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8

If you're trying to simulate a 2D solution of the Laplace equation (which is the only unambiguous reading of your post as currently stated; if that's not what you're doing then you should clarify your question with exactly what it is you're doing and how), then your code is wrong. The reason is that your results don't obey the maximum principle: a harmonic ...


7

You can show this by using perturbation theory (only for suitable small changes in the potential). When you assume, that $\tilde{V}(x) = V(x) + c$ with $c > 0$, then you can write your problem als perturbation: If the unperturbated hamiltonian $\hat{\mathrm H}$ has eigenstates $ | \Psi_n \rangle $ with discrete energies, then perturbation-theory states ...


4

Let there be given a selfadjoint$^1$ operator $H^{(0)}$ and a (semi)positive operator $V\geq 0$ on a Hilbert space ${\cal H}$. Let the basis of normalized eigenvectors for $H^{(0)}$ be $(|i^{(0)}\rangle)_{i\in I}$ with corresponding eigenvalues $(E^{(0)}_i)_{i\in I}$ ordered such that $$ \forall i,j ~\in~I:\quad i~\leq~j\quad\Rightarrow \quad E^{(0)}_i~\leq~...


4

Technically "potential difference" is the difference in electrical potential, i.e. $\Delta V$, not the difference in electrical potential energy, $\Delta U$. Potential difference ($\Delta V$) is also called voltage, in certain contexts. However, many people and sources are sloppy about their terminology, and they will say just "potential" when they really ...


2

You're trying to understand how a point charge is represented in the charge density $\rho(r)$. The concept you need is the Dirac delta function $\delta(r)$, which can describe the density of a finite amount of stuff packed into an infinitesimal point: $$\delta(r) = \left\{ ^{\infty \text{ if } r=0 }_{0 \text{ if } r\ne0} \right\} $$ $$\int_{-\infty}^{+\...


2

Consider $\tilde{V}(x)=V(x)+\Delta(x)$ where $\Delta(x)>0,\forall x$. Within 1st order perturbation, $E_n=E_n^0+\langle \psi_n|\Delta|\psi_n\rangle=E_n^0+\iint dx_1dx_2\psi_n^*(x_1)\delta(x_1-x_2)\Delta(x_2)\psi_n(x_2)=E_n^0+\int dx|\psi_n(x)|^2\Delta(x)>E_n^0$


2

$$A\sin(k_iL)=De^{-qL}$$ $$Ak_i\cos(k_iL)=-Dqe^{-qL}$$ $$k_i\cot(k_iL)=-q$$ Insert the values for $k_i$ and $q$: $$[2m(V_1-E_i)/\hbar^2]^{1/2}\cot[2m(V_1-E_i)/\hbar]^{1/2}L=-[2mE_i/\hbar^2]^{1/2}$$ The allowed energy levels ($E_i<V_2$) for bound states can be determined by numerical solution of that equation. But particles with energy $E>V_2$ can ...


2

Qualitatively, the wave functions of the bound states in a triangular potential well like the one you described, look like this: For $x<-a$, $\psi=0$ because of the infinite potential in that region. Where the wave function crosses the potential line, quantum tunnelling occurs and $\psi \to 0$. For particle energies above $V_0$, no bound states can ...


1

Potential difference (or voltage) is equivalent to a pressure difference in a pipe system of water. If there is equal pressure at two points, then no water has any reason to move. If there is a difference, water flows (is pushed) towards lower pressure. Similarly, charges will want to move towards the point of lowest possible potential. The larger the ...


1

Even with a delta function potential, continuity of the wave function is still required. (Please see comment from ACuriousMind below on this). The derivative of the wavefunction is obviously not continuous, however. You can find the discontinuity by integration about the delta function from +s to -s, where s is a small parameter. You then let s go to 0 ...


1

Generically, the answer is "no". The Zeeman effect is the splitting of degenerate spectral lines in the presence of a static magnetic field. As the field strength increases, some lines move to higher energies and some lines move to lower energies. Example of the splitting of the $5s$ orbitals of Rubidium: (Graph created by: Danski14. Image used under ...


1

When you connect them, charge can flow, and it will do so until there is no net electric field it can move along (the surfaces are the same potential). Imagine I just suddenly destroyed the wire; now the charges are stuck on the spheres. They are still under the condition that they want to equilibrate and put a constant voltage across the surface, but they ...


1

there is an electric field inside the wire, and there is a loss of potential energy, or voltage as they move but this drop in voltage is usually negligible (thought not in some applications) and we only consider that the drop in voltage comes only from the circuits elements o loads. This idealization often fails not with the wires, but within the battery ...


1

This turns out to have a really boring answer. We can find the two circular orbits by finding the maxima and minima of the effective potential, and we get: $$ r = \frac{L^2}{2M}\left(1 \pm \sqrt{1 - \frac{12M^2}{L^2}}\right) \tag{1} $$ where the $+$ gives the outer stable orbit and the $-$ gives the inner unstable orbit. Note that both orbits exist only ...


1

The mathematical definition that John Rennie gave explains it well, but I'd like to give an intuitive answer to your question. Imagine a rubber sheet horizontally stretched. The height of the rubber sheet at a point is equivalent to the electric potential at that point. Now, since as of now there is no disturbance at all, the height (potential) throughout ...



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