Spherical Well: Difference between revisions
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{{Quantum Mechanics A}} | |||
Let's consider spherical well potentials, | Let's consider spherical well potentials, | ||
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Revision as of 16:42, 31 August 2011
Let's consider spherical well potentials,
The Schrödinger equations for these two regions can be written by
for and
for .
The general solutions are
where and .
For the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l=0\!} term, the centrifugal barrier drops out and the equations become the following
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{cases} 0\leq r< a & (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial r^2}-V_0)u_0(r)=Eu_0(r)\\ r>a & (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial r^2})u_0(r)=Eu_0(r) \end{cases} }
The generalized solutions are
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{cases} 0\leq r< a & u_0(r)=Ae^{ikr}+Be^{-ikr}\\ r>a & u_0(r)=Ce^{ik'r}+De^{-ik'r} \end{cases} }
Using the boundary condition, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u(r=0)=0\!} , we find that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=-B\!} . The second equation can then be reduced to sinusoidal function where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha=2iA\!} .
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_0(r)=2iA\sin(kr)=\alpha\sin(kr)=\alpha\sin\left(\frac{r}{\hbar}\sqrt{2m(E+V_0)}\right)}
for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r>a\!} , we know that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D=0\!} since as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r\!} approaches infinity, the wavefunction does not go to zero.
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_0(r)=Ce^{ik'r}+De^{-ik'r}=Ce^{-\frac{r}{\hbar}\sqrt{-2mE}}}
Matching the conditions that at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=a\!} , the wavefunctions and their derivatives must be continuous which results in 2 equations
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha\sin\left(\frac{a}{\hbar}\sqrt{2m(E+V_0)}\right)=Ce^{-\frac{a}{\hbar}\sqrt{-2mE}}}
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha\frac{\sqrt{2m(E+V_0)}}{\hbar}\cos\left(\frac{a}{\hbar}\sqrt{2m(E+V_0)}\right)=-\frac{\sqrt{-2mE}}{\hbar}Ce^{-\frac{a}{\hbar}\sqrt{-2mE}}}
Dividing the above equations, we find
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\cot\left(\sqrt{\frac{2m}{\hbar^2}(V_0-|E|)a^2}\right)=\frac{\sqrt{\frac{2m|E|}{\hbar^2}}}{\sqrt{\frac{2m(V_0-|E|)}{\hbar^2}}}} , which is the solution for the odd state in 1D square well.
Solving for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V_0\!} , we know that there is no bound state for
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V_0<\frac{\pi^2\hbar^2}{8ma^2}} .