Physical Chemistry Third Edition

16.3 The Operator Corresponding to a Given Variable 691

EXAMPLE16.5

For motion in thex–yplane, transform the expression forL̂zto plane polar coordinates.

Solution

Iffis a differentiable function ofxandyand is also expressible as a function ofρandφ,

then

x

∂f

∂y

x

∂f

∂φ

∂φ

∂y

+x

∂f

∂ρ

∂ρ

∂y

x

∂f

∂φ

x

x^2 +y^2

+x

∂f

∂ρ

y

(x^2 +y^2 )^1 /^2

y

∂f

∂x

y

∂f

∂φ

∂φ

∂x

+y

∂f

∂ρ

∂ρ

∂x

y

∂f

∂φ

−y

x^2 +y^2

−y

∂f

∂ρ

x

(x^2 +y^2 )^1 /^2

Since the second terms cancel when these expressions are added,

L̂zfh ̄

i

(

∂f

∂φ

x^2

x^2 +y^2

−

∂f

∂φ

−y^2

x^2 +y^2

)



h ̄

i

∂f

∂φ

(

x^2 +y^2

x^2 +y^2

)



h ̄

i

∂f

∂φ

so that

L̂z ̄h

i

∂

∂φ

(16.3-20)

When we discuss the hydrogen atom in a later chapter, we will use spherical polar

coordinates:

r

√

x^2 +y^2 +z^2 (16.3-21)

θarccos(z/r) (16.3-22)

φarctan(y/x) (16.3-23)

The expression for the operator̂Lzin Eq. (16.3-20) also holds for spherical polar

coordinates. The operatorsL̂x andL̂y are more complicated in spherical polar

coordinates than iŝLz, and we will try to avoid using them. We present the expression

forL̂

2

in spherical polar coordinates without derivation:

L̂^2 −h ̄^2

[

1

sin(θ)

∂

∂θ

sin(θ)

∂

∂θ

+

1

sin^2 (θ)

∂^2

∂φ^2

]

(16.3-24)

Equation (16.3-24) can also be written in the form

L̂^2 −h ̄^2

[

∂^2

∂θ^2

+cot(θ)

∂

∂θ

+

1

sin^2 (θ)

∂^2

∂φ^2

]

(16.3-25)

Exercise 16.5

Show that Eq. (16.3-24) and Eq. (16.3-25) are equivalent.