![\int_{0}^{\infty }f(x)dx=1\\\\\therefore \int_{0}^{\infty }ae^{-kx}dx=1\\\\\Rightarrow a\int_{0}^{\infty }e^{-kx}dx=1\\\\\frac{a}{-k}[\frac{1}{e^{kx}}]_{0}^{\infty }=1\\\\\frac{a}{-k}[0-1]=1\\\\\therefore a=k](/tpl/images/0105/0879/a0b2d.png)
![\int_{0}^{L }f(x)dx=1\\\\\therefore Re(\int_{0}^{L }ae^{-ikx}dx)=1\\\\\Rightarrow Re(a\int_{0}^{L }e^{-ikx}dx)=1\\\\\therefore Re(\frac{a}{-ik}[\frac{1}{e^{ikx}}]_{0}^{L})=1\\\\\Rightarrow Re(\frac{a}{-ik}(e^{-ikL}-1))=1\\\\\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1](/tpl/images/0105/0879/8aad9.png)
Normalize the equations
a) psi(x)=ae^(-kx) from zero to infinity
b) psi(x)=ae^(ikx) from...
Physics, 18.07.2019 19:20 jngonzo1226
Normalize the equations
a) psi(x)=ae^(-kx) from zero to infinity
b) psi(x)=ae^(ikx) from 0 to l
Answers: 2
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