851HW11_09Solutions

# 851HW11_09Solutions - PHYS851 Quantum Mechanics I Fall 2009...

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PHYS851 Quantum Mechanics I, Fall 2009 HOMEWORK ASSIGNMENT 11 Topics Covered: Orbital angular momentum, center-of-mass coordinates Some Key Concepts: angular degrees of freedom, spherical harmonics 1. [20 pts] In order to derive the properties of the spherical harmonics, 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 ) . Cartesian coordinates are related to spherical coordinates 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 ) . Their derivatives can be related via expansions such as x = ∂r ∂x r + ∂θ ∂x θ + ∂φ ∂x φ . Using these relations, and similar expressions for y and z , find expressions for ( rθφ | L x | ψ ) , ( rθφ | L y | ψ ) , and ( rθφ | L z | ψ ) , involving only spherical coordinates and their derivatives. 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 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 1

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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 Now ( rθφ | L x | ψ ) = i planckover2pi1 ( y d dz z d dy ) ( rθφ | ψ ) So we can say L x = i planckover2pi1 parenleftbigg y d dz z d dy parenrightbigg = 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 ) = 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 ) so that ( rθφ | L y | ψ ) = i planckover2pi1 (cos φ∂ θ cot θ sin φ∂ φ ) ( rθφ | ψ ) Lastly, we have L z = i planckover2pi1 ( x d dy y d dx ) = i planckover2pi1 ( r sin 2 θ sin φ cos
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• Fall '09
• M.Moore

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