quantum field theory - Does Peskin & Schroeder Eq. (4.26), $U(t_1,t_2)U(t_2,t_3) = U(t_1,t_3)$ imply $[H_0,H_{int}] = 0$?

Peskin & Schroeder equation (4.17) define the operator, \begin{equation} U(t,t_{0})~=~e^{i(t-t_{0})H_{0}}e^{-i(t-t_{0})H} \tag{4.17} \end{equation} where $$H~=~H_0+H_{\text{int}}\tag{4.12}$$ is the full Hamiltonian and $H_{0}$ is the free Hamilton, both in the Schrodinger picture. In equation (4.26), Peskin and Schroeder state that the operator satisfies the following identity, \begin{equation} U(t_{1},t_{2})U(t_{2},t_{3})~=~U(t_{1},t_{3}) \tag{4.26} \end{equation} where $t_{1}\ge t_{2}\ge t_{3}$. Does this imply that the free Hamiltonian commutes with the interaction $$[H_{0},H_{\text{int}}]~=~0~ ?$$ Here is my argument that it does.

In the condition $t_{1}\ge t_{2}\ge t_{3}$ take $t_{2}=0$. The identity is then, \begin{equation} U(t_{1},0)U(0,t_{3})=U(t_{1},t_{3})\ . \end{equation} Substitute the definition, \begin{equation} e^{it_{1}H_{0}}e^{-it_{1}H}e^{-it_{3}H_{0}}e^{it_{3}H}=e^{i(t_{1}-t_{3})H_{0}}e^{-i(t_{1}-t_{3})H} \end{equation} and simplify to get, \begin{equation} e^{-it_{1}H}e^{-it_{3}H_{0}}=e^{-it_{3}H_{0}}e^{-it_{1}H} \end{equation} with $t_{1}\ge 0\ge t_{3}$ . Put $t_{1}=t$ and $t_{3}=-t$. \begin{equation} e^{-itH}e^{itH_{0}}=e^{itH_{0}}e^{-itH} \end{equation} Expanding to second order in $t$, \begin{equation} (1-itH-\frac{t^{2}}{2}HH)(1+itH_{0}-\frac{t^{2}}{2}H_{0}H_{0})= (1+itH_{0}-\frac{t^{2}}{2}H_{0}H_{0})(1-itH-\frac{t^{2}}{2}HH) \end{equation} results in, \begin{equation} HH_{0}=H_{0}H \end{equation} so that $[H_{0},H]_{-}=0$. Now $H=H_{0}+H_{int}$ so the free Hamiltonian must commute with the interaction. \begin{equation} [H_{0},H_{int}]_{-}=0 \end{equation} In Peskin and Schroeder, the context for this material is the self-interacting scalar field with Hamiltonian, \begin{equation} H=\int d^{3}x \left(\frac{1}{2}\pi(t,x)^{2}+\frac{1}{2}\frac{\partial \phi}{\partial x^{r}}\frac{\partial \phi}{\partial x^{r}}+V(\phi)\right) \ . \end{equation} In the classical theory, the PB is, \begin{equation} [H_{0},H_{int}]_{PB}=-\int d^{3}x\frac{\delta H_{0}}{\delta \pi}\frac{\delta H_{int}}{\delta \phi}=-\int d^{3}x\ \pi\frac{dV}{d\phi}=-\frac{d}{dt}\int d^{3}x\ V(\phi(t,x)) \end{equation} Going over to quantum theory, \begin{equation} [H_{0},H_{int}]_{-}=-i\frac{d}{dt}\int d^{3}x\ V(\phi(t,x)) \end{equation} so that $[H_{0},H_{int}]_{-}=0\ $ implies the integral of $V(\phi)$ is a conserved charge; is this also a correct result?

Answer

Ref. 1 writes the correct formula

$$ U(t,t^{\prime})~=~e^{iH_0(t-t_0)} e^{-iH(t-t^{\prime})}e^{-iH_0(t^{\prime}-t_0)} , \qquad t~\geq~ t^{\prime},\tag{4.25}$$

which satisfies

$$ U(t_1,t_2)U(t_2,t_3)~=~U(t_1,t_3) , \qquad t_1~\geq~ t_2~\geq~ t_3.\tag{4.26}$$

Here $t_0$ is an arbitrary but fixed fiducial initial instant where operators and states in the Schrödinger picture, the Heisenberg picture and the interaction picture all agree. For $t\neq t_0$, the three pictures are no longer the same, although they are still unitary equivalent.

For $t^{\prime}=t_0$, eq. (4.25) simplifies to

$$ U(t,t_0)~=~e^{iH_0(t-t_0)}e^{-iH(t-t_0)}. \tag{4.17}$$

It appears that OP mistakenly replaces $t_0$ in eq. (4.17) with an arbitrary time $t^{\prime} \leq t$. The resulting equation

$$ U(t,t^{\prime})~=~e^{iH_0(t-t^{\prime})}e^{-iH(t-t^{\prime})}. \qquad(\leftarrow \text{Wrong!})$$

is not correct.

References:

1. M.E. Peskin & D.V. Schroeder, An Intro to QFT; Section 4.2.