40.3 The Fourier transform in Minkowski space

40.3 The Fourier transform in Minkowski space

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One can define a Fourier transform with respect to the four space-time variables, which will take functions of $x_0,x_1,x_2,x_3$ to functions of the Fourier transform variables $p_0,p_1,p_2,p_3:$

Definition (Minkowski space Fourier transform)

The Fourier transform of a function $f$ on Minkowski space is given by

$$\begin{array}{rl}\tilde{f}(p)& =\frac{1}{(2\pi {)}^2}{\int }_{M^4}{e}^{-ip\cdot x}f(x)d^{4}x\\ & =\frac{1}{(2\pi {)}^2}{\int }_{M^4}{e}^{-i(-p_0{x}_0+p_1{x}_1+p_2{x}_2+p_3{x}_3)}f(x)d{x}_0{d}^3\mathbf{x}\end{array}$$

In this case the Fourier inversion formula is

$$f(x)=\frac{1}{(2\pi {)}^2}{\int }_{M^4}{e}^{ip\cdot x}\tilde{f}(p)d^{4}p\text{\hspace{1em}(40.3)}$$

Note that our definition puts one factor of $\frac{1}{\sqrt{2\pi }}$ with each Fourier (or inverse Fourier) transform with respect to a single variable. A common alternate convention among physicists is to put all factors of $2\pi$ with the $\mathbf{p}$ integrals (and thus in the inverse Fourier transform), none in the definition of $\tilde{f}(p)$ , 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 measure $d^{4}x$ will be Lorentz invariant and the Fourier transform of a function will behave under Lorentz transformations in the same ways as the function

$$\begin{array}{rl}\tilde{f}({\Lambda }^{-1}p)& =\frac{1}{(2\pi {)}^2}{\int }_{M^4}{e}^{-i({\Lambda }^{-1}p)\cdot x}f(x)d^{4}x\\ & =\frac{1}{(2\pi {)}^2}{\int }_{M^4}{e}^{-ip\cdot \Lambda x}f(x)d^{4}x\\ & =\frac{1}{(2\pi {)}^2}{\int }_{M^4}{e}^{-ip\cdot x}f({\Lambda }^{-1}x)d^{4}x\end{array}$$

The reason why one conventionally defines the Hamiltonian operator as $i\frac{\partial }{\partial t}$ (with eigenvalues $p_0=E)$ but the momentum operator $P_j$ as $\overline{-i\frac{\partial }{\partial x_j}}$ (with eigenvalues $p_j)$ is due to this Lorentz invariant choice of the Fourier transform.

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