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\begin{document} 
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\heading{E1020: Pauli Matrices} 
\auname{Maxim Zatonskikh}


{\bf The problem:}

\The Pauli Matrices, including the Identity Matrix, are defined as:\Dn

${{\sigma }_{0}}=\left( \begin{matrix}
   1 & 0  \\

   0 & 1  \\
\end{matrix} \right);{{\sigma }_{1}}=\left( \begin{matrix}
   0 & 1  \\

   1 & 0  \\
\end{matrix} \right);{{\sigma }_{2}}=\left( \begin{matrix}
   0 & -i  \\

   i & 0  \\
\end{matrix} \right);{{\sigma }_{3}}=\left( \begin{matrix}
   1 & 0  \\

   0 & -1  \\
\end{matrix} \right)$\Dn

The Pauli Matrices and the Identity Matrix together are the complete basis of all $2\times 2$ matrices.\Dn

We can define the representation of the given operator A by $A=\sum\limits_{i}{{{a}_{i}}{{\sigma }_{i}}}$.\Dn

(1) Write these matrices using Dirac notation.\Dn

(2) Express $Tr\left( {{\sigma }_{k}}A \right)$ and $Tr\left( A \right)$ with ${{a}_{i}}$.\Dn

(3) Generalize the formula for general case: calculate $Tr\left( AB \right)$ of two operators.\Dn




\Dn\Dn


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{\bf The solution:}

\Dn

(1) The $2\times 2$ matrices are two dimensional matrices. Therefore, we define two orthogonal states (vectors):

\Dn
$\left| 1 \right\rangle =\left( \begin{matrix}
   0  \\
   1  \\
\end{matrix} \right);\left| 2 \right\rangle =\left( \begin{matrix}
   1  \\
   0  \\
\end{matrix} \right)$\Dn


Let us use the Dirac notation for writing matrices:
\Dn

$A=\sum\limits_{i,j}{\left| i \right\rangle {{A}_{ij}}\left\langle  j \right|}$
\Dn

So we can use this for writing Pauli matrices:
\Dn

$\begin{align}

& {{\sigma }_{0}}=\left| 1 \right\rangle \left\langle  1 \right|+\left| 2 \right\rangle \left\langle  2 \right| \\ 
\Dn

& {{\sigma }_{1}}=\left| 1 \right\rangle \left\langle  2 \right|+\left| 2 \right\rangle \left\langle  1 \right| \\ 
\Dn

 & {{\sigma }_{2}}=-i\left| 1 \right\rangle \left\langle  2 \right|+i\left| 2 \right\rangle \left\langle  1 \right| \\ 
\Dn

 & {{\sigma }_{3}}=\left| 1 \right\rangle \left\langle  1 \right|-\left| 2 \right\rangle \left\langle  2 \right| \\ 
\end{align}$
\Dn

%%%%%%%%%%%%%%%
(2) At the beginning we express $Tr\left( A \right)$:
\Dn

$Tr\left( A \right)=\left\langle  1 \right|A\left| 1 \right\rangle +\left\langle  2 \right|A\left| 2 \right\rangle =\sum\limits_{i}{{{a}_{i}}\left( \left\langle  1 \right|{{\sigma }_{i}}\left| 1 \right\rangle +\left\langle  2 \right|{{\sigma }_{i}}\left| 2 \right\rangle  \right)}={{a}_{0}}+{{a}_{0}}+{{a}_{3}}-{{a}_{3}}=2{{a}_{0}}$
\Dn

Let us represent the multiplication of Pauli matrices by Pauli matrices:
\Dn

${{\sigma }_{i}}{{\sigma }_{j}}=I{{\delta }_{ij}}+i{{\varepsilon }_{ijk}}{{\sigma }_{k}}$
\Dn

Now we can use those multiplications in the next expression:
\Dn

 $Tr\left( {{\sigma }_{k}}A \right)=\sum\limits_{i}{{{a}_{i}}\left( \left\langle  1 \right|{{\sigma }_{k}}{{\sigma }_{i}}\left| 1 \right\rangle +\left\langle  2 \right|{{\sigma }_{k}}{{\sigma }_{i}}\left| 2 \right\rangle  \right)}$
\Dn

So we get:
\Dn

$Tr\left( {{\sigma }_{k}}A \right)=Tr\left( {{\sigma }_{k}}\sum\limits_{i}{{{a}_{i}}{{\sigma }_{i}}} \right)=Tr\left( \sum\limits_{i}{{{a}_{i}}{{\sigma }_{k}}{{\sigma }_{i}}} \right)=\sum\limits_{i}{{{a}_{i}}}Tr\left( {{\sigma }_{k}}{{\sigma }_{i}} \right)={{a}_{k}}Tr\left( {{\sigma }_{k}}{{\sigma }_{k}} \right)=2{{a}_{k}}$
\Dn

%%%%%%%%%%%%%%
(3) With the replacing ${{\sigma }_{k}}$ by the general operator $B$ easily we can see that:
\Dn

$Tr\left( AB \right)=\sum\limits_{i,j}{{{a}_{i}}}{{b}_{j}}Tr\left( {{\sigma }_{i}}{{\sigma }_{j}} \right)=2\cdot \sum\limits_{i}{{{a}_{i}}{{b}_{i}}}$

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\end{document} 
