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solving the spin 1 electron problem. Finding $$S_x,S_y,S_z$$

May 25, 2022   Compiled on May 25, 2022 at 9:36pm

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 {align} |S_{z=1}\rangle & =|1\rangle =\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} \tag {1}\\ |S_{z=0}\rangle & =|2\rangle =\begin {bmatrix} 0\\ 1\\ 0 \end {bmatrix} \nonumber \\ |S_{z=-1}\rangle & =|3\rangle =\begin {bmatrix} 0\\ 0\\ 1 \end {bmatrix} \nonumber \end {align}

Then the following three equations result from writing $$S_{z}|S_{z=\omega _{i}}\rangle =\omega _{i}|S_{z=\omega _{i}}\rangle$$ where $$\omega _{i}$$ is the eigenvalue. These three equation are solved to determine $$S_{z}$$. Let $$S_{z}=\begin {bmatrix} a & b & c\\ d & e & f\\ g & h & m \end {bmatrix}$$. Therefore\begin {align*} \begin {bmatrix} a & b & c\\ d & e & f\\ g & h & m \end {bmatrix}\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} & =1\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} \qquad \\\begin {bmatrix} a & b & c\\ d & e & f\\ g & h & m \end {bmatrix}\begin {bmatrix} 0\\ 1\\ 0 \end {bmatrix} & =0\begin {bmatrix} 0\\ 1\\ 0 \end {bmatrix} \\\begin {bmatrix} a & b & c\\ d & e & f\\ g & h & m \end {bmatrix}\begin {bmatrix} 0\\ 0\\ 1 \end {bmatrix} & =-1\begin {bmatrix} 0\\ 0\\ 1 \end {bmatrix} \end {align*}

Or\begin {align*} \begin {bmatrix} a\\ d\\ g \end {bmatrix} & =\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} \qquad \\\begin {bmatrix} b\\ e\\ h \end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \\\begin {bmatrix} c\\ f\\ m \end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ -1 \end {bmatrix} \end {align*}

Which gives\begin {align} S_{z} & =\begin {bmatrix} a & b & c\\ d & e & f\\ g & h & m \end {bmatrix} \nonumber \\ & =\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} \tag {1A} \end {align}

Now that $$S_{z}$$ is found, the goal is to determine $$S_{x},S_{y}.\$$Let $$S_{-}=S_{x}-iS_{y}$$ and $$S_{+}=S_{x}+iS_{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_{+}$$). We always start with commutator $$\left [ S_{z},S_{+}\right ]$$ \begin {align*} \left [ S_{z},S_{+}\right ] & =\left [ S_{z},S_{x}+iS_{y}\right ] \\ & =\left [ S_{z},S_{x}\right ] +i\left [ S_{z},S_{y}\right ] \end {align*}

But $$\left [ S_{z},S_{x}\right ] =i\sum _{k}\epsilon _{ijk}S_{k}$$. Here $$i=3,j=1$$ since $$z=3,x=1$$. Then $$\left [ S_{z},S_{x}\right ] =i\epsilon _{312}S_{2}=iS_{y}$$ and similarly $$\left [ S_{z}S_{y}\right ] =i\sum _{k}\epsilon _{ijk}S_{k}$$. Here$$\ i=3,j=2$$ since $$z=3,y=1.$$ Then$$\ \left [ S_{z},S_{y}\right ] =i\epsilon _{321}S_{1}=-iS_{x}$$. The above becomes\begin {align*} \left [ S_{z},S_{+}\right ] & =iS_{y}+i\left ( -iS_{x}\right ) \\ & =iS_{y}+S_{x}\\ & =S_{+} \end {align*}

This implies\begin {align} \left [ S_{z},S_{+}\right ] & =S_{+}\nonumber \\ S_{z}S_{+}-S_{+}S_{z} & =S_{+}\nonumber \\ S_{z}S_{+} & =S_{+}S_{z}+S_{+}\tag {2} \end {align}

