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1

Well, without the delta potential the wave function is $$\tag{1} \psi_0(x,t)~=~\exp\left[ -\frac{iE_1 t}{\hbar}\right] \phi(x) ,$$ where $$\tag{2} \phi(x)~:=~\sqrt{\frac{2}{L}}\sin\frac{\pi x}{L}, \qquad E_1~:=~ \frac{\hbar^2}{2m}\frac{\pi^2}{L^2}.$$ Next we are supposed to incorporate the "full" effect of the delta function $\delta(t)$ (as opposed to ...

1

What confused me was the explanation from the tangentbundle homepage (second yellow box in OP). The generalization is straightforward, for simple zeros we have: $$\bigg\vert\frac{\mathrm{d}f(x)}{\mathrm{d}x}\bigg\vert_{x_0}\delta[f(x)] = \delta(x-x_0)$$ integrate $$\int \mathrm{d}x\,\bigg\vert\frac{\mathrm{d}f(x)}{\mathrm{d}x}\bigg\vert_{x_0}\delta[f(x)] ... 2 The notation $$\frac{ \partial f_i}{ \partial x ^i }$$ means the diagonal elements of the matrix: $$J _{ ij} = \frac{ \partial f _i }{ \partial x ^j }$$ where f_i is the component of the vector \vec{f} (x). I found this very confusing a few weeks ago so. Here is the proof I wrote up for the ... 3 Again assuming it only has a zero x^i=x_0^i what you have is$$ \delta(f(x^i)) = \frac{\delta(x^1-x_0^1)}{\left|\frac{\partial f}{\partial x^1}\right|_{x^i=x_0^i}} \frac{\delta(x^2-x_0^2)}{\left|\frac{\partial f}{\partial x^2}\right|_{x^i=x_0^i}}\cdots \frac{\delta(x^n-x_0^n)}{\left|\frac{\partial f}{\partial x^n}\right|_{x^i=x_0^i}} = \prod_{j=1}^n ...

2

A "kosher" way to do this employs test functions. Consider a test function $\phi:\mathbb R^4\to \mathbb R$. Notice that \begin{align} \int_{\mathbb R^4} d^4 x\, \partial_\mu j^\mu(x) \phi(x) &= ec\int_{-\infty}^{\infty} ds\,u^\mu(s)\int_{\mathbb R^4} d^4x\,\partial_\mu\delta^4(x - X(s))\phi(x) \\ &= -ec\int_{-\infty}^{\infty} ds\, ...

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