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What is the parity eigenvalue of the $W^{\pm}$ boson, or is it even an eigenstate? I have not found any source that discusses this. I have seen some lists of particles with their parity eigenvalues, but the $W^{\pm}$ and $Z^{0}$ bosons are always left out.

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    $\begingroup$ Parity is broken in the Electroweak interactions. Only CP is a (n approximate) good symmetry. But $W^{\pm}$ are not eigenstates being charged. $\endgroup$ – TwoBs May 26 '14 at 11:56
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Here is the particle table for exchange bosons. You will see that the massive intermediate bosons are not assigned a parity.

Parity is an operator. To have a definite value the state must be an eigenvalue of this operator. In the case of the massive weak interaction mediating bosons no such eigenvalue exists because in the standard model they carry both an axial vector and a vector component, so the operator cannot be diagonal. This is what induces parity violation in weak interactions. A better formulation is that the observation of parity violating weak interactions forced the model to have massive vector and axial vector exchange bosons. Another source that might help is this one.

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    $\begingroup$ i agree with this answer, but in your second link, top of p3, parity $P_{W^+}=P_{W^-}=-1$. huh? is the link, mark thompson's cambridge lecture notes, wrong? $\endgroup$ – innisfree May 26 '14 at 12:50
  • $\begingroup$ @innisfree thanks for catching it. I believe it is wrong for massive (broken symmetry) bosons, but I will do further checks. It could just be a choice of the standard model. I will edit a new link to vector axial vector. $\endgroup$ – anna v May 26 '14 at 13:50
  • $\begingroup$ I understand that the fermionic currents that couples to the W are V-A (vector - axial vector) currents, but does this imply that W itself is V-A? $\endgroup$ – Ihle May 26 '14 at 14:21
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    $\begingroup$ It is not in an eigen state of parity and that is why it is not given a parity in the data book. $\endgroup$ – anna v May 26 '14 at 14:49
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    $\begingroup$ the original link referred to in my previous comment is hep.phy.cam.ac.uk/~thomson/partIIIparticles/handouts/… $\endgroup$ – innisfree May 27 '14 at 9:39

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