Picking $$|1\rangle$$ 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 {align} S_{z}S_{+}|1\rangle & =\left ( S_{+}S_{z}+S_{+}\right ) |1\rangle \nonumber \\ & =S_{+}S_{z}|1\rangle +S_{+}|1\rangle \tag {3} \end {align}

But $$S_{z}|1\rangle =|1\rangle$$ since the eigenvalue is $$1$$ associated with $$|1\rangle$$ eigenvector. The above becomes\begin {align*} S_{z}S_{+}|1\rangle & =S_{+}|1\rangle +S_{+}|1\rangle \\ & =2S_{+}|1\rangle \end {align*}

The above shows that $$S_{+}|1\rangle$$ is eigenvector of $$S_{z}$$ associated with the eigenvalue $$2$$ which is not compatible with experiments. Therefore the only logical result is that\begin {equation} S_{+}|1\rangle =0|1\rangle \tag {4} \end {equation} Taking the adjoint gives\begin {align*} \left ( S_{+}|1\rangle \right ) ^{\dag } & =0|1\rangle ^{\dag }\\ \langle 1|S_{+}^{\dag } & =0\langle 1| \end {align*}

Hence\begin {align} \langle 1|S_{+}^{\dag }S_{+}|1\rangle & =0\langle 1|1\rangle \nonumber \\ \langle 1|S_{+}^{\dag }S_{+}|1\rangle & =0\tag {4A} \end {align}

The above is used to ﬁnd $$S^{2}$$. Since $$S_{+}=S_{x}+iS_{y}$$ then the above becomes\begin {align*} \langle 1|\left ( S_{x}^{\dag }-iS_{y}^{\dag }\right ) \left ( S_{x}+iS_{y}\right ) |1\rangle & =0\\ \langle 1|S_{x}^{\dag }S_{x}+iS_{x}^{\dag }S_{y}-iS_{y}^{\dag }S_{x}+S_{y}^{\dag }S_{y}|1\rangle & =0 \end {align*}

But $$S_{x},S_{y}$$ are Hermitian. Therefore $$S_{x}^{\dag }=S_{x},S_{y}^{\dag }=S_{y}$$ and the above reduces to\begin {align*} \langle 1|S_{x}^{2}+iS_{x}S_{y}-iS_{y}S_{x}+S_{y}^{2}|1\rangle & =0\\ \langle 1|S_{x}^{2}+i\left ( S_{x}S_{y}-S_{y}S_{x}\right ) +S_{y}^{2}|1\rangle & =0\\ \langle 1|S_{x}^{2}+i\left [ S_{x},S_{y}\right ] +S_{y}^{2}|1\rangle & =0 \end {align*}

But $$\left [ S_{x},S_{y}\right ] =iS_{z}$$, therefore the above becomes$\langle 1|S_{x}^{2}-S_{z}+S_{y}^{2}|1\rangle =0$ 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 {equation} S_{+}^{\dag }S_{+}=S^{2}-S_{z}^{2}-S_{z}\tag {4B} \end {equation} Therefore (4A) becomes$\langle 1|S^{2}-S_{z}^{2}-S_{z}|1\rangle =0$ Expanding the above gives\begin {align*} \langle 1|S^{2}|1\rangle -\langle 1|S_{z}^{2}|1\rangle -\langle 1|S_{z}|1\rangle & =0\\ \langle 1|S^{2}|1\rangle & =\langle 1|S_{z}^{2}|1\rangle +\langle 1|S_{z}|1\rangle \end {align*}

But $$\langle 1|S_{z}|1\rangle =1$$ and $$\langle 1|S_{z}^{2}|1\rangle =1$$, therefore the above becomes\begin {equation} \langle 1|S^{2}|1\rangle =2\tag {5} \end {equation} 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$\begin {bmatrix} a & 0 & 0\\ 0 & a & 0\\ 0 & 0 & a \end {bmatrix}$ Substituting this in (5) gives\begin {align*} \langle 1|\begin {bmatrix} a & 0 & 0\\ 0 & a & 0\\ 0 & 0 & a \end {bmatrix} ^{2}|1\rangle & =2\\ \langle 1|\begin {bmatrix} a^{2} & 0 & 0\\ 0 & a^{2} & 0\\ 0 & 0 & a^{2}\end {bmatrix} |1\rangle & =2=\\ a^{2} & =2 \end {align*}

