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\heading{E114: Representation of Operators} 
\auname{Gal Aviv}


{\bf The problem:}

\Dn
In a certain basis the operators A and B are
represented by the square matrix:


\[
\mathcal{A} 
= \left( \amatrix { a & 0 & 0 \cr 0 & -a & 0 \cr 0 & 0 & -a}\right) 
\]

\[
\mathcal{B} 
= \left( \amatrix { -b & 0 & 0 \cr 0 & 0 & ib \cr 0 & -ib & 0}\right) 
\]

\Dn


(1)	Does B have eigenvalue degeneracy?  
\Dn

(2)	Show that the operators A and B commute.  
\Dn 

(3)	Find a new basis where the operators A and B are diagonal. 

\Dn\Dn 


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{\bf The solution:}

\Dn

(1) 
The operator B after digaonalization is:

\[
\mathcal{B} 
= \left( \amatrix { -b & 0 & 0 \cr 0 & -b & 0 \cr 0 & 0 & b}\right) 
\]
That mean $\mathcal{B}$ has second level of degeneration.

\Dn\Dn\Dn

(2)	To show that the operators A and B commute we need to show that 
\Dn
$ [ ${\widehat{A}}$,${\widehat{B}}$ ]$=$ {\widehat{A}\widehat{B}}$ ${-}$ $ {\widehat{B}\widehat{A}} $ = 0

\Dn
By calculating simple matrix multipcation we get the zero value.

\Dn

Therefore the operators commute.

\Dn\Dn 


(3)	To digonalise first we need to find the eigenvectors. 
 
\Dn\Dn	

 	|1>\mapsto$\left( \amatrix { 1  \cr 0  \cr 0  }\right)$  
\Dn 	
 
 	|2>\mapsto$\left( \amatrix { 0  \cr 1/\sqrt{2}  \cr i/\sqrt{2}  }\right)$ 
\Dn 	
 
 	|3>\mapsto$\left( \amatrix { 0  \cr 1/\sqrt{2}  \cr -i/\sqrt{2}  }\right)$ 
\Dn  

\Dn	Matrix T is Unitary therefor T^{-1} = T^{\dagger}

\Dn	By using the transformation elements:
 
\Dn
 
\Dn	${\widetilde{B}}$ = ${ T^{\dagger}}$  $\mathcal{B}$  ${T}$  

\Dn 

We diagonalize B in the new basis

\Dn 

While 

\[
\mathcal{T} 
= \left( \amatrix { 1 & 0 & 0 \cr 0 & 1/\sqrt{2} & 1/\sqrt{2} \cr 0 & i/\sqrt{2} & -i/\sqrt{2}}\right) 
\]
and
\[
\mathcal T^{\dagger}
= \left( \amatrix { 1 & 0 & 0 \cr 0 & 1/\sqrt{2} & -i/\sqrt{2} \cr 0 & 1/\sqrt{2} & i/\sqrt{2}}\right) 
\]

\Dn
The $\mathcal{T}$ Matrix changes $\mathcal{B}$ base to the diagonal one:
\Dn

$\widetilde{B}$ = $ T^{\dagger}$  $\mathcal{B}$ $ T  $

\Dn
	And due to the degeneracy in $\mathcal{B}$, a linear combination of the eigenvectors of $\mathcal{B}$ will digonalise $\mathcal{A}$.

\Dn
	However, by applying ${ T^{\dagger}}$  $\mathcal{A}$  ${T}$ 
\Dn
	We see that $\mathcal{A}$ Remains the same in the new basis
\Dn

$\widetilde{A}$ = $ T^{\dagger}$  $\mathcal{A}$ $ T $ = $\mathcal{A}$

\Dn 

Consequently we have demonstrated that  $\mathcal{A}$ and  $\mathcal{B}$ are both diagonal in the same basis. 

\Dn\Dn\Dn



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