Tough exponential integral (QM, Variational Principle)

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SUMMARY

The discussion centers on solving a complex integral related to quantum mechanics, specifically using the variational principle. The user struggles with evaluating the second term of the integral involving the potential \( V_{0} \) using tools like Maple and Wolfram. A key hint provided is the integral formula \(\int dx \, x^{n} e^{-\alpha x} = \frac{n!}{\alpha^{n+1}}\), which leads to a rearrangement of terms. The user is advised to integrate over three dimensions and utilize the correct volume element for spherical coordinates to resolve the issue.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly the variational principle.
  • Familiarity with integral calculus, specifically exponential integrals.
  • Proficiency in using computational tools like Maple and Wolfram Alpha for symbolic integration.
  • Knowledge of spherical coordinates and their volume elements in multi-dimensional integrals.
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  • Review the derivation of the integral formula \(\int dx \, x^{n} e^{-\alpha x}\) and its applications in quantum mechanics.
  • Learn how to set up and evaluate integrals in spherical coordinates, focusing on volume elements.
  • Explore advanced features of Maple and Wolfram Alpha for solving complex integrals.
  • Study the variational principle in quantum mechanics and its implications for solving physical problems.
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Students and researchers in quantum mechanics, particularly those tackling variational problems and complex integrals in their studies or research projects.

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Homework Statement


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The Attempt at a Solution


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I've uploaded my work so far since its much faster than typing and I'm stuck on the last line trying to solve the integral.
The first part of the integral is calculable but the second term(containing [itex]V_{0}[/itex]) doesn't evaluate with Maple or Wolfram.
On the question sheet it gives the hint that [itex]\int dx x^{n} e^{-\alpha x}[/itex] = [itex]\frac{n!}{\alpha^{n+1}}[/itex] which leads me to rearrange the 2nd term to -2[itex]V_{0}[/itex][itex]\alpha[/itex]a*[itex]e^{r\frac{(-1-2\alpha a)}{a}}[/itex]*[itex]r^{-1}[/itex]
According to the tip, [itex]\int e^{r\frac{(-1-2\alpha a)}{a}}[/itex]*[itex]r^{-1}[/itex] is equal to [itex]\frac{(-1)!}{\frac{(-1-2\alpha a)}{a}^{-1+1}}[/itex] which is just -1.
This doesn't seem to match up with the given Hamiltonian on the problem set so I'm asking for some extra help on this,
 
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You need to start over from the beginning. You need to integrate over all three dimensions, not just over r. Also, don't forget to use the correct volume element for spherical coordinates.
 

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