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general_considerations [2020/07/16 19:34] – [Odd and Even Operators] admingeneral_considerations [2020/07/21 05:39] (current) – [4.iii.4 Boundary Conditions] admin
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 do not have a continuous derivative at $x=0$ or $x=a$.  For example, 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. In fact, there are no solutions with finite energy that have continuous derivatives at the two ends of the well.
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 \[\int_{x'-\epsilon}^{x'+\epsilon}\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}\,\mathrm{d}x=\frac{2m}{\hbar^2}\int_{x'-\epsilon}^{x'+\epsilon}V(x)\psi(x) \,\mathrm{d}x-\frac{2mE}{\hbar^2}\int_{x'-\epsilon}^{x'+\epsilon} \psi(x)\,\mathrm{d}x\]  \[\int_{x'-\epsilon}^{x'+\epsilon}\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}\,\mathrm{d}x=\frac{2m}{\hbar^2}\int_{x'-\epsilon}^{x'+\epsilon}V(x)\psi(x) \,\mathrm{d}x-\frac{2mE}{\hbar^2}\int_{x'-\epsilon}^{x'+\epsilon} \psi(x)\,\mathrm{d}x\] 
  
-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. We will see how this can be used to determine the boundary condition for a delta function potential in section 4.vi.
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 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. 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.   - Prove that the parity operator $\hat{\mathcal{P}}$ is Hermitian, i.e.
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   HINT: Use the fact that $\hat{\mathcal{P}}$ maps $\psi(x)$ to $\psi(-x)$ and the definition of the inner product   HINT: Use the fact that $\hat{\mathcal{P}}$ maps $\psi(x)$ to $\psi(-x)$ and the definition of the inner product
   \[\langle \phi | \psi \rangle = \int_{-\infty}^{+\infty} \mathrm{d}x\, \phi^*(x) \psi(x).\]   \[\langle \phi | \psi \rangle = \int_{-\infty}^{+\infty} \mathrm{d}x\, \phi^*(x) \psi(x).\]
-  - 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.