Second functional derivative of fermion action

[Ravendark]

Homework Statement

[/B]
Consider the following action:
$$\begin{align}S = \int \mathrm{d}^4 z \; \bar\psi_i(z) \, (\mathrm{i} {\not{\!\partial}} - m)_{ij} \, \psi_j(z)\end{align}$$
where $\psi_i$ is a Dirac spinor with Dirac index $i$ (summation convention for repeated indices). Now I would like to calculate the (inverse) propagator $G_{kl}(x,y)$, i.e., the second functional derivative of the action:
$$\begin{align}G_{kl}(x,y) = \frac{\delta^2 S}{\delta \psi_l(y) \delta \bar\psi_k(x)}\end{align}$$
My actual problem deals with the anticommuting of Grassmann quantities. Since the derivatives are Grassmann, I should find (is this correct at all? It is quote from the lecture notes I'm using)
$$\begin{align}\frac{\delta^2 S}{\delta \psi_l(y) \delta \bar\psi_k(x)} = - \frac{\delta^2 S}{\delta \bar\psi_k(x) \delta \psi_l(y)} \; .\end{align}$$
Unfortunately, my calculation yields something different.

2. Conventions

Integral: $\displaystyle \int_z \equiv \int \mathrm{d}^4 z$
Functional derivative: $\; \dfrac{\delta \psi_i(x)}{\delta \psi_j(y)} = \delta_{ij} \, \delta^{(4)}(x-y) = \dfrac{\delta \bar\psi_i(x)}{\delta \bar\psi_j(y)}$

The Attempt at a Solution

Left hand side:
$$\begin{align}
\frac{\delta^2 S}{\delta \psi_l(y) \delta \bar\psi_k(x)} &= \frac{\delta}{\delta \psi_l(y)} \int_z \, \delta_{ik} \, \delta^{(4)}(z-x) \, (\mathrm{i} {\not{\!\partial}} - m)_{ij} \, \psi_j(z) \\ &= \frac{\delta}{\delta \psi_l(y)} \, (\mathrm{i} {\not{\!\partial}} - m)_{kj} \, \psi_j(x) \\ &= (\mathrm{i} {\not{\!\partial}} - m)_{kj} \, \delta_{jl} \, \delta^{(4)}(x-y) \\ &= (\mathrm{i} {\not{\!\partial}} - m)_{kl} \, \delta^{(4)}(x-y)
\end{align}$$
Looks good so far (?), now the next part...

Right hand side:
$$\begin{align}
- \frac{\delta^2 S}{\delta \bar\psi_k(x) \delta \psi_l(y)} &= \frac{\delta}{\delta \bar\psi_k(x)} \int_z \, \bar\psi_i(z) \, (\mathrm{i} {\not{\!\partial}} - m)_{ij} \, \delta_{jl} \, \delta^{(4)}(z-y) \\ &= \frac{\delta}{\delta \bar\psi_k(x)} \int_z \, \bar\psi_i(z) \, (\mathrm{i} {\not{\!\partial}} - m)_{il} \, \delta^{(4)}(z-y) \\ &= \frac{\delta}{\delta \bar\psi_k(x)} \left\{ \int_z \, \bar\psi_i(z) \, \mathrm{i} \, \gamma^\mu_{il} \, \partial_\mu \, \delta^{(4)}(z-y) - \int_z \, \bar\psi_i(z) \, m_{il} \, \delta^{(4)}(z-y) \right\} \\ &= \frac{\delta}{\delta \bar\psi_k(x)} \left\{ - \int_z \, (\partial_\mu \bar\psi_i(z)) \, \mathrm{i} \, \gamma^\mu_{il} \, \delta^{(4)}(z-y) - \bar\psi_i(y) \, m_{il} \right\} \\ &= \frac{\delta}{\delta \bar\psi_k(x)} \bigl\{ - (\partial_\mu \psi_i(y)) \, \mathrm{i} \, \gamma^\mu_{il} - \bar\psi_i(y) \, m_{il} \bigr\} \\ &= -\delta_{ik} \, \partial_\mu \, \delta^{(4)}(y-x) \, \mathrm{i} \, \gamma^\mu_{il} - \delta_{ik} \, \delta^{(4)}(y-x) \, m_{il} \\ &= (-\mathrm{i} {\not{\!\partial}} - m)_{kl} \, \delta^{(4)}(y-x)
\end{align}$$As you can see, there is an additional minus sign in front of the derivative part...and I have no idea how to "remove" it. Can someone give me a hint please?

