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  Spin tensor from Noether theorem and spin tensor from Pauli-Lubanski vector

+ 4 like - 0 dislike

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?

This post imported from StackExchange Physics at 2014-09-09 21:59 (UCT), posted by SE-user Andrew McAddams
asked Sep 8, 2014 in Theoretical Physics by Andrew McAddams (340 points) [ no revision ]

Note that \(W_\mu\)is a space-like vector : \(W^2 = -m^2 s (s+1)\)

In the rest frame of the particle, we have : \(W^0 = 0, \vec W = m \vec J\)

So, to get the correct \(3\)- dimensional spin \(\vec J\)from \(W_\mu\), you  simply make a Lorentz transformation from the "moving" frame (\(P^\mu\)) , to the rest frame of the particle.

2 Answers

+ 3 like - 0 dislike

They are the same. You will see the orbital angular momentum part drop out of the Pauli-Lubanksi vector since $P\wedge P=0$. Roughly it comes from this (with $J^{\nu\alpha}=L^{\nu\alpha}+S^{\nu\alpha}$)

\(\epsilon_{\mu\nu\alpha\beta} \ L^{\nu\alpha} P^\beta \sim \epsilon_{\mu\nu\alpha\beta} x^\nu P^\alpha P^\beta \sim 0\ .\)

So only the spin part can contribute to the Pauli-Lubanski vector.

answered Sep 10, 2014 by suresh (1,545 points) [ no revision ]
+ 2 like - 0 dislike

If one works in an arbitrary Lorentz frame it is $W^2$ that characterizes the spin as an invariant of the representation; in the rest frame of a particle, it is (as explained by Trimok) equivalently $S^2=S_1^2+S_2^2+S_3^2$.

As a dynamical variable, however, $W$ has not enough components to account for all observables, and one needs the 6-dimensional tensor $S_{\mu\nu}$ that figures in the formula for the angular momentum.

answered Sep 10, 2014 by Arnold Neumaier (15,787 points) [ no revision ]

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