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--- general_considerations [2020/07/16 19:21] – [$E>V_+$: Scattering States] admin
+++ general_considerations [2020/07/21 05:39] (current) – [4.iii.4 Boundary Conditions] admin
@@ Line 78: / Line 78: @@
 do not have a continuous derivative at $x=0$ or $x=a$.  For example,
-\[\lim_{\epsilon \rightarrow 0_+} \left . \frac{\mathrm{d}\psi}{\mathrm{d}x}\right |_{\epsilon} = \lim_{\epsilon \rightarrow 0} \frac{n\pi}{a}\cos \frac{n\pi\epsilon}{a} = \frac{n\pi}{a},\]
+\[\lim_{\epsilon \rightarrow 0_+} \left . \frac{\mathrm{d}\psi}{\mathrm{d}x}\right |_{x=\epsilon} = \lim_{\epsilon \rightarrow 0} \frac{n\pi}{a}\cos \frac{n\pi\epsilon}{a} = \frac{n\pi}{a},\]
-\[\lim_{\epsilon \rightarrow 0_-} \left . \frac{\mathrm{d}\psi}{\mathrm{d}x}\right |_{\epsilon} = \lim_{\epsilon \rightarrow 0} 0 = 0.\]
+\[\lim_{\epsilon \rightarrow 0_-} \left . \frac{\mathrm{d}\psi}{\mathrm{d}x}\right |_{x=\epsilon} = \lim_{\epsilon \rightarrow 0} 0 = 0.\]
 In fact, there are no solutions with finite energy that have continuous derivatives at the two ends of the well.
@@ Line 118: / Line 118: @@
 \[\int_{x'-\epsilon}^{x'+\epsilon}\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}\,\mathrm{d}x=\frac{2m}{\hbar^2}\int_{x'-\epsilon}^{x'+\epsilon}V(x)\psi(x) \,\mathrm{d}x-\frac{2mE}{\hbar^2}\int_{x'-\epsilon}^{x'+\epsilon} \psi(x)\,\mathrm{d}x\]
-Note that $\int_{x'-\epsilon}^{x'+\epsilon}\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}\,\mathrm{d}x= \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x'+\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x'-\epsilon}$, so taking the limit of the right hand side as $\epsilon \rightarrow 0_+$ will give us the boundary condition on the derivative.  Now, if $\psi(x)$ is continuous then the second integral on the right hand side vanishes, so we are left with
+Note that $\int_{x'-\epsilon}^{x'+\epsilon}\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}\,\mathrm{d}x= \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x'+\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x'-\epsilon}$, so taking the limit of the right hand side as $\epsilon \rightarrow 0_+$ will give us the boundary condition on the derivative.  Now, if $\psi(x)$ is continuous then the second integral on the right hand side vanishes, so we are left with
-\[\lim_{\epsilon \rightarrow 0_+} \left [ \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x'+\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x'-\epsilon}\right ]=\frac{2m}{\hbar^2}\int_{x'-\epsilon}^{x'+\epsilon}V(x)\psi(x)\,\mathrm{d}x\]
+\[\lim_{\epsilon \rightarrow 0_+} \left [ \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x'+\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x'-\epsilon}\right ]=\frac{2m}{\hbar^2} \lim_{\epsilon \rightarrow 0_+} \left [ \int_{x'-\epsilon}^{x'+\epsilon}V(x)\psi(x)\,\mathrm{d}x \right ]\]
 We will see how this can be used to determine the boundary condition for a delta function potential in section 4.vi.
@@ Line 202: / Line 202: @@
     * Normalizable
     * Nondegenerate
-    * Have a discrete energy spectrum</span>
+    * Have a discrete energy spectrum
   * The solution will be a //semi-scattering state// if the region in which $E > V(x)$ extends to either $x\rightarrow +\infty$ or $x\rightarrow-\infty$ but not both. Semi-scattering states are:
     * Unnormalizable
@@ Line 218: / Line 218: @@
 A node of a wavefunction is a point at which it crosses the $x$-axis, from negative to positive or vice versa.  In the solutions for an infinite square well potential
-\[\psi(x)=\begin{cases} 0, x\leq 0 \\ \sqrt{\frac{2}{a}}\sin\left ( \frac{n\pi x}{a}\right ), & 0<x<a \\  0, & x\geq a, \end{cases}\]
+\[\psi(x)=\begin{cases} 0, & x\leq 0 \\ \sqrt{\frac{2}{a}}\sin\left ( \frac{n\pi x}{a}\right ), & 0<x<a \\  0, & x\geq a, \end{cases}\]
 if we exclude the end points $x=0$ and $x=a$, the ground state has no nodes, the first excited state has one node, the second excited state has two nodes, etc. This behavior is generally true for the bound states of all potentials.  A proof is beyond the scope of this course, but the intuition is the same as for the infinite square well.  In the regions where $E > V(x)$, the solution is oscillatory. The more times the solution oscillates in this region the higher the energy.  Roughly, more oscillations indicate a smaller "wavelength" and hence a larger momentum.  "Wavelength" is in scare quotes here because the solution need not be periodic, so the concept does not strictly apply, but the general idea that a larger number of nodes gives a larger mean momentum, and hence energy, is correct.  Therefore, the ground state will have the smallest possible number of nodes, i.e. zero, the first excited state the next smallest, i.e. one, and so on.
