Question about Weinberg QFT Vol. 1 Eq. 6.1.14

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.

Consider the case $$(x-y)$$ is space-like, first notice that the fermionic fields anticommute: $$$$\psi(x)\psi(y) = - \psi(y)\psi(x)$$$$ There is a frame with $$x^0 > y^0$$, in this frame: $$$$T\{\psi(x)\psi(y)\} = \psi(x)\psi(y) \label{eq1}$$$$ Since $$(x-y)$$ is space-like, there will be another frame in which $$x^0 < y^0$$. In this frame, if we were to use the definition without the minus sign, then: $$$$T\{\psi(x)\psi(y)\} = \psi(y)\psi(x) = -\psi(x)\psi(y) \label{eq2}$$$$ As you can see, the result would be different between the two frames. In order to resolve that, we need the extra minus sign.
The Hamiltonion is always involves products of an even number of fermions, so the definition of time ordering for products of Hamiltonian $$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.
• 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. May 31 at 17:37