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Problem statement

Determine $S_x,S_y,S_z$ angular momentum spin matrices for the electron using spin 1. The given is that experiments show that $S_z$ has three possible values (eigenvalues). These are $1,0,-1$.

Solution

Using eigenbasis for $S_z$ as the following$$\begin{array}{rl}\text{(1)}& |S_{z=1}⟩& =|1⟩=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]|S_{z=0}⟩& =|2⟩=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]\\ |S_{z=-1}⟩& =|3⟩=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]\end{array}$$

Then the following three equations result from writing $S_z|S_{z={\omega }_i}⟩={\omega }_i|S_{z={\omega }_i}⟩$ where ${\omega }_i$ is the eigenvalue. These three equation are solved to determine $S_z$. Let $S_z=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& m\end{array}\right]$. Therefore$$\begin{array}{rl}\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& m\end{array}\right]\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]& =1\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]\\ \left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& m\end{array}\right]\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]& =0\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]\\ \left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& m\end{array}\right]\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]& =-1\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]\end{array}$$

Or$$\begin{array}{rl}\left[\begin{array}{c}a\\ d\\ g\end{array}\right]& =\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]\\ \left[\begin{array}{c}b\\ e\\ h\end{array}\right]& =\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\\ \left[\begin{array}{c}c\\ f\\ m\end{array}\right]& =\left[\begin{array}{c}0\\ 0\\ -1\end{array}\right]\end{array}$$

Which gives$$\begin{array}{rl}{S}_z& =\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& m\end{array}\right]\\ \text{(1A)}& & =\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& -1\end{array}\right]\end{array}$$

Now that $S_z$ is found, the goal is to determine $S_x,S_y.$Let $S_{-}=S_x-i{S}_y$ and $S_{+}=S_x+i{S}_y$. We start with $S_{+}$ (starting with $S_{-}$ will not work, as it will not be possible to determine $S^2$ that way. So we have to start with $S_{+}$). We always start with commutator $[S_z,S_{+}]$ $$\begin{array}{rl}[S_z,S_{+}]& =[S_z,S_x+i{S}_y]\\ & =[S_z,S_x]+i[S_z,S_y]\end{array}$$

But $[S_z,S_x]=i\sum _k{ϵ}_{ijk}{S}_k$. Here $i=3,j=1$ since $z=3,x=1$. Then $[S_z,S_x]=i{ϵ}_{312}{S}_2=i{S}_y$ and similarly $\left[S_z{S}_y\right]=i\sum _k{ϵ}_{ijk}{S}_k$. Here$i=3,j=2$ since $z=3,y=1.$ Then$[S_z,S_y]=i{ϵ}_{321}{S}_1=-i{S}_x$. The above becomes$$\begin{array}{rl}[S_z,S_{+}]& =i{S}_y+i(-i{S}_x)\\ & =i{S}_y+S_x\\ & =S_{+}\end{array}$$

This implies$$\begin{array}{rl}[S_z,S_{+}]& =S_{+}\\ S_z{S}_{+}-S_{+}{S}_z& =S_{+}\\ \text{(2)}& S_z{S}_{+}& =S_{+}{S}_z+S_{+}\end{array}$$

Picking $|1⟩$ to start with, as it lead to finding $S^2$, which we must find before making any progress. $S^2$ is proportional for the identity matrix $I$. From the above we obtain $$\begin{array}{rl}{S}_z{S}_{+}|1⟩& =(S_{+}{S}_z+S_{+})|1⟩\\ \text{(3)}& & =S_{+}{S}_z|1⟩+S_{+}|1⟩\end{array}$$

But $S_z|1⟩=|1⟩$ since the eigenvalue is $1$ associated with $|1⟩$ eigenvector. The above becomes$$\begin{array}{rl}{S}_z{S}_{+}|1⟩& =S_{+}|1⟩+S_{+}|1⟩\\ & =2S_{+}|1⟩\end{array}$$

The above shows that $S_{+}|1⟩$ is eigenvector of $S_z$ associated with the eigenvalue $2$ which is not compatible with experiments. Therefore the only logical result is that$$\begin{array}{}\text{(4)}& S_{+}|1⟩=0|1⟩\end{array}$$ Taking the adjoint gives$$\begin{array}{rl}{(S_{+}|1⟩)}^{†}& =0|1{⟩}^{†}\\ ⟨1|S_{+}^{†}& =0⟨1|\end{array}$$