Hence $$a=\sqrt {2}$$. Therefore\begin {equation} S=\sqrt {2}\begin {bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end {bmatrix} \tag {6} \end {equation} Now that $$S$$ is found, the next step is to ﬁnd $$S_{+}|2\rangle$$ and $$S_{+}|3\rangle$$.  Starting with (2), but now applying it to $$|2\rangle$$ gives$S_{z}S_{+}|2\rangle =S_{+}S_{z}|2\rangle +S_{+}|2\rangle$ But $$S_{z}|2\rangle =0|2\rangle$$ since $$|2\rangle$$ is the eigenvector associated with $$0\,$$ eigenvalue, the above becomes$S_{z}S_{+}|2\rangle =S_{+}|2\rangle$ Which means $$S_{+}|2\rangle$$ is eigenvector of $$S_{z}$$ associated with eigenvalue $$+1$$ which is compatible with experiment. Hence $S_{+}|2\rangle =c|1\rangle$ Where we used $$|1\rangle$$ 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 {align*} \left ( S_{+}|2\rangle \right ) ^{\dag } & =\left ( c|1\rangle \right ) ^{\dag }\\ \langle 2|S_{+}^{\dag } & =c^{\ast }\langle 1| \end {align*}

Therefore\begin {align} \langle 2|S_{+}^{\dag }S_{+}|2\rangle & =c^{\ast }c\langle 1|1\rangle \tag {7}\\ & =\left \vert c\right \vert ^{2}\nonumber \end {align}

But $$S_{+}^{\dag }S_{+}=S^{2}-S_{z}^{2}-S_{z}$$  which was found earlier above in (4B). Therefore the above equation becomes$\langle 2|\left ( S^{2}-S_{z}^{2}-S_{z}\right ) |2\rangle =\left \vert c\right \vert ^{2}$ Using $$S^{2}$$ found in (6) the above becomes\begin {align*} \langle 2|\left ( 2\begin {bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end {bmatrix} -\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end {bmatrix} -\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} \right ) |2\rangle & =\left \vert c\right \vert ^{2}\\ \langle 2|\begin {bmatrix} 0 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix} |2\rangle & =\left \vert c\right \vert ^{2}\\\begin {bmatrix} 0 & 1 & 0 \end {bmatrix}\begin {bmatrix} 0 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix}\begin {bmatrix} 0\\ 1\\ 0 \end {bmatrix} & =\left \vert c\right \vert ^{2}\\ 2 & =\left \vert c\right \vert ^{2} \end {align*}

which implies $$c=\sqrt {2}$$. Therefore \begin {equation} S_{+}|2\rangle =\sqrt {2}|1\rangle \tag {8} \end {equation} Finally, to ﬁnd $$S_{+}|3\rangle$$,  starting again with (2), but now applying it to $$|3\rangle$$ gives$S_{z}S_{+}|3\rangle =S_{+}S_{z}|3\rangle +S_{+}|3\rangle$ But $$S_{z}|3\rangle =-|3\rangle$$ since $$|3\rangle$$ is the eigenvector associated with $$-1\,$$ eigenvalue, the above becomes\begin {align*} S_{z}S_{+}|3\rangle & =-S_{+}|3\rangle +S_{+}|3\rangle \\ & =0S_{+}|3\rangle \end {align*}

Which means $$S_{+}|3\rangle$$ is eigenvector of $$S_{z}$$ associated with eigenvalue $$0$$ which is compatible with experiment. Hence $S_{+}|3\rangle =b|2\rangle$ Where we used $$|2\rangle$$ 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 {align*} \left ( S_{+}|3\rangle \right ) ^{\dag } & =\left ( b|2\rangle \right ) ^{\dag }\\ \langle 3|S_{+}^{\dag } & =b^{\ast }\langle 2| \end {align*}

