Integrating Fermionic fields by parts

[Muoniex]

Homework Statement

Hello there!
I have been studying path integral for fermionic fields, and I don't understand one detail.
I have to proove that:

$$\int d^4x\int d^4 y\, \overline{\eta (y)}S^+_F(x-y)(i\partial\!\!\!/-m)\Psi (x)=\int d^4x \, \overline{\eta (x)} \Psi (x)$$

Homework Equations

We know that the Feynman propagator and the Dirac operator verify:
$$(i\partial\!\!\!/-m)S_F(x-y)=i \delta (x-y)$$

The Attempt at a Solution

This integral is supposed to do by parts.
It's clear that:
$$-\int d^4x\int d^4 y\, \overline{\eta (y)}S^+_F(x-y)m\Psi (x)=-\int d^4x\int d^4 y\, \overline{\eta (y)}mS^+_F(x-y)\Psi (x)$$
But my problem is with the term:
$$i\int d^4x\int d^4 y\, \overline{\eta (y)}S^+_F(x-y)\partial\!\!\!/\Psi (x)$$
This integral is supposed to do by parts, by I don't know how to do it. It's the first time I saw an integral by parts with operators and not just with functions...

 

[mark726]

Thank you for your question. It seems like you are on the right track with your solution, but let me offer some additional guidance.

When dealing with integrals involving operators, we can use the same techniques as with integrals involving functions. In this case, we can use integration by parts to manipulate the integral involving the derivative.

Let's start by writing the integral in a slightly different form:
i\int d^4x\int d^4y \, \overline{\eta(y)}S_F^+(x-y)\partial\!\!\!/\Psi(x) = i\int d^4x\int d^4y \, \partial\!\!\!/(\overline{\eta(y)}S_F^+(x-y))\Psi(x)

Now, we can apply the product rule for derivatives to the term inside the parentheses:
i\int d^4x\int d^4y \, \partial\!\!\!/(\overline{\eta(y)})S_F^+(x-y)\Psi(x) + i\int d^4x\int d^4y \, \overline{\eta(y)}\partial\!\!\!/(S_F^+(x-y))\Psi(x)

The first term is just the original integral, so we can substitute it back in:
i\int d^4x\int d^4y \, \partial\!\!\!/(\overline{\eta(y)})S_F^+(x-y)\Psi(x) = i\int d^4x\int d^4y \, \overline{\eta(y)}S_F^+(x-y)\partial\!\!\!/\Psi(x)

Now, let's focus on the second term. We know that the Dirac operator and the Feynman propagator satisfy the relation (i\partial\!\!\!/-m)S_F(x-y) = i\delta(x-y), so we can substitute this in:
i\int d^4x\int d^4y \, \overline{\eta(y)}\partial\!\!\!/(S_F^+(x-y))\Psi(x) = i\int d^4x\int d^4y \, \overline{\eta(y)}(i\partial\!\!\!/-m)\delta(x-y)\Psi(x)

Now, we can use the delta function to simplify