QUANTUM OPERATORS
RHS gives
(−i)^2
2 m∂
∂x∂
∂x+(−i)^2
2 m∂
∂y∂
∂y+(−i)^2
2 m∂
∂z∂
∂z.The potential energyV, being a function of position only, becomes a purely
multiplicative operator, thus creating the full expression for the Hamiltonian,
H=−^2
2 m(
∂^2
∂x^2+∂^2
∂y^2+∂^2
∂z^2)
+V(x, y, z),and giving the corresponding Schr ̈odinger equation as
Hψn=−^2
2 m(
∂^2 ψn
∂x^2+∂^2 ψn
∂y^2+∂^2 ψn
∂z^2)
+V(x, y, z)ψn=Enψn.We are not so much concerned in this section with solving such differential
equations, but with the commutation properties of the operators from which they
are constructed. To this end, we now turn our attention to the topic of angular
momentum, the operators for which can be constructed in a straightforward
manner from the two basic sets.
19.2.1 Angular momentum operatorsAs required by the substitution rules, we start by expressing angular momentum
in terms of the classical quantitiesrandp, namelyL=r×pwith Cartesian
components
Lz=xpy−ypx,Lx=ypz−zpy,Ly=zpx−xpz.Making the substitutions (19.22) yields as the corresponding quantum-mechanical
operators
Lz=−i(
x∂
∂y−y∂
∂x)
,Lx=−i(
y∂
∂z−z∂
∂y)
, (19.25)Ly=−i(
z∂
∂x−x∂
∂z)
.It should be noted that forxpy, say,xand∂/∂ycommute, and there is no
ambiguity about the way it is to be carried into its quantum form. Further, since
the operators corresponding to each of its factors commute and are Hermitian,
the operator corresponding to the product is Hermitian. This was shown directly
for matrices in exercise 8.7, and can be verified using equation (17.16).
The first question that arises is whether or not these three operators commute.