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If $S_F(x-y)$ is the Green's function for the Dirac operator $(i\gamma^\mu\partial_\mu-m)$, that is, I assume the following matrix equation holds: $$ (i\gamma^\mu\partial_\mu-m)S_F(x-y)=i\delta(x-y) $$

The adjoint dirac equation is: $$ -i\partial_\mu\bar{\psi}\gamma^\mu -m\bar\psi=0 $$

I am looking for the Green's function of the above equation, in terms of $S_F(x-y)$, that is, if $F[S]$ is some function of $S$ (may include complex conjugation, transposing, etc), then I am looking for such an $F[S]$ such that this equation holds: $$ -i\partial_\mu F[S]\gamma^\mu -mF[S]=i\delta(x-y) $$

What I have done so far:

  1. Dagger this equation: $(i\gamma^\mu\partial_\mu-m)S_F(x-y)=i\delta(x-y)$ and get $$(-i\partial_\mu S_F(x-y)^\dagger \gamma^{\mu\dagger} - S_F(x-y)^\dagger m)=-i\delta(x-y)$$
  2. Multiply by $\gamma^0$ on the right to get: $$ (-i\partial_\mu [-\bar{S_F}(x-y)] \gamma^{\mu} - [-\bar{S_F}(x-y)] m)=i\delta(x-y)\gamma^0 $$
  3. So $[-\bar{S_F}(x-y)]$ almost solves this equation, except you get a factor of $\gamma^0$ on the right which I am not sure how to handle.
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  • $\begingroup$ $(\gamma^0)^2 = 1$, no? $\endgroup$ Apr 6, 2014 at 19:24
  • $\begingroup$ Yes, but if you multiply again by $\gamma^0$ on the right you get on the first term $\gamma^\mu \gamma^0=2\delta^{\mu 0}-\gamma^0 \gamma^\mu$ which is not proportional to $\gamma^\mu$. $\endgroup$
    – PPR
    Apr 6, 2014 at 19:34
  • $\begingroup$ You're missing a $\gamma^0 $ on the left of $\gamma^\mu $ in the adjoint of the latter. Your final expression for $ S $ should absorb this and one on the left to eliminate the one on the right-hand side. $\endgroup$ Apr 6, 2014 at 22:32
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    $\begingroup$ people.math.osu.edu/gerlach.1/math/BVtypset/node81.html $\endgroup$
    – yess
    Apr 6, 2014 at 22:57

3 Answers 3

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The solution to the problem can be found in the nature of the propagator of the Dirac equation: it is a matrix with two spinor indices, i.e.

$$S_F(x-y)=S_F(x-y)_{\alpha\beta}.$$

In step 2, you have treated the propagator as if it was a spinor with a single index, which is not correct. Avoiding this, but multiplying the equation by $\gamma^0$ from the left leaves you with the result

$$F(S)=\gamma^0 S^\dagger \gamma^0=\bar{S}.$$

The last equality sign is consistent with the reference given in one of the comments and with the definition of the adjoint of a matrix discussed in this question.

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  • $\begingroup$ Thanks for your solution. How do you show that $\gamma^0 S^\dagger \gamma^0 = S$ from $ (i\gamma^\mu\partial_\mu-m)S_F(x-y)=i\delta(x-y) $? $\endgroup$
    – PPR
    Apr 6, 2014 at 23:27
  • $\begingroup$ I mixed something up, I have edited my answer. $\endgroup$ Apr 7, 2014 at 2:55
  • $\begingroup$ So, the conclusion is that $F(S)\neq S$? $\endgroup$
    – PPR
    Apr 7, 2014 at 8:11
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    $\begingroup$ Yes, that is my understanding of the answer. $\endgroup$ Apr 7, 2014 at 10:14
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After thinking for a while I think it's easiest to treat this problem in momentum space. Then $$S(p) = \frac{p\!\!\!/ + m}{p^2 - m^2 + i\varepsilon}$$ and satisfies $$(p\!\!\!/ - m)S(p) = 1.$$ Note that $S$ carries spinor indices.

Then from dagger on the above equation, $$S(p)^\dagger (p\!\!\!/ - m)^\dagger =1.$$ Now, $p\!\!\!/ = p_\mu \gamma^\mu$ and $p_\mu$is real, so $$p\!\!\!/^\dagger = p_\mu (\gamma^\mu)^\dagger = p_\mu \gamma^0 \gamma^\mu \gamma^0 = \gamma^0 p\!\!\!/ \gamma^0.$$ Therefore, $$S(p)^\dagger(\gamma^0 p\!\!\!/ \gamma^0 - m) = S(p)^\dagger \gamma^0(p\!\!\!/ +m)\gamma^0 = 1$$ so $S(p)^\dagger\gamma^0$ solves the adjoint Dirac equation with unit source.

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  • $\begingroup$ The last few steps are not clear. What happened to the minus sign? Shouldn't you have a gamma on the left of your final solution? $\endgroup$ Apr 6, 2014 at 22:28
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Can we just multiply from the left with $\gamma^{0}$ then we would get , \begin{equation} \left[-\gamma^{0}S_{F}(x-y)\gamma^{0}\right]\left(-i\partial_{\mu}\gamma^{\mu}-m\right)=i\delta(x-y) \end{equation}

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