Therefore\begin {align} \langle 3|S_{+}^{\dag }S_{+}|3\rangle & =b^{\ast }b\langle 2|2\rangle \tag {9}\\ & =\left \vert b\right \vert ^{2}\nonumber \end {align}

But $$S_{+}^{\dag }S_{+}=S^{2}-S_{z}^{2}-S_{z}$$  which was found earlier in (4B). Therefore the above equation becomes$\langle 3|\left ( S^{2}-S_{z}^{2}-S_{z}\right ) |3\rangle =\left \vert b\right \vert ^{2}$ Using $$S^{2}$$ from eq (6) the above becomes\begin {align*} \langle 3|\left ( 2\begin {bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end {bmatrix} -\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end {bmatrix} -\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} \right ) |3\rangle & =\left \vert b\right \vert ^{2}\\ \langle 3|\begin {bmatrix} 0 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix} |3\rangle & =\left \vert b\right \vert ^{2}\\\begin {bmatrix} 0 & 0 & 1 \end {bmatrix}\begin {bmatrix} 0 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix}\begin {bmatrix} 0\\ 0\\ 1 \end {bmatrix} & =\left \vert b\right \vert ^{2}\\ 2 & =\left \vert b\right \vert ^{2} \end {align*}

which implies $$b=\sqrt {2}$$. Therefore \begin {equation} S_{+}|3\rangle =\sqrt {2}|2\rangle \tag {10} \end {equation} Now that $$S_{+}|1\rangle ,S_{+}|2\rangle ,S_{+}|3\rangle$$ are found, $$S_{+}$$ can be calculated. From (4,6,8) the result is\begin {align*} S_{+}|1\rangle & =0|1\rangle \\ S_{+}|2\rangle & =\sqrt {2}|1\rangle \\ S_{+}|3\rangle & =\sqrt {2}|2\rangle \end {align*}

Hence\begin {align*} S_{+} & =\begin {bmatrix} \langle 1|S_{+}|1\rangle & \langle 1|S_{+}|2\rangle & \langle 1|S_{+}|3\rangle \\ \langle 2|S_{+}|1\rangle & \langle 2|S_{+}|2\rangle & \langle 2|S_{+}|3\rangle \\ \langle 3|S_{+}|1\rangle & \langle 3|S_{+}|2\rangle & \langle 3|S_{+}|3\rangle \end {bmatrix} \\ & =\begin {bmatrix} 0 & \langle 1|\sqrt {2}|1\rangle & \langle 1|\sqrt {2}|2\rangle \\ 0 & \langle 2|\sqrt {2}|1\rangle & \langle 2|\sqrt {2}|2\rangle \\ 0 & \langle 3|\sqrt {2}|1\rangle & \langle 3|\sqrt {2}|2\rangle \end {bmatrix} \\ & =\sqrt {2}\begin {bmatrix} 0 & \langle 1|1\rangle & \langle 1|2\rangle \\ 0 & \langle 2|1\rangle & \langle 2|2\rangle \\ 0 & \langle 3|1\rangle & \langle 3|2\rangle \end {bmatrix} \end {align*}

Therefore\begin {equation} S_{+}=\begin {bmatrix} 0 & \sqrt {2} & 0\\ 0 & 0 & \sqrt {2}\\ 0 & 0 & 0 \end {bmatrix} \tag {11} \end {equation} Now that we know $$S_{+}$$ we turn our attention to ﬁnding $$S_{-}$$. Considering the commutator $$\left [ S_{z},S_{-}\right ]$$ \begin {align*} \left [ S_{z},S_{-}\right ] & =\left [ S_{z},S_{x}-iS_{y}\right ] \\ & =\left [ S_{z},S_{x}\right ] -i\left [ S_{z},S_{y}\right ] \end {align*}

