(a) Apply $\mathit{S}\mathit{\_}$tolocalid="1656131461017" $\overline{)10>}$(Equation$\mathbf{4}\mathbf{.}\mathbf{177}$), and confirm that you getlocalid="1656131442455" $\sqrt{\mathbf{2}}\sout{\mathbf{h}}\overline{)\mathbf{1}\mathbf{-}\mathbf{1}}\mathbf{>}$.

(b) Apply${\mathbf{S}}_{\mathbf{+}}$to$\mathbf{[}\mathbf{00}\mathbf{>}$(Equation$\mathbf{4}\mathbf{.}\mathbf{178}$), and confirm that you get zero.

(c) Show thatlocalid="1656131424007" $\overline{)\mathbf{11}}\mathbf{>}$andlocalid="1656131406083" $\overline{)\mathbf{1}\mathbf{-}\mathbf{1}\mathbf{>}}$(Equation$\mathbf{4}\mathbf{.}\mathbf{177}$) are eigenstates of${\mathit{S}}^{\mathbf{2}}$, with the appropriate eigenvalue

A quantum state whose wave function is an eigenfunction of the linear operator that corresponds to an observable is called an eigenstate. When you measure that observable, the eigenvalue of that wave function is the quantity you see (the eigenvalue could be a vector quantity).

(a) From eq. $4.177$know that$\overline{)01>}=\frac{1}{\sqrt{2}}\left(⇵+⇵\right)$, and the lowering operator if$S\_=S_{\_}^{\left(1\right)}+S_{\_}^{\left(2\right)}$therefore write:

localid="1656132528995" $$\begin{gathered} S\_\overline{)01>=S_{\_}^{\left(1\right)}+S_{\_}^{\left(2\right)}\frac{1}{\sqrt{2}}\left(⇅+⇅\right)} \\ =\frac{1}{\sqrt{2}}{S}_{\_}^{\left(1\right)}+S_{\_}^{\left(2\right)}\left(⇅+⇅\right) \\ =\frac{1}{\sqrt{2}}{S}_{\_}^{\left(1\right)}\left(⇅\right)+S_{\_}^{\left(1\right)}\left(⇅\right)+S_{\_}^{\left(2\right)}\left(⇅\right)+S_{\_}^{\left(2\right)}\left(⇅\right) \end{gathered}$$

Notice here, $S_{\_}^{\left(1\right)}$can only act on the first particle (the first arrow), and $S_{\_}^{\left(2\right)}$can only act on the Second particle (the second arrow), thus,

$S\_\overline{)01>}=\frac{1}{\sqrt{2}}\left(S_{\_}^{\left(1\right)}\uparrow \right)\downarrow +\left(S_{\_}^{\left(1\right)}\downarrow \right)\uparrow +\uparrow \left(S_{\_}^{\left(2\right)}\downarrow \right)+\downarrow \left(S_{\_}^{\left(2\right)}\uparrow \right)$

Here $\left(S_{\_}^{\left(1\right)}\downarrow \right)=\left(S_{\_}^{\left(2\right)}\downarrow \right)=0$because we cannot lower the lowerest state, and$\left(S_{\_}^{\left(1\right)}\uparrow \right)=\left(S_{\_}^{\left(2\right)}\uparrow \right)=\sout{h}\downarrow$, therefore,

$$\begin{gathered} S\_\overline{)01>=\frac{1}{\sqrt{2}}\left(\sout{h}\downarrow \downarrow +\sout{h}\downarrow \downarrow \right)} \\ =\frac{2\sout{h}}{\sqrt{2}}\downarrow \downarrow \end{gathered}$$

Where$\downarrow \downarrow =\overline{)1-1>}$, thus,

$S\_\overline{)01>}=\sqrt{2}\sout{h}\overline{)1-1>}$

The lowerest state value of$S\_\overline{)01>}=\sqrt{2}\sout{h}\overline{)1-1>}$

(b) $S_{\pm }=S_{\pm }^{\left(1\right)}+S_{\pm }^{\left(2\right)}$, and from eq.$4.178\overline{)00>=\frac{1}{\sqrt{2}}\left(⇅+⇵\right)}$, so let us start with $S_{\pm }\overline{)00>}$

