# Tag Info

1

I have found a good explanation in the net to my question, so I am sharing it just in case somebody else wants an answer too. Note that the question about "why it reaches double the amplitude" remains , as well as a new problem on why does the answer say that the ring has momentum(because its massless): The end of the stationary string, figure (a), is ...

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For one dimensional crystal, when we have periodic boundary condition, we may think that the atoms are arranged in a circle, and so each lattice point will be equal to other lattice point and the translation symmetry is conserved. And for three dimensional crystal, due to periodic boundary condition, each atom will feel the same in the lattice. They are in ...

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The holomorphic/coherent state path integral is explained in e.g. Ref. 1. Let us here highlight some of the points. Notation in this answer: In this answer, let $z,z^{\ast}\in \mathbb{C}$ denote two independent complex numbers. Let $\overline{z}$ denote the complex conjugate of $z$. Also Planck's constant $\hbar=1$ is put equal to one. It is customary to ...

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This question is very broad so I will try to answer only its parts. Why do we even have reflection? Impedance of the string is the key. Wave on the string reaches the point with impedance discontinuity and therefore reflection occurs. In your case there is reflection from "zero impedance boundary". The opposite textbook example is the reflection from ...

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I guess the $2$ depends on taking into account spin or not. Spin can be up or down, so just multiply the total number of allowed states by $2$. In http://www.phys.ufl.edu/~pjh/teaching/phy4605/notes/landau.pdf on page 8 they conclude the allowed number of states for a given fixed Landau level, is $\dfrac{eB}{h}$. They do not mention spin. So when ...

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Let's talk about 1d just to keep it as simple as possible. Consider the ground state of the simple harmonic oscillator. The energy eigenstates are each a Gaussian multiplied by a polynomial. The ground state is just a Gaussian. So each of those states goes to zero as $x\rightarrow\pm\infty.$ And each of them is unit norm. And any finite linear combination ...

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Actually it has been observed by experiment that the actual antinode of an air column with closed-open boundary condition is actually just outside the open end, not exactly at the open end. We do not consider the effects of viscosity of air and the radiation acoustic impedance of the open end of the pipe. Hence an end correction is applied to compensate for ...

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Each boundary condition comes from an independent Maxwell equation, so the four boundary conditions are independent. The right-hand side of the third boundary condition should K, the surface current. The first boundary condition could be replaced by $\phi_1=\phi_2$, which is easier to implement

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Fermat's principle says the path with minimum optical path or minimum time is chosen by light. It can be direct or indirect (containing reflections or refractions). But as others said, it's not necessarily unique, because there might be paths all with minimum optical paths, that means, all with equal minimum optical paths. That's exactly what happens in an ...

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So far, the only thing that I can be sure of is that there is no solution of the form $\psi(x,y)=g(y)f(x)$, which would lead to an obvious contradiction: If we have a solution $\psi(x,y)==f(x)g(y)$, denote $\frac{f''}{f}=F(x),\frac{g''}{g}=G(y)$, For any fixed x<-a, F(x)+G(y)=constant requires G(y) is constant, similarly, fixed y<-b, requires F(x) is ...

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To put it another way: typically with this sort of setup we want there to be an incoming wave from the left, $\psi_0 = e^{-i \omega t + i k x},$ which is partially "reflected" into an outgoing wave to the left $\psi_\ell = r e^{- i \omega t - i k x},$ and partly "transmitted" to an outgoing wave to the right, $\psi_r = t e^{- i \omega t + i k x}.$ The ...

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The idea is the following. You have three zones. One before the barrier, $x<0$, with $V(x)=0$ (zone 1), one inside (or above) the barrier $0\leq x\leq a$ with $V(x)=V_0>0$ (zone 2), and one after the barrier with $V(x)=0$ (zone 3). Let's say that a particle comes from the left, with energy $0<E<V_0$, so it will go "inside" the barrier. Then, ...

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