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general_considerations [2020/07/21 05:37] admingeneral_considerations [2020/07/21 05:39] (current) – [4.iii.4 Boundary Conditions] admin
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 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 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
  
-\[\lim_{\epsilon \rightarrow 0_+} \left [ \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x'+\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x'-\epsilon}\right ]=\frac{2m}{\hbar^2}\int_{x'-\epsilon}^{x'+\epsilon}V(x)\psi(x)\,\mathrm{d}x\]+\[\lim_{\epsilon \rightarrow 0_+} \left [ \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x'+\epsilon} - \left . \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |_{x=x'-\epsilon}\right ]=\frac{2m}{\hbar^2} \lim_{\epsilon \rightarrow 0_+} \left [ \int_{x'-\epsilon}^{x'+\epsilon}V(x)\psi(x)\,\mathrm{d}x \right ]\]
  
 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.