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the_delta_function_potential [2020/07/20 23:46] – adminthe_delta_function_potential [2020/07/20 23:52] (current) – admin
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 We still have to deal with the boundary condition on the derivative, which is a little more tricky.  We cannot apply continuity of the derivative because the potential is infinite at $x=0$.  However, recall that in section 4.iii we derived that, if the wavefunction is continuous at $x = x'$ then We still have to deal with the boundary condition on the derivative, which is a little more tricky.  We cannot apply continuity of the derivative because the potential is infinite at $x=0$.  However, recall that in section 4.iii we derived that, if the wavefunction is continuous at $x = x'$ then
  
-\[\lim_{\epsilon \rightarrow 0_+} \left [ \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x'+\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x'-\epsilon}\right ]=\frac{2m}{\hbar^2} \lim_{\epsilon \rightarrow 0_+} \left [ \int_{x'-\epsilon}^{x'+\epsilon}V(x)\psi(x)\,\mathrm{d}x \right ]\]+\[\lim_{\epsilon \rightarrow 0_+} \left [ \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x'+\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x'-\epsilon}\right ]=\frac{2m}{\hbar^2} \lim_{\epsilon \rightarrow 0_+} \left [ \int_{x'-\epsilon}^{x'+\epsilon}V(x)\psi(x)\,\mathrm{d}x \right ]\]
  
-Applying this at $x=0$ gives+Applying this at $x=0$, the left hand side is
  
-\[\lim_{\epsilon \rightarrow 0_+} \left [ \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{-\epsilon}\right ] = \left . \frac{\mathrm{d}\psi_+}{\mathrm{d}x} \right |_{0} - \left . \frac{\mathrm{d}\psi_-}{\mathrm{d}x} \right |_{0} =-\frac{2m}{\hbar^2}\lim_{\epsilon\rightarrow 0_+} \left [\int_{-\epsilon}^{+\epsilon} \alpha \delta(x)\psi(x)\,\mathrm{d}x \right ] = -\frac{2m\alpha}{\hbar^2}\psi(0) = -\frac{2m\alpha}{\hbar^2}\sqrt{k}.\]+\[\lim_{\epsilon \rightarrow 0_+} \left [ \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=-\epsilon}\right ] = \left . \frac{\mathrm{d}\psi_+}{\mathrm{d}x} \right |_{x=0} - \left . \frac{\mathrm{d}\psi_-}{\mathrm{d}x} \right |_{x=0},\]  
 + 
 +and the right hand side is 
 + 
 +\[-\frac{2m}{\hbar^2}\lim_{\epsilon\rightarrow 0_+} \left [\int_{-\epsilon}^{+\epsilon} \alpha \delta(x)\psi(x)\,\mathrm{d}x \right ] = -\frac{2m\alpha}{\hbar^2}\psi(0) = -\frac{2m\alpha}{\hbar^2}\sqrt{k}.\]
  
 Evidently, we must have Evidently, we must have
  
-\[\left . \frac{\mathrm{d}\psi_+}{\mathrm{d}x} \right |_{0} - \left . \frac{\mathrm{d}\psi_-}{\mathrm{d}x} \right |_{0} = -2k^{\frac{3}{2}} = -\frac{2m\alpha}{\hbar^2}\sqrt{k},\]+\[\left . \frac{\mathrm{d}\psi_+}{\mathrm{d}x} \right |_{x=0} - \left . \frac{\mathrm{d}\psi_-}{\mathrm{d}x} \right |_{x=0} = -2k^{\frac{3}{2}} = -\frac{2m\alpha}{\hbar^2}\sqrt{k},\]
  
 which yields which yields
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 With this, we can calculate the transmission and reflection coefficients: With this, we can calculate the transmission and reflection coefficients:
  
-\[T = \frac{|C|^2}{|A|^2} = \frac{k^2}{\beta^2 + k^2}, \qquad R = \frac{\beta^2}{\beta^2 + k^2}.\]+\[T = \frac{|C|^2}{|A|^2} = \frac{k^2}{\beta^2 + k^2}, \qquad R = \frac{|B|^2}{|A|^2} = \frac{\beta^2}{\beta^2 + k^2}.\]
  
 Substituting $k = \sqrt{2m\frac{E}{\hbar^2}}$ and $\beta = \frac{m\alpha}{\hbar^2}$ gives Substituting $k = \sqrt{2m\frac{E}{\hbar^2}}$ and $\beta = \frac{m\alpha}{\hbar^2}$ gives