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?
Best 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: