Quantum Mechanics for Mathematicians

40.3 The Fourier transform in Minkowski space

One can define a Fourier transform with respect to the four space-time variables,
which will take functions ofx 0 ,x 1 ,x 2 ,x 3 to functions of the Fourier transform
variablesp 0 ,p 1 ,p 2 ,p 3 :

Definition(Minkowski space Fourier transform). The Fourier transform of a
functionfon Minkowski space is given by

f ̃(p) =^1

(2π)^2

∫

M^4

e−ip·xf(x)d^4 x

=

1

(2π)^2

∫

M^4

e−i(−p^0 x^0 +p^1 x^1 +p^2 x^2 +p^3 x^3 )f(x)dx 0 d^3 x

In this case the Fourier inversion formula is

f(x) =

1

(2π)^2

∫

M^4

eip·xf ̃(p)d^4 p (40.3)

Note that our definition puts one factor of√^12 πwith each Fourier (or inverse

Fourier) transform with respect to a single variable. A common alternate con-
vention among physicists is to put all factors of 2πwith thepintegrals (and
thus in the inverse Fourier transform), none in the definition off ̃(p), the Fourier
transform itself.
The sign change between the time and space variables that occurs in the
exponent of this definition is there to ensure that this exponent is Lorentz
invariant. Since Lorentz transformations have determinant 1, the measured^4 x
will be Lorentz invariant and the Fourier transform of a function will behave
under Lorentz transformations in the same ways as the function

f ̃(Λ−^1 p) =^1

(2π)^2

∫

M^4

e−i(Λ

− (^1) p)·x
f(x)d^4 x

=

1

(2π)^2

∫

M^4

e−ip·Λxf(x)d^4 x

=

1

(2π)^2

∫

M^4

e−ip·xf(Λ−^1 x)d^4 x

The reason why one conventionally defines the Hamiltonian operator asi∂t∂

(with eigenvaluesp 0 =E) but the momentum operatorPj as−i∂x∂j (with
eigenvaluespj) is due to this Lorentz invariant choice of the Fourier transform.

40.4 Spin and the Lorentz group

Just as the groupsSO(n) have double coversSpin(n), the groupSO(3,1) has
a double coverSpin(3,1), which we will show can be identified with the group
SL(2,C) of 2 by 2 complex matrices with unit determinant. This group will