Why does the time translation operator have a different sign than the time evolution operator?

The time evolution operator is $$\hat U(t)=e^{-i\hat H t}$$ where $$\hat H$$ is the Hamiltonian, $$\hbar=1$$, and the state is at time $$t=0$$. The time translation operator is defined similarly, but with opposite sign $$\mathcal{\hat U} =e^{i\hat H t}$$ Why is it that these two operators have different sign? I am going through the book Quantum Field Theory for The Gifted Amateur by Lancaster and Blundell, and they derive the spatial evolution operator to be $$\hat U_a=e^{i\hat pa}$$ where $$\hat P$$ is the momentum and $$a$$ is the amount evolved by. The book then says that this is not the operator that is needed, and randomly changes the sign to give the spatial translation operator $$\mathcal {\hat U_a}= e^{-i\hat p a}$$ and I have no idea where the sign change comes from. Any help would be greatly appreciated.

• Sounds like bad writing. Consider a better book. Jan 20 '21 at 20:47
• It's an alright introduction, but it's just that, an introduction Jan 20 '21 at 20:48

In this case, there are two possible ways to define the time translation operator. If we want to think about active transformations, we would keep the same coordinate system and evolve the system using $$U(t_0)=e^{-iHt_0}$$. If we want to think about a passive transformation, we would switch to a new coordinate system $$t'=t-t_0$$. In this new coordinate system, $$|\psi(t)\rangle$$ becomes $$|\psi(t')\rangle=|\psi(t-t_0)\rangle=e^{iHt_0}|\psi(t)\rangle$$.