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Heisenberg's uncertainty principle is $$\Delta x \Delta p \geq \hbar/2.$$ Since the well is of width $L$, you have a measure for the uncertainty on the position $\Delta x$. Then assume the lowest possible value for $\Delta p$, i.e. the one for which the above inequality becomes an equality. Lastly, use $E = \dfrac{p^2}{2m}$ to find an expression for $E$. ...

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In Mathematica, using the formula $\sqrt{u_0^2-v^2}=\begin{cases}\\v\tan(v) \\-v\cot(v)\end{cases}$ where $u_0=mL^2V_0/(2\hbar^2)$, we can first compute $u_0$: << PhysicalConstants` u = Convert[ NeutronMass (1 Femto Meter)^2 V0 ElectronVolt/(2 \ PlanckConstantReduced^2), 1] Out: 1.20649*10^-8 V0 Since $u_0$ is typically on the order of an ...

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First: The Schrödinger equation is a equation using functionals. Solutions to this equation are such $\Psi(r,t)$, that fulfill this equation. The finite square well 1.0 fm wide means you have a potential $V(r)$ which is zero for r<0 and r>1fm and -d in between. d is your depth. Now you have to determine d such, that only two of the resulting $\Psi(r,t)$ ...

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your coordinates are pointing up, let's call the coordinates y-axis. at the top your initial coordinate is $y_i=h$, where $h$ is simply a distance, it's positive. at the bottom your location is $y_f=-x$, where $x$ is again a positive value, indicating how far the spring will go down. so, the equations comes up as $mgy_i=mgy_f+kx^2/2$, which is the same as ...

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It is a matter of how you define your variable. Regardless of how you define your coordinate system the result shouldn't change. Consider 2 cases, the first one being how you currently defined your vertical axis, and the second one being with an origin at the foot of the spring. Let $\Delta x$ be a positive value which is the displacement of the spring. 1st ...

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The only thing that's really important is the differential eqaution. The situation is, outside the well, in both cases: $\dfrac{d^2 \psi}{dx^2}= - \frac{2mE}{\hbar^2} \psi$ Now it's foundamental notice that for bound states E<0 so we can write: $E=-|E|$ and Sc. equation become: $\dfrac{d^2 \psi}{dx^2}= + \frac{2m|E|}{\hbar^2} \psi$ So the usual way ...

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I should think that the potential energy would increase by exactly the amount of work being done. Generally in such thought experiments, the background electrical "potential" is considered constant or fixed, but the field (and the implied potential) would obviously be changing in the neighborhood of the point charge. The net potential is the sum of the ...

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Short version: In the infinite potential well, $E \geq 0$ (because $V_{min}=0$, and $E \geq V_{min}$). In your finite potential well, it sounds like you are looking for bound states, in which case $E < 0$, so you absorb the negative into the square root. Long version: When you are tackling a QM problem, first you should figure out the admissibility of ...

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Electromagnetic force is not propagated by electrons, it is propagated by photons. By definition these travel at the speed of light (in the material). Impedance and capacitance play a part in how quickly the system responds to you turning it on / connection a battery, but are generally very small in a plain wire. The electrons are moved by electromagnetism ...

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The information about beginning of the flow of current is transmitted through the propagation of electromagnetic waves and not with drift velocity of electrons. Hence, any electric appliance turns on almost instantly, when the switch is closed.

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Although the electron velocity is very low, which is propagated almost instantaneously is the electric field. This causes the effect that all the electrons in the wire to start moving simultaneously (almost).

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Am I correct that you can rephrase your question to 'electrons move so slow, how come that when I flip the light switch the light comes on basically instantly?'? It's true that the electrons travel very slowly. But these electrons don't have to travel across the wire to power your light bulb. In electromagnetism, we have the continuity equation $\nabla J = ... 1 I've always treated anharmonic oscillators to mean the potential has the form $$V(x)=\gamma x^2 + \beta_ix^i$$ with$i$being any value except 2, including negative values as well. Anharmonicity then follows as the deviation of the eigenvalue of$V(x)$above from the harmonic solution. For example, the paper you link above, Case 1 has an energy eigenvalue ... 2 One has to be careful with the given potential. To start with it must be shown that $$h=-(d/dx)^2+V(x),$$ defines a unique self-adjoint operator$H$, i.e., is essentially self-adjoint. In case $$V(x)=ax^2+bx^3+cx^4$$ with$c>0$this is indeed the case. In fact the resolvent of$H$is compact (these matters are discussed in the books by Reed and Simon), ... 1 This is pretty simple. Consider$V$the electric potential, then an equipotential surface is simply a surface in$\mathbb{R}^3$defined by$V(x,y,z)=c$where$c$is some constant. In that case, consider$S(c)$the equipotential surface defined like that. Consider$\gamma : (-\varepsilon, \varepsilon)\to S(c)$a path in that surface. Consider also the ... 4 Here's one way to think about it (though it isn't mathematically rigorous). From very far away the dipole would appear to have zero charge and thus there wouldn't be an electric field at all. However, you also know that the electric field falls off as$1/r$, so from very far away you'd expect the electric field to be small. The additional charge ... 0 a) Change in potential energy =$(V_f - V_i)q = 70e$mJ where$e$is the charge on one proton. Therefore, potential energy increases. b) The electric field inside the membrane points towards the inside. Since$\vec{E} = -\frac{dV}{dx}$, assuming the electric field to be uniform, we obtain$|\vec{E}| = 70$mV/$8$nm$= 8.75 \times 10^6\$ V/m. Since work is ...

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