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In the traditional Physics literature, we find $\delta$ be defined as $$\delta(x)=\begin{cases}0, & x\neq 0, \\ \infty, & x = 0\end{cases}$$ and then one "derives" that if $f$ is a function we have …

Most of this is taken from analogy with the finite-dimensional integral case. First recall that Fourier theory gives us the representation of the Dirac delta as $$\delta(x-y)=\int\dfrac{d^Nk}{(2\pi)^N …

This is a common abuse of notation when using distributions in Physics in general, so not something specific to operator-valued ones. Given a test function $f(x)$ you can always define a distribution …

First some general comments. The Dirac delta distribution you find merely constrains the values of $p$ that span the space of solutions to the Klein-Gordon equation. When you try to solve the KG equat …

This is notation from Distribution Theory in Functional Analysis. The theory of distributions is meant to make things like the Dirac Delta rigorous. In this context, just to give you one overview, a …

Equation (2.39) is written the way it is because one has chosen the covariant normalization. In this normalization the momentum eigenstates are normalized as $$\langle p'|p\rangle = (2\pi)^3 (2E_p) \ …

The reason why the LHS is nonzero is quite simple. On the one hand, $$\int_{-\epsilon}^{\epsilon}\dfrac{df}{dx}dx=f(\epsilon)-f(-\epsilon).$$ The limit as $\epsilon\to 0$ quantifies the discontinuity …

If the index $k$ is meant to be continuous, then you can replace $\frac{1}{V}\sum_k\to \int dk$ and your equation follows, up to factors of $2\pi$, from the fact that $$\int dk \ e^{ik(x-x')}=2\pi \de …

I want to propose a different approach, as I believe the OP might benefit from comparing different viewpoints on distribution theory. The question is really about the transformation of distributions u …