What is the nature of perturbation theory in QFT?

Question

In the perturbation theory approach to QFT, the total Hamiltonian $H$ is separated into a free part $H_0$ which we can solve exactly and another $H_{\rm int}$ which we cannot such that $$H=H_0+H_{\rm int}.$$ Is this a time-dependent perturbation theory like the one we are used to in ordinary quantum mechanics? I may have a few follow-up questions depending on the answer to this.

Answer 1

It is rough answer withour rigorous mathematical construction. In QFT You always deal with integrals like $$\int_{-\infty}^{+\infty}dx\exp\left(-m^2x^2-\lambda x^4+Jx\right)$$ and we unfortunately cannot calculate this type of integrals explicitly. Hopefully, part of Nature can be described with assumption that $\lambda$ is very small and we can perform expansion by $\lambda$. You can check first two chapters (may be paragraphs) of A. Zee "QFT in a nutshell" to see an anologue between calculation of simple 1D integrals with exp and non-quadratic terms and path integral in QFT. There is lot of criticism of path integral description of QFT but I believe that it is the most simple way to think about QFT.

Answer 2

The perturbation theory approach to QFT assumes the dynamics of a quantum field as determined by a Hamiltonian $H$ via the Heisenberg equations of motion
$i \partial_t \phi(x) = [\phi(x), H(t)]$
The formal solution is
$\phi(\vec x, t) = S(t, t_0)^\dagger \phi(\vec x, t_0) S(t, t_0)$
where $S(t, t_0)$ is the time-evolution operator that satisfies
$i \partial_t S(t, t_0) = S(t, t_0) H(t)$
and
$H(t) = S(t, t_0)^\dagger H(t_0) S(t, t_0)$

The first step in time-dependent perturbation theory is to write the Hamiltonian as
$H(t) = H_0 + V(t)$
where the time evolution by $H_0$ can be solved exactly and $V(t)$ is small. $H_0$ could be the free Hamiltonian which is time independent and $V$ some interaction, for instance a $\phi^3$, written as $V(t) = \int d^3 x \frac{g}{3!} \phi(\vec x, t)^3$. The time and spatial dependence of the perturbation potential is hidden in the fields.
Note that the operators $\phi(\vec x, t), H, H_0, V$ are in the Heisenberg picture, while $\phi(\vec x, t_0)$ is the Schroedinger picture field.

As second step we have to change to the interaction picture where the fields evolve with $H_0$ as
$\phi_0(\vec x, t) = exp(i H_0 (t-t_0)) \phi(\vec x, t_0) exp(-i H_0 (t-t_0))$
$\phi(\vec x, t) = U(t, t_0)^\dagger \phi_0(\vec x ,t) U(t, t_0)$
The operator $U(t, t_0) = exp(i H_0 (t-t_0)) S(t, t_0)$ relates the full Heisenberg picture fields to the free fields at the same time $t$ and obeys the differential equation
$i \partial_t U(t, t_0) = V_I(t) U(t, t_0)$
where $V_I$ is the original $V$ now in the interaction picture
The solution is
$U(t, t_0) = T\{exp(-i \int_{t_0}^t dt' V_I(t'))\}$
where $T\{\}$ is the time ordering operator.

This perturbation series is known as the Dyson series and is similar to the time-dependent perturbation theory in quantum mechanics.

Answer 3

Recalling that in QFT the fields are functions of space-time coordinates $x^\mu =(t, \mathbf x) $ in general the interaction part of the Hamiltonian density will depend on space and time positions.

One usually works in the so called Interaction Picture (somewhere between the Schrödinger and Heisenberg pictures) and in this picture the perturbative expansion often resembles more the series that come out of time independent perturbation theory in quantum mechanics.