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--- general_considerations [2020/07/16 19:27] – [Number of Nodes in a Bound state] 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 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.