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

\texttt{Improved phrasing of the problem by DC + remarks}

\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: 

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

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

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

(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

\texttt{DC: missing $1/\sqrt{2}$ below.}

\Dn

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

\[ P^{n} = \ket{n}\bra{n} \]
%
\[ P^{nm}  = (\ket{n}+\ket{m})(\bra{n}+\bra{m})   %
            = \ket{n}\bra{n} + \ket{m}\bra{m} + \ket{n}\bra{m} + \ket{m}\bra{n} \]
%
\[ Q^{nm} = (\ket{n}+i\ket{m})(\bra{n}-i\bra{m}) %
            = \ket{n}\bra{n} + \ket{m}\bra{m} - i\ket{n}\bra{m} + i\ket{m}\bra{n} \]

\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

\texttt{DC: At this point it is useful to define the matrices $X^{nm}$ and $Y^{nm}$ and $P^{n}$. 
Then it is very easy to express $A$ using thses matrices. By inspection the result should be:}

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


\texttt{DC: The $X$ and $Y$ matrices are easily expressed uisning the projectors. Results in the next page should be corrected.}

\newpage

Due to the orthonormality relations of the $\{ \ket{n}\}$ basis, the matrix elements of the
projection operators are either 1 or $\pm i$, with the diagonal elements always taking
the value 1:  
 
\[
\bra{k}P^{n,im}\ket{k'} = \braket{k}{n}\braket{n}{k'} %
                        + \braket{k}{m}\braket{m}{k'} %
                        - i\braket{k}{n}\braket{m}{k'} %
                        + i\braket{k}{m}\braket{n}{k'}  %
\]

\Dn

Since $\mathcal{A}$ is Hermitian, it's easy to find the representation in the basis of the
projection operators by inspection:
the symmetrical real and imaginary off-diagonal elements are matched to the multiples of
the operators $P^{n,m}$ and $P^{n,im}$, respectively, while the diagonal elements are added
a $-1$ for each overlap with the former operators, and are then matched to multiples of the
operators $P^{n}$.

\[
 \mathcal{A} =
\left( \amatrix {
0 & 0 & 0 & 0 \cr 0 & 1 & 0 & i \cr 0 & 0 & 0 & 0 \cr 0 & -i & 0 & 1
}\right)
+  \left( \amatrix {
0 & 0 & 0 & 0 \cr 0 & 2 & 2 & 0 \cr 0 & 2 & 2 & 0 \cr 0 & 0 & 0 & 0
}\right)
+  \left( \amatrix {
3 & 0 & 0 & 0 \cr 0 & 4 & 0 & 0 \cr 0 & 0 & 3 & 0 \cr 0 & 0 & 0 & 8
}\right)
 \]

\[ \mathcal{A} = P^{4,i2} + 2P^{2,3} + 3P^{1} + 4P^{2} +3P^{3} +8P^{4}  \]

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\Dn

2.  The representation in the projection operators is easily converted to representation by
the generalized Pauli matrices by using the equation which defines the matrices:

\[ \sigma_{n} = 2P^{n} - P^{n} - P^{n} = P^{n} \]
\[ \sigma_{n,m} = 2P^{n,m} - P^{n} - P^{m} \]
\[ \sigma_{n,im} = 2P^{n,im} - P^{n} - P^{m} \]

\Dn

Thus, the representation in the generalized Pauli matrices is calculated from the result of
part 1:

\begin{align*}
\mathcal{A}  & =  P^{4,i2} + 2P^{2,3} + 3P^{1} + 4P^{2} +3P^{3} +8P^{4}  \\
             & = (P^{4,i2} -\frac{1}{2}P^{4} - \frac{1}{2}P^{2}) + (2P^{2,3} -P^{2} - P^{3}) %
               + 3P^{1} + 4P^{2} +3P^{3} +8P^{4}  \\
             & = \frac{1}{2}\sigma_{4,i2} + \sigma_{2,3} + 3\sigma_{1} %
               + \frac{11}{2}\sigma_{2} +4\sigma_{3} +\frac{17}{2}\sigma_{4}
\end{align*}



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