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Given data ${\mathrm{J}}_1=1$ is and ${\mathrm{J}}_2=\frac{3}{2}$.

In order to find the angle between the quantum mechanical angular momentum vectors J₁ and J₂ when the quantum numbers for me two of them are 1 and 3/2 respectively, equations for the angular momentum, the quantum number for the angular momentum, and the law of cosines will be needed.

The magnitude of some angular momentum J.

$\mathbf{J}\mathbf{=}\sqrt{\mathbf{j}\left(j+1\right)}\sout{\sout{\mathbf{h}}}$ ……. (1)

Here data-custom-editor="chemistry" $\sout{\mathbf{h}}$¡s Planck's reduced constant, and data-custom-editor="chemistry" $\mathbf{j}$is the quantum number for some angular momentum.

The quantum number for the total angular momentumdata-custom-editor="chemistry" ${\mathbf{j}}_{\mathbf{T}}$.

data-custom-editor="chemistry" ${\mathbf{j}}_{\mathbf{T}}\mathbf{=}\left|j_1-j_2\right|\mathbf{,}\left|j_1-j_2\right|\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\left|j_1+j_2\right|\mathbf{+}\mathbf{1}\mathbf{,}\left|j_1+j_2\right|$ ……. (2)

Here j₁ and j₂ are the quantum numbers for angular momentum vectors J₁ and J₂ respectively.

The law of cosines relating the lengths of sides a, b and c and angle data-custom-editor="chemistry" $\mathbf{θ}$between sides a and b is, data-custom-editor="chemistry" ${\mathbf{c}}^{\mathbf{2}}\mathbf{=}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{ab}\mathbf{}\mathbf{³¦´Ç²õθ}$.

The general picture for the relation between ${\mathrm{J}}_1,{\mathrm{J}}_2$and data-custom-editor="chemistry" ${\mathrm{J}}_{\mathrm{T}}$looks like in figure 1.

Figure 1

Here, angle between J₁ and J₂ is data-custom-editor="chemistry" $\mathrm{θ}$.

Use law of cosine and magnitudes of the vectors.

$$\begin{gathered} {\mathrm{c}}^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}\mathrm{³¦´Ç²õθ} \\ {\mathrm{J}}_{\mathrm{T}}^2={\mathrm{J}}_1^2+{\mathrm{J}}_2^2-2{\mathrm{J}}_1{\mathrm{J}}_2\mathrm{³¦´Ç²õθ} \end{gathered}$$

Apply angle data-custom-editor="chemistry" $\mathrm{φ}$instead of data-custom-editor="chemistry" $\mathrm{θ}$.

$$\begin{gathered} {\mathrm{J}}_{\mathrm{T}}^2={\mathrm{J}}_1^2+{\mathrm{J}}_2^2-2{\mathrm{J}}_1{\mathrm{J}}_2\mathrm{³¦´Ç²õθ} \\ {\mathrm{J}}_{\mathrm{T}}^2={\mathrm{J}}_1^2+{\mathrm{J}}_2^2-2{\mathrm{J}}_1{\mathrm{J}}_2\cos \left({180}^{\mathrm{o}}-\mathrm{φ}\right) \\ {\mathrm{J}}_{\mathrm{T}}^2={\mathrm{J}}_1^2+{\mathrm{J}}_2^2-2{\mathrm{J}}_1{\mathrm{J}}_2\mathrm{³¦´Ç²õφ} \end{gathered}$$

From the diagram, data-custom-editor="chemistry" $\mathrm{θ}+\mathrm{φ}={180}^{\mathrm{o}}$, solve for data-custom-editor="chemistry" $\mathrm{φ}$.

role="math" localid="1658395808801" $\begin{array}{rcl}{\mathrm{J}}_{\mathrm{T}}^2& =& {\mathrm{J}}_1^2+{\mathrm{J}}_2^2-2{\mathrm{J}}_1{\mathrm{J}}_2\mathrm{³¦´Ç²õφ}\\ {\mathrm{J}}_{\mathrm{T}}^2-{\mathrm{J}}_1^2+{\mathrm{J}}_2^2& =& 2{\mathrm{J}}_1{\mathrm{J}}_2\mathrm{³¦´Ç²õφ}\\ \mathrm{³¦´Ç²õφ}& =& \frac{{\mathrm{J}}_{\mathrm{T}}^2-{\mathrm{J}}_1^2+{\mathrm{J}}_2^2}{2{\mathrm{J}}_1{\mathrm{J}}_2}\\ \mathrm{φ}& =& {\cos }^{-1}\left(\frac{{\mathrm{J}}_{\mathrm{T}}^2-{\mathrm{J}}_1^2+{\mathrm{J}}_2^2}{2{\mathrm{J}}_1{\mathrm{J}}_2}\right)\end{array}$ ……. (3)

