TITLE: Delta Function Identity in Peskin & Schroeder
QUESTION [0 upvotes]: In Notation and Conventions of their QFT textbook (page no. xxi), Peskin and Schroeder mentions the following identity:
$$\int d^4x \, e^{ik\cdot x} = (2\pi)^4 \delta^{(4)}(k).$$
They define the Fourier transforms in four dimensions as follows.
$$f(x) = \int \frac{d^4k}{(2\pi)^4} e^{-ik\cdot x} \, \tilde{f}(k), \quad \tilde{f}(k) = \int d^4 x  \, e^{ik\cdot x} f(x).$$
I am having trouble to see how the identity follows from these definitions. Could you please clarify?

REPLY [0 votes]: Given that $$f(x) = \int \frac{d^4k}{(2\pi)^4} e^{-ik\cdot x} \, \tilde{f}(k) \tag 1$$
and,
$$\tilde{f}(k) = \int d^4 x  \, e^{ik\cdot x} f(x). \tag 2$$
Let $$\tilde{f}(k) = \delta(k) = \left\{ \begin{array}{rl}
                                         \infty &\mbox{ if $x=0$} \\
                                          0 &\mbox{otherwise.}
                                         \end{array} \right.$$
Then the inverse Fourier transformation of $\tilde{f}(k) = \delta(k)$ is $f(x)$ given by
$$f(x) = \int \frac{d^4k}{(2\pi)^4} e^{-ik\cdot x} \, \delta(k) = \frac{1}{(2\pi)^4}.$$
Then it follows that the Fourier transformation of $\frac{1}{(2\pi)^4}$ is $\tilde{f}(k) = \delta(k)$:
$$ \int d^4 x  \, e^{ik\cdot x} \frac{1}{(2\pi)^4} = \tilde{f}(k) = \delta(k) \\
\Rightarrow \boxed{(2\pi)^4 \delta(k) = \int d^4 x  \, e^{ik\cdot x}}.$$
As the Dirac delta $\delta(k)$ is an even function, we get
$$(2\pi)^4 \delta(k) = \int d^4 x  \, e^{-ik\cdot x}.$$
Therefore, $$(2\pi)^4 \delta(k) = \int d^4 x  \, e^{\pm ik\cdot x}.$$
For more discussion, please check here.
{"set_name": "stack_exchange", "score": 0, "question_id": 653287}