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Second functional derivative of fermion action

  1. Aug 11, 2015 #1
    1. The problem statement, all variables and given/known data

    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)}##

    3. 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
     
    Last edited: Aug 11, 2015
  2. jcsd
  3. Aug 13, 2015 #2
    You don't use integration by parts to transfer the spacetime derivative onto PsiBar since [tex]
    \; \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)}
    [/tex] you have the choice over which delta function you integrate first[tex]
    \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) [/tex]

    So integrating over the left delta function will give you the desired result
     
  4. Aug 14, 2015 #3
    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)?
     
  5. Aug 14, 2015 #4
    Forgive me what I should have said is you don't have to transfer the space-time derivative. You are free to do so
     
    Last edited: Aug 14, 2015
  6. Aug 18, 2015 #5
    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)##.
     
  7. Aug 18, 2015 #6
    I'd be careful here because the RHS gives [tex]\frac{\partial}{\partial x^{\mu}}\delta^4(x-y)[/tex] whilst the LHS gives you [tex]\frac{\partial}{\partial y^{\mu}}\delta^4(y-x)[/tex] if you set [tex]t=x-y[/tex] and use the identity you gave above you can see that [tex]\frac{\partial}{\partial x^{\mu}}\delta^4(x-y)=\frac{\partial}{\partial y^{\mu}}\delta^4(y-x)[/tex] 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
     
  8. Aug 19, 2015 #7
    Mhh, it seems you're right...

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

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