any topological vector space , is to think of as a subspace of the space of functions , or equivalently a subspace of the product of copies of ; then the weak-* topology is the subspace topology with respect to the product topology on this product.)
Since morphisms have norm , in fact lies in the unit ball of . We now need the following result, which shows that the weak-* topology is very different from the norm topology on .
Theorem (Banach-Alaoglu): let be a normed space. Then the unit ball of is compact in the weak-* topology.
Proof. Thinking of as a subspace of with the product topology, the unit ball consists of linear functionals which send the unit ball of to the unit ball (disk) , hence can be thought of as a subspace of with the product topology. This is compact by Tychonoff’s theorem. The unit ball of is a closed subspace of (since it is defined by the additional requirement that it consists of linear functions, which impose a collection of closed conditions), hence is also compact.
(Note that is compact Hausdorff, so we only need Tychonoff’s theorem for compact Hausdorff spaces here, hence we only need the ultrafilter lemma and not the full strength of the axiom of choice. In fact Banach-Alaoglu is equivalent to the ultrafilter lemma.)
Corollary: For a commutative Banach algebra, is compact Hausdorff in the weak-* topology.
It follows that the homomorphism above, applied to a commutative Banach algebra, gives us its Gelfand representation
Since is compact Hausdorff, the space of continuous functions on it is itself a Banach algebra with the sup norm. We record the following properties of the Gelfand representation to summarize what we have already proven.
- has range the spectrum of .
- Consequently is the spectral radius .
- Consequently the Gelfand representation is a weak contraction (hence a morphism of Banach algebras).
- The kernel of the Gelfand representation is . It is also the set of elements of with spectral radius zero (the quasi-nilpotent elements).
- Consequently the Gelfand representation is faithful if and only if is semiprimitive, if and only if has n