Question

Using the general rule for adding angular momenta discussed in Section 8.7 and further in Exercise 66, Find the allowed values of${\mathbf{j}}_{\mathbf{T}}$for three spin $\frac{\mathbf{1}}{\mathbf{2}}$ fermions. First add two, then add the third.

Short answer

The values of ${\mathrm{j}}_{\mathrm{T}}$allowed for three spin $\frac{1}{2}$fermions is ${\mathrm{j}}_{\mathrm{T}}=\frac{1}{2},\frac{3}{2}.$

Step-by-step solution

Given data

Three spin is $\frac{1}{2}$fermions.

Formula of spin fermions

To get the allowed values of ${\mathbf{j}}_{\mathbf{T}}$for three fermions, first tile allowed values are gotten for two spin $\frac{\mathbf{1}}{\mathbf{2}}$fermions as follows:

$$\begin{gathered} {\mathbf{j}}_{\mathbf{T}\mathbf{}\mathbf{}\mathbf{max}}\mathbf{=}{\mathbf{j}}_{\mathbf{1}}\mathbf{+}{\mathbf{j}}_{\mathbf{2}} \\ {\mathbf{j}}_{\mathbf{T}\mathbf{}\mathbf{}\mathbf{min}}\mathbf{=}\left|j_1-j_2\right| \end{gathered}$$

Find the values of jT allowed for three spin 12 fermions

Find the values of ${\mathrm{j}}_{\mathrm{T}}$allowed for three spin $\frac{1}{2}$fermions.

role="math" localid="1658389841980" $$\begin{gathered} {\mathrm{j}}_{\mathrm{T}\min }=\left|{\mathrm{j}}_1-{\mathrm{j}}_2\right|,{\mathrm{j}}_{\mathrm{T}\max }={\mathrm{j}}_1+{\mathrm{j}}_2 \\ {\mathrm{j}}_{\mathrm{T}\min }=\left|\left(\frac{1}{2}\right)-\left(\frac{1}{2}\right)\right|,{\mathrm{j}}_{\mathrm{T}\max }=\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right) \\ {\mathrm{j}}_{\mathrm{T}\min }=0,{\mathrm{j}}_{\mathrm{T}\max }=1 \end{gathered}$$

Since those have a difference of just 1, those are the only allowed valued for two fermions.

So the allowed are then checked for three fermions by having the new ${\mathrm{j}}_1$be 0 or 1 with the ${\mathrm{j}}_2$still being 1/2.

$$\begin{gathered} {\mathrm{j}}_{\mathrm{T}\min 1}=\left|{\mathrm{j}}_1-{\mathrm{j}}_2\right|,{\mathrm{j}}_{\mathrm{T}\max 1}={\mathrm{j}}_1+{\mathrm{j}}_2 \\ {\mathrm{j}}_{\mathrm{T}\min 1}=\left|\left(0\right)-\left(\frac{1}{2}\right)\right|,{\mathrm{j}}_{\mathrm{T}\max 1}=\left(0\right)+\left(\frac{1}{2}\right)\$ \\ {\mathrm{j}}_{\mathrm{T}\min 1}=\frac{1}{2},{\mathrm{j}}_{\mathrm{T}\max 1}=\frac{1}{2} \end{gathered}$$

Similarly calculate as shown below.

$$\begin{gathered} {\mathrm{j}}_{\mathrm{T}\min 2}=\left|{\mathrm{j}}_1-{\mathrm{j}}_2\right|,{\mathrm{j}}_{\mathrm{T}\max 2}={\mathrm{j}}_1+{\mathrm{j}}_2 \\ {\mathrm{j}}_{\mathrm{T}\min 2}=\left|\left(1\right)-\left(\frac{1}{2}\right)\right|,{\mathrm{j}}_{\mathrm{T}\max 2}=\left(1\right)+\left(\frac{1}{2}\right) \\ {\mathrm{j}}_{\mathrm{T}\min 2}=\frac{1}{2},{\mathrm{j}}_{\mathrm{T}\max 2}=\frac{3}{2} \end{gathered}$$

And then the two sets can be combined to get the total allowed values $\mathrm{jT}=\frac{1}{2},\frac{3}{2}$.

The two sets can be combined to get the total allowed values $\mathrm{jT}=\frac{1}{2},\frac{3}{2}$.