Use this, we must know magnitudes of vectors J₁ , J₂ and J_T which can be obtained in quantum numbers j₁, j₂ and j_T.

j₁ And j₂ are 1 and 3/2 respectively, so those can be inserted into equation (1) for the magnitudes of J₁ and J₂.

$$\begin{gathered} {\mathrm{J}}_1=\sqrt{{\mathrm{j}}_1\left({\mathrm{j}}_1+1\right)}\sout{\mathrm{h}}{\mathrm{J}}_2=\sqrt{{\mathrm{j}}_2\left({\mathrm{j}}_2+1\right)}\sout{\mathrm{h}} \\ {\mathrm{J}}_1=\sqrt{1\left(1+1\right)}\sout{\mathrm{h}}{\mathrm{J}}_2=\sqrt{\left(\frac{3}{2}\right)\left[\left(\frac{3}{2}\right)+1\right]}\sout{\mathrm{h}} \\ {\mathrm{J}}_1=\sqrt{2}\sout{\mathrm{h}}{\mathrm{J}}_2=\frac{\sqrt{3}}{2}\sout{\mathrm{h}} \end{gathered}$$

Find the magnitude of data-custom-editor="chemistry" ${\mathrm{J}}_{\mathrm{T}}$ , you need to know what the quantum numbers for that are, use equation (2), with 1 for j₁ and 3/2 for j₂.

data-custom-editor="chemistry" $\begin{array}{rcl}{\mathrm{j}}_{\mathrm{T}\min }& =& \left|{\mathrm{j}}_1-{\mathrm{j}}_2\right|\\ & =& \left|\left(1\right)-\left(\frac{3}{2}\right)\right|\\ & =& \frac{1}{2}\end{array}$

Simplify further as shown below.

$\begin{array}{rcl}{\mathrm{j}}_{\mathrm{T}\max }& =& {\mathrm{j}}_1+{\mathrm{j}}_2\\ & =& \left(1\right)+\left(\frac{3}{2}\right)\\ & =& \frac{5}{2}\end{array}$

Given that 1 is added in increments to data-custom-editor="chemistry" ${\mathrm{j}}_{\mathrm{T}\min }$until you reach ${\mathrm{j}}_{\mathrm{T}\max }$(according to equation (2)), all the values for data-custom-editor="chemistry" ${\mathrm{j}}_{\mathrm{T}}$.

data-custom-editor="chemistry" ${\mathrm{j}}_{\mathrm{T}}=\frac{1}{2},\frac{3}{2},\frac{5}{2},\frac{7}{2}$

So the equation (1) can be used to find the values for the various data-custom-editor="chemistry" ${\mathrm{j}}_{\mathrm{T}}$.

role="math" localid="1658397950336" $$\begin{gathered} {\mathrm{J}}_{\mathrm{T}1}=\sqrt{{\mathrm{j}}_{\mathrm{T}1}\left({\mathrm{j}}_{\mathrm{T}1}+1\right)}\sout{\mathrm{h}}{\mathrm{J}}_{\mathrm{T}2}=\sqrt{{\mathrm{j}}_{\mathrm{T}2}\left({\mathrm{j}}_{\mathrm{T}2}+1\right)}\sout{\mathrm{h}}{\mathrm{J}}_{\mathrm{T}3}=\sqrt{{\mathrm{j}}_{\mathrm{T}3}\left({\mathrm{j}}_{\mathrm{T}3}+1\right)}\sout{\mathrm{h}} \\ {\mathrm{J}}_{\mathrm{T}1}=\sqrt{\left(\frac{1}{2}\right)\left[\left(\frac{1}{2}\right)+1\right]}\sout{\mathrm{h}}{\mathrm{J}}_{\mathrm{T}2}=\sqrt{\left(\frac{3}{2}\right)\left[\left(\frac{3}{2}\right)+1\right]}\sout{\mathrm{h}}{\mathrm{J}}_{\mathrm{T}3}=\sqrt{\left(\frac{5}{2}\right)\left[\left(\frac{5}{2}\right)+1\right]}\sout{\mathrm{h}} \\ {\mathrm{J}}_{\mathrm{T}1}=\frac{\sqrt{3}}{2}\sout{\mathrm{h}}{\mathrm{J}}_{\mathrm{T}2}=\frac{\sqrt{15}}{2}\sout{\mathrm{h}}{\mathrm{J}}_{\mathrm{T}3}=\frac{\sqrt{35}}{2}\sout{\mathrm{h}} \end{gathered}$$

Find the first angle ${â‹„}_1$ allowed between J₁ and J₂ , $\sqrt{2}\sout{\mathrm{h}}$ is used for J₁, $\frac{\sqrt{15}}{2}\sout{\mathrm{h}}$ is used for J₂ and $\frac{\sqrt{3}}{2}\sout{\mathrm{h}}$ is used for J_T in equation (3).

