Chapter 7, Problem 7F.11P

Interpretation Introduction

Interpretation:

The operator ${\widehat{l}}_z$ in spherical polar co-ordinates can be expressed as ${\widehat{l}}_z=-\text{i}\hslash \partial /\partial \varphi$ has to be shown.

Concept introduction:

The vector model of angular momentum calculates the value of orbital angular momentum for an electron in an atom.  The vector model is represented by a cone.  The length of the cone should be equal to ${\left\{l\left(l+1\right)\right\}}^{1/2}$ and a projection of $m_l$ along the $z$ axis.

Answer to Problem 7F.11P

It has been proved that the operator ${\widehat{l}}_z$ in spherical polar co-ordinates can be expressed as ${\widehat{l}}_z=-\text{i}\hslash \partial /\partial \varphi$.

Explanation of Solution

The operator ${\widehat{l}}_z$ in cartesian co-ordinates is expressed as shown below.

    ${\widehat{l}}_z=\frac{\hslash }{\text{i}}\left(x\frac{\partial }{\partial y}-y\frac{\partial }{\partial x}\right)$        (1)

The relation between the cartesian co-ordinates and polar-co-ordinates is shown below.s

    $x=r\sin \theta \cos \varphi$, $y=r\sin \theta \sin \varphi$ and $z=r\cos \theta$        (2)

The relation between $r$ and $x,y$ and $z$ is shown below.

    $r^2=x^2+y^2+z^2$        (3)

Differentitate equation (3) with respect to $x$ as shown below.

    $\begin{array}{c}2r\frac{\partial r}{\partial x}=2x\\ 2r\frac{\partial r}{\partial x}=2r\sin \theta \cos \varphi \left(\because x=r\sin \theta \cos \varphi \right)\\ \frac{\partial r}{\partial x}=\sin \theta \cos \varphi \end{array}$

Differentitate equation (3) with respect to $y$ as shown below.

    $\begin{array}{c}2r\frac{\partial r}{\partial y}=2y\\ 2r\frac{\partial r}{\partial y}=2r\sin \theta \sin \varphi \left(\because y=r\sin \theta \sin \varphi \right)\\ \frac{\partial r}{\partial y}=\sin \theta \sin \varphi \end{array}$

The relation of $z$ co-ordinate with spherical co-ordinate is calculated as shown below.

    $\begin{array}{c}z=r\cos \theta \\ \cos \theta =\frac{z}{r}\end{array}$        (4)

Substitute the value of $r$ from equation (3) in equation (4).

    $\cos \theta =\frac{z}{\sqrt{x^2+y^2+z^2}}$        (5)

Differentiate equation (5) with respect to $x$.

    $\begin{array}{c}\frac{\partial \cos \theta }{\partial x}=\frac{\partial }{\partial x}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)\\ -\sin \theta \frac{\partial \theta }{\partial x}=\frac{-1}{2}\frac{z}{{\left(x^2+y^2+z^2\right)}^{3/2}}\frac{\partial x^2}{\partial x}\\ \frac{\partial \theta }{\partial x}=\frac{1}{2\sin \theta }\frac{z}{{\left(r^2\right)}^{3/2}}2x\left(\because r^2=x^2+y^2+z^2\right)\\ =\frac{x}{\sin \theta }\frac{z}{r^3}\end{array}$        (6)

Substitute the value of $x$ and $z$ from equation (2) to equation (6).

    $\begin{array}{c}\frac{\partial \theta }{\partial x}=\frac{r\sin \theta \cos \varphi }{\sin \theta }\frac{r\cos \theta }{r^3}\\ =\frac{\cos \varphi \cos \theta }{r}\end{array}$

Differentiate equation (5) with respect to $y$.

    $\begin{array}{c}\frac{\partial \cos \theta }{\partial y}=\frac{\partial }{\partial y}\left(\frac{z}{\sqrt{x^2+y^2+z^2}}\right)\\ -\sin \theta \frac{\partial \theta }{\partial y}=\frac{-1}{2}\frac{z}{{\left(x^2+y^2+z^2\right)}^{3/2}}\frac{\partial y^2}{\partial y}\\ \frac{\partial \theta }{\partial y}=\frac{1}{2\sin \theta }\frac{z}{{\left(r^2\right)}^{3/2}}2y\left(\because r^2=x^2+y^2+z^2\right)\\ =\frac{y}{\sin \theta }\frac{z}{r^3}\end{array}$        (7)

Substitute the value of $y$ and $z$ from equation (2) to equation (7).

