Here we Pursue the more rigorous approach to the claim that the property quantized according to m_l is L_z,

(a) Starting with a straightforward application of the chain rule,

$\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{\phi }\mathbf{\hspace{0.33em}}}\mathbf{=}\mathbf{\hspace{0.33em}}\frac{\mathbf{\partial }\mathbf{x}}{\mathbf{\partial }\mathbf{\phi }}\mathbf{/}\frac{\mathbf{}\mathbf{\partial }}{\mathbf{\partial }\mathbf{x}}\mathbf{+}\frac{\mathbf{\partial }\mathbf{y}}{\mathbf{\partial }\mathbf{\phi }}\mathbf{}\mathbf{}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{y}}\mathbf{+}\frac{\mathbf{\partial }\mathbf{z}}{\mathbf{\partial }\mathbf{\phi }}\mathbf{}\mathbf{}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{z}}$

Use the transformations given in Table 7.2 to show that

$\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{\phi }}\mathbf{\hspace{0.33em}}\mathbf{=}\mathbf{\hspace{0.33em}}\mathbf{-}\mathit{y}\mathbf{}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{x}}\mathbf{+}\mathit{x}\frac{\mathbf{}\mathbf{\partial }}{\mathbf{\partial }\mathbf{y}}$

(b) Recall that L = r x p. From the z-component of this famous formula and the definition of operators for p_x and p_y, argue that the operator for L_z is $\mathbf{-}\mathit{i}\mathit{h}\mathbf{}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{\phi }}\mathbf{.}$.

(c) What now allows us to say that our azimuthal solution${\mathit{e}}^{\mathbf{i}{\mathbf{m}}_{\mathbf{l}}\mathbf{\phi }}$ has a well-defined z-component of angular momentum and that is value m_lh.

The function given is $\frac{\partial }{\partial \mathrm{\phi }\hspace{0.33em}}=\hspace{0.33em}\frac{\partial \mathrm{x}}{\partial \mathrm{\phi }}/\frac{\partial }{\partial \mathrm{x}}+\frac{\partial \mathrm{y}}{\partial \mathrm{\phi }}\frac{\partial }{\partial \mathrm{y}}+\frac{\partial \mathrm{z}}{\partial \mathrm{\phi }}\frac{\partial }{\partial \mathrm{z}}$.

The chain rule is a technique for finding the differential of composite functions.

Given,

$\frac{\partial }{\partial \mathrm{\phi }\hspace{0.33em}}=\hspace{0.33em}\frac{\partial \mathrm{x}}{\partial \mathrm{\phi }}/\frac{\partial }{\partial \mathrm{x}}+\frac{\partial \mathrm{y}}{\partial \mathrm{\phi }}\frac{\partial }{\partial \mathrm{y}}+\frac{\partial \mathrm{z}}{\partial \mathrm{\phi }}\frac{\partial }{\partial \mathrm{z}}$

From table 7.2, we know that, distances from the origin on all the axes are given by:

$$\begin{gathered} x=rsin\theta cos\varphi \\ y\hspace{0.33em}=\hspace{0.33em}rsin\theta sin\phi \\ z\hspace{0.33em}=\hspace{0.33em}rcos\theta \hspace{0.33em} \end{gathered}$$

Now, if you use the equations given above to solve the given equation

$$\begin{gathered} \hspace{0.33em}\frac{\partial \mathrm{x}}{\partial \mathrm{\phi }}=\frac{\partial \left(r\sin \theta \cos \phi \right)}{\partial \phi }\frac{\partial }{\partial \mathrm{x}}+\frac{\partial \left(r\sin \theta \sin \phi \right)}{\partial \phi }\frac{\partial }{\partial \mathrm{y}}+\frac{\partial \left(r\cos \theta \right)}{\partial \phi }\frac{\partial }{\partial \mathrm{z}} \\ =r\sin \theta \cos \phi \frac{\partial }{\partial y}-r\sin \theta \sin \phi \frac{\partial }{\partial x}+0\frac{\partial }{\partial \mathrm{z}} \\ =-y\frac{\partial }{\partial x}+x\frac{\partial }{\partial y} \end{gathered}$$(1)

Thus we found $\frac{\partial }{\partial \phi }=-y\frac{\partial }{\partial x}+x\frac{\partial }{\partial y}.$.

As you know that, the z-component of L is x.p_y – y.p_x.

You also know that, ${\mathit{p}}_{\mathbf{x}}\mathbf{=}\mathbf{-}\mathit{i}\mathit{h}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{x}}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{p}}_{\mathbf{y}}\mathbf{=}\mathbf{-}\mathit{i}\mathit{h}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{y}}$

Where, h = Plank’s constant

Hence, now the equation (1) becomes,

$\frac{\partial }{\partial \phi }=-y\frac{p_x}{-ih}+x\frac{p_y}{-ih}$

And you can see that L_z is the operator from eq (1) multiplied by -ih orrole="math" localid="1659177997912" $-ih\frac{\partial }{\partial \phi }$.

Thus,L_z is the operator from eq (1) multiplied by -ih or $-ih\frac{\partial }{\partial \phi }$.

From section 5.11, you know that when the operator operates on the function, it gives the product of itself and the well-defined observable.

Hence,

$-ih\frac{\partial }{\partial \phi }{e}^{i{m}_l\phi }=m_{l}h{e}^{i{m}_l\phi }$

Thus, when the operator operates on the function, it gives the product of itself and the well-defined observable