But $$\left [ S_{z},S_{x}\right ] =iS_{y}$$ and $$\left [ S_{z}S_{y}\right ] =-iS_{x}$$. The above becomes\begin {align*} \left [ S_{z},S_{-}\right ] & =iS_{y}-i\left ( -iS_{x}\right ) \\ & =iS_{y}-S_{x}\\ & =-S_{-} \end {align*}

This implies\begin {align} \left [ S_{z},S_{-}\right ] & =-S_{-}\nonumber \\ S_{z}S_{-}-S_{-}S_{z} & =-S_{-}\nonumber \\ S_{z}S_{-} & =S_{-}S_{z}-S_{-}\tag {12} \end {align}

Picking $$|1\rangle$$ to start with, gives \begin {align} S_{z}S_{-}|1\rangle & =\left ( S_{-}S_{z}-S_{-}\right ) |1\rangle \nonumber \\ & =S_{-}S_{z}|1\rangle -S_{-}|1\rangle \tag {13} \end {align}

But $$S_{z}|1\rangle =|1\rangle$$ since the eigenvalue is $$1$$ associated with $$|1\rangle$$ eigenvector from (1). The above now becomes\begin {align*} S_{z}S_{-}|1\rangle & =S_{-}|1\rangle -S_{-}|1\rangle \\ & =0S_{-}|1\rangle \end {align*}

The above shows that $$S_{+}|1\rangle$$ is eigenvector of $$S_{z}$$ associated with the eigenvalue $$0$$ which is compatible with experiments. This implies \begin {equation} S_{-}|1\rangle =c|2\rangle \tag {14} \end {equation} Where $$|2\rangle$$ 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 {align*} \left ( S_{-}|1\rangle \right ) ^{\dag } & =\left ( c|2\rangle \right ) ^{\dag }\\ \langle 1|S_{-}^{\dag } & =c^{\ast }\langle 2| \end {align*}

Therefore\begin {align} \langle 1|S_{-}^{\dag }S_{-}|1\rangle & =c^{\ast }c\langle 2|2\rangle \tag {15}\\ & =\left \vert c\right \vert ^{2}\nonumber \end {align}

But \begin {align*} S_{-}^{\dag }S_{-} & =\left ( S_{x}-iS_{y}\right ) ^{\dag }\left ( S_{x}-iS_{y}\right ) \\ & =\left ( S_{x}^{\dag }+iS_{y}^{\dag }\right ) \left ( S_{x}-iS_{y}\right ) \end {align*}

Since $$S_{x},S_{y}$$ are Hermitian, then $$S_{x}^{\dag }=S_{x}$$ and $$S_{y}^{\dag }=S_{y}$$. The above becomes\begin {align*} S_{-}^{\dag }S_{-} & =\left ( S_{x}+iS_{y}\right ) \left ( S_{x}-iS_{y}\right ) \\ & =S_{x}^{2}-iS_{x}S_{y}+iS_{y}S_{x}+S_{y}^{2}\\ & =S_{x}^{2}+i\left ( S_{y}S_{x}-S_{x}S_{y}\right ) +S_{y}^{2}\\ & =S_{x}^{2}+i\left [ S_{y},S_{x}\right ] +S_{y}^{2} \end {align*}

But $$\left [ S_{y},S_{x}\right ] =-iS_{z}$$. The above becomes\begin {align} S_{-}^{\dag }S_{-} & =S_{x}^{2}+i\left ( -iS_{z}\right ) +S_{y}^{2}\nonumber \\ & =S_{x}^{2}+S_{y}^{2}+S_{z}\tag {16} \end {align}

