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Why is the term $m\delta_2\delta_m\bar{\psi}\psi$ ignored in the QED Lagrangian?

+ 4 like - 0 dislike
250 views

Consider the QED Lagrangian

$$\mathcal{L}=\bar{\psi}_0(i\gamma^{\mu}\partial_{\mu}-e_0\gamma^{\mu}A_{0\mu}-m_0)\psi_0-\frac{1}{4}(\partial_{\mu}A_{0\nu}-\partial_{\nu}A_{0\mu})^2$$

where the 0 subscript denotes bare fields. The bare fields are related with the renormalized fields via

$$\psi_0=\sqrt{Z_2}\psi\qquad{}A_{0\mu}=\sqrt{Z_3}A_{\mu}$$

$$m_0=Z_mm\qquad{}e_0=Z_ee$$

with these redefinitions the Lagrangian takes the form

$$\mathcal{L}=Z_2\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi-eZ_eZ_2\sqrt{Z_3}\bar{\psi}\gamma^{\mu}A_{\mu}\psi-mZ_2Z_m\bar{\psi}\psi-\frac{1}{4}Z_3(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})^2$$


it is customary to define

$$Z_1\equiv{}Z_eZ_2\sqrt{Z_3}$$

leaving

$$\mathcal{L}=Z_2\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi-eZ_1\bar{\psi}\gamma^{\mu}A_{\mu}\psi-mZ_2Z_m\bar{\psi}\psi-\frac{1}{4}Z_3(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})^2.$$

Moreover, the $Z$ renormalization constants are defined to be

$$Z_1=1+\delta_1\qquad{}Z_2=1+\delta_2\qquad{}Z_3=1+\delta_3\qquad{}Z_m=1+\delta_m$$

it is often said that this leaves the QED Lagrangian in the following form (see page 2 of these notes http://isites.harvard.edu/fs/docs/icb.topic1146665.files/III-5-RenormalizedPerturbationTheory.pdf).

$$\mathcal{L}=\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi+\delta_2\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi-e\bar{\psi}\gamma^{\mu}A_{\mu}\psi-e\delta_1\bar{\psi}\gamma^{\mu}A_{\mu}\psi-m\bar{\psi}\psi-m(\delta_m+\delta_2)\bar{\psi}\psi-\frac{1}{4}(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})^2-\frac{1}{4}\delta_3(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})^2.$$

Nonetheless, the carefull reader will notice that one term coming from $mZ_2Z_m\bar{\psi}\psi$ is completely ignored, namely

$$m\delta_2\delta_m\bar{\psi}\psi.$$

Why?

asked Oct 28, 2015 in Theoretical Physics by Dmitry hand me the Kalashnikov (720 points) [ revision history ]
retagged Oct 29, 2015 by Dilaton

2 Answers

+ 4 like - 0 dislike

Because formally we treat the counterterms as small, so anything higher than first order in counterterms is negligible. Alternatively, any contributions that the extra factor of \(\delta_2\)would make to the propagator could be absorbed by a redefinition of \(\delta_2\) and \(\cancel{\delta_m}\). This would add extra complication without changing the end-result.

Let me qualify the first statement in the answer rather than leaving it in a comment. The counterterms are part of a perturbative expansion in the renormalized coupling constant \(e_R\) and we take this to be small, therefore higher order terms in the coupling, like the one this question is asking about, are negligible. This is in contrast to the expansion in bare perturbation theory where the bare coupling, a non-observable and un-physical quantity, is formally infinite.

answered Oct 29, 2015 by holomorphic (70 points) [ revision history ]
edited Oct 29, 2015 by holomorphic

Only the second explanation is correct, as counterterms are large. Only $\delta_m$ must be redefined to get the correct Lagrangian. See my answer.

Maybe I should have been more clear. Yes, obviously counterterms are large, but they are part of a perturbative expansion in the coupling constant. More specifically in QED the 1-loop counterterms are second order in \(e_R\), and therefore the product in question, using the counterterms as commonly defined, is quartic in the charge.

Let me note that divergences are independent of the charge value. It is the structure of interaction who is bad.

+ 3 like - 0 dislike

To define the $Z$ renormalization constants as

$$Z_1=1+\delta_1,~~Z_2=1+\delta_2,~~Z_3=1+\delta_3,~~Z_m=1+\delta_m$$

is just sloppiness. A more careful definition as

$$Z_1=1+\delta_1,~~Z_2=1+\delta_2,~~Z_3=1+\delta_3,~~Z_m=1+\delta_m/Z_2$$

removes the extra term in the renormalized Lagrangian.

Note that the counterterms are not tiny but huge (infinite if the cutoff is removed), hence a product of counterterms cannot be neglected.

answered Oct 29, 2015 by Arnold Neumaier (12,355 points) [ no revision ]

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