Spherical Well: Difference between revisions
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{{Quantum Mechanics A}} | {{Quantum Mechanics A}} | ||
Let us now consider a spherical well potential, given by | Let us now consider a spherical well potential, given by | ||
<math> | <math> | ||
V(\mathbf{r}) = | V(\mathbf{r}) = | ||
\begin{cases} | \begin{cases} | ||
-V_0, & 0\leq r< a \\ | -V_0, & 0\leq r< a \\ | ||
0, & r>a | 0, & r>a. | ||
\end{cases} | \end{cases} | ||
</math> | </math> | ||
Line 28: | Line 29: | ||
</math> | </math> | ||
where <math> k' = \sqrt{\frac{2m(E+V_0)}{\hbar^2}} \!</math> and <math> k = \sqrt{\frac{2mE}{\hbar}} \!</math> | where <math> k' = \sqrt{\frac{2m(E+V_0)}{\hbar^2}} \!</math> and <math> k = \sqrt{\frac{2mE}{\hbar}}.\!</math> | ||
Let us now consider bound states for the special case, <math>l=0.\!</math> In this case, the centrifugal barrier drops out and the equations become | Let us now consider bound states for the special case, <math>l=0.\!</math> In this case, the centrifugal barrier drops out and the equations become | ||
Line 70: | Line 71: | ||
Dividing the second equation by the first, we obtain | Dividing the second equation by the first, we obtain | ||
<math>-k\cot{ka}=\kappa,</math> | <math>-k\cot{ka}=\kappa,\!</math> | ||
which is just the solution for the odd states in a [[One-Dimensional Bound States|one-dimensional square well]]. | which is just the solution for the odd states in a [[One-Dimensional Bound States|one-dimensional square well]]. | ||
This, combined with the fact that | |||
<math>\kappa^2+k^2=\frac{2mV_0}{\hbar^2},</math> | |||
shows that no bound state exists if <math>V_0<\frac{\pi^2\hbar^2}{8ma^2}.</math> |
Latest revision as of 09:43, 26 October 2013
Let us now consider a spherical well potential, given by
The Schrödinger equations for these two regions are
for and
for
The general solutions are
where and
Let us now consider bound states for the special case, In this case, the centrifugal barrier drops out and the equations become
The solution for this case is
where
Using the boundary condition, we find that The wave functions for thus reduces to
where
For , we know that since, as the wavefunction must go to zero. Therefore, for the region in which
Using the conditions that at the wave functions and their derivatives must be continuous yields the following equations:
and
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