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## Hot answers tagged schroedinger-equation

8

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 ...

3

(a) It depends on what you mean by "variational methods". The stationary S. equation it is nothing but an elliptic equation, so you can use all the standard method appropriate for elliptic equations, existence of weak solutions and elliptic regularity (Weyl, Sobolev, Hoermander...). The temporal equation, to some extent is similar to the heat equation. At ...

2

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 $\Psi(r,t)$ ...

1

You might be interested in this "elementary" derivation of the free particle Schroedinger equation from Maxwell's equations. It seems to be in the same spirit as Schroedinger's original reasoning. The niceness of this approach is that if you also include special relativity, it nets you both the free particle Schroedinger equation and its relativistic ...

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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, note that $\hat H_f$ has degenerate spectrum: it has equal eigenvalues for states with $\left|k\right\rangle$ and for $\left|-k\right\rangle$. This in turn means that, in particular, the state $\frac{-i}{\sqrt{2}}\left(\left|k\right\rangle-\left|-k\right\rangle\right)$ is also an eigenvector of $\hat H_f$. But in position representation it will look ...

1

What you are proposing will work, it is essentially what is known as the shooting method for solving the eigenvalue problem. Note that the eigenfunction is defined up to a multiplicative constant so you can just set $\epsilon$=1 and there is only one parameter $\epsilon'$ to vary in order to achieve a properly decaying solution at infinity. The shooting ...

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Well, without the delta potential the wave function is $$\tag{1} \psi_0(x,t)~=~\exp\left[ -\frac{iE_1 t}{\hbar}\right] \phi(x) ,$$ where $$\tag{2} \phi(x)~:=~\sqrt{\frac{2}{L}}\sin\frac{\pi x}{L}, \qquad E_1~:=~ \frac{\hbar^2}{2m}\frac{\pi^2}{L^2}.$$ Next we are supposed to incorporate the "full" effect of the delta function $\delta(t)$ (as opposed to ...

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