Work out the normalization factor for the spherical harmonics, as follows. From Section 4.1.2we know that

$Y_l^m=B_l^m{e}^{im\varphi }{P}_l^m\left(cos\theta \right)$

the problem is to determine the factor (which I quoted, but did not derive, in Equation 4.32). Use Equations 4.120, 4.121, and 4.130to obtain a recursion

relation giving $B_l^{m+1}$ in terms of $B_l^m$. Solve it by induction on to get $B_l^m$ up to an overall constant $C\left(l\right)$, .Finally, use the result of Problem 4.22 to fix the constant. You may find the following formula for the derivative of an associated Legendre function useful:

$\left(1-x^2\right)\frac{d{P}_l^m}{dx}=\sqrt{1-x^2}{P}_l^{m+1}-mx{P}_l^m$ [4.199]

Normalization is the process of scaling wave functions to the point where all probabilities equal one.

The probabilistic description of quantum mechanics only makes sense when all of the probabilities add up to one.

Apply the raising operator to the spherical harmonics and utilizing its coordinate representation as follows,

$L_{+}{Y}_l^m=\hslash \sqrt{l\left(l+1\right)-m\left(m+1\right)}{y}_l^{m+1}$

It is known that $L_{+}=\hslash e^{i\varphi }\left(\frac{\partial }{\partial \theta }+i\cot \theta \frac{\partial }{\partial \varphi }\right)$. Substitute all the values in above expression.

$$\begin{gathered} \hslash e^{i\varphi }\left(\frac{\partial }{\partial \theta }+i\cot \theta \frac{\partial }{\partial \varphi }\right)B_l^m{e}^{im\varphi }{P}_l^{m+}\cos \theta =\hslash \sqrt{l\left(l+1\right)-m\left(m+1\right)}{B}_l^{m+1}{e}^{(m+1)\varphi }{P}_l^{m+1}\left(\cos \theta \right) \\ {B}_l^m\left(\frac{d}{d\theta }-m\cot \theta \right)P_l^m\left(\cos \theta \right)=\sqrt{l\left(l+1\right)-m\left(m+1\right)}{B}_l^{m+1}{P}_l^{m+1}\left(\cos \theta \right) \end{gathered}$$

Assume and find the value of $\frac{d}{d\theta }$.

$\begin{array}{rcl}\cot \theta & =& \frac{x}{\sqrt{1-x^2}}\\ \frac{d}{d\theta }& =& \frac{dx}{d\theta }\times \frac{d}{dx}\\ & =& -\sin \theta \frac{d}{dx}\\ & =& -\sqrt{1-x^2}\frac{d}{dx}\end{array}$

Substitute all the values in above expression.

$\begin{array}{rcl}{B}_l^m\left[-\sqrt{1-x^2}\frac{d}{dx}-m\frac{x}{\sqrt{1-x^2}}\right]P_l^m\left(x\right)& =& \sqrt{l\left(l+1\right)-m\left(m+1\right)}{B}_l^{m+1}{P}_l^{m+1}\left(x\right)\\ -B_l^m\left[\sqrt{1-x^2}\frac{d}{dx}{P}_l^{mm}\left(x\right)+\frac{mx}{\sqrt{1-x^2}}{P}_l^m\right]& =& \sqrt{l\left(l+1\right)-m\left(m+1\right)}{B}_l^{m+1}{P}_l^{m+1}\left(x\right)\\ -B_l^m\left[\frac{\left(1-x^2\right)\frac{d}{dx}{P}_l^m\left(x\right)+mx{P}_l^m\left(x\right)}{\sqrt{1-x^2}}\right]& =& \sqrt{l\left(l+1\right)-m\left(m+1\right)}{B}_l^{m+1}{P}_l^{m+1}\left(x\right)\\ & & \end{array}$

The recursion relation for the Legendre polynomial is given as follows,

$\left(1-x^2\right)\frac{d{P}_l^{m+1}}{dx}=\sqrt{1-x^2}{P}_l^{m+1}-mx{P}_l^m$

Apply the above relation to drive the recursion relation for $B_l^m$

$\begin{array}{rcl}-B_l^{m+}\left[p_l^{m+1}\left(x\right)\right]& =& \sqrt{l\left(l+1\right)-m\left(m+1\right)}{B}_l^{m+1}{P}_l^{m+1}\left(x\right)\\ B_l^{m+1}& =& \frac{-1}{\sqrt{l\left(l+1\right)-m\left(m+1\right)}}{B}_l^m\\ & & \end{array}$

