Physical Chemistry , 1st ed.

Solution

The complete spherical harmonic 3, 3 is given by

3, 3  e^3 isin^3 

Let us consider the total angular momentum first, since we can use the re-

sults of those manipulations to get the total energy. The total angular mo-

mentum can be determined from

Lˆ^2 ^2 







2

 2 cot 







sin

1

2 





2

 (^2) 
Taking derivatives first with respect to , one finds









8

7

2

0



e^3 isin^3 

8

3

2

7



^0 e^3 isin^2 cos 









2

 (^2)  8

7

2

0



e^3 isin^3 

8

3

2

7



0

e^3 i(2 cos^2 sin sin^3 )

The derivative with respect to is simply









2

 (^2)  8

7

2

0



e^3 isin^3  32 i^2 

8

7

2

0



e^3 isin^3 

 9 

8

7

2

0



e^3 isin^3 

Putting all this together, one gets

ˆL^2 ^2 

8

3

2

7



0

e^3 i(2 cos^2 sin sin^3 )

cot 

8

3

2

7



^0 e^3 isin^2 cos 

sin

1

2 ^9  8

7

2

0



e^3 i

This can be simplified by factoring out the wavefunction constants and the

exponential from all of the terms to get

ˆL^2 ^2 3(2 cos^2 sin sin^3 )

3 cot sin^2 cos  9 

sin

1

 (^2) sin
(^3) 
 8

7

2

0



e^3 i

Simplifying all the terms and remembering the definition of the cotangent:

Lˆ^2 ^2 (6 cos^2 sin 3 sin^3 

3 sin cos^2 9 sin )

8

7

2

0



e^3 i

^2 (9 cos^2 sin 3 sin^3 9 sin )

8

7

2

0



e^3 i

Substituting for the trigonometric identity cos^2  1 sin^2 :

ˆL^2 ^2 [9(1 sin^2 ) sin 3 sin^3 9 sin ]

8

7

2

0



e^3 i

^2 (9 sin 9 sin^3 3 sin^3 9 sin )

8

7

2

0



e^3 i

^70

82 

350 CHAPTER 11 Quantum Mechanics: Model Systems and the Hydrogen Atom