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How much functional analysis is needed to understand quantum mechanics?
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A lot, if you want to understand it on a deep level.
However, one needs very little of functional analysis in order to
understand
(a) the foundations of quantum mechanics,
(b) the contents of typical textbooks on quantum mechanics.
One doesn't need anything beyond the definition of a Hilbert space and
the trace of an operator if one is (like most physicists) prepared to
take for granted two quite nontrivial but reasonably intuitive
functional analytic results:
1. The spectral theorem in the following form:
Theorem. (Gelfand & Maurin)
Given an arbitrary set of commuting self-adjoint operators defined on
the same dense subspace of a Hilbert space, there is always an
isomorphic Hilbert space in which these operators are represented by
multiplication with real-valued functions.
2. The Hille-Yosida theorem in the following form:
Theorem. (Hille & Yosida)
The exponential exp(itH) of a Hermitian linear operator H defined on
a (dense subspace of a) Hilbert space exists if and only iff A is
self-adjoint. In this case, exp(itH) is a bounded operator, and hence
defined on the whole Hilbert space.
One may take the latter as a definition of self-adjoint; then there is
hardly anything to prove. (The difficulty is then moved to proving
easily checkable criteria for self-adjointness. On the importance
to distinguish between hermiticity and self-adjointness see
http://arxiv.org/pdf/quant-ph/9907069 (by F. Gieres)
for a gentle introduction and counterexamples. An in depth discussion
is given in Vol. 1 of the math physics treatise by Reed and Simon, or
Vol.3 of the math physics treatise by Thirring.)
The proof of 1. is long, no matter how one does it.
The bounded case is reduced to Gelfand's work
http://en.wikipedia.org/wiki/Gelfand_representation
by noting that bounded commuting operators generate a commutative
C^* algebra. One extends it to a maximal commutative C^* subalgebra B
of the C^* algebra A of all bounded linear operators using Zorn's
lemma (can perhaps be avoided if the Hilbert space is separable?),
and then proceeds to show that A acts already on C_0(Phi_A).
The unbounded case is easily reduced to the bounded case using
Hille-Yosida.
In his book, Maurin probably formulates the theorem not as I do
(I don't know whether my formulation is in the literature - though it
actually might be in Maurin) but in terms of rigged Hilbert spaces
(= Gelfand triples) version, where (non-normalizable) eigenvectors
exist.
But for a rigorous presentation of textbook quantum mechanics, the
rigged Hilbert space extension is not needed - except if one wants to
have a rigorous version of the bra-ket calculus in case of a continuous
spectrum. Indeed, one get the standard bra-ket heuristics for
eigenkets from my formulation of the theorem in precisely the same way
as it is introduced early on in the case of the position representation.
Traditionally, an observable is a self-adjoint operator.
The Gelfand-Maurin theorem proves that there is a diagonal
representation. _This_ is the relevant fact, not the existence of
normalizable eigenvectors, which one hasn't in the continuous part of
the spectrum.
If one takes the components of position, one gets the position
representation.
If one takes the components of momentum, one gets the momentum
representation.
In case of spin, one needs to add in both cases the operators J^2 and
J_3 to get a maximally commuting system and hence up to isomorphism a
unique diagonal representation.