Mathematical Methods for Physics and Engineering - Matematica.NET

19.2 PHYSICAL EXAMPLES OF OPERATORSConsider first L 2 |ψ ′ 〉, recalling that L 2 commutes with both L x and L y and hencewith U:L 2 |ψ ′ 〉 = L 2 U|ψ〉 = UL 2 |ψ〉 = Ua|ψ〉 = aU|ψ〉 = a|ψ ′ 〉.Thus, |ψ ′ 〉 is also an eigenstate of L 2 , corresponding to the same eigenvalue as|ψ〉. Now consider the action of L z :L z |ψ ′ 〉 = L z U|ψ〉=(UL z + U)|ψ〉, using[ L z ,U] = U,= Ub|ψ〉 + U|ψ〉=(b + )U|ψ〉=(b + )|ψ ′ 〉.Thus, |ψ ′ 〉 is also an eigenstate of L z , but with eigenvalue b + .In summary, the effect of U acting upon |ψ〉 is to produce a new state thathas the same eigenvalue for L 2 and is still an eigenstate of L z , though with thateigenvalue increased by . An exactly analogous calculation shows that the effectof D acting upon |ψ〉 is to produce another new state, one that also has the sameeigenvalue for L 2 and is also still an eigenstate of L z , though with the eigenvaluedecreased by in this case. For these reasons, U and D are usually known asladder operators.It is clear that, by starting from any arbitrary eigenstate and repeatedly applyingeither U or D, we could generate a series of eigenstates, all of which have theeigenvalue a for L 2 , but increment in their L z eigenvalues by ±. However, wealso have the physical requirement that, for real values of the z-component, itssquare cannot exceed the square of the total angular momentum, i.e. b 2 ≤ a. Thusb has a maximum value c that satisfiesc 2 ≤ a but (c + ) 2 >a;let the corresponding eigenstate be |ψ u 〉 with L z |ψ u 〉 = c|ψ u 〉. Now it is still truethatL z U|ψ u 〉 =(c + )U|ψ u 〉,and, to make this compatible with the physical constraint, we must have thatU|ψ u 〉 is the zero ket vector |∅〉. Now, using result (19.31), we haveDU|ψ u 〉 =(L 2 − L 2 z − L z )|ψ u 〉,⇒ 0|∅〉= D|∅〉=(a 2 − c 2 − c)|ψ u 〉,⇒ a = c(c + ).This gives the relationship between a and c. We now establish the possible formsfor c.If we start with eigenstate |ψ u 〉, which has the highest eigenvalue c for L z ,and661