Lecture 08

# Modern Physics

• Notes
• davidvictor
• 6

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PHY251 Lecture Notes - Set 8 PHY251 Lecture Notes Set 8 (1 of 6) [4/3/2001 12:30:05 PM]

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PHY251 Lecture Notes Set 8 (2 of 6) [4/3/2001 12:30:05 PM]
PHY251 Lecture Notes Set 8 (3 of 6) [4/3/2001 12:30:05 PM]

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The Schrödinger equation describes the behavior of a particle as function of position and time. The Schrödinger equation with a static potential V = V ( r ) is an example of an eigenvalue problem, with the Operator (called the Hamiltonian or energy operator in this case) - h ² / 2 m ( / x ² + / x ² + / x ² ) + V ( r ) acting on the wave function ψ ( r ), giving the energy E (a constant) times the wavefunction: H op ψ ( r ) = [- h ² / 2 m ( ² / x ² + ² / x ² + ² / x ² ) + V ( r )] ψ ( r ) = E ψ ( r ) (1) For the Hydrogen atom the potential is simply the Coulomb potential, which depends only on distance: V = V ( r ) = - e ²/(4 πε 0 r ) = - α hc / r , where α e ²/(4 πε 0 hc) . Because the potential in equation (1) depends only on the distance r to the atom's nucleus, and not on any direction, we profit from rewriting equation (1) in terms of spherical coordinates r , θ , and φ . That implies that we need to re-express the derivatives in terms of derivatives with respect to these new variables as well, a tedious process as we have seen before! The ultimate result is given below: H op ψ ( r ) = [- h ² / 2 m { ² / r ² + 2 / r r + 1 / r ² ( ² / θ ² + cot θ / θ + ² / sin² θ φ ² )} + V ( r )] ψ ( r , θ , φ ) = E ψ ( r , θ , φ ) (2) This second-order differential equation can be solved using the fact that the solutions factorize, i.e. the wavefunction is a product of three separate functions that depend each on a different variable: ψ ( r , θ , φ ) = R ( r ) T ( θ ) F ( φ ) (3) The equation (2) can be solved; it is technically complicated but not impossible. One finds: ψ ( r , θ , φ ) = R ( r ) T ( θ ) F ( φ ) = R nl Y l m ( θ , φ ) = R nl P l m
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• Spring '01
• Rijssenbeek
• Physics, Atomic orbital, pn, Spin quantum number, Lecture Notes Set, ms. We

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