Known that $$ [Q^a,p^\mu]=0 $$ where Q is a representation for the a-th generator of the algebra of group simmetry and p the 4-momentum, if we consider a single particle state, eigenstate of p(adsbygoogle = window.adsbygoogle || []).push({});

$$ p^\mu | \psi_A (k) > = k^\mu | \psi_A (k) > $$

then also $$ Q^a | \psi_A (k) > $$

is eigenstate of p. My text says that If we suppose that in our theory there aren't massless particle then the result state of the action of Q on the single particle state is in general linear combination of single particle states, because otherwise if i add another particle i change the invariant mass.

How i can derive this?

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# Single particle state and symmetry

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