Lx =1
2
(L++L−) =
̄h
√
2
0 1 0
1 0 1
0 1 0
Ly =1
2 i(L+−L−) =
̄h
√
2 i
0 1 0
−1 0 1
0 −1 0
What is the dimension of the matrices forℓ= 2?
Dimension 5. Derive the matrix operators forℓ= 2.
Just do it.
18.11.2Compute [Lx,Ly] Using Matrices*.
[Lx,Ly] =̄h^2
2 i
0 1 0
1 0 1
0 1 0
0 1 0
−1 0 1
0 −1 0
−
0 1 0
−1 0 1
0 −1 0
0 1 0
1 0 1
0 1 0
̄h^2
2 i
−1 0 1
0 0 0
−1 0 1
−
1 0 1
0 0 0
−1 0 − 1
= ̄h2
2 i
−2 0 0
0 0 0
0 0 2
=i ̄h ̄h
1 0 0
0 0 0
0 0 − 1
=i ̄hLzThe other relations will prove to be correct too, as they must. Itsa reassuring check and a calcula-
tional example.
18.11.3Derive the Expression for Rotation OperatorRz*
The laws of physics do not depend on what axes we choose for our coordinate system- There is
rotational symmetry. If we make an infinitesimal rotation (through and angledφ) about the z-axis,
we get the transformed coordinates
x′ = x−dφy
y′ = y+dφx.We can Taylor expand any functionf,
f(x′,y′) =f(x,y)−
∂f
∂xdφy+
∂f
∂ydφx= (1 +
i
̄hdφLz)f(x,y).So the rotation operator for the function is
Rz(dφ) = (1 +i
̄h
dφLz)A finite rotation can be made by applying the operator for an infinitesimal rotation over and over.
Letθz=ndφ. Then
Rz(θz) = lim
n→∞(1 +
i
̄hθz
n
Lz)n=eiθzLz/ ̄h.The last step, converting the limit to an exponential is a known identity. We can verify it by using
the log of the quantity. First we expandln(x) aboutx= 1:ln(x) =ln(1)+
( 1
x)
x=1(x−1) = (x−1).lim
n→∞
ln(1 +
i
̄hθz
nLz)n=n(
i
̄hθz
nLz) =
i
̄hθzLzSo exponentiating, we get the identity.