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Consider the general diffusion equation of the form, $$ \frac{\partial \psi}{\partial t}=D\frac{\partial\Phi}{\partial x}=D\frac{\partial^2\psi}{\partial x^2}\tag{1} $$ where $\Phi=\partial_x\psi$ is the probability current (flux). In order to replicate a reflecting boundary, we establish an infinite potential at the boundaries, ...


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The point of restricting the string to length $L$ is that we can then construct a periodic function (with wavelength $L$) by imagining repeated copies of the string connected to each other. In this case we can construct the function as a Fourier series with the lowest frequency sine/cosine having the same wavelength $L$. If we have an infinite string then ...


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It doesn't imply that directly, you still have gauge freedom. I think what they mean is that if you set V1=V2 at one point on the boundary, it will hold true for all the boundary because the tangential component is identical.


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Short answer, you can get the result in electrostatics, for simple (path) connected surfaces, if the electric field isn't singular on the surface. How? For simple surfaces were can adjust the whole potential in one surface by a uniform amount and it doesn't change the electric field in that region. So pick a point A on the interface surface and why not ...


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Without the phase change energy conservation would not be satisfied. To see why this is true you can think of a simple Michelson interferometer; without one of the fields having a phase flip you could get constructive (or destructive) interference at both sides of the beam splitter which would result in twice (or none of) the energy which you sent into the ...


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It helps to realize that for the string, "hard boundary" means the displacement dy/dt=0, whereas for a soft boundary it means the force F=0. In the non-extreme cases, neither dy/dt nor F is zero. Their ratio will determine the phase. It's fairly clear that you can't have both F = 0 and dy/dt = 0 at one end, and there aren't any other relevant variables ...


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Every differential equation problem has two parts: the differential equation itself, and the boundary condition. Neither can tell you the other in and of itself. Consider $y'' = -k^2 y$. Is the solution sine or cosine? The answer depends on boundary conditions, and I have to specify them. I can't get them by specifying anything in the equation. So, say your ...


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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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