# How is Lorentz invariance of $S$-matrix related to vanishing of Hamiltonian density commutator at spacelike separations?

In Section 5.1 of the book, 'Quantum Theory of fields Vol-1' by Steven Weinberg, he says that if the Hamiltonian density commutes with itself at spacelike separation then the $$S$$-Matrix satisfies Lorentz invariance.

More specifically, on page 191 it says

"We have seen in chapter 3 that the $$S$$ matrix will be Lorentz invariant if the interaction can be written as $$\tag{5.1.1}V(t)=\int d^3x\, \mathcal{H}(\boldsymbol{x},t)$$ where $$\mathcal{H}$$ is a scalar, in the sense that $$\tag{5.1.2}U_0(\Lambda,a)\mathcal{H}(x)U_0^{-1}(\Lambda,a)=\mathcal{H}(\Lambda x+a)$$ and satisfies the additional condition $$\tag{5.1.3} [\mathcal{H}(x),\mathcal{H}(x')]=0$$ for $$(x-x')^2\geq 0$$."

But, I don't see the connection of the two conditions. The Lorentz invariance of $$S$$-Matrix is satisfied independently of the form of Hamiltonian density. And, $$S$$-operator simply evolves the in-state to one of the possible out-state by performing time evolution from ultimate past to ultimate future. It does not connect with space translation, then how vanishing commutator of Hamiltonian density with itself at spacelike seperation is related with Lorentz invariance of the $$S$$ matrix? I see the intuitive reason that observations at spacelike separations should be independent.

• Actually I saw that bottom of page 144 Weinberg discusses this. Commented Sep 27, 2022 at 3:35

The correct statement is that IF $${\cal H}(x)$$ is scalar and $$[{\cal H}(x),{\cal H}(x')]=0$$ for $$x,x'$$ spacelike related THEN the $$S$$ matrix is invariant under the unitary representation $$U_{(\Lambda, T)}$$ of the orthochronous Poincaré group which implements the above action on $${\cal H}$$.

In other words, if $$U_{(\Lambda, v)} {\cal H}(x) U^{-1}_{(\Lambda, v)} = {\cal H}(v+\Lambda x)$$ and the commutativity requirement is satisfied then

$$U_{(\Lambda, v)} S U^{-1}_{(\Lambda, v)} = S \quad \mbox{for every orthochronous transformation (\Lambda, v)}$$

The converse implication may be false (I am not sure, but dealing with interaction Larangians containing derivatives, everything becomes obscure).

However the proof of Weinberg's statement should be like this (I do not have the book but I guess he follows this route). Formally (i.e., without paying attention to mathematical details), $$S = \sum_{n=0}^{+\infty} \frac{(-i)^n}{n!}\int_{M^4} \cdots \int_{M^4} T({\cal H}(x_1) \cdots {\cal H}(x_n)) d^4x_1\cdots d^4x_n$$ Therefore $$U_{(\Lambda, v)} S U^{-1}_{(\Lambda, v)} = \sum_{n=0}^{+\infty} \frac{(-i)^n}{n!}\int_{M^4} \cdots \int_{M^4} U_{(\Lambda, v)} T({\cal H}(x_1) \cdots {\cal H}(x_n))U^{-1}_{(\Lambda, v)} d^4x_1\cdots dx_n^4\:.$$ If we manage to prove that

$$U_{(\Lambda, v)} T({\cal H}(x_1) \cdots {\cal H}(x_n))U^{-1}_{(\Lambda, v)}= T({\cal H}(x'_1) \cdots {\cal H}(x'_n)) \tag{1}$$ where $$x'_k := v+\Lambda x_k$$ we are done, because $$d^4x_k = d^4x'_k$$ since $$(\Lambda, v)$$ is an isometry.

The only apparent obstruction is that $$T({\cal H}(x_1) \cdots {\cal H}(x_n)) \neq T({\cal H}(x'_1) \cdots {\cal H}(x'_n))$$ because $$\Lambda$$ changes the temporal order of the arguments. This is not the case just because $$\Lambda$$ is orthochronous, so that it does not reverse the chronological order of causal vectors $$x_k-x_h$$, and the Hamiltonias commute when these vectors are spacelike. For instance, with $$n=2$$,

$$T(({\cal H}(x_1) ({\cal H}(x_2)) = {\cal H}(x_1){\cal H}(x_2) \quad \mbox{ if t_1 \geq t_2}$$

$$T(({\cal H}(x_1) ({\cal H}(x_2)) = {\cal H}(x_2){\cal H}(x_1) \quad \mbox{ if t_2 \geq t_1}$$

Therefore

$$UT(({\cal H}(x_1) ({\cal H}(x_2))U^{-1} = {\cal H}(x'_1){\cal H}(x'_2) \quad \mbox{ if t_1 \geq t_2}$$

$$UT(({\cal H}(x_1) ({\cal H}(x_2))U^{-1} = {\cal H}(x'_2){\cal H}(x'_1) \quad \mbox{ if t_2 \geq t_1}$$

If the temporal order of $$t'_1,t'_2$$ were the same as the time order of $$t_1,t_2$$ we would have

$$UT(({\cal H}(x_1) ({\cal H}(x_2))U^{-1}= UT(({\cal H}(x'_1) ({\cal H}(x'_2))U^{-1}$$

just by applying the definition. Actually the order may be reversed only if $$x_1-x_2$$ is spacelike. But this case is irrelevant since $${\cal H}(x_1) {\cal H}(x_2)) ={\cal H}(x_2) {\cal H}(x_1)$$ in that case. This result extends to the case of $$n$$ points establishing that (1) holds true proving the assertion.