whereǫijkis the completely antisymmetric tensor and we assume a sum over repeated indices.
ǫijk=−ǫjik=−ǫikj=−ǫkjiThe tensor is equal to 1 for cyclic permutations of 123, equal to -1for anti-cyclic permutations, and
equal to zero if any index is repeated. It is commonly used for a cross product. For example, if
~L=~r×~pthen
Li=rjpkǫijkwhere we again assume a sum over repeated indices.
Now lets compute commutators of theL^2 operator.
L^2 = L^2 x+L^2 y+L^2 z
[Lz,L^2 ] = [Lz,L^2 x] + [Lz,L^2 y]
= [Lz,Lx]Lx+Lx[Lz,Lx] + [Lz,Ly]Ly+Ly[Lz,Ly]
= i ̄h(LyLx+LxLy−LxLy−LyLx) = 0We can generalize this to
[Li,L^2 ] = 0.L^2 commutes with every component of~L.
14.4.3 Rewritingp
2
2 μUsingL2We wish to use theL^2 andLzoperators to help us solve the central potential problem. If we can
rewriteHin terms of these operators, and remove all the angular derivatives, problems will be greatly
simplified. We will work in Cartesian coordinates for a while, where we know the commutators.
First, write outL^2.
L^2 = (~r×~p)^2= − ̄h^2[(
y∂
∂z−z∂
∂y) 2
+
(
z∂
∂x−x∂
∂z) 2
+
(
x∂
∂y−y∂
∂x) 2 ]
Group the terms.
L^2 = − ̄h^2[
x^2(
∂^2
∂z^2+
∂^2
∂y^2)
+y^2(
∂^2
∂x^2+
∂^2
∂z^2)
+z^2(
∂^2
∂x^2+
∂^2
∂y^2)
−
(
2 xy∂^2
∂x∂y+ 2yz∂^2
∂y∂z+ 2xz∂^2
∂x∂z+ 2x∂
∂x+ 2y∂
∂y+ 2z∂
∂z