Symmetry group of the space time$^1$ on which QFT is defined is usually required to have a representation on the space of states.

Quantum mechanics is just QFT in one dimensions. The spacetime in this case is the time line $\mathbb R$. Fields are $X(t)$, and $P(t)$. Symmetry group is group of translations $t\rightarrow t+b$ of $\mathbb R$. Infinitesimal form of (Hermitian) time translation operator is Hamiltonian $H=i\partial/\partial t$. One **requires** that Hilbert space of states have a representation of this group, which in Heisenberg picture is given as $$exp(iHt)X(0)exp(-iHt)=X(t),\:exp(iHt)P(0)exp(-iHt)=P(t)$$
The case of QFT on Minkowski space is similar. We again require the symmetry group (which is Poincare group in this case) to have a representation on the space of states. In particular for the subgroup of Poincare group generated by four translation operators $P_0,P_1,P_2,P_3$ we **require**
$$exp(iP_0x^0)\psi(0)exp(-iP_0x^0)=\psi(x^0,0,0,0)$$ $$exp(iP_1x^1)\psi(0)exp(-iP_1x^1)=\psi(0,x^1,0,0)$$ $$exp(iP_2x^2)\psi(0)exp(-iP_2x^2)=\psi(0,0,x^2,0)$$ $$exp(iP_3x^3)\psi(0)exp(-iP_3x^3)=\psi(0,0,0,x^3)$$ Since $P_\mu$'s commute with each other these conditions can be collectively written as :
$$exp(iPx)\psi(0)exp(-iPx)=\psi(x)$$ where $P=(P^\mu), \: P^\mu=\eta^{\mu\nu}P_\nu$ and $\eta=diag(1,-1,-1,-1)$. Sredniki's sign convention is different i think.

- For example if space time is a manifold with metric $\eta$ then symmetry group is group of transfomations which preserve $\eta$.

This post imported from StackExchange Physics at 2015-03-30 13:50 (UTC), posted by SE-user user10001