130_notes.dvi

We expect to need to keep the radial derivatives so lets identify those by dotting~rinto~p. This will
also make the units matchL^2.

(~r·~p)^2 = − ̄h^2

(

x

∂

∂x

+y

∂

∂y

+z

∂

∂z

) 2

= − ̄h^2

(

x^2

∂^2

∂x^2

+y^2

∂^2

∂y^2

+z^2

∂^2

∂z^2

+ 2xy

∂^2

∂x∂y

+ 2yz

∂^2

∂y∂z

+ 2xz

∂^2

∂x∂z

+x

∂

∂x

+y

∂

∂y

+z

∂

∂z

)

By adding these two expressions, things simplify a lot.

L^2 + (~r·~p)^2 =r^2 p^2 +i ̄h~r·~p

We can now solve forp^2 and we have something we can use in the Schr ̈odinger equation.

p^2 =

1

r^2

(

L^2 + (~r·~p)^2 −i ̄h~r·~p

)

=

1

r^2

L^2 −

̄h^2

r^2

(

r

∂

∂r

) 2

− ̄h^2

1

r

∂

∂r

The Schr ̈odinger equation now can be written with only radial derivatives andL^2.

− ̄h^2

2 μ

[

1

r^2

(

r

∂

∂r

) 2

+

1

r

∂

∂r

−

L^2

̄h^2 r^2

]

uE(~r) +V(r)uE(~r) =EuE(~r)

14.4.4 Spherical Coordinates and the Angular Momentum Operators.

The transformation from spherical coordinates to Cartesian coordinate is.

x = rsinθcosφ

y = rsinθsinφ

z = rcosθ

The transformation from Cartesian coordinates to spherical coordinates is.

r =

√

x^2 +y^2 +z^2

cosθ =

z

√

x^2 +y^2 +z^2

tanφ =

y

x