the analysis relatively easy, withH =L^2 (S^1 ) and the self-adjoint operator
P =−i~∂φ∂ behaving much the same as in the finite dimensional case: the
eigenvectors ofPgive a countable orthonormal basis ofHandPcan be thought
of as an infinite dimensional matrix.
Unfortunately, in order to understand many aspects of quantum mechanics,
one can’t get away with this trick, but needs to work withRitself. One reason
for this is that the unitary representations ofRare labeled by the same group,
R, and it will turn out (see the discussion of the Heisenberg group in chapter
13) to be important to be able to exploit this and treat positions and momenta
on the same footing. What plays the role then of|n〉=einφ,n∈Zwill be
the|k〉=eikq,k∈R. These are functions onRthat are one dimensional
irreducible representations under the translation action on functions (as usual
using equation 1.3)
π(a)eikq=eik(q−a)=e−ikaeikq
One can try and mimic the Fourier series decomposition, with the coefficients
cnthat depend on the labels of the irreducibles replaced by a functionf ̃(k)
depending on the labelkof the irreducible representation ofR:
Definition(Fourier transform).The Fourier transform of a functionψis given
by a function denotedFψorψ ̃, where
(Fψ)(k) =ψ ̃(k) =1
√
2 π∫∞
−∞e−ikqψ(q)dq (11.3)This integral is not well-defined for all elements ofL^2 (R), so one needs to
specify a subspace ofL^2 (R) to study for which it is well-defined, and then extend
the definition toL^2 (R) by considering limits of sequences. In our case a good
choice of such a subspace is the Schwartz spaceS(R) of functionsψsuch that
the function and its derivatives fall off faster than any power at infinity. We
will not try and give a more precise definition ofS(R) here, but a good class of
examples of elements ofS(R) to keep in mind are products of polynomials and
a Gaussian function. The Schwartz space has the useful property that we can
apply the momentum operatorPan indefinite number of times without leaving
the space.
Just as a function onS^1 can be recovered from its Fourier series coefficients
cnby taking a sum, given the Fourier transformψ ̃(k) ofψ,ψitself can be
recovered by an integral, with the following theorem
Theorem(Fourier Inversion).Forψ∈S(R)one hasψ ̃∈S(R)and
ψ(q) =F ̃ψ ̃=1
√
2 π∫+∞
−∞eikqψ ̃(k)dk (11.4)Note thatF ̃is the same linear operator asF, with a change in sign of the
argument of the function it is applied to. Note also that we are choosing one
of various popular ways of normalizing the definition of the Fourier transform.