    $\begin{array}{c}\frac{\partial \theta }{\partial y}=\frac{r\sin \theta \sin \varphi }{\sin \theta }\frac{r\cos \theta }{r^3}\\ =\frac{\sin \varphi \cos \theta }{r}\end{array}$

The value of $\frac{y}{x}$ is calculated as shown below.

    $\begin{array}{c}\frac{y}{x}=\frac{r\sin \theta \sin \varphi }{r\sin \theta \cos \varphi }\\ =\tan \varphi \end{array}$        (8)

Differentiate equation (8) with respect to $x$.

    $\begin{array}{c}\frac{\partial }{\partial x}\left(\frac{y}{x}\right)=\frac{\partial \tan \varphi }{\partial x}\\ -\frac{y}{x^2}={\sec }^2\varphi \frac{\partial \varphi }{\partial x}\\ -\frac{r\sin \theta \sin \varphi }{{\left(r\sin \theta \cos \varphi \right)}^2}={\sec }^2\varphi \frac{\partial \varphi }{\partial x}\\ -\frac{\sin \varphi }{r\sin \theta {\cos }^2\varphi }={\sec }^2\varphi \frac{\partial \varphi }{\partial x}\end{array}$        (9)

Rearrange and further solve the equation (9) as shown below.

    $\begin{array}{c}\frac{\partial \varphi }{\partial x}=-\frac{\sin \varphi }{r\sin \theta {\cos }^2\varphi {\sec }^2\varphi }\\ =-\frac{\sin \varphi }{r\sin \theta }\left(\because {\cos }^2\varphi {\sec }^2\varphi =1\right)\end{array}$

Differentiate equation (8) with respect to $y$.

    $\begin{array}{c}\frac{\partial }{\partial y}\left(\frac{y}{x}\right)=\frac{\partial \tan \varphi }{\partial y}\\ -\frac{1}{x}={\sec }^2\varphi \frac{\partial \varphi }{\partial y}\\ -\frac{1}{r\sin \theta \cos \varphi }={\sec }^2\varphi \frac{\partial \varphi }{\partial y}\end{array}$        (10)

Rearrange and further solve the equation (10) as shown below.

    $\begin{array}{c}{\sec }^2\varphi \frac{\partial \varphi }{\partial y}=-\frac{1}{r\sin \theta \cos \varphi }\\ \frac{\partial \varphi }{\partial y}=-\frac{1}{r\sin \theta \cos \varphi {\sec }^2\varphi }\left(\because \cos \varphi {\sec }^2\varphi =\sec \varphi \right)\\ =-\frac{1}{r\sin \theta \sec \varphi }\\ =-\frac{\cos \varphi }{r\sin \theta }\left(\because \frac{1}{\sec \varphi }=\cos \varphi \right)\end{array}$

The value of $\frac{\partial }{\partial x}$ is calculated by the following formula.

    $\frac{\partial }{\partial x}=\left(\frac{\partial r}{\partial x}\right)\frac{\partial }{\partial r}+\left(\frac{\partial \theta }{\partial x}\right)\frac{\partial }{\partial \theta }+\left(\frac{\partial \varphi }{\partial x}\right)\frac{\partial }{\partial \varphi }$        (11)

Substitute the value of $\frac{\partial r}{\partial x}$, $\frac{\partial \theta }{\partial x}$ and $\frac{\partial \varphi }{\partial x}$ in equation (11).

    $\frac{\partial }{\partial x}=\left(\sin \theta \cos \varphi \right)\frac{\partial }{\partial r}+\left(\frac{\cos \varphi \cos \theta }{r}\right)\frac{\partial }{\partial \theta }+\left(-\frac{\sin \varphi }{r\sin \theta }\right)\frac{\partial }{\partial \varphi }$

The value of $\frac{\partial }{\partial y}$ is calculated by the following formula.

