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# class42 - Today’s class Schrödinger in 3D •...

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Unformatted text preview: Today’s class: Schrödinger in 3D • Schrödinger in 3D b Hydrogen atom Chapter 8. The 3D Schrodinger Equation In 1D: ) ( ) ( ) ( ) ( 2 2 2 2 x E x x V x x m ψ ψ ψ = + ∂ ∂ − h In 2D: ) , ( ) , ( ) , ( ) , ( 2 2 2 2 2 y x E y x y x V y x ψ ψ ψ = + ∂ + ∂ − h ) , , ( ) , , ( ) , , ( ) , , ( 2 2 2 2 2 2 2 2 z y x E z y x z y x V z y x z y x m ψ ψ ψ = + ∂ ∂ + ∂ ∂ + ∂ ∂ − h In 3D: 2 2 2 y x m ∂ ∂ ) , , ( ) , , ( ) , , ( ) , , ( 2 2 2 2 2 2 2 2 z y x E z y x z y x V z y x z y x m ψ ψ ψ = + ∂ ∂ + ∂ ∂ + ∂ ∂ − h Simplest case: 3D box, infinite wall strength V(x,y,z) = 0 inside, = infinite outside. 3D example: “Particle in a rigid box” c Use mathematics of separation of variables (does not always work, but it works here): Assume we could write the solution as: ψ (x,y,z) = X(x)Y(y)Z(z) Plug it in the Schrödinger eqn. and see what happens! "separated function" a b c ψ (x,y,z) = X(x)Y(y)Z(z) Now, calculate the derivatives for each coordinate: ) ( ) ( ) ( ' ' ) ( ) ( ) ( X ) , , ( 2 2 2 2 z Z y Y x X z Z y Y x x z y x x = ∂ ∂ = ∂ ∂ ψ ) , , ( ) , , ( ) , , ( ) , , ( 2 2 2 2 2 2 2 2 z y x E z y x z y x V z y x z y x m ψ ψ ψ = + ∂ ∂ + ∂ ∂ + ∂ ∂ − h (Do the same for y and z parts) Now put in 3D Schrödinger and see what happens: 2 z y x m ∂ ∂ ∂ ( ) EXYZ VXYZ m = + + + − XYZ" Z XY" YZ X" 2 2 h Divide both sides by XYZ E V Z Y m = + + + − Z" Y" X X" 2 2 h (For simplicity I wrote X instead of X(x) and X" instead of ) 2 2 ) ( X x x ∂ ∂ E V Z Y m = + + + − Z" Y" X X" 2 2 h So we re-wrote the Schrödinger equation as: For the particle in the box we said that V=0 inside and V=∞ outside the box. Therefore, we can write with: ψ (x,y,z) = X(x)Y(y)Z(z) outside the box. Therefore, we can write: 2 2 Z" Y" X X" h mE Z Y − = + + for the particle inside the box....
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class42 - Today’s class Schrödinger in 3D •...

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