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.

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    $\begingroup$ In lorentz invariant quantum field theories the hamiltonian cannot depend on time explicitly . Explicit time depends breaks poincare invariance. Perturbative explanation in QFT is analogous to time independent perturbation theory. $\endgroup$
    – Anonjohn
    Jan 11, 2020 at 17:17
  • $\begingroup$ @Anonjohn But in my exposure to quantum mechanics so far, interaction picture is usually used only when the perturbing hamiltonian is time-dependent. $\endgroup$ Jan 11, 2020 at 17:29
  • $\begingroup$ The perturbative expansion for physical observables is picture independent. Although some books present it in the interaction picture, the observables(green's functions) are all independent of the picture used. If you like, you can develop your perturbative expansion in the Heisenberg picture, which is standard in the derivation of the path integral. In the path integral quantization, it is manifest that the split into interaction and free hamiltonians in artificial and that the quadratic part of the Lagrangian is the "free" bit since Gaussian integrals are doable. $\endgroup$
    – Anonjohn
    Jan 11, 2020 at 18:02
  • $\begingroup$ @Anonjohn there is a PT with Hamiltonian with explicit dependence on time, isn't it? Schwinger-Keldysh PT. $\endgroup$ Jan 12, 2020 at 18:23
  • $\begingroup$ @ArtemAlexandrov. I am not on an expert on Schwinger - Keldysh formalism. But the fact remains that hamiltonians cannot contain explicit time dependance -break poincare symmetry is immediately manifest in observables. However, at finite temperature: which is where this formalism finds use(it appears), the time "coordinate" is an arbitrary choice. $\endgroup$
    – Anonjohn
    Jan 13, 2020 at 2:02

3 Answers 3


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.


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
$i \partial_t S(t, t_0) = S(t, t_0) H(t)$
$H(t) = S(t, t_0)^\dagger H(t_0) S(t, t_0)$

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.

  • $\begingroup$ "$H_0$ could be the free Hamiltonian which is time-independent" In your example both $H_0$ and $V$ are time-independent. So I am not sure why you are writing $V$ as $V(t)$. $\endgroup$ Jan 13, 2020 at 5:27
  • $\begingroup$ @mithusengupta123. Time-independent in the Schroedinger picture, but the perturbation theory is firstly stated in the Heisenberg picture, and secondly in the interaction picture, where the operators are time-dependent. $\endgroup$ Jan 13, 2020 at 17:49

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.

  • $\begingroup$ Follow up questions are best submitted as new questions $\endgroup$
    – lux
    Jan 11, 2020 at 17:20
  • $\begingroup$ Since fields depend on $x^\mu=(t,\vec{x})$ does not make the interaction term explicitly time-dependent. For example, in quantum mechanics, the fact that $H=H(p,x)$ and $x=x(t)$ and $p=p(t)$ does not make the hamiltonian time-dependent. $\endgroup$ Jan 11, 2020 at 17:27
  • $\begingroup$ In quantum mechanics there is a special global time parameter $t$. In quantum field theory this is not the case: time is just the zeroth component of the four vector $x^\mu$ and will differ in different frames. There is no universal time $\endgroup$
    – lux
    Jan 11, 2020 at 17:40
  • $\begingroup$ "There is no universal time." That does not make the Hamiltonian explicitly time-dependent. I think Anonjohn's first comment is reasonable. $\endgroup$ Jan 12, 2020 at 13:18
  • $\begingroup$ I didn't say that it was explicitly time dependent. In fact my point is somewhat the revsere. You want the Lagrangian to be Lorentz invariant so it almost definitely does not depend explicitly on coordinates (unless you form invariant combinations thereof). When going to the Hamiltonian formalism one fixes a ref frame in order to define conjugate momenta as derivatives with respect to the temporal variation of the field. It's unlikely this will lead to explicit time dependence. $\endgroup$
    – lux
    Jan 12, 2020 at 15:04

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