$\begin{array}{rcl}{\mathrm{φ}}_1& =& {\cos }^{-1}\left(\frac{{\mathrm{J}}_{\mathrm{T}}^2-{\mathrm{J}}_1^2+{\mathrm{J}}_2^2}{2{\mathrm{J}}_1{\mathrm{J}}_2}\right)\\ {\mathrm{φ}}_1& =& {\cos }^{-1}\left[\frac{{\left({\displaystyle \frac{\sqrt{3}}{2}}\sout{\mathrm{h}}\right)}^2-{\left(\sqrt{2}\sout{\mathrm{h}}\right)}^2+{\left({\displaystyle \frac{\sqrt{15}}{2}}\sout{\mathrm{h}}\right)}^2}{2\left(\sqrt{2}\sout{\mathrm{h}}\right)\left({\displaystyle \frac{\sqrt{15}}{2}}\sout{\mathrm{h}}\right)}\right]\\ {\mathrm{φ}}_1& =& {\cos }^{-1}\left(-\frac{5}{\sqrt{30}}\right)\\ & =& 155.9^o\end{array}$

So the first angle allowed between J₁ and J₂ is role="math" localid="1658399171396" ${\mathrm{φ}}_1={156}^{\mathrm{o}}$.

Find the second angle${â‹„}_2$ allowed between J₁ and J₂ , $\sqrt{2}\sout{\mathrm{h}}$ is used for J_1, $\frac{\sqrt{15}}{2}\sout{\mathrm{h}}$ is used for J₂ and $\frac{\sqrt{3}}{2}\sout{\mathrm{h}}$ is used for J_T in equation (3).

$\begin{array}{rcl}{\mathrm{φ}}_2& =& {\cos }^{-1}\left(\frac{{\mathrm{J}}_{\mathrm{T}}^2-{\mathrm{J}}_1^2-{\mathrm{J}}_2^2}{2{\mathrm{J}}_1{\mathrm{J}}_2}\right)\\ {\mathrm{φ}}_2& =& {\cos }^{-1}\left[\frac{{\left({\displaystyle \frac{\sqrt{3}}{2}\sout{\mathrm{h}}}\right)}^2-{\left(\sqrt{2}\sout{\mathrm{h}}\right)}^2-{\left({\displaystyle \frac{\sqrt{15}}{2}\sout{\mathrm{h}}}\right)}^2}{2\left(\sqrt{2}\sout{\mathrm{h}}\right)\left({\displaystyle \frac{\sqrt{15}}{2}\sout{\mathrm{h}}}\right)}\right]\\ {\mathrm{φ}}_2& =& {\cos }^{-1}\left(-\frac{2}{\sqrt{30}}\right)\\ & =& 111.4^{\mathrm{o}}\end{array}$

So the second angle allowed between J₁ and J₂ is ${\mathrm{φ}}_2={111}^{\mathrm{o}}$.

Find the third angle ${â‹„}_3$ allowed between J₁ and J₂, data-custom-editor="chemistry" $\sqrt{2}\sout{\mathrm{h}}$ is used for J₁, $\frac{\sqrt{15}}{2}\sout{h}$ is used for J₂ , and $\frac{\sqrt{35}}{2}\sout{h}$is used for J_T in equation (3).

role="math" localid="1658400686867" $\begin{array}{rcl}{\mathrm{φ}}_3& =& {\cos }^{-1}\left(\frac{{\mathrm{J}}_{\mathrm{T}}^2-{\mathrm{J}}_1^2-{\mathrm{J}}_2^2}{2{\mathrm{J}}_1{\mathrm{J}}_2}\right)\\ {\mathrm{φ}}_3& =& {\cos }^{-1}\left[\frac{{\left({\displaystyle \frac{\sqrt{35}}{2}\sout{\mathrm{h}}}\right)}^2-{\left(\sqrt{2}\sout{\mathrm{h}}\right)}^2-{\left({\displaystyle \frac{\sqrt{15}}{2}\sout{\mathrm{h}}}\right)}^2}{2\left(\sqrt{2}\sout{\mathrm{h}}\right)\left({\displaystyle \frac{\sqrt{15}}{2}\sout{\mathrm{h}}}\right)}\right]\\ {\mathrm{φ}}_3& =& {\cos }^{-1}\left(\frac{3}{\sqrt{30}}\right)\\ & =& 56.{79}^o\end{array}$

So the last angle allowed between J₁ and J₂ is 56.8^o.

The first angle allowed between J₁ and J₂ is 156^o.

The second angle allowed between J₁ and J₂ is 111^o.

The last angle allowed between J₁ and J₂ is 56.8^o.