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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.

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    $\begingroup$ I am not sure I completely understand your question, so let me say this as a comment rather than as an answer. How the eigenvalues are ordered just depend on which basis you choose, you can do it in any way you like, the physics is equivalent. So its purely up to convention. The reason people usually choose a basis such that the eigenvalues of the Hamiltonian are ordered such that we start we the lowest and go up, is because the low energy sector of the theory is the most interesting typically from a physical point of view. It tell you the ground state energy, and more importantly, (continued) $\endgroup$
    – Heidar
    Jun 2, 2013 at 1:47
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    $\begingroup$ the energy gap to the lowest excitations. In the case of spin, there is no particular reason to prefer the ordering higher spin->lower, or the other way around. In the situation where your system has $SU(2)$ spin rotation symmetry, the energies of spin up and down are the same. But I think the reason why its more conventional to choose a basis such that the $S_z$ eigenvalues are decreasing, comes from math. The different spins come from representation theory of $\mathfrak{su}(2)$. These type of representations are usually called "highest weight representations". (continued) $\endgroup$
    – Heidar
    Jun 2, 2013 at 1:58
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    $\begingroup$ They are roughly found by diagonalizing the so-called Cartan subalgebra (which in this case is just $S_z$). One starts with the highest eigenvalue (weight), and construct the other states by acting with step down/annihilation operators (which are the other elements of the algebra, besides the Cartan elements). So what I am basically trying to say is that it does not matter which basis you choose, the physics is the same. But the particular convention for spin probably originates from a more general construction in math, the highest weight representations. And there, its just a convention. $\endgroup$
    – Heidar
    Jun 2, 2013 at 2:03
  • $\begingroup$ Woa! yes, I think you got it! My quantum mechanics course is still a basic one, so I guess that's why we didn't got a deeper explanation, indeed I just found out when working on it. But now I know which direction I should look up to, so thanks a lot! $\endgroup$
    – user24999
    Jun 2, 2013 at 2:16
  • $\begingroup$ I'm glad it was useful. $\endgroup$
    – Heidar
    Jun 2, 2013 at 2:58

2 Answers 2

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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.

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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.

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  • $\begingroup$ Thank you! Your answer also helps. Maybe if you could complement Heidar's comments or give detail about what consequences would have if I stick with my basis order, I'd be pleased to mark your answer as accepted. $\endgroup$
    – user24999
    Jun 2, 2013 at 3:19
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    $\begingroup$ There would be no notable consequences, other than the headaches induced from matching up your algebraic signs with the ones found in textbooks and literature :). This would inevitable once you get to raising and lowering operators for AM. $\endgroup$
    – childofsaturn
    Jun 2, 2013 at 3:26

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