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The Feynman propagator is defined as $$i G_F = \theta(t-t')G^+ + \theta(t'-t)G^-. \tag{1}$$ Using $$G^{(1)} = G^+ + G^-,$$ $$G_R = -\theta(t-t')G, $$ $$G_A = \theta(t'-t)G, $$ $$\bar{G} = \frac{1}{2}(G_R + G_A), $$ and $$iG = G^+ - G^- ,$$ it can be shown that $$G_F = - \bar{G} - \frac{1}{2}iG^{(1)}. $$ I tried reproducing the calculation and got the following. First note that $$G_R = i\theta(t-t')G^+ - i\theta(t -t')G^-, $$ and $$G_A =-i\theta(t'-t)G^+ + i\theta(t' -t)G^-. $$ Using this, I get $$G_F = - \bar{G} - \frac{1}{2}iG^{(1)}$$ $$= -\frac{1}{2}(G_R + G_A)- \frac{1}{2}i(G^+ + G^-),$$ $$= -\frac{1}{2}(i\theta(t-t')G^+ - i\theta(t -t')G^- -i\theta(t'-t)G^+ + i\theta(t' -t)G^-) - \frac{1}{2}i(G^+ + G^-)$$ $$ = \frac{-i}{2} ( \theta(t-t') - \theta(t'-t) + 1)G^+ - \frac{i}{2} ( \theta(t'-t) - \theta(t-t') + 1)G^-, $$ thus, $$ iG_F = \frac{1}{2} ( \theta(t-t') - \theta(t'-t) + 1)G^+ + \frac{1}{2} ( \theta(t'-t) - \theta(t-t') + 1)G^-.$$ Nevertheless, this is not equal to equation (1). Independent of the value of the heaviside function, I'll always get a substraction of $G^+$ and $G^-$. Is there a mistake in my process? Or are my last equation and equation (1) equivalent?

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    $\begingroup$ Which conventions? Which reference? Which page? $\endgroup$
    – Qmechanic
    Nov 24, 2023 at 7:31

1 Answer 1

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If you use

$$ \theta(x)+\theta(-x) =1 $$

You will see that they are the same.

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