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I'm just a beginner learning renormalization. I have a short question: In interacting QFT, why the previous Lorentz invariant scaler mass $m$ will be replace by physical mass $m_\lambda$ (or say renormalized mass) once we introduce interaction into theory? Or equally: Why interaction will bring a correction to mass term?

And also, electric charge and field operator (wave function) will do corresponding corrections, are the physical nature of these corrections are same?

Qmechanic
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a Fish in Dirac Sea
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    Why…? Are you looking for something different from “when we do the calculation of the effect of interactions, the mass changes”? – Ghoster Nov 07 '22 at 04:28
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    Even in classical EM, the electric field of a point charge at rest has infinite energy because the field gets infinitely strong infinitely close to the charge. You can think of that energy as classically renormalizing the point charge’s mass. So it should not be surprising that when the EM field is quantized the mass still requires renormalization; the point-iness of the point charge is responsible. – Ghoster Nov 07 '22 at 04:31
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    In an interacting theory, the pole in the propagator (which actually determines the mass we observe) is not the same as the coefficient of the quadratic term in the Lagrangian. The details of this are covered in every QFT book. – Andrew Nov 07 '22 at 12:09

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In QFT one notion of mass is the coefficient of the mass term quadratic in the fields. Coefficients in front of each term in the action are coupling constants, which potentially run. The elementary electric charge $e$, which is the coefficient in front of the $\bar{\psi}\gamma^{\mu}A_{\mu}\psi$ action term, is another running coupling constant.

Perhaps the running is easiest to see in the Wilsonian effective action (WEA) when integrating out heavy modes: Each action term in the WEA gets modified by amputated Feynman diagrams, made from heavy modes and with the same number of external fields as the action term in question, cf. e.g. my Phys.SE answer here.

Qmechanic
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