# 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

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

Thus

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

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

The ladder operators are

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

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

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

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

or

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

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

Therefore

$L_+ Y_{1/2,-1/2}=L_+ 0=0$ and $L_+ Y_{1/2,-1/2}=Y_{1/2,1/2}=A*e^{iφ/2} √(sinθ)$

which is not possible.

Observation. If one takes

$Y_(1/2,-1/2)=Be^{-iφ/2}*√(sinθ )$

Then

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

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

which is trivial.