Hence$$\begin{array}{rl}⟨1|S_{+}^{†}{S}_{+}|1⟩& =0⟨1|1⟩\\ \text{(4A)}& ⟨1|S_{+}^{†}{S}_{+}|1⟩& =0\end{array}$$

The above is used to find $S^2$. Since $S_{+}=S_x+i{S}_y$ then the above becomes$$\begin{array}{rl}⟨1|(S_x^{†}-i{S}_y^{†})(S_x+i{S}_y)|1⟩& =0\\ ⟨1|S_x^{†}{S}_x+i{S}_x^{†}{S}_y-i{S}_y^{†}{S}_x+S_y^{†}{S}_y|1⟩& =0\end{array}$$

But $S_x,S_y$ are Hermitian. Therefore $S_x^{†}=S_x,S_y^{†}=S_y$ and the above reduces to$$\begin{array}{rl}⟨1|S_x^2+i{S}_x{S}_y-i{S}_y{S}_x+S_y^2|1⟩& =0\\ ⟨1|S_x^2+i(S_x{S}_y-S_y{S}_x)+S_y^2|1⟩& =0\\ ⟨1|S_x^2+i[S_x,S_y]+S_y^2|1⟩& =0\end{array}$$

But $[S_x,S_y]=i{S}_z$, therefore the above becomes$$⟨1|S_x^2-S_z+S_y^2|1⟩=0$$ And $S^2=S_x^2+S_y^2+S_z^2$, therefore $S_x^2+S_y^2=S^2-S_z^2$. Hence $$\begin{array}{}\text{(4B)}& S_{+}^{†}{S}_{+}=S^2-S_z^2-S_z\end{array}$$ Therefore (4A) becomes$$⟨1|S^2-S_z^2-S_z|1⟩=0$$ Expanding the above gives$$\begin{array}{rl}⟨1|S^2|1⟩-⟨1|S_z^2|1⟩-⟨1|S_z|1⟩& =0\\ ⟨1|S^2|1⟩& =⟨1|S_z^2|1⟩+⟨1|S_z|1⟩\end{array}$$

But $⟨1|S_z|1⟩=1$ and $⟨1|S_z^2|1⟩=1$, therefore the above becomes$$\begin{array}{}\text{(5)}& ⟨1|S^2|1⟩=2\end{array}$$ It is not possible to use the above to solve for a general $S^2$ which is $3\times 3$ matrix.  But since $S^2$ must be proportional to the Identity matrix for all spin numbers, then it must diagonal matrix with same element on the diagonal, then let $S$ be$$\left[\begin{array}{ccc}a& 0& 0\\ 0& a& 0\\ 0& 0& a\end{array}\right]$$ Substituting this in (5) gives$$\begin{array}{rl}⟨1|{\left[\begin{array}{ccc}a& 0& 0\\ 0& a& 0\\ 0& 0& a\end{array}\right]}^2|1⟩& =2\\ ⟨1|\left[\begin{array}{ccc}{a}^2& 0& 0\\ 0& a^2& 0\\ 0& 0& a^2\end{array}\right]|1⟩& =2=\\ a^2& =2\end{array}$$

Hence $a=\sqrt{2}$. Therefore$$\begin{array}{}\text{(6)}& S=\sqrt{2}\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\end{array}$$ Now that $S$ is found, the next step is to find $S_{+}|2⟩$ and $S_{+}|3⟩$.  Starting with (2), but now applying it to $|2⟩$ gives$$S_z{S}_{+}|2⟩=S_{+}{S}_z|2⟩+S_{+}|2⟩$$ But $S_z|2⟩=0|2⟩$ since $|2⟩$ is the eigenvector associated with $0$ eigenvalue, the above becomes$$S_z{S}_{+}|2⟩=S_{+}|2⟩$$ Which means $S_{+}|2⟩$ is eigenvector of $S_z$ associated with eigenvalue $+1$ which is compatible with experiment. Hence $$S_{+}|2⟩=c|1⟩$$ Where we used $|1⟩$ since that is the eigenvector of $S_z$ associated with $+1$ eigenvalue. Now we need to find $c$. Taking adjoint of both sides of the above gives$$\begin{array}{rl}{(S_{+}|2⟩)}^{†}& ={(c|1⟩)}^{†}\\ ⟨2|S_{+}^{†}& =c^{\ast }⟨1|\end{array}$$

Therefore$$\begin{array}{rl}\text{(7)}& ⟨2|S_{+}^{†}{S}_{+}|2⟩& =c^{\ast }c⟨1|1⟩& ={\left|c\right|}^2\end{array}$$

