Mathematical Methods for Physics and Engineering : A Comprehensive Guide

19.2 PHYSICAL EXAMPLES OF OPERATORS

Consider firstL^2 |ψ′〉, recalling thatL^2 commutes with bothLxandLyand hence

withU:

L^2 |ψ′〉=L^2 U|ψ〉=UL^2 |ψ〉=Ua|ψ〉=aU|ψ〉=a|ψ′〉.

Thus,|ψ′〉is also an eigenstate ofL^2 , corresponding to the same eigenvalue as

|ψ〉. Now consider the action ofLz:

Lz|ψ′〉=LzU|ψ〉

=(ULz+
U)|ψ〉,using[Lz,U]=
U,

=Ub|ψ〉+
U|ψ〉

=(b+
)U|ψ〉

=(b+
)|ψ′〉.

Thus,|ψ′〉is also an eigenstate ofLz, but with eigenvalueb+
.

In summary, the effect ofUacting upon|ψ〉is to produce a new state that

has the same eigenvalue forL^2 and is still an eigenstate ofLz, though with that

eigenvalue increased by
. An exactly analogous calculation shows that the effect

ofDacting upon|ψ〉is to produce another new state, one that also has the same

eigenvalue forL^2 and is also still an eigenstate ofLz, though with the eigenvalue

decreased by
in this case. For these reasons,UandDare usually known as

ladderoperators.

It is clear that, by starting from any arbitrary eigenstate and repeatedly applying

eitherUorD, we could generate a series of eigenstates, all of which have the

eigenvalueaforL^2 , but increment in theirLzeigenvalues by±
. However, we

also have the physical requirement that, for real values of thez-component, its

square cannot exceed the square of the total angular momentum, i.e.b^2 ≤a. Thus

bhas a maximum valuecthat satisfies

c^2 ≤a but (c+
)^2 >a;

let the corresponding eigenstate be|ψu〉withLz|ψu〉=c|ψu〉. Now it is still true

that

LzU|ψu〉=(c+
)U|ψu〉,

and, to make this compatible with the physical constraint, we must have that

U|ψu〉is the zero ket vector|∅〉. Now, using result (19.31), we have

DU|ψu〉=(L^2 −L^2 z−
Lz)|ψu〉,

⇒ 0 |∅〉=D|∅〉=(a^2 −c^2 −
c)|ψu〉,

⇒ a=c(c+
).

This gives the relationship betweenaandc. We now establish the possible forms

forc.

If we start with eigenstate|ψu〉, which has the highest eigenvaluecforLz,and