The recurrence relation for $B_l^m$is found in the above step, compute further terms while taking both the positive and negative values of m into account to arrive at the following formula for $B_l^m$$B_l^{m+1}=\frac{-1}{\sqrt{l-m}\sqrt{l+1+m}}{B}_l^m$$B_l^{m+1}=\frac{-1}{\sqrt{l-m}\sqrt{l+1+m}}{B}_l^m$

$\begin{array}{rcl}{B}_l^1& =& \frac{-1}{\sqrt{l-0}\sqrt{l+1+0}}{B}_l^0\\ & =& \frac{-1}{\sqrt{l}\sqrt{l+1}}{B}_l^0\\ & & \end{array}$

$\begin{array}{rcl}{B}_l^2& =& \frac{-1}{\sqrt{l-1}\sqrt{l+1+1}}{B}_l^1\\ & =& \frac{-1}{\sqrt{l-1}\sqrt{l+2}}\times \frac{-1}{\sqrt{l\left(l+1\right)}}{B}_l^0\\ & =& \frac{1}{\sqrt{l\left(l-1\right)}\sqrt{\left(l+1\right)\left(l+2\right)}}{B}_l^0\\ & & \end{array}$

Write the value of $B_l^m$.

$B_l^m={\left(-1\right)}^{m+1}\sqrt{\frac{\left(l-m\right)!}{\left(l+m\right)!}}C\left(l\right)$

Here,$C\left(l\right)=B_l^0$for $m⩾0$

For $m⩾0$,

$B_l^{-1}=\frac{-B_l^0}{\sqrt{l\left(l+1\right)}}$

$\begin{array}{rcl}{B}_l^{-2}& =& \frac{-1}{\sqrt{\left(l-1\right)\left(l+2\right)}}{B}_l^{-1}\\ & =& \frac{1}{\sqrt{l\left(l-1\right)}\sqrt{\left(l+1\right)\left(l+2\right)}}{B}_l^0\\ & & \end{array}$

It is known that $B_l^{-m}=B_l^m$, so the value of $B_l^m$can be written as follows,

$B_l^m={\left(-1\right)}^m\sqrt{\frac{\left(l-\left|m\right|\right)!}{\left(l+\left|m\right|\right)!}}C\left(l\right)$

Write the expression for $Y_l^l$.

$\begin{array}{rcl}{Y}_l^l& =& \frac{1}{2^{l}l!}\sqrt{\frac{\left(2l+1\right)!}{\pi }}{\left(e^{i\varphi }\sin \theta \right)}^l\\ & =& B_l^l{e}^{il\varphi }·P_l^l\left(\cos \theta \right)\\ & & \end{array}$

The value of $P_l^l\left(x\right)$can be obtained as follows,

$\begin{array}{rcl}{P}_l^l\left(x\right)& =& {\left(1-x^2\right)}^{\frac{1}{2}}{\left(\frac{d}{dx}\right)}^l\frac{1}{2^{l}l!}{\left(\frac{d}{dx}\right)}^l{\left(x^2-1\right)}^l\\ & =& \frac{{\left(1-x^2\right)}^{\frac{l}{2}}}{2^{l}l!}·{\left(\frac{d}{dx}\right)}^{2l}\left(x^{2l}-\dots \right)\\ & =& \frac{\left(2l\right)!}{2^{l}l!}{\left(1-x^2\right)}^{\frac{l}{2}}\\ P_l^l\left(\cos \theta \right)& =& \frac{\left(2l\right)!{\left(\sin \theta \right)}^l}{2^{l}l!}\\ & & \end{array}$

Now,

$\begin{array}{rcl}\frac{1}{2^{l}l!}\sqrt{\frac{\left(2l+1\right)!}{\pi }}{\left(e^{i\varphi }\sin \theta \right)}^l& =& B_l^l{e}^{i\varphi \varphi }\frac{\left(2l\right)}{2^{l}l!}(\sin \theta {)}^{l}n{B}_l^l\\ & =& \frac{1}{\left(2l\right)!}\sqrt{\frac{\left(2l+1\right)!}{\pi }}\\ & =& \sqrt{\frac{\left(2l+1\right)!}{\pi \left(2l\right)!\left(2l\right)!}}n\\ & =& \sqrt{\frac{\left(2l+1\right)}{\pi \left(2l\right)!}}\\ & & \end{array}$

Use the above value in equation, $B_l^l={\left(-1\right)}^l\sqrt{\frac{1}{\left(2l\right)!}}C\left(l\right)$. So, the value of $C\left(l\right)$can be written as follows,

$C\left(l\right)={\left(-1\right)}^l\sqrt{\frac{2l+1}{\pi }}$

Thus, the normalization factor is ${\left(-1\right)}^{l+m}\sqrt{\frac{\left(2l+1\right)}{\pi }\frac{\left(l-\left|m\right|\right)!}{\left(l+\left|m\right|\right)!}}$.