The raising and lowering operators change the value of m by one unit:

L±flm=(Alm)flm+1, (4.120).

Where Almare constant. Question: What is Alm, if the Eigen functions are to be normalized? Hint: First show thatL±is the Hermitian conjugate of L±(Since LxandLyare observables, you may assume they are Hermitian…but prove it if you like); then use Equation 4.112.

Short Answer

Expert verified

If the Eigen functions are to be normalized, they are upper and lower signs.

Step by step solution

01

Given.

The raising and lowering operators are:

L+flm=(Alm)flm+1,L-flm=(Blm)flm-1,

02

Eigen functions to be normalized

We start solving the problem by taking the inner product between a hydro genic set acted upon L±and Las following:

Alm=ħl(l+1)-m(m+1)=ħl(l+1)-m(m+1),Blm=ħl(l+1)-m(m+1)=ħl(l+1)-m(m+1),L2=L±L+Lz2±ħLz(4.112).

Note what happens at the top and bottom of the ladder (i.e. when you apply L+L+tofllorL-tofl-lNow,usingEq.4.112,inthefromL±L=L2-Lz2ħLzLL2=L±L+Lz2ħLzL2=L4.112.

flmlLmL±flm=flmlL2-Lz2mhLzflm=flml[h2ll+l-h2m2mh2m]flm=h2ll+1-mm±1flmlflm=h2ll+1-mm±1=L±flmlL±flmUppersigns:ħ2ll+1-mm+1=L+flmlL+flm=Almflm+1lAlmflm+1=Alm2Alm=ħll+1-mm+1Lowersigns:ħ2ll+1-mM-1=L-flmlL-flm=Blmflm-1lBlmflm-1=Blm2

At the top of the ladderm=lwe get All=0wegetAll=0, so there is no higher rung; at the bottom of the ladder m=l we getBl-l=0 , so there is no lower rung.

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