What is the correct arrangement of the elements of Pauli matrices? I'm dealing with angular momentum, or particularly spin, on my quantum mechanics course; I guess the Pauli matrices thing is a more general one, but I'd like to illustrate my doubt with them (maybe get a deeper answer). Say, for the electron ($s=1/2$), why does
$$S_z=\begin{pmatrix}\left\langle\frac{1}{2},\frac{1}{2}\middle|S_z\middle|\frac{1}{2},\frac{1}{2}\right\rangle&\left\langle\frac{1}{2},\frac{1}{2}\middle|S_z\middle|\frac{1}{2},-\frac{1}{2}\right\rangle\\[0.1in]\left\langle\frac{1}{2},-\frac{1}{2}\middle|S_z\middle|\frac{1}{2},\frac{1}{2}\right\rangle&\left\langle\frac{1}{2},-\frac{1}{2}\middle|S_z\middle|\frac{1}{2},-\frac{1}{2}\right\rangle\end{pmatrix}=\frac{\hbar}{2}\sigma_z$$
instead of
$$S_z=\begin{pmatrix}\left\langle\frac{1}{2},-\frac{1}{2}\middle|S_z\middle|\frac{1}{2},-\frac{1}{2}\right\rangle&\left\langle\frac{1}{2},-\frac{1}{2}\middle|S_z\middle|\frac{1}{2},\frac{1}{2}\right\rangle\\[0.1in]\left\langle\frac{1}{2},\frac{1}{2}\middle|S_z\middle|\frac{1}{2},-\frac{1}{2}\right\rangle&\left\langle\frac{1}{2},\frac{1}{2}\middle|S_z\middle|\frac{1}{2},\frac{1}{2}\right\rangle\end{pmatrix}=\frac{\hbar}{2}\begin{pmatrix}-1&0\\0&1\end{pmatrix}=-\frac{\hbar}{2}\sigma_z$$
What I mean to say is, why does the matrix elements go from $m_s=+s$ to $m_s=-s$ (left-upper corner to right-lower corner) instead of the other way, as usual?
We usually take rows and columns from smaller to bigger value, but why is this not the case? For instance, recently we've seen the matrix representation of the hamiltonian of a simple harmonic oscilator, with elements $H_{mn}=(n+1/2)\hbar\omega\,\delta_{mn}$ and it goes like $\hbar\omega\begin{pmatrix}1/2&0&\ldots\\0&3/2&\ldots\\\vdots&\vdots&\vdots\end{pmatrix}$, not as, say $\hbar\omega\begin{pmatrix}\vdots&\vdots&\vdots\\\ldots&3/2&0\\\ldots&0&1/2\\\end{pmatrix}$, for example. I'm just trying to make clearer my question. Here ($s=1/2$) is only a difference of a minus sign, but when I did it for $s=3/2$, the ladder (spin) operators would interchange ($S_-$ would have non-zero values 'above' the diagonal and $S_+$ 'below' the diagonal). I tried to find the reason but found none; is it a mere convention or definition? Or am I missing something important here? Thank you in advance.
 A: I don't think there will yield any new physics if you use your order of eigenvectors to form Pauli matrices. The reason we are using the basis eigenvectors from $m_s=+s$ to $m_s=-s$(left-upper corner to right-lower corner) is by convention, I guess. 
Now we can figure out what happens if we use your suggestion. Suppose we adopt your suggestion and use the basis eigenvectors from $m_s=-s$ to $m_s=+s$(left-upper corner to right-lower corner), what will the Pauli matrices be under the basis in this order? We should see that the $S_x$, $S_y$, $S_z$ operators do not change, they are still:
$S_x=\frac{\hbar}{2} (|+\rangle \langle -|+ |-\rangle \langle +|)$, 
$S_y=\frac{\hbar}{2} (-i|+\rangle \langle -|+ i|-\rangle \langle +|)$,
$S_z=\frac{\hbar}{2} (|+\rangle \langle +|- |-\rangle \langle -|)$. Here $|+\rangle$ and $|-\rangle$ are eigenvectors of $S_z$. 
Then we can write the Pauli matrices as following:
$S_x=\begin{pmatrix}\left \langle -|S_x | - \right\rangle & \left\langle -|S_x | + \right\rangle\\[0.1in]\left \langle + |S_x| - \right \rangle& \left \langle + |S_x | + \right \rangle \end{pmatrix}=\frac{\hbar}{2} \begin{pmatrix} 0 & 1 \\[0.1in] 1 & 0\end{pmatrix}=\frac{\hbar}{2}\sigma_x$.
Similarly, we have 
$S_y=\begin{pmatrix}\left \langle -|S_y | - \right\rangle & \left\langle -|S_y | + \right\rangle\\[0.1in]\left \langle + |S_y| - \right \rangle& \left \langle + |S_y | + \right \rangle \end{pmatrix}=\frac{\hbar}{2} \begin{pmatrix} 0 & i \\[0.1in] -i & 0\end{pmatrix}=\frac{\hbar}{2}\sigma_y$.
$S_z=\begin{pmatrix}\left \langle -|S_z | - \right\rangle & \left\langle -|S_z | + \right\rangle\\[0.1in]\left \langle + |S_z| - \right \rangle& \left \langle + |S_z | + \right \rangle \end{pmatrix}=\frac{\hbar}{2} \begin{pmatrix} -1 & 0 \\[0.1in] 0 & 1\end{pmatrix}=\frac{\hbar}{2}\sigma_z$.
We can check that the commutation relation $[S_i,S_j]=\epsilon_{ijk}i \hbar S_k$ still holds for the new Pauli matrices. Besides, other properties of Pauli matrices hold as well.
Therefore, it does not matter which order of eigenvectors you use. We use the commonly-accepted order by convention. 
A: There is no deep reason for why this is the case. We could have just as well went from $s= -1/2$ to $s = 1/2$ and nothing of physical or mathematical importance would change. When giving a matrix representation to an operator, we usually have to choose an ordered basis for our vector space. The usual ordered basis is $\{|1/2,1/2 \rangle, |1/2, -1/2\rangle\}$ whereas if we do what you suggest, we would have to choose $\{|1/2,-1/2 \rangle, |1/2, 1/2\rangle\}$. When writing down the matrices for the SHO we went from low energy eigenkets to higher ones, i.e we chose the ordered basis $\{|0\rangle ,|1\rangle, |2\rangle \dots \}$. So you just have to choose an ordered basis and then stick to it.
