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2

I henceforth assume $\hbar =1$. There is no reason to introduce Dirac deltas here, everything is elementary. Moreover as the function $\psi$ is not differentiable, one cannot use the form of the momentum operator $P$ as derivative which is valid only on smooth functions. Forcing this way would introduce unnecessary difficulties as the derivative must be ...

5

I) One problem is that the momentum operator $\hat{p}$ is an unbounded operator, which means that it is only defined on a domain $D(\hat{p}) \subsetneq {\cal H}$ of the Hilbert space ${\cal H}=L^2(\mathbb{R})$. When we apply the differentiation operator $\hat{p}=\frac{\hbar}{i}\frac{d}{dx}$ to OP's wave function $$\tag{1} \psi(x)~=~A(a-x)\theta(a-|x|), ... 3 Well, you can conclude that something is wrong by the following logic: momentum is an observable, which means its allowed values must be things that you could read off a measuring device (assuming you had one that measures momentum). These are necessarily real values, and since the expectation value is some linear combination of possible measurements, it ... 11 The wavefunction has a discontinuity at x=-a, which gives a term -2aA i \hbar \delta(x+a) when you act with p. The contribution from this to the expectation value of momentum exactly cancels the imaginary value you have calculated. Two more-general points: The momentum operator is hermitian, which means its expectation value must be real (provided ... 0 Your last integral is$$ J = \int{d^3r \;\psi_0(r) \left[-2\pi\delta(\vec{r})Res(\psi_0(0)) + \sqrt{2}\pi a\delta(\vec{r})\frac{\partial}{\partial r}(r\psi_0(r))\right]} =\\ = 4\pi \int_0^\infty{dr\; r^2 \psi_0(r) \left[-2\pi\frac{\delta(r)}{2\pi r^2}Res(\psi_0(0)) + \sqrt{2}\pi a\frac{\delta(r)}{2\pi r^2}\frac{\partial}{\partial r}(r\psi_0(r))\right]} \\ = ...

2

Here's how I would come to some intuition for it. I would think about the rate of "probability flow" into a region by integrating the equation over a region in space. For now, let's suppose that no diffusion occurs at all, since that is more complicated (although directly doable and understandable). Then $$\int_a^b dx\frac{\partial p(x,t|x_0)}{\partial ... 0 Note: c=1 in the following. Every time-like worldline can be parametrized by its proper time. If you are given \vec r(t), then the proper time at t_0 is given by$$\tau(t_0) = \int_0^{t}\sqrt{1-\left(\frac{\mathrm{d}\vec r}{\mathrm{d}t}\right)^2}\mathrm{d}t  and inverting this expression to get $t(\tau)$ gives you the worldline \$r^\mu = ...

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