quantum field theory - Why the Hamiltonian and the Lagrangian are used interchangeably in QFT perturbation calculations

Whenever one needs to calculate correlation functions in QFT using perturbations one encounters the following expression:

$\langle 0| some\ operators \times \exp(iS_{(t)}) |0\rangle$

where, depending on the textbook S is either (up to a sign)

1. 
   

   $\int \mathcal{L}dt$ where $\mathcal{L}$ is the interaction Lagrangian

   
   

   or

   
   


2. 
   
   

   $-\int \mathcal{H}dt$ where $\mathcal{H}$ is the interaction Hamiltonian.

   
   

It is straightforward to prove that if you do not have time derivatives in the interaction terms these two expressions are equivalent. However these expressions are derived through different approaches and I can not explain from first principles why (and when) are they giving the same answer.

Result 1 comes from the path-integral approach where we start with a Lagrangian and do perturbation with respect to the action which is the integral of the Lagrangian. Roughly, the exponential is the probability amplitude of the trajectory.

Result 2 comes from the approach taught in QFT 101: Starting from the Schrödinger equation, we guess relativistic generalizations (Dirac and Klein-Gordon) and we guess the commutation relations to be used for second quantization. Then we proceed to the usual perturbation theory in the interaction picture. Roughly, the exponential is the time evolution operator.

Why and when are the results the same? Why and when the probability amplitude from the path integral approach is roughly the same thing as the time evolution operator?

Or restated one more time: Why the point of view where the exponential is a probability amplitude and the point of view where the exponential is the evolution operator give the same results?

Answer

Starting from the Hamiltonian formulation of QM one can derive the path-integral formalism (see chapter 9 in Weinberg's QFT volume 1), where the Hamiltonian action is found to be proportional to $\int \mathrm{d}t (pv - H)$.

For a subclass of theories with "a Hamiltonian that is quadratic in the momenta" (see section "9.3 Lagrangian Version of the Path-Integral formula" in above textbook), the term $(pv - H)$ can be transformed into a Lagrangian $L_H = (pv - H)$. Then the Lagrangian action is proportional to $\int \mathrm{d}t L_H$. Both actions give the same results because one is exactly equivalent to (and derived from) the other.

$$ \int \mathrm{d}t (pv - H) = \int \mathrm{d}t L_H$$

Moreover, when working in the interaction representation you do not use the total Hamiltonian but only the interaction. The derivation of the Hamiltonian action is the same, except that now the total Hamiltonian is substituted by the interaction Hamiltonian $V$. Again you have two equivalent forms of write the action either in Hamiltonian or Lagrangian form.

If you consider Hamiltonians whose interaction $V$ does not depend on the momenta, then the $pv$ term vanishes and the above equivalence between the actions reduces to

$$ - \int \mathrm{d}t V = \int \mathrm{d}t L_V$$

where, evidently, the interaction Lagrangian is $L_V = -V$

This is what happens for instance in QED, where the interaction $V$ depends on both position and Dirac $\alpha$ but not on momenta.

Note: There is a sign mistake in your post. I cannot edit because is less than 10 characters and I have noticed the mistake in a comment to you above, but it remains.