Solving Ladderoperator Problem for c_+ Expressed in j and m

[danja347]

I need help figuring out the expression for the constant c_+
expressed in j and m in the following equation:

\hat J_+|Y_{jm}>=c_+|Y_{jm+1}>

Y is just spherical harmonics and \hat J_+=\hat J_x + i\hat J_y is a ladderoperator.

/Daniel

 

[dextercioby]

I need help figuring out the expression for the constant c_+
expressed in j and m in the following equation:

\hat J_+|Y_{jm}>=c_+|Y_{jm+1}>

Y is just spherical harmonics and \hat J_+=\hat J_x + i\hat J_y is a ladderoperator.

/Daniel

1.Have u tried to look it into your QM book?It's something pretty "classical".Try Cohen-Tanoudji.
2.I would have given u alink,but the server at univ texas at austin is dead.Anyway...I would have actually wanted to upload that chapter from the course,but the server wouldn't accept anything more than 50KB.

Good Luck!

 

[danja347]

Thanks... its all clear now! :-/

 

[RedX]

trying to recall...oh yeah:

J+|jm>=C|j(m+1)>

<jm|adjoint(J+)=<j(m+1)|C*

J+=Jx+iJy
adjoint(J+)=Jx-iJy=J- (since J is Hermitian)

<jm|adjoint(J+)=<jm|J-

So taking the inner product:

<jm|J-J+|jm> = CC*<j(m+1)|j(m+1)> = CC*

J-J+=(Jx-iJy)(Jx+iJy)=JxJx+JyJy+i[Jx,Jy]=J^2 - (Jz)^2 -hJz

<jm|J-J+|jm>=<jm|J^2 - (Jz)^2 -hJz}jm>=j(j+1)h^2 -m^2 h^2 - h^2 m = CC*

So taking the square root:

C=h sqrt(j(j+1) -m^2 -m)