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2

$$δ^3(q⃗ )=\frac{δ(q)δ(\theta)}{2\pi q^2\sin(\theta)}$$ 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.

4

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 $$\tag{1} ... 1 This proof from Griffiths book introduction to electrodynamics Consider the vector function$$\vec{a}=\frac{1}{r^2}\hat{r}$$At every location \vec{a} is directed radially outward ; if ever there was a function that ought to have a large positive divergence, this is it. and yet, when you actually calculate the divergence, you will get ... 4 Indeed the answer is not zero but -4\pi\delta(r) (Dirac delta function). The formula of divergence can be found in any standard textbook on mathematical physics, for example chapter 2 of Mathematical methods for physicists by Arfken. But since this function is singular at r=0 we must be careful. At any other points is easy to calculate it. It is ... 0$$i {{\partial}\over{\partial t}}\pi=[\pi,\int d^3x\tfrac{1}{2}\pi^2+\tfrac{1}{2}\phi()\phi]=[\pi,\int d^3x\tfrac{1}{2}\phi()\phi]=\tfrac{1}{2}\int d^3x[\pi,\phi()\phi]=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+\phi[\pi,()\phi]=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+\phi[\pi,()]\phi+\phi()[\pi,\phi]=\tfrac{1}{2}\int ...

2

I was just playing with things and got this, let me know if you find it a useful idea (or if I'm missing something which makes this useless). Denote $\delta_k(x)=\frac{1}{k\sqrt{\pi}} e^{-(x/ k)^2}$. You get the well known (completely mathematically rigorous) result: $\lim_{k\to 0} \int_{-\infty}^\infty f(x) \delta_k(x) dx=f(0)$ for well behaved enough ...

1

Actually, you got the right result, you just didn't see it. In the sense of distributions we have that $$x\delta(x-x')e^{i(g(x) - g(x'))/\hbar}=x\delta(x-x')$$ These expressions, make no sense if not under an integration sign with a test function $h\in C^\infty_0(\Bbb{R})$ or $h\in\mathcal{S}(\Bbb{R})$, the smooth (infinitely differentiable) functions of ...

2

So I actually was able to figure this out! It turns out I skipped a step. So you begin with: $\langle\tilde{x}'\mid X \mid \tilde{x}\rangle = e^{-ig(x')/\hbar}\langle x'\mid X \mid x\rangle e^{ig(x)/\hbar}$ = $\langle x'\mid X \mid x\rangle e^{i(g(x)-g(x'))/\hbar}$ Next since $X\mid x\rangle = x\mid x\rangle$, the previous formula becomes \$x\langle x'\mid ...

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