Physical Chemistry , 1st ed.

where and are varied (but ris still constant!), the Hamiltonian opera-

tor is

Hˆ



2 I

2









2

 2 cot 







sin

1

2 





2

 (^2) Vˆ (11.43)
where Iis the moment of inertia and Vˆis the potential energy operator. As
with two-dimensional rotations, the Hamiltonian can be written in terms of
the angular momentum, but now it must be written in terms of the total
angular momentum, not just in terms of the angular momentum in a single
dimension. The 3-D rotational Hamiltonian is thus also written as
Hˆ

Lˆ

2 I

2

Vˆ (11.44)

By inspecting the two previous expressions, you can see that

ˆL^2 ^2 







2

 2 cot 







sin

1

2 





2

 (^2)  (11.45)
The square root of the right side of equation 11.45 cannot be taken analytically.
Therefore, an operator for the total angular momentum is not commonly used
in 3-D quantum mechanical systems. Only an operator for the squareof the
total angular momentum is common. In order to find the angular momentum,
one must determine the value (ultimately an eigenvalue) of the square of the
angular momentum, then take the square root of that observable.
Again, the potential energy Vfor 3-D rotational motion can be set to zero,
so acceptable wavefunctions for rotation in three dimensions must satisfy the
Schrödinger equation, which is




2



I

2









2

 2 cot 







sin

1

2 





2

 (^2) E (11.46)
Even though the mass is moving in all three Cartesian coordinates, in spheri-
cal polar coordinates we only need and to define the motion. The detailed
solution of the above differential equation is long, and will not be presented
here. However, several points can be made before the solution is simply pre-
sented. First, it is assumed that the solution is separable.That is, we assume that
the wavefunctions are products of two functions and , each of which de-
pends only on the variables and , respectively:
(,) () ()
If we consider the variables and in equation 11.46 independently, we see
that only one term in the differential contains , the last term. Ifwere held
constant, then the first two differential terms would be identically zero (deriv-
atives are zero if the variable in question is held constant), and the Schrödinger
equation would have the same form as that for 2-D rotational motion.
Therefore, the first part of the solution contains only the variable and is the
same function derived for the 2-D rotating system:

() 

1

2 

eim

The same restrictions on the possible values ofm, the quantum number, hold

true in this case, also:m0,1,2, and so on.

Unfortunately, a similar analysis cannot be done by holding constant and

varying only in order to find the function (). This is because all three of the

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