Question

A simplified approach to the question of how lis related to angular momentum – due to P. W. Milonni and Richard Feynman – can be stated as follows: If can take on only those values ${\mathit{m}}_{\mathbf{l}}\mathit{h}$, where${\mathit{m}}_{\mathbf{l}}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{1}\mathbf{,}\mathbf{\dots }\mathbf{\dots }\mathbf{\pm }\mathit{l}$ , then its square is allowed only values${{\mathit{m}}_{\mathbf{l}}}^{\mathbf{2}}{\mathit{h}}^{\mathbf{2}}$, and the average of localid="1659178449093" ${\mathit{l}}^{\mathbf{2}}$should be the sum of its allowed values divided by the number of values,$\mathbf{2}\mathit{l}\mathbf{+}\mathbf{1}$ , because there really is no preferred direction in space, the averages of $\mathit{L}{\mathit{x}}^{\mathbf{2}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{L}{\mathit{y}}^{\mathbf{2}}$should be the same, and sum of all three should give the average of role="math" localid="1659178641655" ${\mathit{L}}^{\mathbf{2}}$. Given the sumrole="math" localid="1659178770040" $\mathbf{\sum }_{\mathbf{1}}^{\mathbf{S}}{\mathit{n}}^{\mathbf{2}}\mathbf{=}\mathit{N}\mathbf{(}\mathit{N}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathit{N}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{/}\mathbf{6}$, show that these arguments, the average of ${\mathit{L}}^{\mathbf{2}}$ should be $\mathit{l}\mathbf{(}\mathit{l}\mathbf{+}\mathbf{1}\mathbf{)}{\mathit{h}}^{\mathbf{2}}$.

Short answer

Given the sum $\sum _1^{\mathrm{S}}{n}^2=N(N+1)(2N+1)/6$the average of $L^{2}isl(l+1)h^2$.

Step-by-step solution

Average of Lz2 :

The Azimuthal quantum number specifies the shape and angular momentum of the orbital in the space.

Given that,

If $L_2$can take on only those values $m_{l}h$, where $m_l=0,\pm 1,\dots \dots \pm l$, then its square is allowed only values $m{l}^2{h}^2$, and the average of $l^2$ should be the sum of its allowed values divided by the number of values, $2l+1$, because there really is no preferred direction in space, the averages of $L_x^{2}and{L}_y^2$ should be the same, and sum of all three should give the average of$L^2$ .

Where, L is the Orbital angular momentum, $L_x$ is the component of orbital angular momentum along x-axis, $L_y$ is the component of orbital angular momentum along y-axis, $L_z$ is the component of orbital angular momentum along z-axis, l is theAzimuthal quantum number,$m_l$ is theMagnetic quantum number.

role="math" localid="1659179529670" $L_z^2\hspace{0.33em}=\hspace{0.33em}\frac{1}{2l+1}\sum _{ml=-l}^{l}m{l}^2{h}^2$

Where, h is Planck’s constant.

$$\begin{gathered} L_z^2\hspace{0.33em}=\hspace{0.33em}\frac{2}{2l+1}\sum _{ml=-l}^{l}m{l}^2{h}^2 \\ =\frac{2}{2l+1}{h}^{2}l\left(l+1\right)\left(2l+1\right)/6 \\ =\frac{1}{3}{h}^{2}l\left(l+1\right) \end{gathered}$$

Average of  :

It is also given that,

$Avg.of{L}_x^2=Avg.of{L}_y^2=Avg.of{L}_z^2$ ….. (1)

Hence,

$$\begin{gathered} Avg.of{L}^2=Avg.of{L}_y^2=Avg.of{L}_z^2 \\ =3\times Avg.of{L}_x^2+Avg.of{L}_y^2+Avg.of{L}_z^2 \\ =3\times Avgof{L}_z^2 \\ =3\times \frac{1}{3}{h}^{2}l\left(l+1\right) \\ =h^{2}l\left(l+1\right) \end{gathered}$$