Tag Info

New answers tagged

0

electric potential of a body depend on the surface area of the body, that is as the surface area increases potential decreases  Exactly. Potential is the "will" for the charges to move away. Like the pressure in a water pipe, potential is what pushes charges to move faster through a conductive path (a wire). It should be quite simple intuitively that ...


0

The suggestion of Kevin Zhou, on other hand, has the advantage that we could try to control all the poles of the original well and the halved one. The polology of the square well was studied in Nussenzveig "The poles of the S-matrix of a rectangular potential well or barrier" Nuclear Physics, 11:409–521, 1959. and revisited in his book "Causality and ...


0

I'd say that the non adiabatic problem is ill-defined. Lets see. All the initial solutions $\Psi_n(x) = A_n \sin(n \pi x / 2a)$ have a probability current $\Psi'^* \Psi - \Psi^* \Psi'$ equal to zero everywhere, so in principle you can cut them at any point without having a leak of probability. The problem is that the boundary conditions for the domain of ...


0

As all the answers are oriented in the theoretical side let me remind everybody of the Balmer series, an experimental observation fitted with a mathematical series, which was first modeled with the Bohr model of the hydrogen atom, and then was the cornerstone of building quantum mechanics, as it came out from the Schrodinger equation. So in this sense a ...


0

The article of D J Griffiths assumes that the delta interaction can be approximated by a sequence of even functions and then infers two boundary conditions: $$ \Psi'(0^+)-\Psi'(0^-)= (-1)^n {m c \over \hbar^2} (\Psi'^{(n)}(0^+)+\Psi'^{(n)}(0^-))$$ $$ \Psi(0^+)-\Psi(0^-)= (-1)^{n-1} {m c \over \hbar^2} n (\Psi'^{(n-1)}(0^+)+\Psi'^{(n-1)}(0^-))$$ My own ...


0

The problems solved by the method of images are, at their heart, boundary value problem involving the divergence of a scalar field, and those problem have uniqueness theorems that says there is only one configuration of fields in the volume that generates a particular set of boundary conditions. So if you find any method of generating a field that meets the ...


4

Okay, this seems to me correct since the force imparted by $q'$ on $\left(Q- q'\right)$ is cancelled by the field of $q$ since it is a zero equipotential surface of $q-q'$ system. I have no idea why you think a zero potential surface has anything to so with anything. While the field of $q$ and the field of $q'$ together make a zero potential surface on ...


4

Let's simplify things, $\hbar=m=1$ and we put in the $\delta^{(n)}$ as $V(x)=V_0\delta^{(n)}/2$. Then the problem is $$-\psi'' - V_0\delta^{(n)}\psi = E \psi$$ Apart from $x=0$ the equation gives $$\psi'' = -E \psi $$ we want a bound state $E<0$ which falls off at infinity, so we get a solution $\psi_+ = A \exp(-k x)$ for $x>0$ and $\psi_- = A ...


5

Ok, I have a solution for $\delta'(x)$ based on a very crude limit. I'm going to neglect factors of $\hbar$, $m$, etc for the sake of eliminating clutter. Let $V_\epsilon(x)=\frac{\delta(x+\epsilon)-\delta(x)}{\epsilon}$. Then $\lim_{\epsilon\rightarrow 0}V_e(x)=\delta'(x)$. We'll solve the Schrodinger equation for finite $\epsilon$ and then take the limit ...


0

I wanted to bring up the same question as you raised, namely question 2). I agree with the answers to 1) and 3), but am unsatisfied with the proposed answer to 2) of buzhidao and levitopher (the argument given is also used in well-known textbooks such as Shankar, p.176). It is too sloppy in my opinion to say that "at infinity, the function is zero". This is ...


1

In terms of radiography / radiology, kVp is the tube voltage / tube potential between the cathode and the anode, set by the operator. The unit eV (or keV as used in the range of general radiography, MeV as used by Radiation Therapy) describes the energy of the particles - in this case, the electrons in the x-ray tube, and the x-ray photons coming off the ...


1

If we consider a satellite in orbit around the earth, the satellite has an infinite number of different potential energies: one for each possible frame of reference we could choose. Correct. And the kinetic energy is also different for different frames. A conserved quantity is one that is the same at two moments. An invariant quantity is something that ...


0

If you consider the viral theorem then you can relate the expectation value of the kinetic energy to the potential. Just make sure to use a fully general form, like one that involves the derivative of the potential. After you have the kinetic energy you should be able to get what you want easily.


0

This would not work for an arbitrary smooth central potential $V: \mathbb{R}_+ \to \mathbb{R}$ with $$V~<~0, \qquad V^{\prime}~>~0,\qquad V(0)~=~-\infty,\qquad V(\infty)~=~0,$$ because there might only be a finite number of bound states, and hence no continuous limit of bound states. For instance, one can use WKB methods to argue that if $V(r)$ ...


0

A summarised derivation for the 1 D triangular potential well can be found here. Interesting graphs of the wave functions $\psi_n(x)$ are provided. Also treated at [this *.pdf] (p.5, section 1.2.8.2)2.


0

It would be better to specify that $V_0 < 0$ as otherwise your problem is that of a potential barrier (no bound states) and not that of a finite potential well. The case of a finite potential well is fully developed here (Wikipedia entry), the case of a rectangular potential barrier you can find here (Wiki).


1

Assume that the free space is vacuum and the battery is perfectly insulated as well as all the electrical contacts. In this case there are two possibilities for an electronic device to fail: 1. The induced emf in the inductive elements of electronics device because of the changing electrical field caused by the addition of electrons. 2. Most of the ...


0

You are correct that for $n = 3$ there are $2$ non-boundary zero points. Also, the modulus of $\psi(x)$ is highest where $V(x)$ is lowest. Where you are wrong is that $\psi(A)$ is not zero and $\psi(B)$ is not zero either, as you indicate in your schematic. For $x < A$ and $x > B$, $V(x)$ is not $\infty$ (your well is a finite, not infinite well) and ...


0

This problem is somewhat similar to the ammonia inversion. In that problem the probability densities $\psi_n(x)^2$ are similar for $n =1$ and $2$, for $n = 3$ and $4$ etc. As a result the Hamiltonian levels $E_1$ and $E_2$ are close together, as are $E_3$ and $E_4$, etcetera. See for example here and here. See also Energy levels for $NH_3$.


1

The wavefunction $\psi(x)$ satisfies $$ -\frac{\hbar^2}{2m}\psi'' + V_0\left(\frac{x}{L} - 1\right) \psi = E\psi, \quad 0 \leq x \leq L\\ -\frac{\hbar^2}{2m}\psi'' = E\psi, \quad x > L $$ Since the bound states have $E < 0$ let's introduce $$ k = \frac{\sqrt{-2mE}}{\hbar}\\ \varkappa = \frac{\sqrt{2mV_0}}{\hbar} $$ Then $$ \psi'' - ...


1

Disclaimer: In this answer, we will just derive a rough semiclassical estimate for the threshold between the existence of zero and one bound state. Of course, one should keep in mind that the semiclassical WKB method is not reliable$^1$ for predicting the ground state. We leave it to others to perform a full numerical analysis of the problem using Airy ...



Top 50 recent answers are included