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Unformatted text preview: PHY251 Lecture Notes  Set 8 PHY251 Lecture Notes Set 8 file:///C/Documents/phy_courses/PHY251/PHY251_lecture_08.HTML (1 of 6) [4/3/2001 12:30:05 PM] PHY251 Lecture Notes Set 8 file:///C/Documents/phy_courses/PHY251/PHY251_lecture_08.HTML (2 of 6) [4/3/2001 12:30:05 PM] PHY251 Lecture Notes Set 8 file:///C/Documents/phy_courses/PHY251/PHY251_lecture_08.HTML (3 of 6) [4/3/2001 12:30:05 PM] The Schrdinger equation describes the behavior of a particle as function of position and time. The Schrdinger 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 m ( / x + / x + / x ) + V ( r ) acting on the wave function ( r ), giving the energy E (a constant) times the wavefunction: H op ( r ) = [ h / 2 m ( / x + / x + / x ) + V ( r )] ( r ) = E ( r ) (1) For the Hydrogen atom the potential is simply the Coulomb potential, which depends only on distance: V = V ( r ) =  e /(4 r ) = hc / r , where e /(4 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 , , and . That implies that we need to reexpress 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 ( r ) = [ h / 2 m { / r + 2 / r r + 1 / r ( / + cot / + / sin )} + V ( r )] ( r , , ) = E ( r , , ) (2) This secondorder 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: ( r , , ) = R ( r ) T ( ) F ( ) (3) The equation (2) can be solved; it is technically complicated but not impossible. One finds:The equation (2) can be solved; it is technically complicated but not impossible....
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 Spring '01
 Rijssenbeek
 Physics

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