Quantum Mechanics for Mathematicians

In others, the factor of 2πmay appear instead in the exponent of the complex
exponential, or just in one ofForF ̃and not the other.
The operatorsFandF ̃are thus inverses of each other onS(R). One has

Theorem(Plancherel). FandF ̃extend to unitary isomorphisms ofL^2 (R)
with itself. In particular
∫∞

−∞

|ψ(q)|^2 dq=

∫∞

−∞

|ψ ̃(k)|^2 dk (11.5)

Note that we will be using the same inner product on functions onR

〈ψ 1 ,ψ 2 〉=

∫∞

−∞

ψ 1 (q)ψ 2 (q)dq

both for functions ofqand their Fourier transforms, functions ofk, with our
normalizations chosen so that the Fourier transform is a unitary transformation.
An important example is the case of Gaussian functions where

Fe−α

q 22

=

1

√

2 π

∫+∞

−∞

e−ikqe−α

q 22

dq

=

1

√

2 π

∫+∞

−∞

e−

α 2 ((q+ikα)^2 −(ikα)^2 )

dq

=

1

√

2 π

e−

k 22 α

∫+∞

−∞

e−

α 2 q′^2

dq′

=

1

√

α

e−

k 2 α^2

(11.6)

A crucial property of the unitary operatorFonHis that it diagonalizes the
differentiation operator and thus the momentum operatorP. Under the Fourier
transform, constant coefficient differential operators become just multiplication
by a polynomial, giving a powerful technique for solving differential equations.
Computing the Fourier transform of the differentiation operator using integra-
tion by parts, we find

̃dψ

dq

=

1

√

2 π

∫+∞

−∞

e−ikq

dψ

dq

dq

=

1

√

2 π

∫+∞

−∞

(

d

dq

(e−ikqψ)−

(

d

dq

e−ikq

)

ψ

)

dq

=ik

1

√

2 π

∫+∞

−∞

e−ikqψdq

=ikψ ̃(k) (11.7)

So, under Fourier transform, differentiation byqbecomes multiplication byik.
This is the infinitesimal version of the fact that translation becomes multiplica-
tion by a phase under the Fourier transform, which can be seen as follows. If