Particle in a Ring. Wells

1. The wave function of a particle in a ring is $\psi(\phi,t)=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2}}e^{-i\phi}e^{i\hbar t/2I}-\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2}}e^{-i\phi}e^{i2\hbar t/2I}$. Please find the expectation value of the energy.

$\psi=\frac{1}{\sqrt{2\pi}}(C_1\phi_1+C_2\phi_2)$ is a mix of elementary states.
For first state $C_1=1/\sqrt{2}$ and $\phi_1=\exp(-i\phi)\exp(-i\hbar t/2I)$
For second state $C_1=-i\sqrt{2}$ and $\phi_2=\exp(i\phi\exp(-2i\hbar t/2I)$

By comparying the elementary states  with the general type $\phi=\exp(im\phi)\exp(-i\omega t)$
we obtain two energies in the spectrum
$E_1=\hbar\omega_1=\hbar^2/2I$ and $e_2=\hbar\omega_2=2\hbar^2/I=4\hbar^2/2I$
For the coefficients we have $|C_1|^2+|C_2|^2=1$  so that the expectation energy is
$<E>=E_1*|C_1|^2+E_2*|C_2|^2=[(1/2)*1+(1/2)*4](\hbar^2/2I)=5\hbar^2/4I$

2.
Rank for the penetration distance $d$ of the wavefunction for the energy levels in the figure. The penetration distance is defined from $\psi\sim e^{-x/d}$.

Three Wells

The decay of the wavefunction in the well wall is of the type $\psi \sim \exp (-kx)$ so that $d=1/k=\frac{\hbar}{\sqrt{2m(U-E)}}$ and thus the penetration depth depends only on difference \sqrt{U-E}.
For wells in figures a) and b) the energy difference is the same $U-E=5 eV$. therefore the penetration depth is the same in a) and b) cases. For case c) $U-E=16-10=6 eV >5eV$ so that the penetration depth is in this case smaller than in the first two cases (a and b).

3.
Specify the features of the wavefunctions for the indicated energies.

One Well

Features:
-The energy level n has  n-1 nodes inside the well.
-The amplitude of a wave inside a well is proportional to the $\sqrt{1/L}$ where $L$ is the width of the well. So lower energy levels will have a bigger amplitude in the case given.
-The penetration depth is inversely proportional to the difference $\sqrt{U-E}$ so that the lowest energy level will have a smaller penetration depth into the wall.
-All wavefunctions have a a zero at the left wall (it is infinite).
The frequency of the wavefunction is increasing with the energy level, according to the relation $E_n=\hbar\omega_n$.


valentin68

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