HW11_Solutions - PHYS851 Quantum Mechanics I, Fall 2008...

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Unformatted text preview: PHYS851 Quantum Mechanics I, Fall 2008 HOMEWORK ASSIGNMENT 11 1. [20 pts] In order to derive the properties of the angular momentum eigenstate wavefunctions, we need to determine the action of the angular momentum operator in spherical coordinates. Just as we have ( x | P x | ψ ) = − i planckover2pi1 d dx ( x | ψ ) , we should find a similar expression for ( rθφ | vector L | ψ ) . From vector L = vector R × vector P and our knowledge of momentum operators, it follows that ( rθφ | vector L | ψ ) = − ı planckover2pi1 parenleftbigg vectore x parenleftbigg y d dz − z d dy parenrightbigg + vectore y parenleftbigg z d dx − x d dz parenrightbigg + vectore z parenleftbigg x d dy − y d dx parenrightbiggparenrightbigg ( rθφ | ψ ) . The coordinates are defined via the transformations x = r sin θ cos φ y = r sin θ sin φ z = r cos θ and the inverse transformations r = radicalbig x 2 + y 2 + z 2 θ = arctan( radicalbig x 2 + y 2 z ) φ = arctan( y x ) , while their derivatives can be related via expansions such as d dx = ∂r ∂x ∂ ∂r + ∂θ ∂x ∂ ∂θ + ∂φ ∂x ∂ ∂φ . Using these relations (and similar relations for d/dy and d/dz ) find expressions for ( rθφ | L x | ψ ) , ( rθφ | L y | ψ ) , and ( rθφ | L z | ψ ) , involving only spherical coordinates and their derivatives. This is an alternate approach to the one described in lecture. Answer: ∂ x r = x r = sin θ cos φ ∂ x θ = z 2 r 2 x z √ x 2 + y 2 = cos θ cos φ r ∂ x φ = − x 2 x 2 + y 2 y x 2 = − csc θ sin φ r So d dx = sin θ cos φ∂ r + cos θ cos φ r ∂ θ − csc θ sin φ r ∂ φ ∂ y r = y r = sin θ sin φ ∂ y θ = z 2 r 2 y z √ x 2 + y 2 = cos θ sin φ r 1 ∂ y φ = x 2 x 2 + y 2 1 x = csc θ cos φ r So d dy = sin θ sin φ∂ r + cos θ sin φ r ∂ θ + csc θ cos φ r ∂ φ ∂ z r = z r = cos θ ∂ z θ = − z 2 r 2 √ x 2 + y 2 z 2 = − sin θ r ∂ z φ = 0 So d dz = cos θ∂ r − sin θ r ∂ θ ( rθφ | L x | ψ ) = − i planckover2pi1 ( y d dz − z d dy ) ( rθφ | ψ ) So we can say L x = i planckover2pi1 parenleftBig y d dz − z d dy parenrightBig L x = − i planckover2pi1 ( r sin θ cos θ sin φ∂ r − sin 2 θ sin φ∂ θ − r sin θ cos θ sin φ∂ r − cos 2 θ sin φ∂ θ + cot θ cos φ∂ φ ) Which means ( rθφ | L x | ψ ) = − i planckover2pi1 ( − sin φ∂ θ − cot θ cos φ∂ φ ) ( rθφ | ψ ) Similarly we can say L y = − i planckover2pi1 ( z d dx − x d dz ) L y = − i planckover2pi1 ( r sin θ cos θ cos φ∂ r + cos 2 θ cos φ∂ θ − cot θ sin φ∂ φ − r sin θ cos θ cos φ∂ r + sin 2 θ cos φ∂ θ ) ( rθφ | L y | ψ ) = − i planckover2pi1 (cos φ∂ θ − cot θ sin φ∂ φ ) ( rθφ | ψ ) And L z = − i planckover2pi1 ( x d dy − y d dx ) L z = − i planckover2pi1 ( r sin 2 θ sin φ cos φ∂ r + sin θ cos θ...
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This note was uploaded on 11/26/2010 for the course PHYSICS PHYS 851 taught by Professor Michaelmoore during the Spring '08 term at Michigan State University.

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HW11_Solutions - PHYS851 Quantum Mechanics I, Fall 2008...

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