$S_{\pm }\overline{)00>=S_{\pm }^{\left(1\right)}+}{S}_{\pm }^{\left(2\right)}\frac{1}{\sqrt{2}}\left(⇅-⇵\right)$

$$\begin{gathered} =\frac{1}{\sqrt{2}}{S}_{\pm }^{\left(1\right)}\left(⇅\right)-S_{\pm }^{\left(1\right)}\left(⇵\right)+S_{\pm }^{\left(2\right)}\left(⇅\right)-S_{\pm }^{\left(2\right)}\left(⇵\right) \\ =\frac{1}{\sqrt{2}}\left(S_{\pm }^{\left(1\right)}\uparrow \right)\downarrow -\left(S_{\pm }^{\left(1\right)}\downarrow \right)\uparrow +\uparrow \left(S_{\pm }^{\left(2\right)}\downarrow \right)-\downarrow \left(S_{\pm }^{\left(2\right)}\uparrow \right) \end{gathered}$$

Where $\left(S_{\pm }^{\left(1\right)}\uparrow \right)=\left(S_{\pm }^{\left(2\right)}\uparrow \right)=0$because we cannot rais the higher possible state, and$\left(S_{\pm }^{\left(1\right)}\downarrow \right)=\left(S_{\pm }^{\left(2\right)}\downarrow \right)=\sout{h}\uparrow$, thus,

$$\begin{gathered} S_{\pm }^{}\overline{)00>=\frac{1}{\sqrt{2}}\left(-\sout{h}\uparrow \uparrow +\sout{h}\uparrow \uparrow \right)} \\ =0 \end{gathered}$$

Then we will work with$S\_\overline{)00>}$.

$$\begin{gathered} S\_\overline{)00>=S_{\_}^{\left(1\right)}+S_{\_}^{\left(2\right)}\frac{1}{\sqrt{2}}\left(⇅-⇵\right)} \\ =\frac{1}{\sqrt{2}}{S}_{\_}^{\left(1\right)}\left(⇅\right)-S_{\_}^{\left(1\right)}\left(⇵\right)+S_{\_}^{\left(2\right)}\left(⇅\right)-S_{\_}^{\left(2\right)}\left(⇵\right) \\ =\frac{1}{\sqrt{2}}\left(S_{\_}^{\left(1\right)}\uparrow \right)\downarrow -\left(S_{\_}^{\left(1\right)}\downarrow \right)\uparrow +\uparrow \left(S_{\_}^{\left(2\right)}\downarrow \right)-\downarrow \left(S_{\_}^{\left(2\right)}\uparrow \right) \end{gathered}$$

Where$\left(S_{\_}^{\left(1\right)}\uparrow \right)=\left(S_{\_}^{\left(2\right)}\uparrow \right)=0$, and$\left(S_{\_}^{\left(1\right)}\downarrow \right)=\left(S_{\_}^{\left(2\right)}\downarrow \right)=\sout{h}\uparrow$, thus,

$$\begin{gathered} S\_\overline{)00>}=\frac{1}{\sqrt{2}}\left(-\sout{h}\uparrow \uparrow +\sout{h}\uparrow \uparrow \right) \\ =0 \end{gathered}$$

The higher possible state gets the value of$S\_\overline{)00>=S\_\overline{)00>=0}}$

(c) First, define the operator $S^2$for two combined states as

$S^2=S^{\left(1\right)}+S^{\left(2\right)}.S^{\left(1\right)}+S^{\left(2\right)}=S^{{\left(1\right)}^2}+S^{{\left(2\right)}^2}+2S^{\left(1\right)}.S^{\left(2\right)}$

Where

$S^{\left(1\right)}{S}^{\left(2\right)}=S_x^{\left(1\right)}{S}_x^{\left(2\right)}+S_y^{\left(1\right)}{S}_y^{\left(2\right)}+S_z^{\left(1\right)}{S}_z^{\left(2\right)}$

And can show that $S^2=\overline{)11>}$is as eiginstate as follow: (remebmber from eq$4.177$.$\overline{)11>=\uparrow \uparrow }$)

$S^2\left(\uparrow \uparrow \right)=S^{{\left(1\right)}^2}\left(\uparrow \uparrow \right)+S^{{\left(2\right)}^2}\left(\uparrow \uparrow \right)+2S^{\left(2\right)}\left(\uparrow \uparrow \right)$

Let's break it down term by term:

First term:$S^{\left(1\right)2}\left(\uparrow \uparrow \right)=S^{\left(1\right)2}\uparrow \uparrow =\frac{3{\sout{h}}^2}{4}\uparrow \uparrow$

Second term:$S^{\left(2\right)2}\left(\uparrow \uparrow \right)=\uparrow S^{\left(2\right)2}\uparrow =\frac{3{\sout{h}}^2}{4}\uparrow \uparrow$

