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  Why is the quadratic term in the Lagrangian a mass term and not the propagator?

+ 2 like - 2 dislike

Typically we interpret the propagator to be associated with the kinetic piece of the the Lagrangian... but I am confused about why the $m^2 \phi^2$ term is mass, not also a propagator? That is, we interpret higher order terms as interactions, but if we draw the quadratic term, it looks like a propagator.


Closed as per community consensus as the post is undergraduate level
asked Sep 13, 2014 in Closed Questions by anonymous [ no revision ]
recategorized Jun 22, 2015 by dimension10

The quadratic term is either a propagator, if you choose to make it part of the propagator, or else if you choose to make it part of the interaction, it is a two prong interaction which when you add up all of them on any line, it sums to the same thing. That is the geometric series identity

$${1 \over k^2 + m^2} = {1\over k^2} - {1\over k^2} m^2 {1\over k^2} + {1\over k^2} m^2 {1\over k^2} m^2 {1\over k^2} + ...$$

the terms of which represents the contribution to a line from term with 0,1,2,3 mass-term vertices.

This question is not clear, and I am voting to close. The complete answer is the identity above.

Things are generally interconnected in mathematics, but insofar as there is a single reason for the \(m^2\hat\phi(x)^2\) term being taken to be part of the Feynman propagator \((k^2-m^2+\mathrm{i}\epsilon)^{-1}\) instead of being taken to be an interaction, it is because there is an exact solution of the Klein-Gordon differential equation \(\left(\frac{\partial^2}{\partial x^\mu\partial x_\mu}+m^2\right)\phi(x)=0\), and there are, correspondingly, various exact Green's functions, including the mass \(m\) Feynman propagator.

1 Answer

+ 3 like - 0 dislike

The bare propagator

$(p^2+m^2)^{-1}$ for a scalar field, $(\gamma\cdot p+m)^{-1}$ for a Dirac field

is the inverse of the Fourier transformed differential operator inside the full quadratic part of the Lagrangian density. It includes the kinetic term ($p^2$ for a scalar field, $\gamma\cdot p$ for a Dirac field) and the mass term ($m^2$ for a scalar field, $m$ for a Dirac field). 

In renormalization theory, the bare mass term is replaced by the physical mass tern and the difference is added to the interaction as a mass counterterm. This is the first step needed to make the perturbation theory finite. In a second step, one adds momentum-dependent self-energy terms to get the renormalized propagator. Only the latter is a physical propagator with observable information.

answered Sep 14, 2014 by Arnold Neumaier (15,787 points) [ revision history ]

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