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 {equation} S_{-}^{\dag }S_{-}=S^{2}-S_{z}^{2}+S_{z}\tag {16A} \end {equation} Substituting (16A) in (15) gives\begin {equation} \langle 1|\left ( S^{2}-S_{z}^{2}+S_{z}\right ) |1\rangle =\left \vert c\right \vert ^{2}\tag {17} \end {equation} But \begin {equation} S_{z}^{2}=\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix}\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} =\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end {bmatrix} \tag {18} \end {equation} And $$S^{2}$$ was found in (6). Substituting (6,18) back in (17) gives an equation to solve for $$c$$\begin {align*} \langle 1|\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix} -\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end {bmatrix} +\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} |1\rangle & =\left \vert c\right \vert ^{2}\\ \langle 1|\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 0 \end {bmatrix} |1\rangle & =\left \vert c\right \vert ^{2}\\\begin {bmatrix} 1 & 0 & 0 \end {bmatrix}\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 0 \end {bmatrix}\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} & =\left \vert c\right \vert ^{2}\\ 2 & =\left \vert c\right \vert ^{2} \end {align*}

Hence $$c=\sqrt {2}$$. From (14) this implies\begin {equation} S_{-}|1\rangle =\sqrt {2}|2\rangle \tag {19} \end {equation} Now we pick $$|2\rangle$$ and using (12) gives \begin {align} S_{z}S_{-}|2\rangle & =\left ( S_{-}S_{z}-S_{-}\right ) |2\rangle \nonumber \\ & =S_{-}S_{z}|2\rangle -S_{-}|2\rangle \tag {20} \end {align}

But $$S_{z}|2\rangle =0|2\rangle$$ since the eigenvalue is $$0$$ associated with $$|2\rangle$$ eigenvector. The above now becomes$S_{z}S_{-}|2\rangle =-S_{-}|2\rangle$ The above shows that $$S_{-}|2\rangle$$ is eigenvector of $$S_{z}$$ associated with the eigenvalue $$-1$$ which is compatible with experiments. This implies \begin {equation} S_{-}|2\rangle =b|3\rangle \tag {21} \end {equation} Where $$|3\rangle$$ 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 {align*} \left ( S_{-}|2\rangle \right ) ^{\dag } & =\left ( b|3\rangle \right ) ^{\dag }\\ \langle 2|S_{-}^{\dag } & =b^{\ast }\langle 3| \end {align*}

Therefore\begin {align*} \langle 2|S_{-}^{\dag }S_{-}|2\rangle & =b^{\ast }b\langle 3|3\rangle \\ & =\left \vert b\right \vert ^{2} \end {align*}

But $$S_{-}^{\dag }S_{-}=S^{2}-S_{z}^{2}+S_{z}$$ as calculated earlier in (16A). Hence the above becomes $\langle 2|\left ( S^{2}-S_{z}^{2}+S_{z}\right ) |2\rangle =\left \vert b\right \vert ^{2}$ Using $$S_{z}^{2},S^{2}$$ calculated earlier in the above gives\begin {align*} \langle 2|\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix} -\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end {bmatrix} +\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} |2\rangle & =\left \vert b\right \vert ^{2}\\ \langle 2|\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 0 \end {bmatrix} |2\rangle & =\left \vert b\right \vert ^{2}\\\begin {bmatrix} 0 & 1 & 0 \end {bmatrix}\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 0 \end {bmatrix}\begin {bmatrix} 0\\ 1\\ 0 \end {bmatrix} & =\left \vert b\right \vert ^{2}\\ 2 & =\left \vert b\right \vert ^{2} \end {align*}

Hence $$b=\sqrt {2}$$. From (21) this implies\begin {equation} S_{-}|2\rangle =\sqrt {2}|3\rangle \tag {22} \end {equation} And ﬁnally using $$|3\rangle$$ in (12) results in\begin {align*} S_{z}S_{-}|3\rangle & =\left ( S_{-}S_{z}-S_{-}\right ) |3\rangle \\ & =S_{-}S_{z}|3\rangle -S_{-}|3\rangle \end {align*}