@@ Line 268: / Line 268: @@
 At the end of section 4.ii we considered the symmetric infinite potential well
-\[V(x) = \begin{cases} \infty, & x<-\frac{a}{2}, \\ 0, & -\frac{a}{2} \leq x \leq +\frac{a}{2}, \\ \infty, x>+\frac{a}{2}, \end{cases}\]
+\[V(x) = \begin{cases} \infty, & x<-\frac{a}{2}, \\ 0, & -\frac{a}{2} \leq x \leq +\frac{a}{2}, \\ \infty, & x>+\frac{a}{2}, \end{cases}\]
 as illustrated below.
@@ Line 276: / Line 276: @@
 We noted that the eigenstates
-\[\psi_n(x) = \begin{cases} \sqrt{\frac{2}{a}} \cos \left ( \frac{n\pi x}{a}\right ), & n=1,3,5,7,\cdots \sqrt{\frac{2}{a}} \sin \left ( \frac{n\pi x}{a} \right ), & n=2,4,6,8,\cdots,\\ \end{cases}\]
+\[\psi_n(x) = \begin{cases} \sqrt{\frac{2}{a}} \cos \left ( \frac{n\pi x}{a}\right ), & n=1,3,5,7,\cdots \\ \sqrt{\frac{2}{a}} \sin \left ( \frac{n\pi x}{a} \right ), & n=2,4,6,8,\cdots,\\ \end{cases}\]
 were a sequence of alternating odd and even functions.  The ground state is odd, the first excited state is even, the second excited state is odd, etc.
@@ Line 307: / Line 307: @@
 \[\hat{\mathcal{P}}|\psi\rangle = -\int_{+\infty}^{-\infty} \mathrm{d}x'\, \psi(-x') |-x'\rangle = \int_{-\infty}^{+\infty} \mathrm{d}x\, \psi(-x) |x\rangle, \]
-and hence $\hat{\mathcal{P}}$ maps a wavefunction $\psi(x)$ to $\psi(-x)$.
+where, in the second step we got rid of the minus sign by switching the limits and we replaced $x'$ by $x$ because $x'$ is a dummy variable. Hence, $\hat{\mathcal{P}}$ maps a wavefunction $\psi(x)$ to $\psi(-x)$.
 In the in class activities, you will show that $\hat{\mathcal{P}}$ is a Hermitian operator, which implies that it has real eigenvalues.  Note also that
@@ Line 335: / Line 335: @@
 \[\hat{\mathcal{P}} \hat{p}^2 = -\hat{p} \hat{\mathcal{P}} \hat{p} = +\hat{p}^2 \hat{\mathcal{P}}.\]
-Similarly, if the potential is even, $V(-\hat{x}) = V(\hat{x})$ the its Taylor expansion only includes even powers of $\hat{x}$,
+Similarly, if the potential is even, $V(-\hat{x}) = V(\hat{x})$, its Taylor expansion only includes even powers of $\hat{x}$,
 \[V(\hat{x}) = a_0 + a_2 \hat{x}^2 + a_3 \hat{x}^4 + \cdots.\]
@@ Line 355: / Line 355: @@
 The same type of argument works for the rest of the excited states.  If they have an even number of nodes then there cannot be a node at $x=0$ and hence the wavefunction must be even.  If there are an odd number of nodes then one of them must be at $x=0$ meaning that the wavefunction must be odd.  Since the number of nodes increases by one each time we increase the energy to the next allowed value in the spectrum, we get an alternating sequence of even and odd wavefunctions.
-{{:question-mark.png?nolink&50 |}}====== In Class Activities ======
+{{:question-mark.png?nolink&50 |}}
+====== In Class Activities ======
   - Prove that the parity operator $\hat{\mathcal{P}}$ is Hermitian, i.e.
@@ Line 361: / Line 362: @@
   HINT: Use the fact that $\hat{\mathcal{P}}$ maps $\psi(x)$ to $\psi(-x)$ and the definition of the inner product
   \[\langle \phi | \psi \rangle = \int_{-\infty}^{+\infty} \mathrm{d}x\, \phi^*(x) \psi(x).\]
-  - Show that $\hat{x}$ is an odd operator.  It is sufficient to show that $\hat{x}\hat{\mathcal{P}}|x\rangle = -\hat{\mathcal{P}}\hat{x}|x\rangle$.  Why?
+  -
-  - Show that $\hat{\mathcal{P}}|p\rangle = |-p\rangle$ using $\langle x | p \rangle = \frac{1}{\sqrt{2\pi \hbar}} e^{ipx/\hbar}$.
+    - Show that $\hat{x}$ is an odd operator.  It is sufficient to show that $\hat{x}\hat{\mathcal{P}}|x\rangle = -
-  - Show that $\hat{p}$ is an odd operator.
+    \hat{\mathcal{P}}\hat{x}|x\rangle$.  Why?
+    - Show that $\hat{\mathcal{P}}|p\rangle = |-p\rangle$ using $\langle x | p \rangle = \frac{1}{\sqrt{2\pi \hbar}} e^{ipx/\hbar}$.
+    - Show that $\hat{p}$ is an odd operator.