Time

For a spin 1 system l = 1 and m = -1 , 0 , 1. For the operator $S_{z}$ we have

$S_{z}|l,m\rangle$ $\Rightarrow$ $\langle l,n|S_{z}|l,m\rangle =m\hbar \langle l,n|l,m\rangle$ $\Rightarrow (S_{z})_{nm}=m\hbar \delta _{nm}$

So

$S_{z}=\hbar {\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\\\end{pmatrix}}$ $\Rightarrow$ $S_{z}^{2}={\hbar }^{2}{\begin{pmatrix}1&0&0\\0&0&0\\0&0&1\\\end{pmatrix}}$

For the $S_{x}$ operator we have

$S_{x}={\frac {\hbar }{2}}{\begin{pmatrix}0&{\sqrt {2}}&0\\{\sqrt {2}}&0&{\sqrt {2}}\\0&{\sqrt {2}}&0\\\end{pmatrix}}$ $\Rightarrow$ $S_{x}^{2}={\hbar }^{2}{\begin{pmatrix}{\frac {1}{2}}&0&{\frac {1}{2}}\\0&1&0\\{\frac {1}{2}}&0&{\frac {1}{2}}\\\end{pmatrix}}$ $\Rightarrow$ $S_{y}={\frac {\hbar }{2i}}{\begin{pmatrix}0&{\sqrt {2}}&0\\-{\sqrt {2}}&0&{\sqrt {2}}\\0&-{\sqrt {2}}&0\\\end{pmatrix}}$ $S_{y}^{2}={\hbar }^{2}{\begin{pmatrix}{\frac {1}{2}}&0&-{\frac {1}{2}}\\0&1&0\\-{\frac {1}{2}}&0&{\frac {1}{2}}\\\end{pmatrix}}$

Thus the Hamiltonian can be represented by the matrix

$H={\hbar }^{2}{\begin{pmatrix}A&0&B\\0&0&0\\B&0&A\\\end{pmatrix}}$

To find the energy eigenvalues we have to solve the secular equation

$det(H-\lambda I)=0$ $\Rightarrow$ $det{\begin{pmatrix}A{\hbar }^{2}-\lambda &0&B{\hbar }^{2}\\0&-\lambda &0\\B{\hbar }^{2}&0&A{\hbar }^{2}-\lambda \\\end{pmatrix}}=0$

$\Rightarrow$ $\lambda _{1}$ = 0, $\lambda _{2}={\hbar }^{2}(A+B)$, $\lambda _{3}={\hbar }^{2}(A-B)$

To find the eigenstate $|n_{(}\lambda )$ that corresponds to the eigenvalue $\lambda$ we have to solve the following equation:

${\hbar }^{2}{\begin{pmatrix}A&0&B\\0&0&0\\B&0&A\\\end{pmatrix}}{\begin{pmatrix}a\\b\\c\end{pmatrix}}\Rightarrow \lambda _{c}{\begin{pmatrix}a\\b\\c\end{pmatrix}}$

For $\lambda _{1}=0$

${\hbar }^{2}{\begin{pmatrix}A&0&B\\0&0&0\\B&0&A\\\end{pmatrix}}{\begin{pmatrix}a\\b\\c\end{pmatrix}}=0\Rightarrow a=0,c=0$

$|n_{0}\rangle ={\begin{pmatrix}0\\b\\0\end{pmatrix}}(normalizing)\Rightarrow |n_{0}\rangle ={\begin{pmatrix}0\\1\\0\end{pmatrix}}\Rightarrow |n_{0}\rangle =|1,0\rangle$

In the same way for $\lambda _{2}={\hbar }^{2}(A+B)$

${\begin{pmatrix}A&0&B\\0&0&0\\B&0&A\\\end{pmatrix}}{\begin{pmatrix}a\\b\\c\end{pmatrix}}=(A+B){\begin{pmatrix}a\\b\\c\end{pmatrix}}\Rightarrow a=c,b=0$

$|n_{A+B}\rangle ={\begin{pmatrix}c\\0\\c\end{pmatrix}}(normalizing)\Rightarrow |n_{0}\rangle ={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1\\0\\1\end{pmatrix}}\Rightarrow |n_{A+B}\rangle ={\frac {1}{\sqrt {2}}}|1,+1\rangle +{\frac {1}{\sqrt {2}}}|1,-1\rangle$

In the same way for $\lambda _{2}={\hbar }^{2}(A+B)$

${\begin{pmatrix}A&0&B\\0&0&0\\B&0&A\\\end{pmatrix}}{\begin{pmatrix}a\\b\\c\end{pmatrix}}=(A+B){\begin{pmatrix}a\\b\\c\end{pmatrix}}\Rightarrow a=c,b=0$

$|n_{A+B}\rangle ={\begin{pmatrix}c\\0\\c\end{pmatrix}}(normalizing)\Rightarrow |n_{A+B}\rangle ={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1\\0\\1\end{pmatrix}}\Rightarrow |n_{A+B}\rangle ={\frac {1}{\sqrt {2}}}|1,+1\rangle +{\frac {1}{\sqrt {2}}}|1,-1\rangle$

For $\lambda _{3}={\hbar }^{2}(A-B)$

${\begin{pmatrix}A&0&B\\0&0&0\\B&0&A\\\end{pmatrix}}{\begin{pmatrix}a\\b\\c\end{pmatrix}}=(A-B){\begin{pmatrix}a\\b\\c\end{pmatrix}}\Rightarrow a=-c,b=0$

$|n_{A-B}\rangle ={\begin{pmatrix}c\\0\\-c\end{pmatrix}}(normalizing)\Rightarrow |n_{A-B}\rangle ={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1\\0\\-1\end{pmatrix}}\Rightarrow |n_{A-B}\rangle ={\frac {1}{\sqrt {2}}}|1,+1\rangle -{\frac {1}{\sqrt {2}}}|1,-1\rangle$

Now we are going to check if the Hamiltonian is invariant under time reversal

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