Question about Weinberg QFT Vol. 1 Eq. 6.1.14

Question

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
$$langle 0 | T{psi(x),psi^dagger(y)}|0rangle$$
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.

mike stone · Answer

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
$$
langle0| T {psi(x)psi(y)}|0rangle= theta(x_0-y_0)langle0|  psi(x)psi(y)|0rangle -  theta(y_0-x_0)langle0|  psi(y)psi(x)|0rangle,
$$
it makes subsequent equations nicer.

JF132 · Answer

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.

Question about Weinberg QFT Vol. 1 Eq. 6.1.14

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