But $S_{+}^{†}{S}_{+}=S^2-S_z^2-S_z$  which was found earlier above in (4B). Therefore the above equation becomes$$⟨2|(S^2-S_z^2-S_z)|2⟩={\left|c\right|}^2$$ Using $S^2$ found in (6) the above becomes$$\begin{array}{rl}⟨2|(2\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]-\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right]-\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& -1\end{array}\right])|2⟩& ={\left|c\right|}^2\\ ⟨2|\left[\begin{array}{ccc}0& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]|2⟩& ={\left|c\right|}^2\\ \left[\begin{array}{ccc}0& 1& 0\end{array}\right]\left[\begin{array}{ccc}0& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]& ={\left|c\right|}^2\\ 2& ={\left|c\right|}^2\end{array}$$

which implies $c=\sqrt{2}$. Therefore $$\begin{array}{}\text{(8)}& S_{+}|2⟩=\sqrt{2}|1⟩\end{array}$$ Finally, to find $S_{+}|3⟩$,  starting again with (2), but now applying it to $|3⟩$ gives$$S_z{S}_{+}|3⟩=S_{+}{S}_z|3⟩+S_{+}|3⟩$$ But $S_z|3⟩=-|3⟩$ since $|3⟩$ is the eigenvector associated with $-1$ eigenvalue, the above becomes$$\begin{array}{rl}{S}_z{S}_{+}|3⟩& =-S_{+}|3⟩+S_{+}|3⟩\\ & =0S_{+}|3⟩\end{array}$$

Which means $S_{+}|3⟩$ is eigenvector of $S_z$ associated with eigenvalue $0$ which is compatible with experiment. Hence $$S_{+}|3⟩=b|2⟩$$ Where we used $|2⟩$ since that is the eigenvector of $S_z$ associated with $0$ eigenvalue. Now we need to find $b$. Taking adjoint of both sides of the above gives$$\begin{array}{rl}{(S_{+}|3⟩)}^{†}& ={(b|2⟩)}^{†}\\ ⟨3|S_{+}^{†}& =b^{\ast }⟨2|\end{array}$$

Therefore$$\begin{array}{rl}\text{(9)}& ⟨3|S_{+}^{†}{S}_{+}|3⟩& =b^{\ast }b⟨2|2⟩& ={\left|b\right|}^2\end{array}$$

But $S_{+}^{†}{S}_{+}=S^2-S_z^2-S_z$  which was found earlier in (4B). Therefore the above equation becomes$$⟨3|(S^2-S_z^2-S_z)|3⟩={\left|b\right|}^2$$ Using $S^2$ from eq (6) the above becomes$$\begin{array}{rl}⟨3|(2\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]-\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right]-\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& -1\end{array}\right])|3⟩& ={\left|b\right|}^2\\ ⟨3|\left[\begin{array}{ccc}0& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]|3⟩& ={\left|b\right|}^2\\ \left[\begin{array}{ccc}0& 0& 1\end{array}\right]\left[\begin{array}{ccc}0& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]& ={\left|b\right|}^2\\ 2& ={\left|b\right|}^2\end{array}$$

which implies $b=\sqrt{2}$. Therefore $$\begin{array}{}\text{(10)}& S_{+}|3⟩=\sqrt{2}|2⟩\end{array}$$ Now that $S_{+}|1⟩,S_{+}|2⟩,S_{+}|3⟩$ are found, $S_{+}$ can be calculated. From (4,6,8) the result is$$\begin{array}{rl}{S}_{+}|1⟩& =0|1⟩\\ S_{+}|2⟩& =\sqrt{2}|1⟩\\ S_{+}|3⟩& =\sqrt{2}|2⟩\end{array}$$

Hence$$\begin{array}{rl}{S}_{+}& =\left[\begin{array}{ccc}⟨1|S_{+}|1⟩& ⟨1|S_{+}|2⟩& ⟨1|S_{+}|3⟩\\ ⟨2|S_{+}|1⟩& ⟨2|S_{+}|2⟩& ⟨2|S_{+}|3⟩\\ ⟨3|S_{+}|1⟩& ⟨3|S_{+}|2⟩& ⟨3|S_{+}|3⟩\end{array}\right]\\ & =\left[\begin{array}{ccc}0& ⟨1|\sqrt{2}|1⟩& ⟨1|\sqrt{2}|2⟩\\ 0& ⟨2|\sqrt{2}|1⟩& ⟨2|\sqrt{2}|2⟩\\ 0& ⟨3|\sqrt{2}|1⟩& ⟨3|\sqrt{2}|2⟩\end{array}\right]\\ & =\sqrt{2}\left[\begin{array}{ccc}0& ⟨1|1⟩& ⟨1|2⟩\\ 0& ⟨2|1⟩& ⟨2|2⟩\\ 0& ⟨3|1⟩& ⟨3|2⟩\end{array}\right]\end{array}$$

