Codex @codex  0

The Clifford algebra gives $S^{\mu \nu }=\frac{1}{2}{\gamma }^{\mu }{\gamma }^{\nu }-\frac{1}{2}{\eta }^{\mu \nu }I$. Reordering one more gamma matrix gives $[S^{\mu \nu },{\gamma }^{\rho }]={\eta }^{\nu \rho }{\gamma }^{\mu }-{\eta }^{\mu \rho }{\gamma }^{\nu }$. The commutator derivation identity then givesso these matrices represent the Lorentz algebra. This convention has no extra factor of $i$; the spatial generators are $-\frac{i}{2}{ϵ}^{jkl}\mathrm{diag}({\sigma }^l,{\sigma }^l)$.

Use the metric in the question and fix the Levi-Civita symbol convention ${ϵ}_{0123}=+1$. A consistent choice of rotation Lie algebra generators isSubstituting the indices into the printed Poincare algebra givesThe other cyclic brackets follow the same way. The negative spatial metric and antisymmetry of $M^{\rho \sigma }$ both enter this sign.

The printed brackets are real Lie algebra brackets, without the factor $i$ used for ordinary commutators of Hermitian quantum observables. We use those brackets for the algebraic verifications. For physical eigenvalues below, angular momentum is Hermitian, its spin projection is the real number $j_3$, and $\hslash =1$. In that quantum convention the ordinary operator commutators are $i$ times the displayed brackets. This distinction prevents identifying a real spin projection with an anti-Hermitian matrix eigenvalue.

In the universal enveloping algebra, translations commute. The Pauli-Lubanski pseudovector therefore satisfiesFor fixed $\nu ,\rho$, the product of momenta is symmetric in $\mu ,\tau$ while the epsilon coefficient is antisymmetric in them. There is no need to commute the Lorentz generator through the momenta.

Using the commutator derivation identity and the mixed bracket,The two terms become equal after swapping $\nu ,\rho$, and the last expression vanishes by antisymmetry in $\nu ,\tau$. In the Hermitian observable convention, the calculation has one overall extra $i$ and still vanishes. Thus $W_{\mu }$ preserves each momentum eigenspace.

For the specified epsilon orientation, two useful component identities areThe order displayed matters: the momentum operator is on the right, so it can act first on the momentum eigenstate. At rest, $p^{\mu }=(m,0,0,0)$, and on a spin state with $J_3|j,j_3⟩=j_3|j,j_3⟩$ these giveThese are massive rest-frame Pauli-Lubanski eigenvalues. The raised component would be $W^3=+m{j}_3$; confusing $W_3$ with $W^3$ reverses the answer.

Helicity is the projection of spin angular momentum, or equivalently the rotation generator acting internally, along the momentum direction:For the momentum $p^{\mu }=(k,0,0,k)$ with $k>0$, the helicity operator is $J_3$. Hence a helicity-$j_3$ state hasThe result uses only the two longitudinal components and does not need a separate assumption about the transverse little-group generators. These massless longitudinal Pauli-Lubanski eigenvalues agree with $W^{\mu }=j_3{P}^{\mu }$ for ordinary finite-helicity representations.

Finally, at rest the contraction is $m{W}_0=0$. For the chosen null momentum it is $k(W_0+W_3)$, whose two eigenvalues cancel. Thus both results obey $W_{\mu }{P}^{\mu }=0$. Reversing the epsilon orientation reverses all $W$ eigenvalues together; it leaves both identities and both consistency checks intact. For a literal anti-Hermitian derived representation $\rho$ of the printed real algebra, write $H_X=i\rho (X)$ for the Hermitian observable. Since $W_{\mu }$ is bilinear in generators, its abstract enveloping-algebra image is $\rho (W_{\mu })=-W_{\mu ,\mathrm{phys}}$. Thus the corresponding formal-image eigenvalues, if that convention is intended, areHere $\rho (P^{\mu })$ has eigenvalue $-i{p}^{\mu }$, while $p^{\mu }$ and $j_3$ themselves remain real physical labels. This is the same result after the Hermitian quantum generator convention is applied to both factors, not an inconsistent choice of spin sign. Both the Hermitian-observable and literal anti-Hermitian interpretations are consequently specified.

Apply the commutator derivation identity $[S,AB]=[S,A]B+A[S,B]$ and the preceding gamma matrix transformation identity. Keeping the matrix order unchanged givesThusThe order of the gamma matrices matters, since they generally do not commute. The PDF has the four ordinary metric contractions shown here; the conversion's stray powers such as ${\eta }^{\nu }{\rho }^{\sigma }$ are not mathematical expressions to retain.

Since $S^{\rho \sigma }=\frac{1}{4}[{\gamma }^{\rho },{\gamma }^{\sigma }]$, apply the commutator derivation identity to this commutator and insert part (ii):This is exactly the printed Lorentz algebra relation, including its final two signs; their alternative appearance in other formulas comes from $S^{\rho \mu }=-S^{\mu \rho }$. Therefore the matrices S form a representation of the Lorentz algebra. Their definition contains no extra factor of $i$, so the spatial rotation generators below are represented by skew-Hermitian matrices. Adding a factor of $i$ would change the convention for the Lie algebra structure constants and cannot be done silently.