# Dirac operator Feynman propagator

+ 3 like - 0 dislike
214 views

Is it true that the following identity holds for the Feynman prescription Dirac propagator: $$S_F(x) \stackrel{?}{=} \gamma^0[S_F(-x)]^\dagger\gamma^0$$

where $S_F$ is defined as the Green's function: $$(i\gamma^\mu\partial_\mu-m)S_F(x-y)=i\delta(x-y)$$

This is somewhat related to a previous question of mine: Green's function for adjoint Dirac Equation

If the statement is true, how do you prove it?

This post imported from StackExchange Physics at 2014-04-13 11:29 (UCT), posted by SE-user Psycho_pr

(bump)

+ 3 like - 0 dislike

I guess, it's true, although I never heard of it. Let's denote $S_F^{-1}(x)=(i\gamma_\mu\partial^\mu-m)(x)$ First let's compute $[S_F^{-1}(−x)]^\dagger$. Assuming that $(\partial_\mu)^\dagger=\partial_\mu$ we get

$$[S_F^{-1}(−x)]^\dagger=(-i\gamma_\mu^\dagger\partial^\mu-m)(-x)=(i\gamma_\mu^\dagger\partial^\mu-m)(x)$$

Using $\gamma^0\gamma^0=1$ and $\gamma^0\gamma_\mu\gamma^0=\gamma_\mu^\dagger$ we get

$$\gamma^0[S_F^{-1}(−x)]^\dagger\gamma^0=(i\gamma^0\gamma_\mu^\dagger\gamma^0\partial^\mu-m\gamma^0\gamma^0)(x)=(i\gamma_\mu\partial^\mu-m)(x)=S_F^{-1}(x)$$

But how do we prove, that it also holds for $S_F(x)$?

Let's apply this transformation on the defining equation $S_F^{-1}(x)S_F(x)=\delta(x)$. Then we get:

$$\gamma^0\left(S_F^{-1}(-x)S_F(-x)\right)^\dagger\gamma^0=\gamma^0\delta(-x)^\dagger\gamma^0$$

The l.h.s. gives: $$\gamma^0\left(S_F^{-1}(-x)S_F(-x)\right)^\dagger\gamma^0=\gamma^0S_F^\dagger(-x)[\overleftarrow{S_F^{-1}}(-x)]^\dagger\gamma^0=\gamma^0S_F^\dagger(-x)\gamma^0\gamma^0[\overleftarrow{S_F^{-1}}(-x)]^\dagger\gamma^0=\gamma^0S_F^\dagger(-x)\gamma^0\overleftarrow{S_F^{-1}}(x)$$

where in the last step we used the relation for $S_F^{-1}$ derived in the beginning.

The r.h.s. simply gives $\gamma^0\delta(-x)^\dagger\gamma^0=\delta(x)$ since the delta function is symmetric. And yeah, we need to assume that it is self-adjoint.

From this we see that $$\gamma^0S_F^\dagger(-x)\gamma^0\overleftarrow{S_F^{-1}}(x)=\delta(x)$$ and therefore $\gamma^0S_F^\dagger(-x)\gamma^0$ is the Green's function of $S_F^{-1}$ which is unique. Therefore it must hold $$\gamma^0S_F^\dagger(-x)\gamma^0=S_F(x)$$

This post imported from StackExchange Physics at 2014-04-13 11:29 (UCT), posted by SE-user Photon
answered Apr 12, 2014 by (70 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOv$\varnothing$rflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.