19.2 PHYSICAL EXAMPLES OF OPERATORS
Consider first
LxLy=−^2(
y∂
∂z−z∂
∂y)(
z∂
∂x−x∂
∂z)=−^2(
y∂
∂x+yz∂^2
∂z∂x−yx∂^2
∂z^2−z^2∂^2
∂y∂x+zx∂^2
∂y∂z)
.Now consider
LyLx=−^2(
z∂
∂x−x∂
∂z)(
y∂
∂z−z∂
∂y)=−^2(
zy∂^2
∂x∂z−z^2∂^2
∂x∂y−xy∂^2
∂z^2+x∂
∂y+xz∂^2
∂z∂y)
.These two expressions arenotthe same. The difference between them, i.e. the
commutator ofLxandLy, is given by
[
Lx,Ly]
=LxLy−LyLx=^2(
x∂
∂y−y∂
∂x)
=iLz. (19.26)This, and two similar results obtained by permuttingx,yandzcyclically,
summarise the commutation relationships between the quantum operators corre-
sponding to the three Cartesian components of angular momentum:
[
Lx,Ly]
=iLz,
[
Ly,Lz]
=iLx, (19.27)[Lz,Lx]=iLy.As well as its separate components of angular momentum, the total angularmomentum associated with a particular state|ψ〉is a physical quantity of interest.
This is measured by the operator corresponding to the sum of squares of its
components,
L^2 =L^2 x+L^2 y+L^2 z. (19.28)This is an Hermitian operator, as each term in it is the product of two Hermitian
operators that (trivially) commute. It might seem natural to want to ‘take the
square root’ of this operator, but such a process is undefined and we will not
pursue the matter.
We next show that, although no two of its components commute, the totalangular momentum operator does commute with each of its components. In the
proof we use some of the properties (19.17) to (19.20) and result (19.27). We begin