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

Therefore\begin {align*} S_{-} & =\begin {bmatrix} \langle 1|S_{-}|1\rangle & \langle 1|S_{-}|2\rangle & \langle 1|S_{-}|3\rangle \\ \langle 2|S_{-}|1\rangle & \langle 2|S_{-}|2\rangle & \langle 2|S_{-}|3\rangle \\ \langle 3|S_{-}|1\rangle & \langle 3|S_{-}|2\rangle & \langle 3|S_{-}|3\rangle \end {bmatrix} \\ & =\begin {bmatrix} \langle 1|\sqrt {2}|2\rangle & \langle 1|\sqrt {2}|3\rangle & \langle 1|0|1\rangle \\ \langle 2|\sqrt {2}|2\rangle & \langle 2|\sqrt {2}|3\rangle & \langle 2|0|1\rangle \\ \langle 3|\sqrt {2}|2\rangle & \langle 3|\sqrt {2}|3\rangle & \langle 3|0|1\rangle \end {bmatrix} \\ & =\sqrt {2}\begin {bmatrix} 0 & 0 & 0\\ 1 & 0 & 0\\ 0 & 1 & 0 \end {bmatrix} \end {align*}

Therefore\begin {equation} S_{-}=\begin {bmatrix} 0 & 0 & 0\\ \sqrt {2} & 0 & 0\\ 0 & \sqrt {2} & 0 \end {bmatrix} \tag {24} \end {equation} Now that $$S_{+},S_{-}$$ are known, $$S_{x},S_{y}$$ can be found. Using\begin {align} S_{-} & =S_{x}-iS_{y}\tag {25}\\ S_{+} & =S_{x}+iS_{y}\tag {26} \end {align}

Adding the above two equations gives, and using (11,24)\begin {align*} S_{-}+S_{+} & =2S_{x}\\\begin {bmatrix} 0 & 0 & 0\\ \sqrt {2} & 0 & 0\\ 0 & \sqrt {2} & 0 \end {bmatrix} +\begin {bmatrix} 0 & \sqrt {2} & 0\\ 0 & 0 & \sqrt {2}\\ 0 & 0 & 0 \end {bmatrix} & =2S_{x}\\ S_{x} & =\frac {1}{2}\begin {bmatrix} 0 & \sqrt {2} & 0\\ \sqrt {2} & 0 & \sqrt {2}\\ 0 & \sqrt {2} & 0 \end {bmatrix} \\ & =\frac {\sqrt {2}}{2}\begin {bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end {bmatrix} \end {align*}

Hence\begin {equation} S_{x}=\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end {bmatrix} \tag {27} \end {equation} And subtracting (25,26) gives\begin {align*} S_{-}-S_{x} & =-2iS_{y}\\\begin {bmatrix} 0 & 0 & 0\\ \sqrt {2} & 0 & 0\\ 0 & \sqrt {2} & 0 \end {bmatrix} -\begin {bmatrix} 0 & \sqrt {2} & 0\\ 0 & 0 & \sqrt {2}\\ 0 & 0 & 0 \end {bmatrix} & =-2iS_{y}\\\begin {bmatrix} 0 & -\sqrt {2} & 0\\ \sqrt {2} & 0 & -\sqrt {2}\\ 0 & \sqrt {2} & 0 \end {bmatrix} & =-2iS_{y}\\ S_{y} & =\frac {-1}{2i}\begin {bmatrix} 0 & -\sqrt {2} & 0\\ \sqrt {2} & 0 & -\sqrt {2}\\ 0 & \sqrt {2} & 0 \end {bmatrix} \\ & =\frac {i}{2}\begin {bmatrix} 0 & -\sqrt {2} & 0\\ \sqrt {2} & 0 & -\sqrt {2}\\ 0 & \sqrt {2} & 0 \end {bmatrix} \\ & =\frac {\sqrt {2}}{2}\begin {bmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end {bmatrix} \end {align*}

Hence \begin {equation} S_{y}=\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end {bmatrix} \tag {28} \end {equation} This completes the solution. We have found $$S_{x},S_{y}$$, starting from just knowing the eigenvalues of $$S_{z}$$.