Schwartz book: Heisenberg equation of motion for quantum fields (equation 7.28)

Question

In his book "Quantum field theory and the standard model", section 7.2 "Hamiltonian derivation" (of the Feynman rules), Schwartz states that the equations of motion $i\partial_t\phi(x)=[\phi,H]$ have the formal solution (eq 7.27)

$$ \phi(\vec x, t) = S(t,t_0)^\dagger\phi(\vec x)S(t, t_0)\tag{7.27} $$ where $S$ is the time evolution operator for the momentum eigenstates, and it satisfies $$ i\partial_t S(t, t_0)=H(t)S(t,t_0).\tag{7.28} $$

I tried to justify this expression, but I didn't manage. Here what I tried.

From $i\partial_t\phi(x)=[\phi(x),H(t)]$ I get $$ i\partial_t\phi(x) = i\partial_t[S^\dagger(t,t_0) \phi(\vec x)S(t,t_0)] = (i\partial_t S^\dagger(t,t_0)) \phi(\vec x)S(t,t_0) + S^\dagger(t,t_0)\phi(\vec x)(i\partial_t S(t,t_0)) = [S^\dagger(t,t_0)\phi(\vec x)S(t,t_0), H(t)] $$

Using the commutator identity, I expand the commutator on the rhs: $$ (i\partial_t S^\dagger) \phi(\vec x)S + S^\dagger\phi(\vec x)(i\partial_t S) = S^\dagger\phi(\vec x)[S,H] + [S^\dagger\phi(\vec x), H]S = S^\dagger\phi SH - S^\dagger\phi HS + S^\dagger\phi HS - HS^\dagger\phi S = S^\dagger\phi SH - HS^\dagger\phi S $$

For the lhs to be equal to the rhs I have to require: $$ i\partial_t S = SH $$ $$ i\partial_t S^\dagger = -HS^\dagger $$

But Schwartz claim is $i\partial_t S = HS$ and in general $S$ does not commute with $H$. Where am I wrong?

Answer 1

I found the source of the confusion.

In the derivation, I always assumed that the Hamiltonian is in the Heisenberg representation. If I now express my last equation, $i\partial_t S = SH_H$ (the $H$ subscript stands for Heisenberg), in the Schrödinger representation I get $$ \partial_t S = SH_H = SS^\dagger H_S S = H_S S $$

So the time evolution given by Schwartz is the time evolution of the S operator using the Hamiltonian in the Schrödinger representation.