handrea2009
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- TL;DR Summary
- QED: replacing photon field with current in the definition of the renormalized 3-point function vertex
I am self-studying QFT in the Schwartz book "Quantum Field Theory and the Standard Model", currently I am struggling to understand the all-orders proof that ##Z_1=Z_2## using Ward-Takahashi identity (page 352).
He states that ## -ie_R\Gamma^\mu ##, which is the sum of the 1PI contributions to matrix elements for the 3-point function ##\langle \psi(x_1)A_\nu(x)\bar\psi(x_2)\rangle## with externa legs amputated , can be formally defined as follow:
$$-ie_R\Gamma^\mu(p,q_1,q_2)(2\pi)^4\delta^4(p+q_1-q_2) \\
\equiv -ie_R\int d^4x d^4x_1 d^4x_2 e^{ipx}e^{iq_1x_1}e^{-iq_2x_2}\\(iG)^{-1}(\not q_1) \langle j^\mu(x) \psi(x_1) \bar\psi(x_2)\rangle (iG)^{-1}(\not p+\not q_1) \tag {19.78}
$$
I don't understand how he gets to replace in ##\langle \psi(x_1)A_\nu(x)\bar\psi(x_2)\rangle## the ##A_v## photon field with the current ##j^\mu = \bar\psi\gamma^\mu \psi##.
The only clue I could find is on page 281 where we have the following formula which comes from Schwinger-Dyson equation:
$$
\square_{\alpha\beta}^k \square_{\mu\nu} \langle A_\nu(x)...A_\beta(x_k)...\rangle = \langle j_\mu(x)...j_\alpha(x_k)...\rangle
$$
so basically you can remove ##A_\mu## field and insert current ##j_\mu##.
However, even using that result it seems to me that in the formula (19.78) in the right-hand-side we have a wrong extra ##ie_R## factor, since we already have one which comes from exploding the expectation value ##\langle ... \rangle ## ( see (7.77) ):
$$
\langle j^\mu(x) \psi(x_1) \bar\psi(x_2)\rangle \equiv \langle \Omega|T\{j^\mu(x)\psi(x_1) \bar\psi(x_2)\}|\Omega\rangle = \langle 0|T\{j^\mu_0(x)\psi_0(x_1) \bar\psi_0(x_2)e^{-ie_R\int \bar\psi_0\gamma^\mu\psi_0 A_\mu}\}|0\rangle_{no bubbles}
$$
where ## -ie_R\bar\psi\gamma^\mu\psi A_\mu ## is the interaction term in the QED Lagrangian density
He states that ## -ie_R\Gamma^\mu ##, which is the sum of the 1PI contributions to matrix elements for the 3-point function ##\langle \psi(x_1)A_\nu(x)\bar\psi(x_2)\rangle## with externa legs amputated , can be formally defined as follow:
$$-ie_R\Gamma^\mu(p,q_1,q_2)(2\pi)^4\delta^4(p+q_1-q_2) \\
\equiv -ie_R\int d^4x d^4x_1 d^4x_2 e^{ipx}e^{iq_1x_1}e^{-iq_2x_2}\\(iG)^{-1}(\not q_1) \langle j^\mu(x) \psi(x_1) \bar\psi(x_2)\rangle (iG)^{-1}(\not p+\not q_1) \tag {19.78}
$$
I don't understand how he gets to replace in ##\langle \psi(x_1)A_\nu(x)\bar\psi(x_2)\rangle## the ##A_v## photon field with the current ##j^\mu = \bar\psi\gamma^\mu \psi##.
The only clue I could find is on page 281 where we have the following formula which comes from Schwinger-Dyson equation:
$$
\square_{\alpha\beta}^k \square_{\mu\nu} \langle A_\nu(x)...A_\beta(x_k)...\rangle = \langle j_\mu(x)...j_\alpha(x_k)...\rangle
$$
so basically you can remove ##A_\mu## field and insert current ##j_\mu##.
However, even using that result it seems to me that in the formula (19.78) in the right-hand-side we have a wrong extra ##ie_R## factor, since we already have one which comes from exploding the expectation value ##\langle ... \rangle ## ( see (7.77) ):
$$
\langle j^\mu(x) \psi(x_1) \bar\psi(x_2)\rangle \equiv \langle \Omega|T\{j^\mu(x)\psi(x_1) \bar\psi(x_2)\}|\Omega\rangle = \langle 0|T\{j^\mu_0(x)\psi_0(x_1) \bar\psi_0(x_2)e^{-ie_R\int \bar\psi_0\gamma^\mu\psi_0 A_\mu}\}|0\rangle_{no bubbles}
$$
where ## -ie_R\bar\psi\gamma^\mu\psi A_\mu ## is the interaction term in the QED Lagrangian density
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