# Tag Info

## Hot answers tagged semiclassical

8

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 ...

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 ...

5

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

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 ...

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 ...

4

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 ...

4

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 ...

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. ...

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 ...

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 = ... 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 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 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 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 ... 2 There are results that are mathematically rigorous concerning the semiclassical limit of quantum theories. It is in fact an ongoing and interesting theme of research in mathematical physics. However you need to be rather well versed in analysis to understand the results. The bibliography is quite huge, but I would like to mention the following (some quite ... 2 Unfortunately I cannot tell you what went wrong on your first try, since I don't exactly know what you did. However, I sat down and tried to solve the system you describe: We are looking for Eigenstates of the Hamiltonian$$ H = \frac{p^2}{2} + V(x),\qquad \text{where }V(x)=\left\{\begin{array}{ll}\infty & \text{if } x<0 \\ x & ...

2

I) Let us for simplicity put the physical constants $\hbar=1=m$ to one. OP is considering the usual transcription $u(r)\equiv rR(r)$ of the 3D radial TISE into a 1D TISE, $$\tag{A} - \frac{1}{2} u^{\prime\prime}(r)+U_{\ell}(r)u(r) ~=~E u(r),$$ where the total potential energy $$\tag{49.8b} U_{\ell}(r)~:=~ U(r) + \frac{C_{\ell}}{2r^2}$$ is a sum of a ...

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 ...

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 $$\psi(x=0)~=~0 \qquad \text{and} \qquad ... 2 The (finite) imaginary part comes solely from the singularity at r = 2(M-\omega^\prime) in the integration over r away from this singularity the integral is finite and real. Performing a change of variables: r = 2(M-\omega^\prime) + u, since only the singularity contributes to the imaginary part , we can approximate the integrand by by his ... 1 I agree with QMechanic that the velocity in question is the quantity$$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

This was explained very successfully the work that Paul Drude began in 1900 and extended by Hendrik Lorentz in 1905, culminating in the Drude–Lorentz model which is a classical model. I will not go through the derivation but the main points. Inside a metal, the electrons do not feel a restoring force from the lattice when they interact with an impinging EM ...

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