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\heading{E1025: Projector Operators as a Representation Basis}
\auname{Christine Khripkov}

\Dn

{\bf The problem:}

\Dn

The projection operators $P^{n}$ and $P^{nm}$ and $Q^{nm}$ 
on the $N^2$ states $\ket{n}$, and $\ket{n}+\ket{m}$, and $\ket{n}+i\ket{m}$ respectively,  
form a complete basis for the representations of operators in the $N$-dimensional Hilbert space.
We also define the generalized Pauli matrices: 

\vspace{-15pt}
\[ X^{nm} = 2P^{nm} - P^{n} - P^{m} \]
\[ Y^{nm} = 2Q^{nm} - P^{n} - P^{m} \]

Consider the matrix $\mathcal{A}$ which is written in the standard basis ${\ket{n}}$  as:

\vspace{-10pt}
\[
  \mathcal{A} =
\left( \amatrix {
3 & 0 & 0 & 0 \cr 0 & 7 & 2 & i \cr 0 & 2 & 5 & 0 \cr 0 & -i & 0 & 9
}\right)
 \]

(1) Explain how the $P^{n}$ projectors, together with the generalized Pauli matrices 
can be used as a {\em convenient} basis. Express $\mathcal{A}$ in this basis.

(2) Express $\mathcal{A}$ as a linear combination of the projection operators as defined above.


\Dn\Dn

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{\bf The solution:}

\Dn

1.  The projection operators are represented in the standard basis $\{ \ket{n}\}$ by:

\vspace{-10pt}
\[ P^{n} = \ket{n}\bra{n} \]

\vspace{-15pt}
\[ P^{nm}  = \frac{1}{\sqrt{2}}(\ket{n}+\ket{m})\frac{1}{\sqrt{2}}(\bra{n}+\bra{m})   %
            = \frac{1}{2}(\ket{n}\bra{n} + \ket{m}\bra{m} + \ket{n}\bra{m} + \ket{m}\bra{n}) \]
            
\vspace{-10pt}
\[ Q^{nm} = \frac{1}{\sqrt{2}}(\ket{n}+i\ket{m})\frac{1}{\sqrt{2}}(\bra{n}-i\bra{m}) %
            = \frac{1}{2}(\ket{n}\bra{n} + \ket{m}\bra{m} - i\ket{n}\bra{m} + i\ket{m}\bra{n}) \]

\Dn

For the 4-dimentional space, there are 4 different $P^{n}$ operators, 6 different $P^{nm}$, 
and 6 different $Q^{nm}$, making a total of $N^2 = 16$ basis operators.

\Dn\Dn

While the $P^{n}$ operators have a single diagonal element, the $P^{nm}$ and the $Q^{nm}$
operators have both diagonal and off-diagonal elements. Thus, it is convenient to define a 
new basis for the $N^2$ representation:

\vspace{-10pt}
\[ P^{n} = \ket{n}\bra{n} \]

\vspace{-15pt}
\[ X^{nm} = 2P^{n,m} - P^{n} - P^{m} = \ket{n}\bra{m} + \ket{m}\bra{n}\]

\vspace{-15pt}
\[ Y^{nm} = 2Q^{n,m} - P^{n} - P^{m} = - i\ket{n}\bra{m} + i\ket{m}\bra{n}\]

\pagebreak
In this basis the diagonal and the off-diagonal elements are separated; hence, the matrix
 $\mathcal{A}$ can be easily decomposed as:

\vspace{-10pt}
\[ \mathcal{A} = 3P^{1} + 7P^{2} + 5P^{3} + 9P^{4} + 2X^{2,3} - Y^{2,4} \]

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\Dn

2.  The representation in the $\left\lbrace  P^{n}, X^{nm}, Y^{nm} \right\rbrace $ basis can be converted to the
$\left\lbrace P^{n},  P^{n,m}, Q^{n,m}\right\rbrace $ basis by substituting back the definitions of $X^{nm}$ and
 $Y^{nm}$:

\begin{align*}
\mathcal{A} &= 3P^{1} + 7P^{2} + 5P^{3} + 9P^{4} + 2(2P^{2,3} - P^{2} - P^{3}) - (2Q^{2,4} - P^{2} - P^{4}) \\
   & = 3P^{1} + 6P^{2} + 3P^{3} + 10P^{4} + 4P^{2,3} - 2Q^{2,4} 
   \end{align*}  




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