3dOrbitalAngMomn

3dOrbitalAngMomn - Orbital Angular Momentum in Three...

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Orbital Angular Momentum in Three Dimensions Michael Fowler 11/05/07 The Angular Momentum Operators in Spherical Polar Coordinates The angular momentum operator Lrp i r =× = − × GG G = . In spherical polar coordinates, 22 2 2 2 2 sin cos sin sin cos sin xr yr zr ds dr r d r d 2 θ φ θφ = = = =+ + the gradient operator is 11 ˆˆ ˆ sin r rr r ∂∂ ∇= + + G where now the little hats denote unit vectors : is radially outwards, ˆ r ˆ points along a line of longitude away from the north pole (and therefore in the direction of increasing ) and ˆ points along a line of latitude in an anticlockwise direction as seen looking down on the north pole (that is, in the direction of increasing ). ˆ r G : perpendicular to surface ˆ ϕ G : along line of latitude ˆ G : along line of longitude Top View: Pole ( z -axis) ˆ G : along line of longitude ˆ G : along line of latitude The three unit vectors in the spherical polar system
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2 Here ˆˆ ˆ ,, r θφ form an orthonormal local basis, and ˆ , rr ˆ θ φφ × = , as should be clear from the diagram. So 1 sin r φθ ×∇= G G . (Explicitly, and () ˆ sin , cos , 0 φ =− ( ) ˆ cos cos , cos sin , sin .) The vector ˆ has zero component in the z -direction, the vector ˆ has component sin in the z -direction, so we can immediately conclude that z z z Lr p i r i = × ×∇ =− G G GG == just as in the two-dimensional case. The operator 22 11 . sin sin L ⎛⎞ ∂∂ ⎜⎟ ⎝⎠ = To evaluate this expression, we use ˆˆˆ ˆ 1, 0 φθφ = =⋅ = but we must also check the effects of the differential operators in the first expression on the variables in the second, including the unit vectors . From the explicit coordinate expressions for the unit vectors, or by staring at the diagram, you should be able to establish the following: ˆ /0 ,/ ˆ θθ ∂= ∂ ∂ is in the r -direction, ˆ / is a horizontal unit vector pointing inwards perpendicular to ˆ , and having component cos in the ˆ -direction, /c o s φφ θ ∂∂= . Therefore, the only “differentiation of a unit vector” term that contributes to L 2 is ˆ 1 ˆ cot sin ⋅= . The ˆ acting on the sin in 1 ˆ sin contributes nothing because . 0 Hence
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3 22 2 2 1 cot sin 11 sin sin sin L θ θθ φ 2 θθθ ⎛⎞ ∂∂∂ =− + + ⎜⎟ ⎝⎠ ∂∂ + = = Now, we know that L 2 and L z have a common set of eigenkets (since they commute) and we’ve already established that those of L z are ( ) /2 im m e π Φ= , with m an integer, so the eigenkets of L 2 must have this same dependence, so they must be of the form ( )( ) m l ΘΦ , where () m l Θ is a (suitably normalized) solution of the equation ( ) () ( ) () 2 2 1 sin 1 sin sin m l mm ll d dm dd Θ −Θ = + Θ more conveniently written ( ) sin sin 1 sin 0. m l m l d d m Θ + +− Θ = To summarize : the solutions to this differential equation, with integer ,, , lm m l will (together with m Φ ) give the complete set of eigenstates of L 2 , L z in the coordinate representation.
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3dOrbitalAngMomn - Orbital Angular Momentum in Three...

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