Therefore$$\begin{array}{}\text{(11)}& S_{+}=\left[\begin{array}{ccc}0& \sqrt{2}& 0\\ 0& 0& \sqrt{2}\\ 0& 0& 0\end{array}\right]\end{array}$$ Now that we know $S_{+}$ we turn our attention to finding $S_{-}$. Considering the commutator $[S_z,S_{-}]$ $$\begin{array}{rl}[S_z,S_{-}]& =[S_z,S_x-i{S}_y]\\ & =[S_z,S_x]-i[S_z,S_y]\end{array}$$

But $[S_z,S_x]=i{S}_y$ and $\left[S_z{S}_y\right]=-i{S}_x$. The above becomes$$\begin{array}{rl}[S_z,S_{-}]& =i{S}_y-i(-i{S}_x)\\ & =i{S}_y-S_x\\ & =-S_{-}\end{array}$$

This implies$$\begin{array}{rl}[S_z,S_{-}]& =-S_{-}\\ S_z{S}_{-}-S_{-}{S}_z& =-S_{-}\\ \text{(12)}& S_z{S}_{-}& =S_{-}{S}_z-S_{-}\end{array}$$

Picking $|1⟩$ to start with, gives $$\begin{array}{rl}{S}_z{S}_{-}|1⟩& =(S_{-}{S}_z-S_{-})|1⟩\\ \text{(13)}& & =S_{-}{S}_z|1⟩-S_{-}|1⟩\end{array}$$

But $S_z|1⟩=|1⟩$ since the eigenvalue is $1$ associated with $|1⟩$ eigenvector from (1). The above now becomes$$\begin{array}{rl}{S}_z{S}_{-}|1⟩& =S_{-}|1⟩-S_{-}|1⟩\\ & =0S_{-}|1⟩\end{array}$$

The above shows that $S_{+}|1⟩$ is eigenvector of $S_z$ associated with the eigenvalue $0$ which is compatible with experiments. This implies $$\begin{array}{}\text{(14)}& S_{-}|1⟩=c|2⟩\end{array}$$ Where $|2⟩$ was used above, since that is the eigenvector with $0$ eigenvalue. $c$ is constant to be found. Taking adjoint of both sides of the above gives$$\begin{array}{rl}{(S_{-}|1⟩)}^{†}& ={(c|2⟩)}^{†}\\ ⟨1|S_{-}^{†}& =c^{\ast }⟨2|\end{array}$$

Therefore$$\begin{array}{rl}\text{(15)}& ⟨1|S_{-}^{†}{S}_{-}|1⟩& =c^{\ast }c⟨2|2⟩& ={\left|c\right|}^2\end{array}$$

But $$\begin{array}{rl}{S}_{-}^{†}{S}_{-}& ={(S_x-i{S}_y)}^{†}(S_x-i{S}_y)\\ & =(S_x^{†}+i{S}_y^{†})(S_x-i{S}_y)\end{array}$$

Since $S_x,S_y$ are Hermitian, then $S_x^{†}=S_x$ and $S_y^{†}=S_y$. The above becomes$$\begin{array}{rl}{S}_{-}^{†}{S}_{-}& =(S_x+i{S}_y)(S_x-i{S}_y)\\ & =S_x^2-i{S}_x{S}_y+i{S}_y{S}_x+S_y^2\\ & =S_x^2+i(S_y{S}_x-S_x{S}_y)+S_y^2\\ & =S_x^2+i[S_y,S_x]+S_y^2\end{array}$$

But $[S_y,S_x]=-i{S}_z$. The above becomes$$\begin{array}{rl}{S}_{-}^{†}{S}_{-}& =S_x^2+i(-i{S}_z)+S_y^2\\ \text{(16)}& & =S_x^2+S_y^2+S_z\end{array}$$