Third term:

$$\begin{gathered} 2S^{\left(1\right)}.S^{\left(2\right)}\left(\uparrow \uparrow \right)=2S_x^{\left(1\right)}{S}_x^{\left(2\right)}\left(\uparrow \uparrow \right)+S_y^{\left(1\right)}{S}_y^{\left(2\right)}\left(\uparrow \uparrow \right)+S_z^{\left(1\right)}{S}_z^{\left(2\right)}\left(\uparrow \uparrow \right) \\ =2S_x^{\left(1\right)}\uparrow S_x^{\left(2\right)}\uparrow +S_y^{\left(1\right)}\uparrow S_y^{\left(2\right)}\uparrow +S_z^{\left(1\right)}\uparrow S_z^{\left(2\right)}\uparrow \\ =2\frac{\sout{h}}{2}\downarrow \frac{\sout{h}}{2}\downarrow \frac{i\sout{h}}{2}+\frac{i\sout{h}}{2}\downarrow \frac{\sout{h}}{2}\downarrow \frac{\sout{h}}{2}\uparrow \\ =2\frac{{\sout{h}}^2}{4}\downarrow \downarrow +\frac{-{\sout{h}}^2}{4}\downarrow \downarrow +\frac{{\sout{h}}^2}{4}\downarrow \downarrow \\ =\frac{{\sout{h}}^2}{2}\uparrow \uparrow \end{gathered}$$

Now, combine the terms,

$$\begin{gathered} S^2\left(\uparrow \uparrow \right)=\frac{3{\sout{h}}^2}{4}+\frac{3{\sout{h}}^2}{4}+\frac{{\sout{h}}^2}{2}\uparrow \uparrow \\ =2{\sout{h}}^2\uparrow \uparrow \end{gathered}$$

Which is

$S^2\overline{)11>=2{\sout{h}}^2\overline{)11>}}$

Now, show that $S^2=\overline{)1-1}$is as eiginstate as follow: (remebmber from eq. $4.177\overline{)1-1>=\downarrow \downarrow )}$

$S^2\left(\downarrow \downarrow \right)=S^{\left(1\right)2}\left(\downarrow \downarrow \right)+S^{\left(2\right)2}\left(\downarrow \downarrow \right)+2S^{\left(1\right)}.S^{\left(2\right)}\left(\downarrow \downarrow \right)$

Let's break it down term by term:

First term: $S^{\left(1\right)2}\left(\downarrow \downarrow \right)=S^{\left(1\right)2}\downarrow \downarrow =\frac{3{\sout{h}}^2}{4}\downarrow \downarrow$

Second term: $S^{\left(2\right)2}\left(\downarrow \downarrow \right)=\downarrow S^{\left(2\right)2}\downarrow =\frac{3{\sout{h}}^2}{4}\downarrow \downarrow$

Third term:

$$\begin{gathered} 2S^{\left(1\right)}.S^{\left(2\right)}\left(\downarrow \downarrow \right)=2S_x^{\left(1\right)}{S}_x^{\left(2\right)}\left(\downarrow \downarrow \right)+S_y^{\left(1\right)}{S}_y^{\left(2\right)}\left(\downarrow \downarrow \right)+S_z^{\left(1\right)}{S}_z^{\left(2\right)}\left(\downarrow \downarrow \right) \\ =2S_x^{\left(1\right)}\downarrow S_x^{\left(2\right)}\downarrow +S_y^{\left(1\right)}\downarrow S_y^{\left(2\right)}\downarrow +S_z^{\left(1\right)}\downarrow S_z^{\left(2\right)}\downarrow \\ =2\frac{\sout{h}}{2}\uparrow \frac{\sout{h}}{2}\uparrow \frac{-i\sout{h}}{2}+\frac{-i\sout{h}}{2}\uparrow \frac{\sout{-h}}{2}\uparrow \frac{-\sout{h}}{2}\downarrow \\ =2\frac{{\sout{h}}^2}{4}\uparrow \uparrow +\frac{-{\sout{h}}^2}{4}\uparrow \uparrow +\frac{{\sout{h}}^2}{4}\uparrow \uparrow \\ =\frac{{\sout{h}}^2}{2}\downarrow \downarrow \end{gathered}$$

Now, combine the terms,

$$\begin{gathered} S^2\left(\downarrow \downarrow \right)=\frac{3{\sout{h}}^2}{4}+\frac{3{\sout{h}}^2}{4}+\frac{{\sout{h}}^2}{2}\downarrow \downarrow \\ =2{\sout{h}}^2\downarrow \downarrow \end{gathered}$$

Which is

$S^2\overline{)1-1>=2{\sout{h}}^2\overline{)1-1>}}$

The eiginstate is $S^2$value of $S^2\overline{)11>=2{\sout{h}}^2\overline{)11>}}$and$S^2\overline{)1-1>=2{\sout{h}}^2\overline{)1-1>}}$