Intiutive approach to Green's function for SE

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SUMMARY

The discussion focuses on the derivation of Green's function for the Helmholtz equation as presented in Griffiths' Quantum Mechanics book. The equation is defined as (k^2 + ∇^2) G(𝑟) = δ³(𝑟). Participants explore the independence of the Green's function G(𝑟) from the direction of 𝑟, concluding that G(𝑟) can be expressed as G(𝑟) = G(r). The significance of this result lies in its simplicity, which arises from the space-invariance of the potential.

PREREQUISITES
  • Understanding of the Schrödinger equation in quantum mechanics
  • Familiarity with Green's functions and their applications
  • Knowledge of Fourier transforms and residue integrals
  • Concept of space-invariance in physical systems
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  • Study the derivation of Green's functions in quantum mechanics
  • Learn about the application of Fourier transforms in solving differential equations
  • Research the concept of space-invariance in potential fields
  • Explore advanced techniques in residue calculus for complex integrals
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Students and professionals in quantum mechanics, physicists working with differential equations, and researchers interested in mathematical physics and integral equations.

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Griffiths develops an intelgral equation for Scrödinger equation in his QM book. As doing so, he requires Green's function for Helmholtz equation

[tex] (k^2 + \nabla^2) G( \mathbf r) = \delta^3(\mathbf r)[/tex]

A rigourious series of steps, including Fourier transforms and residue integrals follow immidiately. As it turns out, the Green's function is independent of direction of [itex]\mathbf r[/itex], i.e. [itex]G(\mathbf r) = G(r)[/itex]. Does anyone know a trick deducing this condition without actually finding the Green's function. The importance to prove is that, it turns out that there's a relatively simple solution if we assume [itex]G(\mathbf r) = G(r)[/itex])
 
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This simpler form of the GF is caused by a space-invariance of the potential.
 

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