# Tag Info

6

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 ...

6

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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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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Always include units Always. No exception. Ever. If your Nobel prize winning professor writes down equations like this without units, you look him in the eye and tell him he's not doing physics. Your analysis suffers from the fatal flaw that you are comparing quantities with different units. $gh/c^2$ is dimensionless. It is the fractional energy change in ...

3

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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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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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 ...

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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 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) 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 I believe it is valid without a continuity assumption, but (as always) only up to higher orders in$\hbar$. 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 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 No. The differences between the eigenvalues (giving the spectral frequencies) are highly specific for each chemical substance, so knowledge of E_0 tells very little about the remainder of the spectrum. 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 Not quite. the correspondence principle tells us that a theory should reproduce the older theory that it aims to supersede in an appropriate limit, e.g. \hbar \rightarrow 0 to reproduce classical behavior from quantum mechanics or c\rightarrow \infty to reproduce Newtonian kinematics from special relativity. As you can see, this is not restricted to ... 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 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 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 The first one is regularized the second one isn't. 1 The bounds for r should still be the classical turning points, as you mentioned for the harmonic oscillator. Presumably you're in a bound state of Hydrogen, i.e. have an energy of the form \frac{-13.6 eV}{n^2} for some integer n. The problem then reduces to finding the zeros of the equation$$\frac{-13.6 eV}{n^2} = -\frac{e}{r^2} - \frac{l(l+1) ... 1 (Not really an answer, but as one should not state such things in comments, I'm putting it here) You commented: "This seems to boil down to the relationship between the phase space and the Hilbert space." That's a deep question. I recommend reading Urs Schreiber's excellent post on how one gets from the phase space to the operators on a Hilbert space in a ... 1 Well, first of all, it is important to realize that the integrals (1) and (3) are not merely ordinary double integrals over a single$x$- and a single$p$-variable. Instead they are (Wick-rotated) path integrals containing, heuristically speaking, infinitely many integrations. The path integral derivation of the free particle and the harmonic oscillator ... 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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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 ...

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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 ...

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