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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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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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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 ...

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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: $\Psi(x) = \sum_n \hbar^n a_n(x) exp(\frac{i}{\hbar} \int^x \sqrt{2m(E-V(y)) ... 3 It may be a reference to the fact that you can reproduce the characteristics of the photoelectron production in a model which treats the incident light classically, but treats the matter in the target quantum mechanically. This is explained in Mandel and Wolf's book (chapter 9), which explains how a simple semiclassical calculation can be used to derive the ... 2 The (finite) imaginary part comes solely from the singularity at$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 ... 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 $$\psi(x=0)~=~0 \qquad \text{and} \qquad ... 2 I) Schrödinger is apparently talking about the speed$$ v ~=~\sqrt{\frac{2(E-V)}{m}} $$of a non-relativistic particle with energy E in a potential V. II) If the energy E<V_0 is less than the asymptotic value of the potential$$V_0~:=~\lim_{|{\bf r}|\to \infty} V({\bf r})$$(corresponding to a bound state), then the wave function solution ... 2 The semiclassical/WKB quantization rule is discussed in numerous textbooks. The discrete quantization condition follows from requiring single-valueness of the wavefunction. Note that quantization formula gets modified by the metaplectic correction/Maslov index because of turning points. For an elementary treatment, see e.g. Refs 1 and 2 below. For a ... 2 I think you're very close to the right understanding with the following statement: Currently the only "solution" I can think of is to treat the light field quantum mechanically - in that case photon creation/annihilation operators will take care of these matrix elements. However I would like to treat my problem semiclassically. You have not ... 2 Inside vs outside, there is a sign change inside the square root, so that changes the nature of the "phase" \phi(r). Normally, when you match wave functions you require that \psi_\mathrm{left}(x) = \psi_\mathrm{right}(x) (continuity) and that the derivative changes according to what you get when you integrate the Schrodinger equation: ... 2 For first question, you must simly take l = 0(in your notations, l is curved). It's because angular momentum is conserved in radial-symmetry field problem, and you can simply take L^2 = \hbar^2 l(l+1). So, for l = 0 states you simply ommit \frac{L^2}{2mr^2} term. Potential is given in problem. It behaves like constant for some r values, and ... 2 I recently updated the wikipedia article on statistical ensembles which might be relevant. Basically, in classical physics the probability distribution for the state of a system is written as an integral over position and momentum as in your equation. It turns out to be necessary to choose an arbitrary unit of action (energy times time) in order to define ... 1 I agree with QMechanic that the velocity in question is the quantity$$v=\sqrt\frac{2(E-V)}{m}.$$Schrödinger is using terminology that's quite convoluted by modern standards so it is understandable to be confused by this passage. There are two possible regimes for this velocity, and they both affect how the wavefunction behaves. Here Schrödinger is working ... 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 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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