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9D Chapter 8,9,13-15 Notes

# 9D Chapter 8,9,13-15 Notes - Class Notes for Modern Physics...

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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 three-dimensional situation. In the 3 D case, Ψ = Ψ( ~ r,t ) = Ψ( x,y,z,t ) , P ( ~ r,t ) = | Ψ( ~ 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 Ψ 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 ∂ Ψ ∂t =- ¯ h 2 2 m ~ ∇ 2 Ψ + U Ψ (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 Ψ( ~ r,t ) = ψ ( ~ r ) φ ( t ) = ψ ( ~ r ) e- iωt , (5) where i ¯ h ∂ ∂t Ψ( ~ r,t ) = ¯ hω Ψ( ~ r,t ) = E Ψ( ~ r,t ) and consistency with the SE requires the time-independent SE:- ¯ h 2 2 m ~ ∇ 2 ψ ( ~ r ) + U ( ~ r ) ψ ( ~ r ) = Eψ ( ~ 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 time-independent SE reduces to- ¯ h 2 2 m ~ ∇ 2 ψ ( ~ r ) = Eψ ( ~ r ) , (7) J. Gunion 9D, Spring Quarter 3 while outside the box U = ∞ and ψ = 0 . Given the symmetry of the situation, it is appropriate to further separate variables by writing ψ ( ~ r ) = ψ 1 ( x ) ψ 2 ( y ) ψ 3 ( z ) . (8) If we substitute this form into the 3 D SE, we get ( ∂ → d is ok)- ¯ h 2 2 m d 2 ψ 1 ( x ) dx 2 ψ 2 ψ 3 + ψ 1 d 2 ψ 2 ( y ) dy 2 ψ 3 + ψ 1 ψ 2 d 2 ψ 3 ( z ) dz 2 = Eψ 1 ψ 2 ψ 3 . (9) We may then divide both sides of this equation by ψ 1 ψ 2 ψ 3 to obtain:- ¯ h 2 2 m d 2 ψ 1 ( x ) dx 2 ψ 1 + d 2 ψ 2 ( y ) dy 2 ψ 2 + d 2 ψ 3 ( z ) dz 2 ψ 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 ψ 1 dx 2 ψ 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 ψ 1 at x = 0 and x = L , and so forth, implies ψ 1 ( x ) ∝ sin k 1 x, with k 1 = n 1 π L ψ 2 ( y ) ∝ sin k 2 y , with k 2 = n 2 π L ψ 3 ( z ) ∝ sin k 3 z , with k 3 =...
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9D Chapter 8,9,13-15 Notes - Class Notes for Modern Physics...

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