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Expert-verified Found in: Page 1247 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # If orbital angular momentum is measured along, say, a z-axis to obtain a value for ${{\mathbf{L}}}_{{\mathbf{z}}}$, show that role="math" localid="1661497092782" ${\mathbf{\left(}{\mathbf{L}}_{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{L}}_{\mathbf{y}}^{\mathbf{2}}\mathbf{\right)}}^{\mathbf{1}\mathbf{/}\mathbf{2}}{\mathbf{=}}{\mathbf{\left[}\mathbf{I}\left(I+1\right)\mathbf{-}{\mathbf{m}}_{\mathbf{I}}^{\mathbf{2}}\mathbf{\right]}}^{\mathbf{1}\mathbf{/}\mathbf{2}}{\mathbit{ħ}}$ is the most that can be said about the other two components of the orbital angular momentum.

It is proved for the other two components of the orbital angular momentum is

${\left({\mathrm{L}}_{\mathrm{x}}^{2}+{\mathrm{L}}_{\mathrm{y}}^{2}\right)}^{1/2}={\left[\mathrm{I}\left(\mathrm{I}+1\right)-{\mathrm{m}}_{\mathrm{I}}^{2}\right]}^{1/2}ħ$.

See the step by step solution

## Step 1: The given data:

The orbital angular momentum $\stackrel{\to }{\mathrm{L}}$ is measured along a z-axis to get a value ${\mathrm{L}}_{\mathrm{z}}$.

## Step 2: Understanding the concept of angular momentum:

In quantum mechanics, angular momentum is a vector operator with well-defined commutation relations between its three components.

Using the concept of orbital angular momentum and the value of the momentum in the z-axis in the resultant value of the momentum, define the required expression of the x- and y-component of the angular momentum.

Formulas:

The magnitude of the orbital angular momentum in terms $ħ$ of is,

$\stackrel{\to }{\mathrm{L}}=\sqrt{I\left(I+1\right)ħ}$ ….. (1)

The z-component of the orbital angular momentum is,

${\mathrm{L}}_{\mathrm{z}}={\mathrm{m}}_{\mathrm{I}}\mathrm{ħ}$ ….. (2)

The resultant value of the angular momentum is,

${\left|\stackrel{\to }{\mathrm{L}}\right|}^{2}={\mathrm{L}}_{\mathrm{x}}^{2}+{\mathrm{L}}_{\mathrm{y}}^{2}+{\mathrm{L}}_{\mathrm{z}}^{2}$ ….. (3)

## Step 3: Calculation to get the required equation:

Substituting the values of equations (1) and (2) in equation (3), the required equation to be proved as follows:

${\left(\sqrt{\mathrm{I}\left(\mathrm{I}+1\right)}ħ\right)}^{2}={\mathrm{L}}_{\mathrm{x}}^{2}+{\mathrm{L}}_{\mathrm{y}}^{2}+{\left({\mathrm{m}}_{\mathrm{I}}^{}\mathrm{ħ}\right)}^{2}\phantom{\rule{0ex}{0ex}}{\mathrm{L}}_{\mathrm{x}}^{2}+{\mathrm{L}}_{\mathrm{y}}^{2}={\left(\sqrt{\mathrm{I}\left(\mathrm{I}+1\right)}\mathrm{ħ}\right)}^{2}-{\left({\mathrm{m}}_{\mathrm{I}}^{}\mathrm{ħ}\right)}^{2}\phantom{\rule{0ex}{0ex}}\left({\mathrm{L}}_{\mathrm{x}}^{2}+{\mathrm{L}}_{\mathrm{y}}^{2}\right)={\left[\mathrm{I}\left(\mathrm{I}+1\right)-{\mathrm{m}}_{\mathrm{I}}^{2}\right]}^{1/2}\mathrm{ħ}$

For a given value of I, the greatest value of ${\mathrm{m}}_{1}$ is , so the smallest value of $\sqrt{{\mathrm{L}}_{\mathrm{x}}^{2}+{\mathrm{L}}_{\mathrm{y}}^{2}}\mathrm{is}\mathrm{ħ}\sqrt{\mathrm{I}}$.

Now, the smallest possible value of ${\mathrm{m}}_{1}$ is zero, thus the largest possible value of is .

$\sqrt{{\mathrm{L}}_{\mathrm{x}}^{2}+{\mathrm{L}}_{\mathrm{y}}^{2}}\mathrm{is}\mathrm{ħ}\sqrt{\mathrm{I}\left(\mathrm{I}+1\right)}$

So, the range is given by:

$\mathrm{ħ}\sqrt{\mathrm{I}}\le \sqrt{{\mathrm{L}}_{\mathrm{x}}^{2}+{\mathrm{L}}_{\mathrm{y}}^{2}}\le \sqrt{\mathrm{I}\left(\mathrm{I}+1\right)}$

Hence, the given equation is proved. ### Want to see more solutions like these? 