# Single particle state and symmetry

1. Feb 23, 2016

### andrex904

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

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

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?

2. Feb 28, 2016