Robben said:
I can't see it. Can you elaborate please?
For example let us take ##S_z## and systematically write down it's representation in the ##\{ |1,-1\rangle, |1,0\rangle, |1,1\rangle \}## basis. First of all we need to know the action of the operator ##S_z## on this basis. Since these states are eigenstates of the ##S_z## operator this is simple to write down as,
## S_z |1,-1\rangle = -\hbar |1 -1\rangle ##
## S_z |1,0\rangle = 0 ##
## S_z |1,1\rangle = \hbar |1 1\rangle ##
Now that we know this we can construct the matrix representation of ##S_z## in this basis by making the following assignements
## |1,1\rangle = \left( \begin{array}{c}
1\\
0 \\
0 \end{array} \right),
|1,0\rangle =
\left( \begin{array}{c}
0\\
1 \\
0 \end{array} \right),
|1,-1\rangle = \left( \begin{array}{c}
0\\
0 \\
1 \end{array} \right)
##
and the matrix representation of the operator is
##S_z = \left( \begin{array}{ccc}
\langle 1,1 | S_z | 1,1 \rangle &\langle 1,1 | S_z |1,0\rangle &\langle 1,1| S_z | 1,-1 \rangle \\
\langle 1,0 | S_z | 1,1 \rangle &\langle 1,0 | S_z |1,0\rangle &\langle 1,0| S_z | 1,-1 \rangle \\
\langle 1,-1 | S_z | 1,1 \rangle &\langle 1,-1 | S_z |1,0\rangle &\langle 1,-1| S_z | 1,-1 \rangle
\end{array} \right)
##
which you can verify is the definition that you have given.
Robben said:
In order to find the eigenvectors I must find the eigenvalues. Now from this, what exactly are we writing the eigenvectors as a linear combination of?
##S_x## is also given in the same basis, so if you find the eigenvectors of ##S_x ## you will get some vector
##
|\phi\rangle = \left( \begin{array}{c}
a\\
b \\
c \end{array} \right) =
a\left( \begin{array}{c}
1\\
0 \\
0 \end{array} \right) +
b\left( \begin{array}{c}
0\\
1 \\
0 \end{array} \right) +
c\left( \begin{array}{c}
0\\
0 \\
1 \end{array} \right)
##
or
##
| \phi \rangle = a|1,1\rangle + b|1,0\rangle + c|1,-1\rangle
##
So essentially you need to find the eigenvectors of the matrix representation of ##S_x## you have given.