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--- the_delta_function_potential [2020/07/20 23:45] – admin
+++ the_delta_function_potential [2020/07/20 23:52] (current) – admin
@@ Line 9: / Line 9: @@
 ====== 4.vi.1 $E<0$: Bound States ======
-Our strategy is to solve the TISE for the regions $x < 0$ and $x > 0$ and apply the boundary conditions at $x=0$ where the delta function resides.  Since $E$ is negative, we define the positive constant $k = \sqrt{-2m\frac{E}{\hbar^2}}$ and the general solution to the TISE will then be
+Our strategy is to solve the TISE for the regions $x < 0$ and $x > 0$ and apply the boundary conditions at $x=0$ where the delta function resides.  Since $E$ is negative, we define the positive constant $k = \sqrt{-\frac{2mE}{\hbar^2}}$ and the general solution to the TISE will then be
 \[\psi(x) = \begin{cases} \psi_-(x) = Ae^{kx} + Be^{-kx}, & x < 0,  \\ \psi_+(x) = C e^{kx} + D e^{-kx}, & x>0. \end{cases}\]
@@ Line 27: / Line 27: @@
 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
-\[\left . \frac{\mathrm{d}\psi_+}{\mathrm{d}x} \right |_{0} - \left . \frac{\mathrm{d}\psi_-}{\mathrm{d}x} \right |_{0} = -2k^{\frac{3}{2}} = -\frac{2m\alpha}{\hbar^2}\sqrt{k},\]
+\[\left . \frac{\mathrm{d}\psi_+}{\mathrm{d}x} \right |_{x=0} - \left . \frac{\mathrm{d}\psi_-}{\mathrm{d}x} \right |_{x=0} = -2k^{\frac{3}{2}} = -\frac{2m\alpha}{\hbar^2}\sqrt{k},\]
 which yields
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 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