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The $J$ in the question is not a constant. It is a phasor. It has an inherent $e^{jwt}$ with it. In the book, phasor form is used in these aspects. [Found an answer after researching a bit. Now it seems was really an easy concept...]


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The answer is yes: Obviously it is possible to solve radially symmetric Schroedinger equations, including the specialized treatment (radial component only) that you ask about. However, you will need a well-defined problem to meaningfully begin something as rigorous as an analytical (or numerical) treatment. The differential equation you have given for the ...


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Let us suppress (world-sheet) time $\tau$ in what follows, i.e. consider a fixed time $\tau$. Let there be given a continuous map $\phi:\Sigma\to M$, where the world-space $\Sigma$ and the target space $M$ are both 1D manifolds. We will assume that such a 1D manifold is either a real line $\mathbb{R}$ or a circle $S^1\cong\mathbb{R}/\mathbb{Z}$. That gives ...


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Notice first that even before restricting the domain of $\phi$, we are considering the theory on the cylinder and identifying the boundary condition $\phi(x + L,t) = \phi(x,t)$. Now to explain the restriction, let's take this example. Consider a field configuration at some fixed time $\phi(x,0)$, we only have to study this in the domain $[0,L]$. Now pick ...


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Maxwell's equations and Lorentz force law do indeed summarize all electrodynamics. However there are physical situations in which you do not know the charge distribution a priori, but instead you specify some surfaces on which the electric field is always normal, which happens when you have metals around. In such situations you can in principle you can ...


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At least three different quantities in physics are customary called an action and denoted with the letter $S$: The off-shell action functional $S[q;t_i,t_f]$, The (Dirichlet) on-shell action function $S(q_f,t_f;q_i,t_i)$, and Hamilton's principal function$^1$ $S(q,\alpha, t).$ For their definitions and how they are related, see e.g. my Phys.SE answer ...


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Well, if I'm right, then this problem has no solution - at least no finite analytical one. And btw. the problem is more general than one might think. Look at the upper right quadrant ($x\gt0$ and $y\gt0$). Then the boundary conditions are symmetry boundary conditions (scalar product of normal vector and gradient of the solution is zero at $x=0$ and $y=0$). ...



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