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In Weinberg QFT Vol. 1 Eq. 6.1.14, he defines the propagator for a fermion to be $$ \theta(x-y)\{\psi^+(x),\psi^+{}^\dagger(y)\} - \theta(x-y)\{\psi^-{}^\dagger(x),\psi^-(y)\}$$ where $\psi^+$ annihilates particles, and $\psi^-$ creates antiparticles. He claims (cf. 6.1.1) that this comes from a term $$\langle 0 | T\{\psi(x),\psi^\dagger(y)\}|0\rangle$$ in the S-matrix. My question is, I don't see how the relative minus sign between the two terms in 6.1.14 (my first equation) follows from the time ordering in the second equation. Peskin and Schroeder say that it's just a definition of time ordering for fermions, but that can't be true because the time ordering is already defined in the definition of the S-matrix in terms of time-ordered products of the Hamiltonian.

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In addition to the other answer, the reason we need the extra minus sign is so that the time-order is Lorentz invariant.

Consider the case $(x-y)$ is space-like, first notice that the fermionic fields anticommute: \begin{equation} \psi(x)\psi(y) = - \psi(y)\psi(x) \end{equation} There is a frame with $x^0 > y^0$, in this frame: \begin{equation} T\{\psi(x)\psi(y)\} = \psi(x)\psi(y) \label{eq1} \end{equation} Since $(x-y)$ is space-like, there will be another frame in which $x^0 < y^0$. In this frame, if we were to use the definition without the minus sign, then: \begin{equation} T\{\psi(x)\psi(y)\} = \psi(y)\psi(x) = -\psi(x)\psi(y) \label{eq2} \end{equation} As you can see, the result would be different between the two frames. In order to resolve that, we need the extra minus sign.

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The Hamiltonion is always involves products of an even number of fermions, so the definition of time ordering for products of Hamiltonian $H(t)$'s does not dictate the time time ordering rule for individual Fermi fields. It turns out that if we define $$ \langle0| T \{\psi(x)\psi(y)\}|0\rangle= \theta(x_0-y_0)\langle0| \psi(x)\psi(y)|0\rangle - \theta(y_0-x_0)\langle0| \psi(y)\psi(x)|0\rangle, $$ it makes subsequent equations nicer.

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  • $\begingroup$ You're saying that if we define the fermion propagator with a + sign, we get the same answer at the end? $\endgroup$ May 31 at 15:46
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    $\begingroup$ @Eric David Kramer. No! I just meant that the definition with the minus sign is useful, and that the one without is not. $\endgroup$
    – mike stone
    May 31 at 15:55
  • $\begingroup$ Then I'm still confused. The thing I want to calculate is the time-ordered product of Hamiltonians, which is well-defined. By Wick's theorem, this should be equal to a product of propagators. I just wanted a proof (or a hand-wavy argument) that this only works with the minus sign. $\endgroup$ May 31 at 16:12
  • $\begingroup$ The Hamiltonians are even in fermion number, so no minus sign is needed in the T product of $H$'s. You need minus signs when using Wick's theorem to evaluate the contractions of out-of-order fermions in the T product of $H$'s to get propagators. $\endgroup$
    – mike stone
    May 31 at 17:37
  • $\begingroup$ Right, that's the claim. (It's also what we use and seems to work.) I just want a proof of that point exactly. I.e. a proof that you need the minus sign when using Wick's theorem. But OK I agree it somehow has to be true. $\endgroup$ May 31 at 18:50

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