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This is really straight forward, once you get used to the notation. (Don't you hate it when people say that?) $$[\pi (\vec{x},t), (-\nabla^{2} +m^{2})\phi (\vec{y},t)] ,$$ Here you need to remember that $\nabla^2$ acts on the $\phi(\vec{y},t)$ only, so $\pi$ can pass right through this wave operator. Now when you evaluate the commutator you'll end up with ...


@Suresh Doesn't the normalization condition follow directly from the fourier transform relation for delta function? I'm not clear about the need for box normalization and PBC. Sorry I'm writing this as an answer, I don't have enough reputation to comment. Thanks.


$$δ^3(q⃗ )=\frac{δ(q)δ(\theta)}{2\pi q^2\sin(\theta)}$$ is wrong. The delta function is spherically symmetric, and thus has no θ dependence. Simply use: $$d^3(q⃗ )=\frac{δ(q)}{2\pi q^2}$$ instead. Use the Jacobian when you switch coordinate systems (from Cartesian to spherical) ($r^2 \sin(\theta)$), and you should get the result.


Hints: In mathematics, a distribution is usually only defined wrt. smooth testfunctions. However the function ${\bf q}\mapsto({\bf q}\cdot{\bf p})^2/q^2$ is not continuous at the origin ${\bf q}={\bf 0}$. Nevertheless, we can e.g. try to evaluate the triple integral using the following representation of the 3D Dirac delta distribution $$\tag{1} ...


This proof from Griffiths book introduction to electrodynamics Consider the vector function $$\vec{a}=\frac{1}{r^2}\hat{r}$$ At every location $\vec{a}$ is directed radially outward ; if ever there was a function that ought to have a large positive divergence, this is it. and yet, when you actually calculate the divergence, you will get ...


Indeed the answer is not zero but $-4\pi\delta(r)$ (Dirac delta function). The formula of divergence can be found in any standard textbook on mathematical physics, for example chapter 2 of Mathematical methods for physicists by Arfken. But since this function is singular at $r=0$ we must be careful. At any other points is easy to calculate it. It is ...

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