QUANTUM OPERATORS
with
[
L^2 ,Lz]
=[
L^2 x+L^2 y+L^2 z,Lz]=Lx[Lx,Lz]+[Lx,Lz]Lx+Ly[
Ly,Lz]
+[
Ly,Lz]
Ly+[
L^2 z,Lz]=Lx(−i)Ly+(−i)LyLx+Ly(i)Lx+(i)LxLy+0=0.Thus operatorsL^2 andLzcommute and, continuing in the same way, it can be
shown that
[
L^2 ,Lx]
=[
L^2 ,Ly]
=[
L^2 ,Lz]
=0. (19.29)Eigenvalues of the angular momentum operatorsWe will now use the commutation relations forL^2 and its components to find
the eigenvalues ofL^2 andLz, without reference to any specific wavefunction. In
other words, the eigenvalues of the operators follow from the structure of their
commutators. There is nothing particular aboutLz,andLxorLycould equally
well have been chosen, though, in general, it is not possible to find states that are
simultaneously eigenstates of two or more ofLx,LyandLz.
To help with the calculation, it is convenient to define the two operatorsU≡Lx+iLy and D≡Lx−iLy.These operators are not Hermitian; they are in fact Hermitian conjugates, in that
U†=DandD†=U, but they do not represent measurable physical quantities.
We first note their multiplication and commutation properties:
UD=(Lx+iLy)(Lx−iLy)=L^2 x+L^2 y+i[
Ly,Lx]=L^2 −L^2 z+Lz, (19.30)DU=(Lx−iLy)(Lx+iLy)=L^2 x+L^2 y−i[
Ly,Lx]=L^2 −L^2 z−Lz, (19.31)[Lz,U]=[Lz,Lx]+i[
Lz,Ly]
=iLy+Lx=U, (19.32)
[Lz,D]=[Lz,Lx]−i[
Lz,Ly]
=iLy−Lx=−D. (19.33)In the same way as was shown for matrices, it can be demonstrated that if two
operators commute they have a common set of eigenstates. SinceL^2 andLz
commute they possess such a set; let one of the set be|ψ〉with
L^2 |ψ〉=a|ψ〉 and Lz|ψ〉=b|ψ〉.Now consider the state|ψ′〉=U|ψ〉and the actions ofL^2 andLzupon it.