130_notes.dvi

Now we Taylor expand this equation.

Hψ(x,y,z) +Hdθ

(

∂ψ

∂x

y−

∂ψ

∂y

x

)

=Eψ(x,y,z) +Edθ

(

∂ψ

∂x

y−

∂ψ

∂y

x

)

Subtract off the original equation.

H

(

∂

∂x

y−

∂

∂y

x

)

ψ=

(

∂

∂x

y−

∂

∂y

x

)

Hψ

We find an operator that commutes with the Hamiltonian.

[

H,

̄h

i

(

x

∂

∂y

−y

∂

∂x

)]

= 0

Note that we have inserted the constant ̄hi in anticipation of identifying this operator as thez
component of angular momentum.
~L=~r×~p

Lz=

̄h

i

(

x

∂

∂y

−y

∂

∂x

)

=xpy−ypx

We could have done infinitesimal rotations about thexoryaxes and shown that all the components
of the angular momentum operator commute with the Hamiltonian.

[H,Lz] = [H,Lx] = [H,Ly] = 0

Remember that operators that commute with the Hamiltonian imply physical quantities that are
conserved.

14.4.2 The Commutators of the Angular Momentum Operators

[Lx,Ly] 6 = 0,

however, the square of the angular momentum vector commutes with all the components.

[L^2 ,Lz] = 0

This will give us the operators we need to label states in 3D central potentials.

Lets just compute the commutator.

[Lx,Ly] = [ypz−zpy,zpx−xpz] =y[pz,z]px+ [z,pz]pyx

=

̄h

i

[ypx−xpy] =i ̄hLz

Since there is no difference betweenx,yandz, we can generalize this to

[Li,Lj] =i ̄hǫijkLk