Will difference of creation operators in free and interacting theories affect the construction of $\phi_I$ similar to the free-field theory?

Question

In Peskin and Schroeder's book p 83, the authors said

At any fixed time $t_0$, we can of course expand $\phi$ in terms of ladder operators: $$\phi(t_0, \textbf{x})=\displaystyle\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf{p}} }}(a_{\mathbf{p}} e^{i\textbf{p.x}}+a_{\mathbf{p}} ^\dagger e^{-i\textbf{p.x}}) (1) $$ ... $e^{iH_0 (t-t_0)} \phi (t_0, \mathbf{x}) e^{-iH_0 (t-t_0)} \equiv \phi_I(t,\mathbf{x}).$...

Since we can diagonize $H_0$, it is easy to construct $\phi_I$ explicitly:

$\phi_I(t, \mathbf{x} ) = \int \frac{d^3p}{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}} } \left( a_{\mathbf{p}} e^{-ipx} + a^{\dagger}_{\mathbf{p}} e^{ipx} \right) |_{x^0 = t - t_0} $ (4.15)

To derive (4.15), I thought about $$ e^{i H_0 (t-t_0)} a_{\mathbf{p}} e^{-i H_0 (t-t_0)} $$ use BCH expansion in terms of $e^{-B} A e^B$ to work out (4.15), similar to the free field theories.

As answered in Interacting Fields in QFT $a_{\mathbf{p}}$ is time-indenpendent for free theory (I thought about $(\partial^2 + m^2) \phi(x) =0 $ in Eq. (2) in the aforementioned link for free theories) but time-dependent on interacting theories. Will this difference affect commutation relation $[a^{int}_{\mathbf{p}},H_0]$? More specifically, $$ [a^{int}_{\mathbf{p}},H_0 = \int \frac{d^3p }{(2\pi)^3} \omega_p a_{\mathbf{p}}^{free, \dagger} a_{\mathbf{p}}^{free} ] =? $$

Answer 1

By considering the Heisenberg-picture, we can deduce that given a free Hamiltonian $H_0$ and the annihilation operator $a_{\bf{k}}$, we have that $$[a_{{\bf{k}}},H_0]=\omega_{{\bf{k}}}a_{{\bf{k}}}\implies a_{\bf{k}}(t)=e^{iH_0t}a_{{\bf{k}}}e^{-iH_0t}=e^{-i\omega_{\bf{k}}t}a_{\bf{k}}.$$Here $a_{\bf{k}}(t)$ is the time dependent annihilation operator. Since $\omega_{\bf{k}}\in\mathbb{R}$, we have that $e^{-i\omega_{\bf{k}}t}\in\mathbb{C}$. We can use this to compute the commutation relation between $a_{\bf{k}}(t)$ and $H_0$. We thus have$$[a_{\bf{k}}(t),H_0]=[e^{-i\omega_{\bf{k}}t}a_{\bf{k}},H_0]=e^{-i\omega_{\bf{k}}t}[a_{\bf{k}},H_0]=\omega_{\bf{k}}e^{-i\omega_{\bf{k}}t}a_{\bf{k}}.$$But we have already established that $e^{-i\omega_{\bf{k}}t}a_{\bf{k}}=a_{\bf{k}}(t)$. Thus we have that$$[a_{\bf{k}}(t),H_0]=\omega_{\bf{k}}a_{\bf{k}}(t).$$ So we can see that the time-dependent annihilation (for example) commutates with $H_0$ in the same manner as the time-independent annihilation operator. Intuitively, this makes sense; if the Hamiltonian doesn't change with time, then at all times, the creation and annihilation operators will create/annihilate the same eigenstates $|k\rangle$ with corresponding energy eigenvalues $\omega_{\bf{k}}$.

It should be noted that this is not true in general. For a general Hamiltonian $H$, the commutation relations might be far more complicated. Furthermore, equal-time commutation relations can be shown to preserve the unitary structure of the operators and unequal-time commutation relations may vary.