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# HW8sol1 - Physics 731 Assignment#8 Solutions 1(a Using the...

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Physics 731 Assignment #8, Solutions 1. (a) Using the result that Π a ( A - a ) = 0 for an operator A and its set of eigenvalues a , we have ( J - 1) J ( J + 1) = 0 , and thus J 3 = J . Another approach is to use spectral decomposition: J 3 = (1) 3 | 11 11 | + ( - 1) 3 | 1 - 1 1 - 1 | = | 11 11 | + | 1 - 1 1 - 1 | = J . Note that the | 10 state does not contribute because of its zero eigenvalue. (b) Expanding D (1) using the result from (a), one obtains D (1) = n =0 ( - ) n J n n ! = 1 + J ( - ) + ( - ) 3 3! + . . . + J 2 ( - ) 2 2! + ( - ) 4 4! + . . . = 1 - i sin φ J + (cos φ - 1) J 2 . (c) Choose ˆ n = ˆ y . Then J = J y ¯ h = i 2 0 - 1 0 1 0 - 1 0 1 0 , J 2 = - 1 2 - 1 - 0 1 0 - 2 0 1 0 - 1 , and d (1) ( β ) = 1 - i sin β J + (cos β - 1) J 2 = 1+cos β 2 - sin β 2 1 - cos β 2 sin β 2 cos β - sin β 2 1 - cos β 2 sin β 2 1+cos β 2 . 2. (a) The wavefunction is ψ ( x ) = ( x + y + 3 z ) f ( r ) = rf ( r )(sin θ cos φ + sin θ sin φ + 3 cos θ ) , and hence ψ ( x ) = 4 π 3 (1 - i ) 2 Y 11 - (1 + i ) 2 Y 1 - 1 + 3 Y 10 . Therefore, ψ ( x ) is an eigenstate of L 2 with eigenvalue l = 1 .

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HW8sol1 - Physics 731 Assignment#8 Solutions 1(a Using the...

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