In Schwartz's "Quantum Field Theory and the Standard Model", section 3.5 Green's Functions, he gives a technique for solving for explicit solutions of interacting Classical Field Theories perturbatively. To do this, we plug in a perturbation series in the coupling $\lambda$ constant $\phi _0 +\lambda \phi _1+\lambda ^2 \phi _2+....$ into the Euler Lagrange equations, and solve the equations upto each order. E.g. for the real $\phi ^4$ theory, we can write the general solution as :
$$\phi (x,t)=\phi_0 (x,t)+\lambda \phi_1(x,t)+\lambda ^2 \phi _2(x,t)+...$$
e.g. solving for the first two terms, we get (where $G$ is the Green's function of the Klein Gordon operator) :
$$\phi _0=\int dk \ a_k e^{ikx-i\omega _kt}+a^*_k e^{-ikx +i \omega _k t}$$
$$\phi _1=\int dx'dt'\ G(x,t;x',t')\ (\int dk \ a_k e^{-i\omega _k t'+ikx'}+a^*_k e^{i\omega_k t' -ikx'})^3$$
This perturbation series seems analogous to the free field solution in terms of mode operators. My question is, in the quantum theory, since $\phi (x,t)$ becomes becomes an operator, what happens to the operator equivalents of $a_k$? Is the corresponding operator analogous to the mode operators in free field theories?
P.S. I did some analysis with these operators which I am adding here:
I think that the $\hat{a_k}$ satisfies $[\hat{a_k}, H]=\omega _k \hat{a_k}$, where $H$ is the full Hamiltonian of the interacting theory. This is because I substituted $t=0$ in the perturbation series to get a series for $\hat{\phi} (x,0)$. Let's say $\hat{\phi} (x,0)= f[\hat{a_k}]$, where $f[\hat{a_k}]$ is a functional in $\hat{a_k}$ that we obtain by substituting $t=0$ in the series.
Then I computed $f[\hat{a_k} e^{-i\omega _k t}]$ upto two terms and this gave back the first two terms of $\hat{\phi}(x,t)$. Now, we independently know that:
$$\hat{\phi}(x,t)=e^{-iHt}\hat{\phi}(x,0)e^{iHt}$$ $$=e^{-iHt} f[\hat{a_k}]e^{iHt}$$ $$=f[e^{-iHt}\hat{a_k}e^{iHt}]$$
Now, if it holds independently that $\hat{\phi}(x,t)= f[\hat{a_k} e^{-i\omega _k t}]$ (I verified that this holds upto the first two terms), then
$$f[\hat{a_k} e^{-i\omega _kt}]=f[e^{-iHt}\hat{a_k}e^{iHt}]$$
which gives
$$e^{-iHt}\hat{a_k}e^{iHt}=\hat{a_k}e^{-i\omega _kt}$$
Differentiating gives $[\hat{a_k}, H]=\omega _k \hat{a_k}$
These properties seem analogous to the properties of mode operators from free field theories.
P.S. I am adding my computation for the first two terms.
$\phi(x,0)$ upto the first two terms is:
$$f[a_k]:=\phi(x,0)=\int dk \ a_k e^{ikx}+a^*_k e^{-ikx}+\lambda\int dx'dt'\ G(x,0;x',t')\ (\int dk \ a_k e^{-i\omega _k t'+ikx'}+a^*_k e^{i\omega_k t' -ikx'})^3$$
Computing $f[a_k e^{-i\omega _k t}]$ gives:
$$f[a_ke^{-i\omega _k t}]=\int dk \ a_k e^{ikx-i\omega _k t}+a^*_k e^{-ikx +i \omega _k t}+\lambda\int dx'dt'\ G(x,0;x',t')\ (\int dk \ a_k e^{-i\omega _k (t'+t)+ikx'}+a^*_k e^{i\omega_k (t'+t) -ikx'})^3$$
In the second term, we use $G(x,0;x',t')=G(x,t;x',t'+t)$ to get :
$$=\int dk \ a_k e^{ikx-i\omega _k t}+a^*_k e^{-ikx +i \omega _k t}+\lambda \int dx'dt' G(x,t;x',t'+t) (\int dk a_k e^{-i\omega _k (t'+t)+ikx'}+a^*_k e^{i\omega_k (t'+t) -ikx'})^3$$
In the second term, we substitute $t'+t=t''$ and $dt'=dt''$ to get:
$$=\int dk \ a_k e^{ikx-i\omega _k t}+a^*_k e^{-ikx +i \omega _k t}+\lambda \int dx'dt'' G(x,t;x',t'') (\int dk a_k e^{-i\omega _k t''+ikx'}+a^*_k e^{i\omega_k t'' -ikx'})^3$$
The above is equal to the first two terms of $\phi(x,t)$.
P.S. The operators $\hat{a_k}$ are equal to the free theory annihilation operators when $\lambda =0$. This is because when we expand the field and the canonical momentum at the $t=0$ space slice: $$\hat{\phi}(x,0)=\hat{\phi_0}(x,0)+\lambda \hat{\phi_1}(x,0)+...$$
$$\hat{\pi}(x,0)=\dot{\hat{\phi_0}}(x,0)+\lambda \dot {\hat{\phi _1}}(x,0)+..$$
and set $\lambda =0$, the two equations can be explicitly inverted to get the expressions of the free mode operators. This inversion is done e.g. in this post. But when $\lambda \neq 0$, I think it would be hard to get the explicit expression of $\hat{a_k}$
