About $\phi^4$ theory

Question

it's literally days I'm struggling with apparently simple problems concerning $\phi^4$ theory. I think the main issues are 2, that appear to me related one to another

1. Is this type of field actually a field?

I know it may seem as a silly question, but for how I learned QFT a field state is a carrier space of an irreducible unitary realization of the Poincaré group: this ultimately means that a field can be only separated in modes on the mass-shell and the fact that every interacting field is not on the mass-shell anymore, despite for $t=\pm\infty$, really confuses me.

2. Why in the interaction term of the hamiltonian operator do we encounter the free field?

In fact $\hat{H}=\hat{H}_0+\delta\hat{H}$ where the hamiltonian density change is $\delta\hat{\mathscr{H}}=\lambda\phi^4/4!$ and the field is the free one! I really can't go over this, because doing all the derivation from the Heisenberg picture to the interaction one doesn't seem that way at all; in fact, just reporting the main results, we should have $$ \hat{\phi}_{\text{H}}(t,\boldsymbol{r}) = \mathcal{U}^\dagger(t,\tilde{t}) \hat{\phi}_{\text{I}}(t,\boldsymbol{r}) \mathcal{U}(t,\tilde{t}) \\ \mathcal{U}(t,\tilde{t}) = T\left( \exp\left( -\frac{i}{\hbar}\int_{\tilde{t}}^t \text{d}{\underline{t}}\, e^{\frac{i}{\hbar}\hat{H}_0(\underline{t}-\tilde{t})} \delta\hat{H}(\underline{t}) e^{-\frac{i}{\hbar}\hat{H}_0(\underline{t}-\tilde{t})} \right)\right) $$ where $\tilde{t}$ is defined such that $\hat{\phi}_{\text{H}}(\tilde{t},\boldsymbol{r})=\hat{\phi}_{\text{I}}(\tilde{t},\boldsymbol{r})$. In other words how is possible that the fourth power of the field carried by the term $$ e^{\frac{i}{\hbar}\hat{H}_0(\underline{t}-\tilde{t})} \delta\hat{H}(\underline{t}) e^{-\frac{i}{\hbar}\hat{H}_0(\underline{t}-\tilde{t})} $$ is the fourth power of the free field? Thank you in advance.

Answer 1

Why in the interaction term of the hamiltonian operator do we encounter the free field?

Because we work in the Interaction picture. You stated this in your question, so I think you already know about the Interaction picture.

The interaction picture evolves the operators with the "free" hamiltonian $H_0$. Whether the operators are "free" or not just refers to their time dependence. For example, you seem to call the free field, the operator in the interaction picture: $$ \hat \phi_I(x,t) = e^{iH_0t}\hat\phi(x)e^{-iH_0t}\;. $$

In the interaction picture, the time dependence of the phi-fourth perturbation is: $$ \delta H_I(t) = e^{iH_0 t}\delta He^{-iH_0 t} $$ $$ =e^{iH_0 t}\frac{\lambda}{4!}\int d^3x \hat\phi(x)\hat\phi(x)\hat\phi(x)\hat\phi(x) e^{-iH_0 t} $$ $$ =\frac{\lambda}{4!}\int d^3x e^{iH_0 t}\hat\phi(x)\hat\phi(x)\hat\phi(x)\hat\phi(x)e^{-iH_0 t} $$ $$ =\frac{\lambda}{4!}\int d^3x e^{iH_0 t}\hat\phi(x)e^{-iH_0 t}e^{iH_0 t}\hat\phi(x)e^{-iH_0 t}e^{iH_0 t}\hat\phi(x)e^{-iH_0 t}e^{iH_0 t}\hat\phi(x)e^{-iH_0 t} $$ $$ =\frac{\lambda}{4!}\int d^3x \hat\phi_I(x,t)\hat\phi_I(x,t)\hat\phi_I(x,t)\hat\phi_I(x,t) $$

Answer 2

Sometimes a complicated interacting system can be described with so called normal modes - collective motions that are approximately linearly independent. The real evolution includes interaction terms, naturally. Unfortunately, we often choose bad initial approximations to begin the perturbation theory. I wrote a methodological paper about it. Maybe it will help.