Quantum Mechanics for Mathematicians

ψa(q) =ψ(q+a) then

ψ ̃a(k) =√^1

2 π

∫+∞

−∞

e−ikqψ(q+a)dq

=

1

√

2 π

∫+∞

−∞

e−ik(q

′−a)

ψ(q′)dq′

=eikaψ ̃(k)

Sincep=~k, one can easily change variables and work withpinstead ofk.
As with the factors of 2π, there’s a choice of where to put the factors of~in the
normalization of the Fourier transform. A common choice preserving symmetry
between the formulas for Fourier transform and inverse Fourier transform is

ψ ̃(p) =√^1

2 π~

∫+∞

−∞

e−i

pq~

ψ(q)dq

ψ(q) =

1

√

2 π~

∫+∞

−∞

ei

pq

~ψ ̃(p)dp

We will however mostly continue to set~= 1, in which case the distinction
betweenkandpvanishes.

11.3 Distributions

While the use of the subspaceS(R)⊂L^2 (R) as state space gives a well-behaved
momentum operatorPand a formalism symmetric between positions and mo-
menta, it still has the problem that eigenfunctions ofP are not in the state
space. Another problem is that, unlike the case ofL^2 (R) where the Riesz rep-
resentation theorem provides an isomorphism between this space and its dual
just like in the finite dimensional case (see 4.3), such a duality no longer holds
forS(R).
For a space dual toS(R) one can take the space of linear functionals on
S(R) called the Schwartz space of tempered distributions (a certain continuity
condition on the functionals is needed, see for instance [88]), which is denoted
byS′(R). An element of this space is a linear map

T:f∈S(R)→T[f]∈C

S(R) can be identified with a subspace ofS′(R), by takingψ∈ S(R) to the
linear functionalTψgiven by

Tψ:f∈S(R)→Tψ[f] =

∫+∞

−∞

ψ(q)f(q)dq (11.8)

Note that takingψ∈S(R) toTψ∈S′(R) is a complex linear map.
There are however elements ofS′(R) that are not of this form, with three
important examples