Forbidden Angular Momentum

Assume that the angular momentum is allowed to have $l=1/2$. This means that the following equations must be satisfied simultaneously:

$L_+Y_{1/2,1/2}=0$;   $L_-Y_{1/2,-1/2}=0$;  $L_+Y_{1/2,-1/2}=A_{1/2,-1/2}Y_{1/2,1/2}$
Using representation of ladder operators for orbital angular momentum in the coordinate representation demonstrate that this is not possible. (Hint: Write down first two equations as differential equations and solve them. then using found $Y_{1/2,-1/2}$ compute $L_+Y_{1/2,-1/2}$ and prove that the result is not consistent with found $Y_{1/2,1/2}$  ).

The spherical harmonics functions are

$Y_{l m} (θ.φ)=C_{l m} e^{imφ}P_l^m (\cos⁡ θ)=C_{l m}e^{imφ}*\frac{1}{\sin^m ⁡θ} \frac{d^{l-m}}{d(\cos⁡θ)^{l-m}} (\sin⁡ θ)^{2l}$


$Y_{1/2,1/2}=A e^{iφ/2}*\frac{1}{√(sin⁡θ)}\sin⁡ θ=A e^{iφ/2}* √(sin⁡θ)$

$Y_{1/2,-1/2}=Be^{-iφ/2}*(√\sinθ)*\frac{d}{d(cos⁡ θ)} (\sin⁡ θ)=Be^{-iφ/2}*√(sin⁡θ )(-\cot ⁡θ)$

The ladder operators are

$L_±=±ℏe^{±iφ} [\frac{∂}{∂θ}±i\cot⁡θ\frac{∂}{∂φ}]$

$L_+ Y_{1/2,1/2}=Aℏe^{iφ}*[e^{iφ/2}*\frac{\cos⁡θ}{[2√(sin⁡θ )]}+i \cot⁡ θ√(sin⁡θ )(i/2)e^{iφ/2}]=$

$=Aℏe^{3iφ/2}*\frac{\cos⁡ θ}{√(\sin⁡θ)}[1/2-1/2]=0$

$L_- Y_{1/2,-1/2}=-Bℏe^{-iφ}*[e^{-iφ/2}(\frac{1}{\sin^{3/2}⁡θ} -\frac{cos^2⁡θ}{2\sin^{3/2}⁡θ} )-i(-i/2)√(sin⁡θ )(-\cot^2⁡θ)e^{-iφ/2}]$


$L_- Y_{1/2,-1/2}=-Bℏe^{-3iφ/2}*\frac{1}{\sin^{3/2}⁡θ} (2-cos^2⁡θ+cos^2⁡ θ)=-Bℏe^{-3iφ/2}2\sin^{3/2}⁡θ =0$ 

which means $φ→∞$  so that $Y_{1/2,-1/2}=0$


$L_+ Y_{1/2,-1/2}=L_+ 0=0$    and $L_+ Y_{1/2,-1/2}=Y_{1/2,1/2}=A*e^{iφ/2} √(sin⁡θ)$
which is not possible.

Observation. If one takes

$Y_(1/2,-1/2)=Be^{-iφ/2}*√(sin⁡θ )$


$L_- Y_{1/2,-1/2}=-Bℏe^{-iφ} [e^{-iφ/2}\frac{\cos⁡ θ}{(2√(\sin⁡θ)}-i \cot⁡ θ√(\sin⁡θ )(-i/2) e^{-iφ/2} ]=$

$=-Bℏe^{-3iφ/2} \frac{\cos⁡ θ}{√(\sin⁡θ)} [1/2-1/2]=0$

which is trivial.