- #1
andrex904
- 15
- 1
Known that $$ [Q^a,p^\mu]=0 $$ where Q is a representation for the a-th generator of the algebra of group symmetry 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?
$$ p^\mu | \psi_A (k) > = k^\mu | \psi_A (k) > $$
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?