Question 4.37: (a) Apply S- to |10〉 (Equation 4.175), and confirm that you...

(a) Apply S_{-} \text {to }|10\rangle (Equation 4.175), and confirm that you get \sqrt{2} \hbar|1-1\rangle .

\left\{\begin{array}{l} |11\rangle=|\uparrow \uparrow\rangle \\ |10\rangle=\frac{1}{\sqrt{2}}(|\uparrow \downarrow\rangle+|\downarrow \uparrow\rangle) \\ |1-1\rangle=|\downarrow \downarrow\rangle \end{array}\right\} \quad s=1 \text { (triplet) } .      (4.175).

(b) Apply S_{±} \text {to }|100\rangle (Equation 4.176), and confirm that you get zero.

\left\{|00\rangle=\frac{1}{\sqrt{2}}(|\uparrow \downarrow\rangle-|\downarrow \uparrow\rangle)\right\} \quad s=0 \text { (singlet) } .      (4.176).

(c) Show that | 11 〉 and |1-1〉 (Equation 4.175) are eigenstates of S² , with the appropriate eigenvalue.

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(a)

S_{-}|10\rangle=\left(S_{-}^{(1)}+S_{-}^{(2)}\right) \frac{1}{\sqrt{2}}(\uparrow \downarrow+\downarrow \uparrow)=\frac{1}{\sqrt{2}}\left[\left(S_{-} \uparrow\right) \downarrow+\left(S_{-} \downarrow\right) \uparrow+\uparrow\left(S_{-} \downarrow\right)+\downarrow\left(S_{-} \uparrow\right)\right] .

But S_{-} \uparrow=\hbar \downarrow, S_{-} \downarrow=0   (line above Eq. 4.146), so S_{-}|10\rangle=\frac{1}{\sqrt{2}}[\hbar \downarrow \downarrow+0+0+\hbar \downarrow \downarrow]=\sqrt{2} \hbar \downarrow \downarrow=\sqrt{2} \hbar|1-1\rangle .

S _{+}=\hbar\left(\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right), \quad S _{-}=\hbar\left(\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right)       (4.146).

(b)

S_{\pm}|00\rangle=\left(S_{\pm}^{(1)}+S_{\pm}^{(2)}\right) \frac{1}{\sqrt{2}}(\uparrow \downarrow-\downarrow \uparrow)=\frac{1}{\sqrt{2}}\left[\left(S_{\pm} \uparrow\right) \downarrow-\left(S_{\pm} \downarrow\right) \uparrow+\uparrow\left(S_{\pm} \downarrow\right)-\downarrow\left(S_{\pm} \uparrow\right)\right] .

S_{+}|00\rangle=\frac{1}{\sqrt{2}}(0-\hbar \uparrow \uparrow+\hbar \uparrow \uparrow-0)=0 ; S_{-}|00\rangle=\frac{1}{\sqrt{2}}(\hbar \downarrow-0+0-\hbar \downarrow \downarrow)=0 .

(c)

S^{2}|11\rangle=\left[\left(S^{(1)}\right)^{2}+\left(S^{(2)}\right)^{2}+2 S ^{(1)} \cdot S ^{(2)}\right] \uparrow \uparrow .

=\left(S^{2} \uparrow\right) \uparrow+\uparrow\left(S^{2} \uparrow\right)+2\left[\left(S_{x} \uparrow\right)\left(S_{x} \uparrow\right)+\left(S_{y} \uparrow\right)\left(S_{y} \uparrow\right)+\left(S_{z} \uparrow\right)\left(S_{z} \uparrow\right)\right] .

=\frac{3}{4} \hbar^{2} \uparrow \uparrow+\frac{3}{4} \hbar^{2} \uparrow \uparrow+2\left[\frac{\hbar}{2} \downarrow \frac{\hbar}{2} \downarrow+\frac{i \hbar}{2} \downarrow \frac{i \hbar}{2} \downarrow+\frac{\hbar}{2} \uparrow \frac{\hbar}{2} \uparrow\right] .

=\frac{3}{2} \hbar^{2} \uparrow \uparrow+2\left(\frac{\hbar^{2}}{4} \uparrow \uparrow\right)=2 \hbar^{2} \uparrow \uparrow=2 \hbar^{2}|11\rangle=(1)(1+1) \hbar^{2}|11\rangle , as it should be.

S^{2}|1-1\rangle=\left[\left(S^{(1)}\right)^{2}+\left(S^{(2)}\right)^{2}+2 S ^{(1)} \cdot S ^{(2)}\right] \downarrow \downarrow .

=\frac{3 \hbar^{2}}{4} \downarrow \downarrow+\frac{3 \hbar^{2}}{4} \downarrow \downarrow+2\left[\left(S_{x} \downarrow\right)\left(S_{x} \downarrow\right)+\left(S_{y} \downarrow\right)\left(S_{y} \downarrow\right)+\left(S_{z} \downarrow\right)\left(S_{z} \downarrow\right)\right] .

=\frac{3}{2} \hbar^{2} \downarrow \downarrow+2\left[\left(\frac{\hbar}{2} \uparrow\right)\left(\frac{\hbar}{2} \uparrow\right)+\left(-\frac{i \hbar}{2} \uparrow\right)\left(-\frac{i \hbar}{2} \uparrow\right)+\left(-\frac{\hbar}{2} \downarrow\right)\left(-\frac{\hbar}{2} \downarrow\right)\right] .

=\frac{3}{2} \hbar^{2} \downarrow \downarrow+2 \frac{\hbar^{2}}{4} \downarrow \downarrow=2 \hbar^{2} \downarrow \downarrow=2 \hbar^{2}|1-1\rangle .

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