Hot answers tagged semiclassical
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 ...
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 ...
4
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 ...
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
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
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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