Hamiltonian in the Heisenberg picture with explicit time dependence My question is about something I read in Zwiebach's "A First Course in String Theory", 2nd edition.
On page 220, just below eqn (11.24) he says $$\alpha(t) = e^{iHt}\alpha e^{-iHt}.$$ Here, $\alpha$ is an operator in the Schroedinger picture and $\alpha(t)$ is that same operator in the Heisenberg picture. On page 219, just below eqn (11.18) he says $$H(p(t), q(t); t)$$ is the Heisenberg Hamiltonian. Here $H(p, q)$ is the Schroedinger Hamiltonian. I assume he means $$H(p(t), q(t); t) = e^{iHt}H(p, q) e^{-iHt}.$$
But
$$H(p(t), q(t); t) = H(e^{iHt}p e^{-iHt}, e^{iHt}q e^{-iHt}; t).$$
In other words, $$e^{iHt}H(p, q) e^{-iHt} = H(e^{iHt}p e^{-iHt}, e^{iHt}q e^{-iHt}; t).$$
How do the exponentials pass from surrounding H, to surrounding $p$ and $q$? Is it possible that the Schroedinger Hamiltonian is $H(p(t), q(t))$ and the Heisenberg Hamiltonian is $H(p(t), q(t); t)$? 3 lines below eqn (11.20) on page 219 he specifies that the Schroedinger Hamiltonian is $H(p, q)$. Am I correct in assuming that whenever $H$ appears in the exponent it is the Schroedinger Hamiltonian?
 A: That's a good question.

*

*If the Hamiltonian $H_S(t)$ in the Schrödinger picture has explicit$^1$ time-dependence, the Hamiltonian at different times might not commute. This means that the formula $e^{-\frac{i}{\hbar}Ht}$ no longer apply! The unitary time-evolution operator is more generally given by the (anti)time-ordered exponentiated Hamiltonian
$$ U(t)~=~\left\{\begin{array}{rcl}
T\exp\left[-\frac{i}{\hbar}\int_0^t\! dt^{\prime}~H_S(t^{\prime})\right]
&\text{for}& t  ~\geq~0 \cr\cr
AT\exp\left[-\frac{i}{\hbar}\int_0^t\! dt^{\prime}~H_S(t^{\prime})\right]
&\text{for}& t  ~\leq~0, \end{array}\right.\tag{A}$$
cf. e.g. this related Phys.SE post. It satisfies its own TDSE:
$$ i\hbar \frac{d U(t)}{d t}
~=~H_S(t)U(t),\qquad U(t\!=\!0)~=~{\bf 1}. \tag{B} $$


*Operators in the Heisenberg picture are given as
$$ A_H(t)~=~U(t)^{-1} A_S(t) U(t). \tag{C}$$
They satisfies the Heisenberg EOM
$$ i\hbar \frac{d A_H(t)}{d t}~=~[A_H(t),H_H(t)]+  i\hbar\frac{\partial A_H(t)}{\partial t}, \tag{D}$$
see the Wikipedia page for details. In particular, eq. (C) holds
if we put the operator $A_H$ equal to the Heisenberg Hamiltonian $H_H$, which OP asked about.
--
$^1$ In section 11.2 Zwiebach assumes after eqs. (11.19) that there is no explicit time dependence.