Since $S^2=S_x^2+S_y^2+S_z^2$, then $S_x^2+S_y^2=S^2-S_z^2$. This implies$$\begin{array}{}\text{(16A)}& S_{-}^{†}{S}_{-}=S^2-S_z^2+S_z\end{array}$$ Substituting (16A) in (15) gives$$\begin{array}{}\text{(17)}& ⟨1|(S^2-S_z^2+S_z)|1⟩={\left|c\right|}^2\end{array}$$ But $$\begin{array}{}\text{(18)}& S_z^2=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& -1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& -1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right]\end{array}$$ And $S^2$ was found in (6). Substituting (6,18) back in (17) gives an equation to solve for $c$$$\begin{array}{rl}⟨1|\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]-\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right]+\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& -1\end{array}\right]|1⟩& ={\left|c\right|}^2\\ ⟨1|\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 0\end{array}\right]|1⟩& ={\left|c\right|}^2\\ \left[\begin{array}{ccc}1& 0& 0\end{array}\right]\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]& ={\left|c\right|}^2\\ 2& ={\left|c\right|}^2\end{array}$$

Hence $c=\sqrt{2}$. From (14) this implies$$\begin{array}{}\text{(19)}& S_{-}|1⟩=\sqrt{2}|2⟩\end{array}$$ Now we pick $|2⟩$ and using (12) gives $$\begin{array}{rl}{S}_z{S}_{-}|2⟩& =(S_{-}{S}_z-S_{-})|2⟩\\ \text{(20)}& & =S_{-}{S}_z|2⟩-S_{-}|2⟩\end{array}$$

But $S_z|2⟩=0|2⟩$ since the eigenvalue is $0$ associated with $|2⟩$ eigenvector. The above now becomes$$S_z{S}_{-}|2⟩=-S_{-}|2⟩$$ The above shows that $S_{-}|2⟩$ is eigenvector of $S_z$ associated with the eigenvalue $-1$ which is compatible with experiments. This implies $$\begin{array}{}\text{(21)}& S_{-}|2⟩=b|3⟩\end{array}$$ Where $|3⟩$ was used above, since that is the eigenvector with $-1$ eigenvalue. $b$ is constant to be found. Taking adjoint of both sides of the above gives$$\begin{array}{rl}{(S_{-}|2⟩)}^{†}& ={(b|3⟩)}^{†}\\ ⟨2|S_{-}^{†}& =b^{\ast }⟨3|\end{array}$$

Therefore$$\begin{array}{rl}⟨2|S_{-}^{†}{S}_{-}|2⟩& =b^{\ast }b⟨3|3⟩\\ & ={\left|b\right|}^2\end{array}$$

But $S_{-}^{†}{S}_{-}=S^2-S_z^2+S_z$ as calculated earlier in (16A). Hence the above becomes $$⟨2|(S^2-S_z^2+S_z)|2⟩={\left|b\right|}^2$$ Using $S_z^2,S^2$ calculated earlier in the above gives$$\begin{array}{rl}⟨2|\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]-\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right]+\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& -1\end{array}\right]|2⟩& ={\left|b\right|}^2\\ ⟨2|\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 0\end{array}\right]|2⟩& ={\left|b\right|}^2\\ \left[\begin{array}{ccc}0& 1& 0\end{array}\right]\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]& ={\left|b\right|}^2\\ 2& ={\left|b\right|}^2\end{array}$$

Hence $b=\sqrt{2}$. From (21) this implies$$\begin{array}{}\text{(22)}& S_{-}|2⟩=\sqrt{2}|3⟩\end{array}$$ And finally using $|3⟩$ in (12) results in$$\begin{array}{rl}{S}_z{S}_{-}|3⟩& =(S_{-}{S}_z-S_{-})|3⟩\\ & =S_{-}{S}_z|3⟩-S_{-}|3⟩\end{array}$$

But $S_z|3⟩=-|3⟩$ since the eigenvalue is $-1$ associated with $|3⟩$ eigenvector. The above now becomes$$S_z{S}_{-}|3⟩=-2S_{-}|3⟩$$ The above shows that $S_{-}|3⟩$ is eigenvector of $S_z$ associated with the eigenvalue $-2$ which is not compatible with experiments. This implies $$\begin{array}{}\text{(23)}& S_{-}|3⟩=0|1⟩\end{array}$$ Now that $S_{-}|1⟩,S_{-}|2⟩,S_{-}|3⟩$ are all known, we are ready to determine $S_{-}$. From (19,22,23)$$\begin{array}{rl}{S}_{-}|1⟩& =\sqrt{2}|2⟩\\ S_{-}|2⟩& =\sqrt{2}|3⟩\\ S_{-}|3⟩& =0|1⟩\end{array}$$

