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I) OP's potential $$V(x)~=~a^x~=~e^{bx}, \qquad b~:=~\ln a ~\in~ \mathbb{R},$$ is the so-called Liouville potential. There are no (discrete) bound states. In scattering theory, an incoming wave at $x=-\infty$ gets reflected by the so-called "Liouville wall", and returns to $x=-\infty$. This potential is used in e.g. Liouville theory, which is ...

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First, one must appreciate that the phase space is classically parameterized by $x,p$ and coordinates on an ordinary plane commute with each other, $xp=px$. However, in quantum mechanics, this ain't the case. Instead, we have the Heisenberg commutator $$xp - px = i\hbar.$$ This means that quantum mechanically, the phase space is not an ordinary plane (or ...

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Don't be intimidated, semiclassical quantization is very simple, and it can be straightforwardly understood from a few examples which lead to the general case. Consider a particle in a box. The classical motions are reflections off the wall. These make a box in phase space, as the particle goes left, hits the wall, goes right, and hits the other wall. If ...

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Many symplectic manifolds (phase spaces of mechanical systems) admit a coordinate system where the symplectic two form can be written locally as: $\omega = \sum_i dp_i \wedge dq_i + \sum_j dI_j \wedge d\theta_j$ Where $p_i, q_i$ are linear coordinates $I_j$ are radial coordinates and $\theta_j$ are angular coordinates. The submanifold parameterized by ...

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If you use tight-binding Hamiltonian, it is reasonable to start not from semiclassical, but one-particle approximation. In that case, you have an amplitude (complex number) at each site, the state is complex vector of length $n$, Hamiltonian is $n\times n$ (sparse) matrix and the problem of time evolution and/or eigenstates (for one particle state) is ...

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In the following, I'll describe to you how in principle one can compute higher order corrections to the Bohr-Sommerfeld condition. In order to find higher order corrections to the Bohr-Sommerfeld formula, we need to include higher order corrections of the wavefunction of the form: $\Psi(x) = \sum_n \hbar^n a_n(x) exp(\frac{i}{\hbar} \int^x \sqrt{2m(E-V(y)) ... 3 It may be a reference to the fact that you can reproduce the characteristics of the photoelectron production in a model which treats the incident light classically, but treats the matter in the target quantum mechanically. This is explained in Mandel and Wolf's book (chapter 9), which explains how a simple semiclassical calculation can be used to derive the ... 2 Landau diamagnetism takes place because of the noncommutativity of$\mathbf{p}$and$\mathbf{A}(\mathbf{r})$. In the classical treatment no such noncommutativity exists, thus the magnetic susceptibility is identically zero. One way to appreciate the dependence of the result on$\hbar$and at the same time perform the full quantum treatment is to perform ... 2 I think you're very close to the right understanding with the following statement: Currently the only "solution" I can think of is to treat the light field quantum mechanically - in that case photon creation/annihilation operators will take care of these matrix elements. However I would like to treat my problem semiclassically. You have not ... 2 To find the bound states for the potential $$V(x) ~=~\left\{\begin{array}{ccc}ae^{cx} &\text{for}& x>0, \\ \infty&\text{for}& x\leq 0, \end{array} \right.$$ where$a,c>0$are two positive constants, one should solve the time-independent Schrödinger eq. with the two boundary conditions$$\psi(x=0)~=~0 \qquad \text{and} \qquad ... 1 Perhaps not a totally satisfactory answer, but a partial clarification of one of the things I was confused about: In the semiclassical treatment of the Hawking radiation process, there is no need to have an interacting quantum field theory. Therefore the vacuum-vacuum bubble diagrams of interacting perturbation theory are completely irrelevant to the basic ... 1 The (finite) imaginary part comes solely from the singularity at$r = 2(M-\omega^\prime)$in the integration over$r$away from this singularity the integral is finite and real. Performing a change of variables:$r = 2(M-\omega^\prime) + u$, since only the singularity contributes to the imaginary part , we can approximate the integrand by by his ... 1 As$x \rightarrow -\infty$the potential$V(x) = a^x$will go to 0 so if you start with a particle as a wavepacket anywhere on that potential, it will eventually end up travelling to$x \rightarrow -\infty$. Even if the packet started traveling in the positive$x$direction, it will bounce off the potential and go to$x \rightarrow -\infty\$. So the only ...

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