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  1. Can someone explain to me what is a pseudoscalar particle?

  2. And how do experiments figure out that what they're dealing with is a scalar or pseudoscalar?

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up vote 4 down vote accepted

As the Wikipedia page you linked to, pseudoscalar particles are just like scalar particles, but their associated field (or wave function) is assigned an additional sign flip if we decide to study how objects behave under the parity, i.e. under $$ (x,y,z) \to (-x,-y,-z) $$ In general, we may study whether some equations of physics are invariant under this mirror reflection (which changes the left hand to the right hand). They may fail to be symmetric altogether. But if they happen to be symmetric, we are still free to transform scalar fields either as $$ s(x,y,z)\to s'(x',y',z') =+s(-x,-y,-z) $$ which are (true) scalars or $$ p(x,y,z)\to p'(x',y',z') =-p(-x,-y,-z) $$ which are pseudoscalars. Scalars and pseudoscalars have $P=+1$ and $P=-1$ where $P$ is the parity and the sign is correlated with the sign in the transformation law above.

If parity is conserved and we may determine the parity of the decay products, the parity of the original state is simply equal to the parity of the products – which is the product of individual parities in the simplest cases (parity is a multiplicative quantum number). So particles that are quanta of scalar fields or pseudoscalar fields behave in the same way "separately" but they may have different decays. If you want me to mention objects that are scalars or pseudoscalars, let me say that $\vec E\cdot \vec E$ is a scalar (the sign, an energy density, doesn't depend on the mirroring) but $\vec \cdot \vec B$ is a pseudoscalar (because the magnetic field is a pseudovector, so its sign is a matter of conventions and one must revert it if we switch to a mirror-symmetric configuration).

In the case of the Higgs-like particle discovered by the LHC, you may figure out what the parity is e.g. by methods described in today's paper by John Ellis et al.

In general, the relative motion of the final products can compensate the parity and allow the decay of the initial particle regardless of the parity. However, the parity will still influence the angular distributions of the final particles as well as the distribution of their relative energies.

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