Therefore$$\begin{array}{rl}{S}_{-}& =\left[\begin{array}{ccc}⟨1|S_{-}|1⟩& ⟨1|S_{-}|2⟩& ⟨1|S_{-}|3⟩\\ ⟨2|S_{-}|1⟩& ⟨2|S_{-}|2⟩& ⟨2|S_{-}|3⟩\\ ⟨3|S_{-}|1⟩& ⟨3|S_{-}|2⟩& ⟨3|S_{-}|3⟩\end{array}\right]\\ & =\left[\begin{array}{ccc}⟨1|\sqrt{2}|2⟩& ⟨1|\sqrt{2}|3⟩& ⟨1|0|1⟩\\ ⟨2|\sqrt{2}|2⟩& ⟨2|\sqrt{2}|3⟩& ⟨2|0|1⟩\\ ⟨3|\sqrt{2}|2⟩& ⟨3|\sqrt{2}|3⟩& ⟨3|0|1⟩\end{array}\right]\\ & =\sqrt{2}\left[\begin{array}{ccc}0& 0& 0\\ 1& 0& 0\\ 0& 1& 0\end{array}\right]\end{array}$$

Therefore$$\begin{array}{}\text{(24)}& S_{-}=\left[\begin{array}{ccc}0& 0& 0\\ \sqrt{2}& 0& 0\\ 0& \sqrt{2}& 0\end{array}\right]\end{array}$$ Now that $S_{+},S_{-}$ are known, $S_x,S_y$ can be found. Using$$\begin{array}{}\text{(25)}& S_{-}& =S_x-i{S}_y\text{(26)}& S_{+}& =S_x+i{S}_y\end{array}$$

Adding the above two equations gives, and using (11,24)$$\begin{array}{rl}{S}_{-}+S_{+}& =2S_x\\ \left[\begin{array}{ccc}0& 0& 0\\ \sqrt{2}& 0& 0\\ 0& \sqrt{2}& 0\end{array}\right]+\left[\begin{array}{ccc}0& \sqrt{2}& 0\\ 0& 0& \sqrt{2}\\ 0& 0& 0\end{array}\right]& =2S_x\\ S_x& =\frac{1}{2}\left[\begin{array}{ccc}0& \sqrt{2}& 0\\ \sqrt{2}& 0& \sqrt{2}\\ 0& \sqrt{2}& 0\end{array}\right]\\ & =\frac{\sqrt{2}}{2}\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right]\end{array}$$

Hence$$\begin{array}{}\text{(27)}& S_x=\frac{1}{\sqrt{2}}\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right]\end{array}$$ And subtracting (25,26) gives$$\begin{array}{rl}{S}_{-}-S_x& =-2i{S}_y\\ \left[\begin{array}{ccc}0& 0& 0\\ \sqrt{2}& 0& 0\\ 0& \sqrt{2}& 0\end{array}\right]-\left[\begin{array}{ccc}0& \sqrt{2}& 0\\ 0& 0& \sqrt{2}\\ 0& 0& 0\end{array}\right]& =-2i{S}_y\\ \left[\begin{array}{ccc}0& -\sqrt{2}& 0\\ \sqrt{2}& 0& -\sqrt{2}\\ 0& \sqrt{2}& 0\end{array}\right]& =-2i{S}_y\\ S_y& =\frac{-1}{2i}\left[\begin{array}{ccc}0& -\sqrt{2}& 0\\ \sqrt{2}& 0& -\sqrt{2}\\ 0& \sqrt{2}& 0\end{array}\right]\\ & =\frac{i}{2}\left[\begin{array}{ccc}0& -\sqrt{2}& 0\\ \sqrt{2}& 0& -\sqrt{2}\\ 0& \sqrt{2}& 0\end{array}\right]\\ & =\frac{\sqrt{2}}{2}\left[\begin{array}{ccc}0& -i& 0\\ i& 0& -i\\ 0& i& 0\end{array}\right]\end{array}$$

Hence $$\begin{array}{}\text{(28)}& S_y=\frac{1}{\sqrt{2}}\left[\begin{array}{ccc}0& -i& 0\\ i& 0& -i\\ 0& i& 0\end{array}\right]\end{array}$$ This completes the solution. We have found $S_x,S_y$, starting from just knowing the eigenvalues of $S_z$.