I'm studying from here: Roberto Soldati - Field Theory 2. Intermediate Quantum Field Theory (A Next-to-Basic Course for Primary Education)

I'm trying to understand and prove an equality at page 52, between equations $(2.36)$ and $(2.37)$. The equality is the following

$$ \langle x|(i\not\!{\partial} - M )^{-1}|y\rangle = - i S_F(x - y) \tag{1} $$

Where $S_F$ is a fermion propagator and satisfies

$$ (i\not\!{\partial} - M ) S_F(x - y) = i \delta^{(4)}(x-y) \tag{2} $$

and whose explicit expression is the following

$$ S_F(x - y) = \int \frac{d^4p}{(2\pi)^4} \frac{i}{\not\! p - m + i\epsilon} e^{-ip\cdot(x-y)}\\ = \int \frac{d^4p}{(2\pi)^4} \frac{i(\not\! p + m)}{p^2 - m^2 + i\epsilon} e^{-ip\cdot(x-y)} \tag{3} $$

I tried to prove the equality, but

  1. I am not sure my proof is correct, therefore I want to ask you: is it?
  2. If it is correct there is a passage I don't really understand a passage, I don't know why I did it, beside from the fact that it gives me the correct result.

My attempt:

Starting from $(2)$ I multiply both sides for $(i\not\!{\partial_x} - M )^{-1}$ obtaining

$$ -iS_F(x-y)= (i\not\!{\partial_x} - M )^{-1} \delta^{(4)}(x-y) = (i\not\!{\partial_x} - M )^{-1} \int \frac{d^4p}{(2\pi)^4} e^{-ip\cdot(x-y)} $$

Now, like in standard QM: $e^{-ip\cdot x} = \langle x|p\rangle$ and $e^{ip \cdot y}=\langle p|y\rangle$ and I have

$$ -iS_F(x-y)= \int \frac{d^4p}{(2\pi)^4} \langle x|(i\not\!{\partial_x} - M )^{-1}|p\rangle \langle p|y\rangle $$

Now I solve the integral in $d^4p$, and since $\int \frac{d^4p}{(2\pi)^4}\langle p|y\rangle= \delta^{(4)}(x-y)$, I obtain

$$ -iS_F(x-y) = \langle x|(i\not\!{\partial} - M )^{-1}|y\rangle $$

So is this correct? If it is why I can put the operator $(i\not\!{\partial_x} - M )^{-1}$ inside $\langle x|p\rangle$ ? I thought that since we have $i\not\!{\partial_x}$ it can be that the operator acts on the bra or ket with $x$, is it right?

I'm pretty sure I'm making some mistakes and confusion, could you clarify why that equality is true?


1 Answer 1


The proof is unnecessarily confusing. It is hard to follow, and so it is hard to tell whether it is correct or not.

But anyway, here is the simplest approach. In fact, let us do something slightly more general. Let $D=D(i\partial)$ be an arbitrary differential operator, and consider the expression $$ \mathscr D:=\langle x|D|y\rangle $$

We will ultimately take $D(z)=1/(\not\! z-M+i\epsilon)$, but the approach works for more general differential operators.

One can insert a complete set of momentum states in the form $$ 1=\int\frac{\mathrm d^np}{(2\pi)^n}|p\rangle\langle p| $$ such that $$ \begin{aligned} \mathscr D&=\langle x| D\,1|y\rangle\\ &=\int\frac{\mathrm d^np}{(2\pi)^n}\langle x|D|p\rangle\langle p|y\rangle \end{aligned} $$

Now use $$ D(i\partial)|p\rangle=D(p)|p\rangle,\qquad \langle p|z\rangle=\mathrm e^{ipz} $$ to write $$ \begin{aligned} \color{red}{\mathscr D}&=\int\frac{\mathrm d^np}{(2\pi)^n}D(p)\overbrace{\langle x|p\rangle}^{\mathrm e^{-ipx}}\overbrace{\langle p|y\rangle}^{\mathrm e^{ipy}}\\ &=\color{red}{\int\frac{\mathrm d^np}{(2\pi)^n}D(p)\mathrm e^{ip(y-x)}} \end{aligned} $$

In the case in the OP, we have $D(z)=1/(\not\! z-M+i\epsilon)$, and so $$ \langle x|(i\not\!\partial-M)^{-1}|y\rangle\equiv \int\frac{\mathrm d^np}{(2\pi)^n}\frac{1}{\not\! p-M+i\epsilon}\mathrm e^{ip(y-x)} $$ which is precisely the fermion propagator, $S(x-y)$, as required.

  • $\begingroup$ Thanks, I think it should be $(i\not\!\partial-M)^{-1}$ in the last equality. I'll check the answer and award the bounty as soon as it's possible $\endgroup$
    – AnOrAn
    Sep 29, 2019 at 19:27
  • $\begingroup$ @AnOrAn Yes, I meant to write $1/(\cdots)$, but missed the slash. Anyway, we typically encourage users to wait a few days before accepting an answer, or awarding a bounty. That way, new answers may be posted, or the current answers improved. There is no rush, it's best if you unaccept it for now, and just leave it be. In a few days come back, and if this is still the best answer, you can accept it back. You never know, maybe you get a better one :-) But anyway, take your time, and if there is anything unclear or you have any further questions let me know $\endgroup$ Sep 29, 2019 at 19:40
  • $\begingroup$ Ok thanks, I'll wait then $\endgroup$
    – AnOrAn
    Sep 29, 2019 at 19:49
  • $\begingroup$ I have a curiosity: if instead of $\langle p|z\rangle=\mathrm e^{ipz}$ I choose the other normalisation: $\langle p|z\rangle=\mathrm e^{-ipz}$ does this simply gives $S(y-x)$ right? Is it all so arbitrary? $\endgroup$
    – AnOrAn
    Sep 30, 2019 at 12:59
  • 1
    $\begingroup$ @AnOrAn Both definitions are acceptable, $\langle p|z\rangle=e^{\pm i pz}$. But they differ by $D(i\partial)|p\rangle=D(\pm p)|p\rangle$. So the sign is in principle arbitrary, but you have to be consistent in your choice. In the end it is just a matter of convention. $\endgroup$ Oct 1, 2019 at 23:52

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