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Spin 3-vector directly from Noether theorem

Let's have one of applications of Noether theorem: the invariance of action under Lorentz group transformations leads to conservation of tensor $$ \tag 1 J_{\mu , \alpha \beta} = x_{\alpha}T_{\mu \beta} - x_{\beta}T_{\mu \alpha} + \frac{\partial L}{\partial (\partial^{\mu}\Psi_{k})}Y_{k, \alpha \beta} = L_{\mu , \alpha \beta} + S_{\mu , \alpha \beta}. $$ Here the second summand is called spin tensor.

The conservation law $\partial^{\mu}J_{\mu , \alpha \beta} = 0$ leads to conservation in time the following tensor: $$ \tag 2 J_{\alpha \beta} = \int d^{3}\mathbf r J_{0, \alpha \beta}. $$ The second summand of $(1)$ after integration $(2)$ gives spin vector. For example, in Dirac theory we have $\hat{\mathbf S}_{i} = \frac{1}{2}\varepsilon_{ijk}S^{jk} = \frac{1}{2}\int d^{3}\mathbf r\Psi^{\dagger} \Sigma \Psi $.

The value $S_{\alpha \beta} = \int S_{0 , \alpha \beta}d^{3}\mathbf r$ isn't conserved in general.

Spin 4-vector (Pauli-Lubanski vector)

It can be shown that quantity $$ W_{\mu} = \frac{1}{2}\varepsilon_{\mu \nu \alpha \beta}J^{\nu \alpha} P^{\beta} $$ refers to eigen angular momentum, and also it is translational invariant. It is conserved in time if $J_{\mu \nu}, P_{\alpha}$ are also conserved (while $S_{i}$ isn't).

The question

Whichever characterizes the spin truly?

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    $\begingroup$ They are the same, as explained here. $\endgroup$ – Dilaton Sep 10 '14 at 9:23

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