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--- the_delta_function_potential [2020/07/20 08:27] – [4.vi.1 $E<0$: Bound States] admin
+++ the_delta_function_potential [2020/07/20 23:52] (current) – admin
@@ Line 3: / Line 3: @@
 where $\alpha$ is a positive constant.
-Now, obviously a potential that dips down to $-\infty$ at just a single point is highly idealized.  If it makes sense to compare the level of idealization of unrealistic potentials then it is even more ridiculous than the infinite square well potential.  However, solving this system is a good stepping stone from the infinite square well to the finite square well as, unlike the infinite square well but like the finite square well, it has both bound and scattering states.  It is much easier to solve than the finite square well and will give us some intuition as to what to expect there.
+Now, obviously a potential that dips down to $-\infty$ at just a single point is highly idealized.  However, solving this system is a good stepping stone from the infinite square well to the finite square well.
 Since the potential is zero everywhere except the origin, bound states occur for $E < 0$ and scattering states for $E > 0$.  Classically, if $E < 0$ then the particle is confined to $x=0$ as it does not have enough kinetic energy to escape the delta function.  Its momentum is horribly divergent.  If you think about the delta function as the limit of an infinite square well where the length of the well goes to zero then its momentum must change direction infinitely fast as the particle hits the "walls" of the delta function.  For $E>0$ a particle moving to the right from $x < 0$ goes along its merry way with kinetic energy $E$ until it hits the delta function.  When it reaches $x=0$, it receives an infinitely large impulse towards the right while, at the same time, receiving an infinitely large impulse towards the left.  This will again cause the momentum to be horribly divergent at the instant the particle passes $x=0$, but since the two impulses cancel each other we would not notice.  The particle will just continue to the right with kinetic energy $E$ for $x>0$.
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 ====== 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 in the present case at $x=0$ gives
+Applying this at $x=0$, the left hand side is
-\[\lim_{\epsilon \rightarrow 0_+} \left [ \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{-\epsilon}\right ] = \frac{\mathrm{d}\psi_+}{\mathrm{d}x} \right |_{0} - \left . \frac{\mathrm{d}\psi_-}{\mathrm{d}x} \right |_{0} =-\frac{2m}{\hbar^2}\lim_{\epsilon\rightarrow 0_+} \left [\int_{-\epsilon}^{+\epsilon} \alpha \delta(x)\psi(x)\,\mathrm{d}x \right ] = -\frac{2m\alpha}{\hbar^2}\psi(0) = -\frac{2m\alpha}{\hbar^2}\sqrt{k}.\]
+\[\lim_{\epsilon \rightarrow 0_+} \left [ \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=-\epsilon}\right ] = \left . \frac{\mathrm{d}\psi_+}{\mathrm{d}x} \right |_{x=0} - \left . \frac{\mathrm{d}\psi_-}{\mathrm{d}x} \right |_{x=0},\]
+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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 ====== 4.vi.2 $E > 0$: Scattering states ======
-For the scattering case we define the positive constant $k = \sqrt{2m\frac{E}{\hbar^2}}$ and then the general solution to the TISE is
+For the scattering case we define the positive constant $k = \sqrt{\frac{2mE}{\hbar^2}}$ and then the general solution to the TISE is
 \[\psi(x) = \begin{cases} \psi_-(x) = Ae^{ikx} + Be^{-ikx}, & x < 0,  \\ \psi_+(x) = C e^{ikx} + D e^{-ikx}, & x>0. \end{cases}\]
@@ Line 75: / Line 79: @@
 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
@@ Line 81: / Line 85: @@
 \[T = \frac{2E\hbar^2}{m\alpha^2 + 2E\hbar^2}, \qquad R = \frac{m\alpha^2}{m\alpha + 2E\hbar^2}.\]
-Classically, there would be no reflection from the delta function potential, so this is an interference effect.  However, for large $E$, when $2E\hbar^2$ is large compared to $m\alpha$, we have $T \approx 1$ and $R \approx 0$, as expected.
+Classically, there would be no reflection from the delta function potential, so this is an interference effect.  However, for large $E$, when $2E\hbar^2$ is large compared to $m\alpha^2$, we have $T \approx 1$ and $R \approx 0$, as expected.
 ====== 4.vi.3 The Delta Function Barrier ======
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 In this case, there are no bound states, and the only difference this will make to the scattering states is that, when we evaluate the boundary condition on the derivative we will get
-\[\lim_{\epsilon \rightarrow 0_+} \left [ \int_{-\epsilon}^{+\epsilon}V(x)\psi(x)\,\mathrm{d}x = +\frac{2m\alpha}{\hbar^2}(A+B),\]
+\[\lim_{\epsilon \rightarrow 0_+} \left [ \int_{-\epsilon}^{+\epsilon}V(x)\psi(x)\,\mathrm{d}x \right ] = +\frac{2m\alpha}{\hbar^2}(A+B),\]
+instead of $-\frac{2m\alpha}{\hbar^2}(A+B)$, which has the effect of changing $B$ and $C$ to
+\[B = \frac{\beta}{-\beta + ik}A, \qquad C = (A+B) = \frac{ik}{-\beta + ik}A.\]
+Because the moduli squared of $B$ and $C$ are the same as for the attractive delta function potential, the transmission and reflection coefficients will be the same.
+A classical particle in this potential would never have enough kinetic energy to pass the delta function barrier, so it would always bounce off it elastically.  The fact that the transmission coefficient is nonzero in the quantum case is another example of quantum tunneling.
+{{:question-mark.png?nolink&50 |}}
+====== In Class Activities ======
-instead of $-\frac{2m\alpha}{\hbar^2}(A+B)$
+  - For the wavefunction
+  \[\psi(x) = \begin{cases} \psi_-(x) = Ae^{kx} , & x < 0,  \\  \psi_+(x) = A e^{-kx}, & x>0, \end{cases}\]
+  show that, if it is normalized, $\int_{-\infty}^{+\infty}\mathrm{d}x\, |\psi(x)|^2$, then the constant $A$ is $\sqrt{k}$.
+  - For the wavefunction
+  \[\psi(x) = \begin{cases} \psi_-(x) = Ae^{ikx} + Be^{-ikx}, & x < 0,  \\ \psi_+(x) = (A+B) e^{ikx}, & x>0, \end{cases}\]
+  show that the boundary condition
+  \[\left . \frac{\mathrm{d}\psi_+}{\mathrm{d}x} \right |_{0} - \left . \frac{\mathrm{d}\psi_-}{\mathrm{d}x} \right |_{0} = -\frac{2m\alpha}{\hbar^2}(A+B)\]
+  implies that
+  \[B = \frac{-\beta}{\beta + ik}A, \qquad C = (A+B) = \frac{ik}{\beta + ik},\]
+  where $\beta = \frac{m\alpha}{\hbar^2}$.