Heuristically (ignoring problems of interchange of integrals that don’t make
sense), the Fourier inversion formula can be written as follows
ψ(q) =1
√
2 π∫+∞
−∞eikqψ ̃(k)dk=
1
√
2 π∫+∞
−∞eikq(
1
√
2 π∫+∞
−∞e−ikq′
ψ(q′)dq′)
dk=
1
2 π∫+∞
−∞(∫+∞
−∞eik(q−q′)
ψ(q′)dk)
dq′=
∫+∞
−∞δ(q−q′)ψ(q′)dq′Physicists interpret the above calculation as justifying the formula
δ(q−q′) =1
2 π∫+∞
−∞eik(q−q′)
dkand then go on to consider the eigenvectors
|k〉=1
√
2 πeikqof the momentum operator as satisfying a replacement for the Fourier series
orthonormality relation (equation 11.1), with theδ-function replacing theδnm:
〈k′|k〉=∫+∞
−∞(
1
√
2 πeik′q)(
1
√
2 πeikq)
dq=1
2 π∫+∞
−∞ei(k−k′)q
dq=δ(k−k′)11.4 Linear transformations and distributions
The definition of distributions as linear functionals on the vector spaceS(R)
means that for any linear transformationAacting onS(R), we can get a linear
transformation onS′(R) as the transpose ofA(see equation 4.1), which takes
Tto
AtT:f∈S(R)→(AtT)[f] =T[Af]∈C
This gives a definition of the Fourier transform onS′(R) as
FtT[f]≡T[Ff]and one can show that, as forS(R) andL^2 (R), the Fourier transform provides
an isomorphism ofS′(R) with itself. Identifying functionsψwith distributions