Spherical Well

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Quantum Mechanics A
Schrödinger Equation
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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Brief Derivation of Schrödinger Equation
Relation Between the Wave Function and Probability Density
Stationary States
Heisenberg Uncertainty Principle
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Linear Vector Spaces and Operators
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One-Dimensional Bound States
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Summary of One-Dimensional Systems
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Commutation Relations
Angular Momentum as a Generator of Rotations in 3D
Spherical Coordinates
Eigenvalue Quantization
Orbital Angular Momentum Eigenfunctions
General Formalism
Free Particle in Spherical Coordinates
Spherical Well
Isotropic Harmonic Oscillator
Hydrogen Atom
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}) = 
-V_0, & 0\leq r< a \\
0, & r>a.

The Schrödinger equations for these two regions are


for  0\leq r< a \! and


for  r>a. \!

The general solutions are

&\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, 

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

\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.

The solution for this case is

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

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


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}\!


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

Dividing the second equation by the first, we obtain


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

This, combined with the fact that


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

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