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Weinberg claims that it is obvious that the $\sigma = 0$ component of $u^\mu$ at zero spatial momentum points in the 3-direction. This is supposed to follow from (5.3.6). Unfortunately I am not seeing it. I thought that the $J=1, m = 0$ component is conventionally aligned in the 2-direction?

Any help would be appreciated.


$u^\mu(0,0) = (2m)^{-1/2} (0,0,1,0)$


$\sum_{\bar{\sigma}} u^\mu(0,\bar{\sigma}) \mathbf{J}^{(j)}_{\bar{\sigma}\sigma} = {\mathbf{\mathcal{J}}^\mu}_\nu u^\nu (0,\sigma)$


$(\mathcal{J}_k)^0_0 = (\mathcal{J}_k)^0_i = (\mathcal{J}_k)^i_0 = 0$


$(\mathcal{J}_k)^i_j = - i \epsilon_{ijk}$

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By the most universal rules and conventions for the angular momentum in quantum mechanics, $J=1$, $m=0$ is always associated with the third direction i.e. with the $z$-axis. It's because $m$ (and, in this Weinberg notation, I suppose that $\sigma$ as well) conventionally denotes the eigenvalue of $J_z$, the generator of rotations around the third axis.

The matrices ${\mathbf J}^{(1)}$ and ${\mathcal J}_3$ are intrinsically the same, representing the angular momentum generators for the spin $j=1$. In the case of ${\mathcal J}$, it's by definition because this is the matrix in the "preferred" vector representation; in the case of ${\mathbf J}$, it was depicted that we talk about the same spin-one representation by the subscript $(1)$.

The last thing one needs to know is what are the eigenvalues of $J_z$ in the three-dimensional space generated by $e_x,e_y,e_z$. Well, the eigenvalues are $\sigma=-1$ and $\sigma=0$ and $\sigma=+1$ and the corresponding eigenstates are proportional to $e_x+i e_y$, $e_z$, $e_x-i e_y$. In particular, $e_z$ itself corresponds to the vanishing eigenvalue $\sigma=0$ of $J_z$. It's because if you rotate the vector in the $z$-axis around the same $z$-axis, it doesn't change at all. So the infinitesimal generator remembering the change is zero.

Alternatively, you may use your last equation implying that $$({\mathcal J}_z)^z_j = 0$$ because the epsilon-symbol is antisymmetric. It means that the matrix elements of $J_z$ in the column corresponding to $e_z$ (and similarly in the row corresponding to $e_z$) are equal to zero.

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I'm sorry but I thought that $\mathbf{J}^{(1)}$ denotes 3-vector of matrices which have the same components as $(\mathcal{J}_k)^i_j$ where $k$ denotes the 3-vector index. I don't see exactly what about (5.3.6) singles out the 3-direction – user11881 Sep 3 '12 at 10:50
Hi, one must choose a basis in the rep of SU(2) and it's a convention that we're choosing eigenvectors of $J_3$, the third component of the angular momentum. In particular, the $J_3=0$ basis vector of the 3-dimensional vector representation is the 3rd basis vector itself. The formula $\sigma=0$ is just meant to denote $J_3=0$. – Luboš Motl Sep 3 '12 at 14:05

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