    $\frac{\partial }{\partial y}=\left(\frac{\partial r}{\partial y}\right)\frac{\partial }{\partial r}+\left(\frac{\partial \theta }{\partial y}\right)\frac{\partial }{\partial \theta }+\left(\frac{\partial \varphi }{\partial y}\right)\frac{\partial }{\partial \varphi }$        (12)

Substitute the value of $\frac{\partial r}{\partial y}$, $\frac{\partial \theta }{\partial y}$ and $\frac{\partial \varphi }{\partial y}$ in equation (12).

    $\frac{\partial }{\partial y}=\left(\sin \theta \sin \varphi \right)\frac{\partial }{\partial r}+\left(\frac{\sin \varphi \cos \theta }{r}\right)\frac{\partial }{\partial \theta }-\left(\frac{\cos \varphi }{r\sin \theta }\right)\frac{\partial }{\partial \varphi }$

Substitute the value of $x$, $y$, $\frac{\partial }{\partial x}$ and $\frac{\partial }{\partial y}$ in equation (1).

    $\begin{array}{c}{\widehat{l}}_z=\frac{\hslash }{\text{i}}\left(\begin{array}{l}r\sin \theta \cos \varphi \left(\left(\sin \theta \sin \varphi \right)\frac{\partial }{\partial r}+\left(\frac{\sin \varphi \cos \theta }{r}\right)\frac{\partial }{\partial \theta }-\left(\frac{\cos \varphi }{r\sin \theta }\right)\frac{\partial }{\partial \varphi }\right)\\ -r\sin \theta \sin \varphi \left(\left(\sin \theta \cos \varphi \right)\frac{\partial }{\partial r}+\left(\frac{\cos \varphi \cos \theta }{r}\right)\frac{\partial }{\partial \theta }+\left(-\frac{\sin \varphi }{r\sin \theta }\right)\frac{\partial }{\partial \varphi }\right)\end{array}\right)\\ =\frac{\hslash }{\text{i}}\left(\begin{array}{l}\left(\left(r{\sin }^2\theta \cos \varphi \sin \varphi \right)\frac{\partial }{\partial r}+\left(\sin \theta \cos \varphi \sin \varphi \cos \theta \right)\frac{\partial }{\partial \theta }-\left({\cos }^2\varphi \right)\frac{\partial }{\partial \varphi }\right)\\ -\left(\left(r{\sin }^2\theta \sin \varphi \cos \varphi \right)\frac{\partial }{\partial r}+\left(\sin \theta \sin \varphi \cos \varphi \cos \theta \right)\frac{\partial }{\partial \theta }-\left({\sin }^2\varphi \right)\frac{\partial }{\partial \varphi }\right)\end{array}\right)\\ =\frac{\hslash }{\text{i}}\left(\begin{array}{l}\left(r{\sin }^2\theta \cos \varphi \sin \varphi \right)\frac{\partial }{\partial r}+\left(\sin \theta \cos \varphi \sin \varphi \cos \theta \right)\frac{\partial }{\partial \theta }-\left({\cos }^2\varphi \right)\frac{\partial }{\partial \varphi }\\ -\left(r{\sin }^2\theta \sin \varphi \cos \varphi \right)\frac{\partial }{\partial r}-\left(\sin \theta \cos \varphi \sin \varphi \cos \theta \right)\frac{\partial }{\partial \theta }-\left({\sin }^2\varphi \right)\frac{\partial }{\partial \varphi }\end{array}\right)\\ =-\frac{\hslash }{\text{i}}\left(\left({\cos }^2\varphi \right)\frac{\partial }{\partial \varphi }+\left({\sin }^2\varphi \right)\frac{\partial }{\partial \varphi }\right)\end{array}$

Further solving the above equation using identity $\left({\cos }^2\varphi \right)+\left({\sin }^2\varphi \right)=1$.

    ${\widehat{l}}_z=-\frac{\hslash }{\text{i}}\left(\frac{\partial }{\partial \varphi }\right)$

Hence, it has been proved that the value of operator ${\widehat{l}}_z$ in spherical polar co-ordinates can be expressed as ${\widehat{l}}_z=-\text{i}\hslash \partial /\partial \varphi$.