Spherical Well

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Quantum Mechanics A
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The most fundamental equation of quantum mechanics; given a Hamiltonian \mathcal{H}, it describes how a state |\Psi\rangle evolves in time.
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Spherical Coordinates
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Free Particle in Spherical Coordinates
Spherical Well
Isotropic Harmonic Oscillator
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WKB in Spherical Coordinates
Feynman Path Integrals
The Free-Particle Propagator
Propagator for the Harmonic Oscillator
Differential Cross Section and the Green's Function Formulation of Scattering
Central Potential Scattering and Phase Shifts
Coulomb Potential Scattering

Let us now consider a spherical well potential, given by


V(\mathbf{r}) = 
\begin{cases}
-V_0, & 0\leq r< a \\
0, & r>a.
\end{cases}

The Schrödinger equations for these two regions are

 \left(\frac{-\hbar^2}{2m}\frac{d^2}{dr^2}+\frac{\hbar^2l(l+1)}{2mr^2}-V_0\right)u_l=Eu_l

for  0\leq r< a \! and

\left(\frac{-\hbar^2}{2m}\frac{d^2}{dr^2}+\frac{\hbar^2l(l+1)}{2mr^2}\right)u_l=Eu_l

for  r>a. \!

The general solutions are


\begin{align}
&\frac{u_{l}(r)}{r} = C_{l} j_{l} (k'r), & r \leq a, \\
&\frac{u_{l}(r)}{r} = A_{l}j_{l}(kr) +Bn_{l}(kr), & r > a, 
\end{align}

where  k' = \sqrt{\frac{2m(E+V_0)}{\hbar^2}} \! and  k = \sqrt{\frac{2mE}{\hbar}}.\!

Let us now consider bound states for the special case, l=0.\! In this case, the centrifugal barrier drops out and the equations become


\begin{cases}
\left (\frac{-\hbar^2}{2m}\frac{d^2}{dr^2}-V_0\right )u_0=Eu_0, & 0\leq r< a \\
\frac{-\hbar^2}{2m}\frac{d^2u_0}{dr^2}=Eu_0, & r>a.
\end{cases}

The solution for this case is


\begin{cases}
u_0(r)=Ae^{ikr}+Be^{-ikr}, & 0\leq r< a \\
u_0(r)=Ce^{\kappa r}+De^{-\kappa r}, & r>a,
\end{cases}

where \kappa=\frac{\sqrt{-2mE}}{\hbar}.

Using the boundary condition, u(0)=0,\! we find that A=-B.\! The wave functions for 0\leq r<a\! thus reduces to

u_0(r)=2iA\sin(kr)=\alpha\sin(kr),\!

where \alpha=2iA.\!

For r>a\!, we know that D=0\! since, as r\rightarrow\infty,\! the wavefunction must go to zero. Therefore, for the region in which r>a,\!

u_0(r)=Ce^{-\kappa r}.\!

Using the conditions that at r=a,\! the wave functions and their derivatives must be continuous yields the following equations:

\alpha\sin{ka}=Ce^{-\kappa a}\!

and

k\alpha\cos{ka}=-\kappa Ce^{-\kappa a}.\!

Dividing the second equation by the first, we obtain

-k\cot{ka}=\kappa,\!

which is just the solution for the odd states in a one-dimensional square well.

This, combined with the fact that

\kappa^2+k^2=\frac{2mV_0}{\hbar^2},

shows that no bound state exists if V_0<\frac{\pi^2\hbar^2}{8ma^2}.

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