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I am reading a paper (arXiv version) in QFT. I am stuck at this point,

$$S [\phi (x)]={1\over 2}\int\phi (x_1)D (x_1 -x_2)\phi (x_2)dx_1dx_2 $$ $$+{\lambda \over 4!}\int V (x_1,x_2,x_3,x_4) \phi(x_1)\phi (x_2)\phi (x_3) \phi (x_4) dx_1 dx_2dx_3dx_4.\tag{9}$$

Here $D (x_1-x_2)$ is an inverse propagator.

I have 3 doubts:

  1. In LHS $\phi$ is a function of $x$ but in RHS there is no $x$. Why?

  2. How the inverse propagator comes?

  3. Here we are studying $\phi^4$ theory so, how $V (x_1,x_2,x_3,x_4)$ comes in second term of RHS?

Qmechanic
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ROBIN RAJ
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1 Answers1

2

  1. $S[\phi(x)]$ is a common misleading notation often used in physics. The symbol $x$ is here a half-baked attempt to indicate that the function $\phi$ is a function of $x$. The author is not saying that the action $S$ only depends on the value $\phi(x)$ at the point $x\in\mathbb{R}^4$. In fact the action functional $S[\phi]$ depends on the function $\phi:\mathbb{R}^4\to \mathbb{R}$. A better notation is perhaps $S[x\!\mapsto\!\phi(x)]$.

  2. The fact that the quadratic action term is the inverse propagator is e.g. explained in Refs. 1-2; this duplicate Phys.SE question; or in my Phys.SE answer here.

  3. The author writes down the most general (possibly non-local) quartic interaction term.

References:

  1. L.H. Ryder, QFT, 2nd eds., 1996; eq. (6.13).

  2. L.S. Brown, QFT, 1992; eq. (3.2.10).

Qmechanic
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