EDIT:
I noticed right now that the arguments of the delta functions are still swapped. I can fix the mass part by using that the delta function is even, i.e., $m_{kl} \, \delta^{(4)}(y-x) = m_{kl} \, \delta^{(4)}(x-y)$. But the derivative part confuses me a bit because I never dealt with a derivative of a delta function in such a way.Best Regards,
Ravendark

 

[sgd37]

You don't use integration by parts to transfer the spacetime derivative onto PsiBar since <br /> \; \dfrac{\delta \psi_i(x)}{\delta \psi_j(y)} = \delta_{ij} \, \delta^{(4)}(x-y) = \dfrac{\delta \bar\psi_i(x)}{\delta \bar\psi_j(y)}<br /> you have the choice over which delta function you integrate first<br /> \frac{\delta}{\delta \bar\psi_k(x)} \int_z \, \bar\psi_i(z) \, (\mathrm{i} {\not{\!\partial}} - m)_{il} \, \delta^{(4)}(z-y)=\int_z \, \delta^{(4)}(z-x) \, (\mathrm{i} {\not{\!\partial}} - m)_{kl} \, \delta^{(4)}(z-y)

So integrating over the left delta function will give you the desired result

 

[Ravendark]

You don't use integration by parts to transfer the spacetime derivative onto PsiBar since <br /> \; \dfrac{\delta \psi_i(x)}{\delta \psi_j(y)} = \delta_{ij} \, \delta^{(4)}(x-y) = \dfrac{\delta \bar\psi_i(x)}{\delta \bar\psi_j(y)}<br /> you have the choice over which delta function you integrate first<br /> \frac{\delta}{\delta \bar\psi_k(x)} \int_z \, \bar\psi_i(z) \, (\mathrm{i} {\not{\!\partial}} - m)_{il} \, \delta^{(4)}(z-y)=\int_z \, \delta^{(4)}(z-x) \, (\mathrm{i} {\not{\!\partial}} - m)_{kl} \, \delta^{(4)}(z-y)

So integrating over the left delta function will give you the desired result

Right...instead of transfer the derivative to $\bar\psi$ I can perform the derivative directly since I know how the derivative w.r.t. $\bar\psi$ looks like.

But why is it "forbidden" to use partial integration ((11) to (12) in my first post) at this point (apart from the fact that this yields a wrong result)?

 

[sgd37]

Forgive me what I should have said is you don't have to transfer the space-time derivative. You are free to do so

 

[Ravendark]

Forgive me what I should have said is you don't have to transfer the space-time derivative. You are free to do so

Thank you very much, I got it...I totally forgot, that there exists an identity for the derivative of the delta function: $\partial_\mu \delta^{(4)}(y-x)=-\partial_\mu \delta^{(4)}(x-y)$.

 

[sgd37]

I'd be careful here because the RHS gives \frac{\partial}{\partial x^{\mu}}\delta^4(x-y) whilst the LHS gives you \frac{\partial}{\partial y^{\mu}}\delta^4(y-x) if you set t=x-y and use the identity you gave above you can see that \frac{\partial}{\partial x^{\mu}}\delta^4(x-y)=\frac{\partial}{\partial y^{\mu}}\delta^4(y-x) so this wouldn't get rid of your minus sign. The reason why you find these inequalities is because on the RHS you're integrating over a delta function whilst on the LHS you're integrating over the derivative of a delta function. If you are consistent on both sides you get the right answer

 

[Ravendark]

I'd be careful here because the RHS gives \frac{\partial}{\partial x^{\mu}}\delta^4(x-y) whilst the LHS gives you \frac{\partial}{\partial y^{\mu}}\delta^4(y-x) if you set t=x-y and use the identity you gave above you can see that \frac{\partial}{\partial x^{\mu}}\delta^4(x-y)=\frac{\partial}{\partial y^{\mu}}\delta^4(y-x) so this wouldn't get rid of your minus sign.

Mhh, it seems you're right...

The reason why you find these inequalities is because on the RHS you're integrating over a delta function whilst on the LHS you're integrating over the derivative of a delta function.

Now I'm a bit confused...about which RHS/LHS are you talking about?

If you are consistent on both sides you get the right answer

You mean consistent on which argument $\partial_\mu$ actually acts?