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The article of D J Griffiths assumes that the delta interaction can be approximated by a sequence of even functions and then infers two boundary conditions: $$\Psi'(0^+)-\Psi'(0^-)= (-1)^n {m c \over \hbar^2} (\Psi'^{(n)}(0^+)+\Psi'^{(n)}(0^-))$$ $$\Psi(0^+)-\Psi(0^-)= (-1)^{n-1} {m c \over \hbar^2} n (\Psi'^{(n-1)}(0^+)+\Psi'^{(n-1)}(0^-))$$ My own ...
Let's simplify things, $\hbar=m=1$ and we put in the $\delta^{(n)}$ as $V(x)=V_0\delta^{(n)}/2$. Then the problem is $$-\psi'' - V_0\delta^{(n)}\psi = E \psi$$ Apart from $x=0$ the equation gives $$\psi'' = -E \psi$$ we want a bound state $E<0$ which falls off at infinity, so we get a solution $\psi_+ = A \exp(-k x)$ for $x>0$ and $\psi_- = A ... 1$\delta'(x)$is a scale invariant barrier, where the S-matrix and phase shifts do not depend of the momentum. A recent divulgative article is Point interactions: boundary conditions or potentials with the Dirac delta function (De Vincenzo - Sánchez) Canadian Journal of Physics 10/2010; 88(11):809-815. DOI: 10.1139/P10-060 Another interesting reference can ... 5 Ok, I have a solution for$\delta'(x)$based on a very crude limit. I'm going to neglect factors of$\hbar$,$m$, etc for the sake of eliminating clutter. Let$V_\epsilon(x)=\frac{\delta(x+\epsilon)-\delta(x)}{\epsilon}$. Then$\lim_{\epsilon\rightarrow 0}V_e(x)=\delta'(x)$. We'll solve the Schrodinger equation for finite$\epsilon$and then take the limit ... 1 Ref.1 is already in eq. (3.1) considering a functional integral over the scalar field$\phi:M\to \mathbb{R}$. Here$M$is spacetime. For a rigorous treatment of functional integrals, Ref. 1 points in the beginning of Section 3 to its Ref. [3.2]. In this answer we will just take an intuitive heuristic approach, and try to construct the functional integral as ... 1 TL;DR: Substitution inside the delta function yields a Jacobian factor $$\tag{1} \delta(f(v))~=~ \sum_{v_{(0)},f(v_{(0)})=0 }\frac{1}{| f^{\prime}(v_{(0)})|} \delta(v-v_{(0)}).$$ Here the sum is over the zeroes$v_{(0)}$of the function$f(v)$. Let us for simplicity consider velocity$v$rather than momentum$p=mv$. So energy conservation $$\tag{2} ... 1 For the second case you want to divide by the force, not the velocity. You are basically computing what fraction of the time you spend at a particular point in phase space. However what you have is a probability density. So P(z,p) is something you multiply by a volume in phase space to get a probability. A correct way to get it would be to consider a ... 1 I am not sure if you mean x one or two dimensional variable. I don't have the article at hand but I guess we probably mean D=1?? The argument goes as follows: you have to use Bochner's theorem, which says that W(x-y):=W_2(x,y) is positive distribution if and only if the Fourier transformation yields is a positive measure \tilde W(p) d^Dp. In this ... 4 Yes, they can. It's not great form, but they can. For a simple example, consider the hamiltonian for a 1D particle in a potential V(x),$$ \hat H=\frac1{2m}\hat{ p}^2+V(\hat{x}). $$This holds equally well if the potential is a delta function at x_0, V(x)=\kappa \delta(x-x_0), in which case$$ \hat H=\frac1{2m}\hat{ p}^2+\kappa\delta(\hat{x}-x_0).$$... 0 I think the point is that if you took the limits to infinity without doing anything else, you'd be implicitly redefining the matrix element in order to make the equations consistent. So he just calls the redefined matrix element$T_{ni}$instead of$V_{ni}$. Later he solves for$T_{ni}$in terms of$V_{nj}\$ (see the section "Solving for the T matrix").