91Ó°ÊÓ

The function given is $\frac{∂}{∂\mathrm{φ} }= \frac{∂\mathrm{x}}{∂\mathrm{φ}}/\frac{∂}{∂\mathrm{x}}+\frac{∂\mathrm{y}}{∂\mathrm{φ}}\frac{∂}{∂\mathrm{y}}+\frac{∂\mathrm{z}}{∂\mathrm{φ}}\frac{∂}{∂\mathrm{z}}$.

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

Given,

$\frac{∂}{∂\mathrm{φ} }= \frac{∂\mathrm{x}}{∂\mathrm{φ}}/\frac{∂}{∂\mathrm{x}}+\frac{∂\mathrm{y}}{∂\mathrm{φ}}\frac{∂}{∂\mathrm{y}}+\frac{∂\mathrm{z}}{∂\mathrm{φ}}\frac{∂}{∂\mathrm{z}}$

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

$$\begin{gathered} x=rsin\mathrm{θ}cos\mathrm{ϕ} \\ y = rsin\mathrm{θ}sin\mathrm{φ} \\ z = rcos\mathrm{θ}  \end{gathered}$$

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

$$\begin{gathered}  \frac{∂\mathrm{x}}{∂\mathrm{φ}}=\frac{∂\left(r\sin \mathrm{θ}\cos \mathrm{φ}\right)}{∂\mathrm{φ}}\frac{∂}{∂\mathrm{x}}+\frac{∂\left(r\sin \mathrm{θ}\sin \mathrm{φ}\right)}{∂\mathrm{φ}}\frac{∂}{∂\mathrm{y}}+\frac{∂\left(r\cos \mathrm{θ}\right)}{∂\mathrm{φ}}\frac{∂}{∂\mathrm{z}} \\ =r\sin \mathrm{θ}\cos \mathrm{φ}\frac{∂}{∂y}-r\sin \mathrm{θ}\sin \mathrm{φ}\frac{∂}{∂x}+0\frac{∂}{∂\mathrm{z}} \\ =-y\frac{∂}{∂x}+x\frac{∂}{∂y} \end{gathered}$$(1)

Thus we found $\frac{∂}{∂\mathrm{φ}}=-y\frac{∂}{∂x}+x\frac{∂}{∂y}.$.

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

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

Where, h = Plank’s constant

Hence, now the equation (1) becomes,

$\frac{∂}{∂\mathrm{φ}}=-y\frac{p_x}{-ih}+x\frac{p_y}{-ih}$

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

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

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,

$-ih\frac{∂}{∂\mathrm{φ}}{e}^{i{m}_l\mathrm{φ}}=m_{l}h{e}^{i{m}_l\mathrm{φ}}$

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