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There is a certain sign mismatch between the time translation operator and time evolution operator in quantum field theory which I hope someone can illuminate.

From my understanding, a Poincaré transformation is a transformation: $$x \to x' = \Lambda x + a,$$ for some Lorentz $\Lambda$ and vector $a$ . The transformed fields can be defined via active or passive transformation. For simplicity let us consider scalar fields only. In the passive formulation we have:

$$ \phi'(x) = \phi(\Lambda^{-1} x - \Lambda^{-1} a),$$ so that $\phi'(x') = \phi(x) $. In the active formulation we have:

$$ \phi'(x) = \phi(\Lambda x + a),$$ so that $$\phi'(x) = \phi(x').$$ Let us stick with the active formulation and postulate a unitary operator $U(\Lambda,a)$ such that the state $| \psi \rangle$ as seen in the frame $S$ corresponds to the state $| \psi' \rangle$ in the frame $S'$, i.e. we have:

$$ |\psi' \rangle = U(\Lambda,a) | \psi \rangle $$

Since we want $\langle \chi' | \phi'(x) | \psi' \rangle = \langle \chi | U^{\dagger} \phi'(x) U | \psi \rangle = \langle \chi | \phi(x) | \psi \rangle$

This gives us the identity: $$\phi'(x) = U \phi(x) U^{\dagger} = \phi(\Lambda x + a )$$

To recover Heisenberg's Equations we must postulate that $P_\mu U(\mathbb{1},a) = -i \frac{1}{\partial a^\mu} U (\mathbb{1},a)$. Thus expanding the previous equation for small $a$ and $\Lambda = \mathbb{1}$ we obtain:

$$ ( 1 + i a^\mu P_\mu) \phi(x) (1 - i a^\mu P_\mu) = \phi(x) + a^\mu \partial_\mu \phi(x)$$

Or:

$$ i [P_\mu, \phi(x) ] = \partial_\mu \phi(x).$$

In particular we have for $\mu =0$, $$i [ H,\phi(x)] = \partial_t \phi(x)$$ which is Heisenberg's equation where we identity $P_0 = H$.

Now, time translation $t \to t' = t + \Delta t$ is also a Poincaré transformation. In the active formulation which we are working now, $$\phi'(t ) = \phi( t + \Delta t ).$$ Thus in our formulation we can identify the time evolution operator $\hat{U}(t)$ via:

$$ \hat{U}(t) = U( \Lambda = \mathbb{1}, a = (t,0,0,0) ) $$

If we take this identification however, then we have by our definition of $P_\mu$, we must have:

$H \hat{U}(t) = -i \frac{d}{d t} \hat{U}(t)$, which gives the wrong sign for Schrodinger's equation, which is $$H \hat{U}(t) = +i \frac{d}{d t} \hat{U}(t).$$

Where did I go wrong?

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I think I have figured it out. Everything I wrote is correct up till I identified:

$$\hat{U}(t) = U(\Lambda = 1, a^\mu = (\Delta t,0,0,0))$$

While is is true that we have:

$$U(\Lambda = 1, a^\mu = (\Delta t,0,0,0)) \phi(x) U^{\dagger}(\Lambda = 1, a^\mu = (\Delta t,0,0,0)) = \phi(x, t_0 + \Delta t )$$

Meanwhile $\hat{U}(t)$ satisfies something different. Recall from quantum mechanics that $\hat{U}(t,t_0)$, the time-evolution operator in the Shrodinger picture, is also the map between heisenberg and Shrodinger states:

$$ | \psi(t) \rangle = \hat{U}(t,t_0) | \psi(t_0) \rangle$$ (I'm taking the reference point to be $t_0$). $\ket{\psi}(t)$ is the Shrodinger state while $| \psi(t_0) \rangle$ is the heisenberg state.

The shrodinger $\phi(x,t_0)$ and heisenberg operators $\phi(x,t)$ are then related by:

$$ \phi(x,t) = \hat{U}^\dagger(t,t_0) \phi(x,t_0) U(t,t_0) $$

Now choose $\Delta t = t - t_0$. Then we have:

$$U(\Lambda = 1, a^\mu = (t-t_0,0,0,0)) \phi(x,t_0) U^{\dagger}(\Lambda = 1, a^\mu = (t-t_0,0,0,0)) = \phi(x, t )$$

Comparing the above two formulas we get:

$$ U(t,t_0) = U^{\dagger}(\Lambda = 1, a^\mu = (t-t_0,0,0,0))$$

Hence we obtain:

$$ HU(t,t_0) = i \frac{d}{dt} U(t,t_0)$$

The Shrodinger equation in the shrodinger picture. So my main confusion in my original question lies in the fact that I was working in the Heisenberg picture all along and needed to transform to the Shrodinger picture in order to reproduce Shrodinger's equation with the correct sign.

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