91影视

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

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$魔$ of is,

$\stackrel{鈫�}{\mathrm{L}}=\sqrt{I\left(I+1\right)魔}$ 鈥�.. (1)

The z-component of the orbital angular momentum is,

${\mathrm{L}}_{\mathrm{z}}={\mathrm{m}}_{\mathrm{I}}\mathrm{魔}$ 鈥�.. (2)

The resultant value of the angular momentum is,

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

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

$$\begin{gathered} {\left(\sqrt{\mathrm{I}(\mathrm{I}+1)}魔\right)}^2={\mathrm{L}}_{\mathrm{x}}^2+{\mathrm{L}}_{\mathrm{y}}^2+{\left({\mathrm{m}}_{\mathrm{I}}^{}\mathrm{魔}\right)}^2 \\ {\mathrm{L}}_{\mathrm{x}}^2+{\mathrm{L}}_{\mathrm{y}}^2={\left(\sqrt{\mathrm{I}(\mathrm{I}+1)}\mathrm{魔}\right)}^2-{\left({\mathrm{m}}_{\mathrm{I}}^{}\mathrm{魔}\right)}^2 \\ \left({\mathrm{L}}_{\mathrm{x}}^2+{\mathrm{L}}_{\mathrm{y}}^2\right)={\left[\mathrm{I}(\mathrm{I}+1)-{\mathrm{m}}_{\mathrm{I}}^2\right]}^{1/2}\mathrm{魔} \end{gathered}$$

For a given value of I, the greatest value of ${\mathrm{m}}_1$is , so the smallest value of $\sqrt{{\mathrm{L}}_{\mathrm{x}}^2+{\mathrm{L}}_{\mathrm{y}}^2}\mathrm{is}\mathrm{魔}\sqrt{\mathrm{I}}$.

Now, the smallest possible value of${\mathrm{m}}_1$ is zero, thus the largest possible value of is .

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

So, the range is given by:

$\mathrm{魔}\sqrt{\mathrm{I}}鈮�\sqrt{{\mathrm{L}}_{\mathrm{x}}^2+{\mathrm{L}}_{\mathrm{y}}^2}鈮�\sqrt{\mathrm{I}\left(\mathrm{I}+1\right)}$

Hence, the given equation is proved.