The Dirac Delta Function Potential: Difference between revisions
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Revision as of 15:28, 15 February 2013
A delta potential, eg. 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\delta(x-a)\!} , is a special case of the finite square well, where the width of the well goes to zero and the depth of the well goes to infinity, while the produce of the height and depth remains constant. For a delta potential, the wavefunction is still continuous across the potential, ie. 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 x=a\!} . However, the first derivative of the wavefunction is discontinuous across the potential.
For a particle subject to an attractive delta potential 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(x) = -V_0\delta(x)\!} the Schrödinger equation is
- 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 \frac{-\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2}-V_0\delta(x)\psi(x)=E\psi(x)}
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 x \neq \!0 } the potential term vanishes, and all that is left is
- 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 \frac{d^2 \psi(x)}{dx^2} + \frac{2mE}{\hbar^2}\psi(x) = 0}
A bound state(s) may exist when 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 E < 0 \! } , and 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 \psi(x) \! } vanishes 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 x = \pm \infty \!} . The bound state solutions are therefore given by:
- 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 \psi_{1}(x) = Ae^{kx} x < 0 \!}
where
The first boundary condition, the continuity of at , yields .
The second boundary condition, the discontinuity of at , can be obtained by integrating the Schrödinger equation from to and then letting
Integrating the whole equation across the potential gives
In the limit , we have
which yields the relation: .
Since we defined , we have . Then, the energy is
Finally, we normalize :
so,
Evidently, the delta function well, regardless of its "strength" , has one bound state:
Similarly, for a delta potential of the form , the discontinuity of the first derivative can be shown as follows:
The Schrödinger equation is
Integrating the whole equation across the potential gives
In the limit , we have
Hence the first derivative of the wave function across a delta potential is discontinuous by an amount: