Will the fermion propagator change if instead of deriving it from the Lagrangian $$\mathcal{L}=i\bar{\Psi}\gamma^{\mu}\partial_{\mu}\Psi -m\bar{\Psi}\Psi\tag{1}$$ I derive it from $$\mathcal{L}'=\frac{i}{2}(\bar{\Psi}\gamma^{\mu}\partial_{\mu}\Psi- \partial_{\mu}\bar{\Psi}\gamma^{\mu}\Psi) -m\bar{\Psi}\Psi\tag{2}$$ and if yes, what will the new propagator be?

The fermion propagator derived from the first Lagrangian, $\mathcal{L}$, is $$(-i)\frac{(-\gamma^{\mu}p_{\mu}+m)}{p^2+m^2-i\epsilon}.\tag{3}$$

EDIT: I will lay out the blueprint for deriving the expression for the Feynman propagator

  1. Define the Feynman propagator as $$S_F(x-y)_{\alpha\beta}= \langle0|T\bar{\Psi}(x)\Psi(y)|0\rangle_{\alpha\beta}\tag{4}$$

  2. and then substitute the mode expansions for $\Psi(y)$ and $\bar{\Psi}(x)$.

  3. The result will yield the expression for the Feynman propagator in position space.

  4. Invert that and get the Feynman propagator in momentum space

So my question reduces to "whether or not the propagator is determined by the Lagrangian." And I guess the answer is yes through the mode expansion, which needs to satisfy the classical equations of motion, right?

Since the equations of motion are the same for $\mathcal{L}$ and $\mathcal{L}'$, I would suggest that the propagator is the same. But at the very beginning I hadn't write that to get a feeling on what other people think... So now, I would just appreciate for some comments/confirmation if any...

  • $\begingroup$ Where's your attempt? If you know how to derive the first propagator, what problem do you have applying this to your second Lagrangian? $\endgroup$
    – ACuriousMind
    Commented Oct 3, 2023 at 12:00
  • $\begingroup$ @ACuriousMind I edited the post $\endgroup$
    – schris38
    Commented Oct 3, 2023 at 12:55

1 Answer 1


Yes they give the same propagator. The bulk Lagrangian is defined up to a total derivative or equivalently a boundary term. Your second formulation is the symmetric version (people sometimes write the action of the partial derivative with \rightleftarrow). The two versions differ by the boundary term: $$ \mathcal L=\mathcal L’+\partial_\mu \left(\frac{i}{2}\bar \psi\gamma^\mu\psi\right) $$ This is why the bulk classical equations of motion are the same. If the domain is finite, even classically these boundary terms are important (they capture the boundary conditions). For topological theories, the boundary terms are important even for the bulk as they capture how the field is twisted. This is not the case for a spin 1/2. Anyway, the topological considerations do not affect the propagator.

Hope this helps.

  • $\begingroup$ Yes, it does indeed help. I know that the two Lagrangians differ by a boundary term. I guess what I was asking is that "Does the fact that they differ by a boundary term ensure that the propagator is going to be the same (because they have the same EoMs)?" $\endgroup$
    – schris38
    Commented Oct 3, 2023 at 15:31

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