Differential equation for the operator relating Heisenberg and interaction-picture fields

Question

I'm following Schwartz's QFT textbook and he finds a differential equation for the operator $$U(t,t_0) \equiv e^{iH_0(t-t_0)}S(t,t_0)\tag{p.85}$$ via the following steps: $$ i\partial_tU(t,t_0) = i(\partial_t e^{iH_0(t-t_0)})S(t,t_0) + e^{iH_0(t-t_0)}i\partial_t S(t,t_0) \\= -e^{iH_0(t-t_0)}H_0 S(t,t_0) + e^{iH_0(t-t_0)}i\partial_tS(t,t_0) \\= e^{iH_0(t-t_0)}[-H_0 + H(t_0)]e^{-iH_0(t-t_0)}e^{iH_0(t-t_0)} S(t,t_0) \ =\ V_I(t)U(t,t_0) .\tag{7.33}$$

where $$ V_I(t) \equiv e^{iH_0(t-t_0)}V(t_0)e^{-iH_0(t-t_0)}\ \tag{p.85}$$ and $\ S(t,t_0)$ is the evolution operator in Heisenberg picture, i.e. $$\phi(\vec{x},t) = S(t,t_0)^\dagger \phi(\vec{x},t_0) S(t,t_0),\tag{7.27}$$ that satisfies the equation $$ i\partial_t S(t,t_0) = H(t)S(t,t_0)\tag{7.28}$$ with $$H(t) = H_0 + V(t).\tag{7.29}$$

However, it seems to me that in the third equality he's suggesting that $$i\partial_t S(t,t_0) = H(t_0)S(t,t_0).$$ Is this relation true? If so, why?

In order to get to the same differential equation, Peskin and Schroeder's book states $$ \phi(\vec{x},t) = e^{iH(t-t_0)} \phi(\vec{x},t_0) e^{iH(t-t_0)}\tag{PS p.83} $$ so that $$U(t,t_0) = e^{iH_0(t-t_0)}e^{-iH(t-t_0)}\tag{4.17}$$ and the following is obvious. Does it mean that $$S(t,t_0) = e^{iH(t-t_0)} $$ also for $H=H(t)$? But then, does this $S(t,t_0)$ satisfy its differential equation?

Answer 1

It seems to me that OP either has a version of the mentioned book with a typo, or miss-copied the derivation. I will highlight here the main steps for reference. According to Schwartz's book, we have \begin{align} i\partial_t S(t, t_0) &= H(t)S(t, t_0)&&\text{eq. (7.28)}\\ H(t) &= H_0 + V(t) &&\text{eq. (7.29)}\\ U(t, t_0) &= \exp\{iH_0(t-t_0)\}S(t, t_0). && \text{definition}\\ \end{align} Therefore, \begin{align} i\partial_t U(t, t_0) &= -\exp\{iH_0(t-t_0)\}H_0S(t, t_0) + \exp\{iH_0(t-t_0)\}H(t)S(t, t_0) \\ &= \exp\{iH_0(t-t_0)\}[H(t)-H_0]S(t, t_0) = \exp\{iH_0(t-t_0)\}V(t)S(t, t_0)\\ &= \exp\{iH_0(t-t_0)\}V(t)\exp\{-iH_0(t-t_0)\}\exp\{iH_0(t-t_0)\}S(t, t_0) \\ &= V_I(t)U(t, t_0), \end{align} where I have used eq. (7.28) in the first step, eq. (7.29) in the third, and the definition in the last. At no point do I need to assume $i\partial_t S(t,t_0) = H(t_0)S(t,t_0)$, which would anyway be incorrect in general.

Concerning the question about Peskin and Schroeder's book, the expression $S(t,t_0) = e^{iH(t-t_0)} $ is only valid for time-independent Hamiltonians. If the Hamiltonian is time-dependent, as assumed in Schwartz's book, we have to consider a time-ordered exponential [see eq. (7.34) for $U(t, t_0)$, which follows formally the same differential equation as $S(t, t_0)$, replacing $H(t)$ with $V_I(t)$]