# Time evolution vs. Time translation in QFT

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?

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.