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First order perturbation energy correction to H-like atom
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[QUOTE="Ed Sheeran Fan, post: 6016782, member: 647587"] [h2]Homework Statement [/h2] Real atomic nuclei are not point charges, but can be approximated as a spherical distribution with radius ##R##, giving the potential $$ \phi(r) = \begin{cases} \frac{Ze}{R}(\frac{3}{2}-\frac{1}{2}\frac{r^2}{R^2}) &\quad r<R\\ \frac{Ze}{r} &\quad r>R \\ \end{cases}$$ We can view the modified Hamiltonian as having a perturbation $$H^\prime = \begin{cases} -\frac{Ze^2}{R}(\frac{3}{2}-\frac{1}{2}\frac{r^2}{R^2}) + \frac{Ze^2}{r} &\quad r<R\\ 0 &\quad r>R\\ \end{cases} $$ Calculate the first order correction to the energy of the ##1s## state. Given answer: ##\frac{2}{5}\frac{Z^4e^2R^2}{a^3}##, where ##a## is the Bohr radius. [h2]Homework Equations[/h2] Nondegenerate case, first order energy correction to the ##n##'th unperturbed energy $$E_n^1 = H^\prime_{nn}= \langle\psi_n^0 | H^\prime| \psi_n^0 \rangle$$ Radial wavefunction of lowest state of hydrogen/hydrogen like ions $$\psi(r) = \sqrt{\frac{Z^3}{\pi a^3}}e^{-Zr}$$ *think there may be a typo in the radial wavefunction, as the exponential should have a factor of ##1/a##, making it be ##e^{-Zr/a}## [h2]The Attempt at a Solution[/h2] Tried to evaluate $$E^1_{nlm=100}=\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{R} \frac{Z^3}{\pi a^3}e^{-2Zr/a}\times (-\frac{Ze^2}{R}(\frac{3}{2}-\frac{1}{2}\frac{r^2}{R^2}) + \frac{Ze^2}{r}) \times r^2sin\theta \,dr \,d\theta \,d\phi$$ but the integral seemed to fall apart. Wondering 1) if this gives the right answer and 2) is there a better way, e.g. some way to evaluate this without using integration by parts 4 times.***EDIT*** Found the/a solution in another book They use the approximation ##e^{-2Zr/a} = 1## since ##a \gg R## after which the answer falls out Feel a bit cheated after having done 3 pages of algebra [/QUOTE]
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First order perturbation energy correction to H-like atom
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