Question

In Problem4.3 you showed that ${\mathit{Y}}_{\mathbf{2}}^{\mathbf{1}}\left(\theta ,\varphi \right)\mathbf{=}\mathbf{-}\sqrt{\mathbf{15}\mathbf{/}\mathbf{8}\mathbf{\pi }}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }{\mathit{e}}^{\mathbf{i}\mathbf{\varphi }}$. Apply the raising operator to find localid="1656065252558" ${\mathit{Y}}_{\mathbf{2}}^{\mathbf{2}}\mathbf{(}\mathbf{\theta }\mathbf{,}\mathbf{\varphi }\mathbf{)}$. Use Equation 4.121to get the normalization.

Short answer

The value of$Y_2^2\left(\theta ,\varphi \right)\mathrm{is}\frac{1}{4}\sqrt{\frac{15}{2\mathrm{\pi }}}\sout{h}{\left(e^{i\varphi }\sin \theta \right)}^2$.

Step-by-step solution

Define Raising Operator

An operator that raises or lowers the eigenvalue of another operator (collectively called as ladder operators) is referred to as a raising or lowering operator.

Find the value of Y22(θ,ϕ)

In this step evaluate first the spherical harmonic is $Y_2^1\left(\theta ,\varphi \right)$.

From problem 4.3, the value of $Y_2^1$

$$\begin{gathered} Y_2^1=-\sqrt{\frac{15}{8\tau \tau }}\sin \theta \cos \theta e^{io}{L}_{+} \\ =\sout{h}{e}^{i\varphi }\left[\frac{\partial }{\partial \theta }+icot\theta \frac{\partial }{\partial \varphi }\right]L_{+}{Y}_2^1 \\ =Y_2^2 \end{gathered}$$

Now, determine the spherical harmonic $Y_2^2\left(\theta ,\varphi \right)$as follows,

localid="1658464225727" $$\begin{gathered} Y_2^1=-\sqrt{\frac{15}{8\mathrm{\pi }}}\sout{h}{e}^{i\varphi }\left[e^{i\varphi }\frac{\partial \left)\right(\sin \theta \cos \theta )}{\partial \theta }+i\cot \theta \sin \theta \cos \theta \frac{\partial e^{i\varphi }}{\partial \varphi }\right] \\ =-\sqrt{\frac{15}{8\mathrm{\pi }}}\sout{h}{e}^{i\varphi }\left[e^{i\varphi }({\cos }^2\theta -{\sin }^2\theta )+i\cot \theta \sin \theta \cos \theta \left(i{e}^{i\varphi }\right)\right] \\ =-\sqrt{\frac{15}{8\mathrm{\pi }}}\sout{h}\left[e^{i\varphi }\cos 2\theta -{\cos }^2\theta e^{i\varphi }\right] \\ =\sqrt{\frac{15}{2\mathrm{\pi }}}\sout{h}{e}^{2i\varphi }{\sin }^2\theta \\ =\frac{1}{2}\sqrt{\frac{15}{2\mathrm{\pi }}}\sout{h}{(e^{2i\varphi }\sin \theta )}^2 \end{gathered}$$

Thus, the value of$Y_2^2\left(\theta ,\varphi \right)$islocalid="1658464244244" $\frac{1}{2}\sqrt{\frac{15}{2\mathrm{\pi }}}\sout{h}{\left(e^{i\varphi }{\sin }^2\theta \right)}^2$.