Is there any relevance to these perturbatively defined mode operators in interacting Quantum Field Theories?

Question

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) :

$$\phi _0=\int dk \ a_k e^{ikx-i\omega _kt}+a^*_k e^{-ikx +i \omega _k t}$$

$$\phi _1=\lambda\int dx'dt'\ G(x,t;x',t')\ (\int dk \ a_k e^{-i\omega _k t'+ikx'}+a^*_k e^{i\omega_k t' -ikx'})^3$$

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 $[\hat{a_k}, H]=\omega _k \hat{a_k}$. 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. 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 $\hat{a_k}$

Answer 1

Yes the field $\phi_0$ is the standard free field. $\phi_1$ solves $(\Box + m^2 )\phi_1=\phi_0^{3}$ and $\phi_2$ for example is a solution of $(\Box + m^2 )\phi_2=\phi_0^{2}\phi_1$. Graphically $\phi_1$ is interpreted as three of the $\phi_0$ meeting at a point $y$ and propagating to point $x$ expressed by $G(y-x)$. What this approach does not see is instantons, which are invisible because they scale like $\frac{1}{\lambda}$.

EDIT: The annihilation operators you have written above remain free field operators even when $\lambda>0$. Since you can express every $\phi_n$ through products of free-fields $\phi_0$, upon quantization you get products of operator-valued distributions, which are mathematically ill-defined. The inability to properly handle such products is then dealt with in renormalization. It makes no sense to express $\hat\phi$ in terms of annihilation and creation operators since it is defined as a sum of integrals over prodcuts of operator valued distributions. The reason for the expressablity of the free field in terms of creation and annihilation operators lies in the fact that classically it is a solution to a linear wave-equation. This is not the case for $\phi$. As soon as you go non-linear in the fields it makes no sense to talk about superposition of waves. Even if you have an analytic classical solution to the non-linear equations of motion, you still quantize them using a linear fluctuation operator in the background of the solution were the fluctuations can again be expressed in terms of familiar superpositions of waves. The requirement is just that the fluctuations asymptote to some vacuum fluctuations at infinity. This is only a naive approach to qunatization, which does not even sratch the surface of whats really going on.