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Be aware that a "mathematical derivation" of a physical principle is, in general, not possible. Mathematics does not concern the real world, we always need empirical input to decide which mathematical frameworks correspond to the real world. However, the Schrödinger equation can be seen arising naturally from classical mechanics through the process of ...

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Small addition to ACuriousMind's great answer, in reply to some of the comments asking for a derivation of Schrödinger wave equation, using the results of Feynman's path integral formalism: (Note: not all steps can be included here, it would be too long to remain in the context of a forum-discussion-answer.) In the path integral formalism, each path is ...

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The momentum operator $P$ in the infinite well can be defined as a self-adjoint operator by infinitely many ways with respect to the boundary conditions by: $$P_\theta=-i\hbar\frac{d}{dx}\\ \mathcal{D}(P_\theta)=\left\{\psi\in \mathcal{H}^1[0,a]:\psi(a)=e^{i\theta}\psi(0)\right\},$$ where $\mathcal{H}^1[0,a]$ is the Sobolev space, on the interval $[0,a]$. In ...

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Good question. You're right, you can't really throw out one of the solutions. The author of this PDF is using a poorly explained shortcut, which goes like this: if you expand out the full complex solution, you get \begin{align} \Psi(\theta) &= c_1 e^{i\omega t} + c_2 e^{-i\omega t} \\ &= (a_1 + b_1 i)(\cos\omega t + i\sin\omega t) + (a_2 + b_2 ... 4 The probability density of the ground state is time independent, so there is no motion in this sense. Yet the expectation value of the kinetic energy is non-zero, so there is movement in this sense. How are these notions of movement reconciled? First off, classically, if we had a particle in a 1/r potential and released it from rest, it would indeed bob ... 4 To explicitly verify this, one solves the problem for a box of finite depth V_0. If you additionally assume the wavefunction and its first derivative to be continuous across the potential step, the solution becomes a matter of Solve the Schrödinger equation in the distinct regions in- and outside of the box. Match \phi and \phi' at the potential ... 4 According to Richard Feynman in his lectures on Physics, volume 3, and paraphrased "The Schrodinger Equation Cannot be Derived". According to Feynman it was imagined by Schrodinger, and it just happens to provide the predictions of quantum behavior. 4 Fundamental laws of physics cannot be derived (turtles all the way down and all that). However, they can be motivated in various ways. Direct experimental evidence aside, you can argue by analogy - in case of the Schrödinger equation, comparisons to Hamiltonian mechanics and the Hamilton-Jacobi equation, fluid dynamics, Brownian motion and optics have been ... 3 Answer to part 1: This is a common method of solving differential equations, employed by physicists to quickly extract a solution without having to make slow progress using more rigorous methods. The first step is to look for an asymptotic solution (i.e. the limit of large \xi in this case), where the equation is easily solvable. Now, we know that this ... 2 The potential energy function is the same for both. The energy level solutions are the same for both. The key difference is that in (most modern interpretations of) the Schrodinger model the electron of a one-electron atom, rather than traveling in fixed orbits about the nucleus, has a probablity distribution permitting the electron to be at almost all ... 2 This definition of bound and scattering states is not quite correct, although it holds for many potentials. There are counterexamples to this fact that have roots in a paper by von Neumann and Wigner. One is the spherical potentialV(r)=32\sin r \frac{\sin r-(1+r)\cos r}{\left(2+2r-\sin 2r\right)^2}It is not hard to check that V(r) is a bounded ... 2 If you assume separability of the wave function, i.e., \psi(\mathbf x)=u(x)v(y)w(z), you can solve the individual components separately: \begin{align} -\frac{\hbar^2}{2\mu}\frac{d^2u(x)}{dx^2}+V_1(x)u(x)&=E_1u(x)\\ -\frac{\hbar^2}{2\mu}\frac{d^2v(y)}{dy^2}+V_2(y)v(y)&=E_2v(y)\tag{1}\\ ... 1 One of the problems with Bohr's theory to describe the hydrogen atom, was that the electron orbiting around the nucleus has an acceleration. Therefore it radiates and loses energy, until it would collapse with the nucleus. This is a common error in physicist's history of physics. The problem was not with Bohr's model, but (as Bohr thought) with ... 1 I also don't quite understand what you mean by "imaginary \psi", but let me give you some general, more mathematically correct view. In general, what we measure is the spectrum of a self-adjoint operator. For the energy, this operator is of course the Hamiltonian. Now, since it is self-adjoint, its spectrum will lie on the real line. To have stability, we ... 1 In the 1d particle in the box the energy of the particle should be completely determined by the momentum of the particle that you observe correct? The Hamiltonian in the position basis is\hat H = \begin{cases}-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}, & 0\lt x \lt a\\\infty, & \text{otherwise} \end{cases}$$and the energy ... 1 Since k_E \propto \hat{p}^2 and \hat{p} is Hermitian you may see that this makes k_E positive semidefinite, that is all of its eigenvalues are larger or equal to 0. In other words when you measure this operator you will always get results which are larger or greater than zero. This "contradiction" is resolved by the fact that the potential is a ... 1 We all know that E=k_E + V(x); E = total energy, k_E = kinetic energy, V(x) = potential One reason why you may be confused by this is that the equation$$ E = \frac{1}{2}m\dot{x}^2 + V(x) $$comes from classical mechanics. When we are solving the Schroedinger equation$$ H\Phi = E\Phi  with Hamiltonian operator $H$, the meaning of the ...

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The paper addresses the case of a circular, finite quantum well with different electron effective masses inside and outside the well. The equation you point to is the result of a derivation from the time-independent Schrodinger equation, effective mass approximation and BenDaniel-Duke boundary conditions. It is also, however, dependent on the statement ...

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