This preview shows pages 1–7. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Class Notes for Modern Physics, Part 3 J. Gunion U.C. Davis 9D, Spring Quarter J. Gunion Quantum Mechanics in 3 D There are a lot of issues regarding coordinate systems and spatial symmetries of the potential or physical configuration that enter into the threedimensional situation. In the 3 D case, = ( ~ r,t ) = ( x,y,z,t ) , P ( ~ r,t ) =  ( ~ r,t )  2 (1) where P ( ~ r,t ) dV = the probability to find the particle in a volume element of size dV about ~ r at time t . What is the 3 D version of the SE? Start from E = p 2 x + p 2 y + p 2 z 2 m + U ( x,y,z ) . (2) Then multiply by and use operator substitutions of b E = i h t , b p x = h i x b p y = h i y , b p z = h i z . (3) J. Gunion 9D, Spring Quarter 1 The resulting SE is i h t = h 2 2 m ~ 2 + U (4) where ~ 2 = 2 x 2 + 2 y 2 + 2 z 2 . For the situation considered, where U has no explicit time dependence, we again use a separation of variables approach by writing ( ~ r,t ) = ( ~ r ) ( t ) = ( ~ r ) e it , (5) where i h t ( ~ r,t ) = h ( ~ r,t ) = E ( ~ r,t ) and consistency with the SE requires the timeindependent SE: h 2 2 m ~ 2 ( ~ r ) + U ( ~ r ) ( ~ r ) = E ( ~ r ) . (6) Cubical symmetry case The simplest case is that of a cube with infinite potential at the sides of the cube. This is analogous to our infinite well in one dimension. The picture is below. J. Gunion 9D, Spring Quarter 2 Inside the box where U = 0 , the timeindependent SE reduces to h 2 2 m ~ 2 ( ~ r ) = E ( ~ r ) , (7) J. Gunion 9D, Spring Quarter 3 while outside the box U = and = 0 . Given the symmetry of the situation, it is appropriate to further separate variables by writing ( ~ r ) = 1 ( x ) 2 ( y ) 3 ( z ) . (8) If we substitute this form into the 3 D SE, we get ( d is ok) h 2 2 m d 2 1 ( x ) dx 2 2 3 + 1 d 2 2 ( y ) dy 2 3 + 1 2 d 2 3 ( z ) dz 2 = E 1 2 3 . (9) We may then divide both sides of this equation by 1 2 3 to obtain: h 2 2 m d 2 1 ( x ) dx 2 1 + d 2 2 ( y ) dy 2 2 + d 2 3 ( z ) dz 2 3 = E . (10) At this point, we note that each of the terms depends only on one coordinate x , y or z . The only way the above equation can hold for all x,y,z is if h 2 2 m d 2 1 dx 2 1 ( x ) = E 1 ... (11) J. Gunion 9D, Spring Quarter 4 where the E 1 , E 2 , E 3 are all constants and E 1 + E 2 + E 3 = E . Each of these individual equations is then solved just like in the 1 D case. Using a coordinate set up where the cubical box runs from to L in each of the three spatial coordinates, it should be clear from the one dimensional case that vanishing of 1 at x = 0 and x = L , and so forth, implies 1 ( x ) sin k 1 x, with k 1 = n 1 L 2 ( y ) sin k 2 y , with k 2 = n 2 L 3 ( z ) sin k 3 z , with k 3 =...
View
Full
Document
This note was uploaded on 04/11/2009 for the course ENG 100 taught by Professor Delwiche during the Spring '08 term at UC Davis.
 Spring '08
 Delwiche
 Spherical Harmonics, Ode

Click to edit the document details