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A classical turning point is a point at which the system's total energy $E$ equals the potential energy $V.$ Past this point, i.e. for $E<V$ the potential is greater than the total energy, such cases we denote as classically forbidden regions, because from a purely classical point of view, the system has 0 chance of being in a state where its potential ...

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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 ...

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 ...

5

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 ...

5

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 ...

5

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 ...

5

First, the classical and semiclassical adjectives are not quite synonyma. "Semiclassical" means a treatment of a quantum system whose part is described classically, and another part quantum mechanically. Fields may be classical, particle positions' inside the fields quantum mechanical; metric field may be classical and other matter fields are quantum ...

5

Before addressing your question, there is a point where I kind of disagree with Orca's answer that I'd like to discuss: I will begin with part 2 of your question about plane waves. The use of this Ansatz is the first clue that you are actually treating the situation quantum-mechanically, but ending up with a result that exactly matches the classical ...

4

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 $\... 4 The semiclassical/WKB quantization rule and connection formulas are discussed in numerous textbooks. The discrete quantization condition follows from requiring single-valueness of the wavefunction. Note that quantization formula gets modified by the metaplectic correction/Maslov index because of turning points. For an elementary treatment, see e.g. Refs. 1-... 4 Not quite. the correspondence principle tells us that a theory should reproduce the older theory that it aims to supersede in an appropriate limit, e.g.$\hbar \rightarrow 0$to reproduce classical behavior from quantum mechanics or$c\rightarrow \infty$to reproduce Newtonian kinematics from special relativity. As you can see, this is not restricted to ... 4 What would somebody take h-->0 to take the classical limit in the first place? Think for a minute about what the presence of Planck's constant in quantum theories does. It creates a granularity to the amount of energy that can be stored in a mode (but not in general, to the amount of energy that can be stored in most systems). Why does that matter? ... 3 The question of how quantum mechanics reduces to classical mechanics in the limit$\hbar\to 0$has been asked several times before, see. e.g. this, this and this Phys.SE posts, and links therein. I) Let$\Delta t:=t_f-t_i$and$\Delta x:=x_f-x_i$. The key fact is now that the Feynman propagator/kernel/amplitude$K(x_f,t_f;x_i,t_i)$localizes to a delta ... 3 If you use tight-binding Hamiltonian, it is reasonable to start not from semiclassical, but one-particle approximation. In that case, you have an amplitude (complex number) at each site, the state is complex vector of length$n$, Hamiltonian is$n\times n$(sparse) matrix and the problem of time evolution and/or eigenstates (for one particle state) is ... 3 In the following, I'll describe to you how in principle one can compute higher order corrections to the Bohr-Sommerfeld condition. In order to find higher order corrections to the Bohr-Sommerfeld formula, we need to include higher order corrections of the wavefunction of the form:$\Psi(x) = \sum_n \hbar^n a_n(x) exp(\frac{i}{\hbar} \int^x \sqrt{2m(E-V(y)) }...

3

I see this as a partial trace on the $\psi$ space, that is : Let $H$ be the hamiltonian applying on the $(\psi, \chi)$ space, and $H_{eff}$ the reduced hamiltonian applying on the $\chi$ space. Take some density matrix $\rho$ applying on the $\psi$ space. Then : $H_{eff} = Tr_\psi(\rho H)$ Here one is taking $\rho = \int d^3p f(E_p) |\psi(p,s)\rangle \... 3 I'm not altogether sure what you are asking, but I suspect the following may help. To represent rotations, spins and vectors in$SU(2)$we work as follows. Rotations live in$SU(2)$. Vectors (in the physicist's sense) live in the algebra$\mathfrak{su}(2)$. The position vector$(x,\,y,z)$is: $$X =x\,\hat{s}_x+y\,\hat{s}_y+z\,\hat{s}_z = \left(\begin{... 3 I) The common starting point is the CCR$$\tag{1} [\hat{Q},\hat{P}]~=~i\hbar~{\bf 1}.$$For a general irreducible representation of the CCR (1), see the Stone-von Neumann theorem. The standard Schrödinger position representation reads$$ \tag{2} \hat{Q}~=~q, \qquad \hat{P}~=~-i\hbar\frac{\partial}{\partial q}. $$There is a similar Schrödinger momentum ... 3 We can treat this system classically because it is one of those nice situations in which the quantum-mechanical treatment produces the same results! I will begin with part 2 of your question about plane waves. The use of this Ansatz is the first clue that you are actually treating the situation quantum-mechanically, but ending up with a result that exactly ... 2 The (finite) imaginary part comes solely from the singularity at r = 2(M-\omega^\prime) in the integration over r away from this singularity the integral is finite and real. Performing a change of variables: r = 2(M-\omega^\prime) + u, since only the singularity contributes to the imaginary part , we can approximate the integrand by by his ... 2 To find the bound states for the potential$$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 \lim_{... 2 I believe it is valid without a continuity assumption, but (as always) only up to higher orders in$\hbar$. 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 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$\...

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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: $\int_\mathrm{... 2 For first question, you must simly take$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$values, and like ... 2 I recently updated the wikipedia article on statistical ensembles which might be relevant. Basically, in classical physics the probability distribution for the state of a system is written as an integral over position and momentum as in your equation. It turns out to be necessary to choose an arbitrary unit of action (energy times time) in order to define "... 2 The bounds for r should still be the classical turning points, as you mentioned for the harmonic oscillator. Presumably you're in a bound state of Hydrogen, i.e. have an energy of the form$\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) \hbar^2}{2 \...

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