I am interested in the ambiguities which exist in defining the composite free field operators--i.e., operators corresponding to monomials of the fundamental field operator (and their derivatives). In particular, I want to know once one has chosen a "renormalized" quadratic operator for the free theory: Are all other composite operators completely fixed or are there additional ambiguities which arise at higher powers of the field?

Consider, for concreteness, the Klein-Gordon scalar field. Products of the field operator at the same spacetime point usually diverge:

\begin{align} \lim_{x_1\to x_2} D(x_1-x_2)& \equiv \lim_{x_1\to x_2}\langle 0|\phi(x_1)\phi(x_2) |0 \rangle \\ &=\lim_{x_1\to x_2} \int \frac{d^3p}{(2\pi)^3} \frac{e^{-ip\cdot(x_1-x_2)}}{2\sqrt{\vec{p}^2+m^2}}\\ &\propto\int_0^\infty d\rho \frac{\rho^2}{\sqrt{\rho^2+m^2}}\\ &\sim\infty. \end{align}

Hence, to define a quadratic field operator with finite expectation value, one needs to remove the singular behavior at the coincidence limit. The conventional choice is simply to define "normal ordered operators" with vanishing expectation value:

\begin{align} :\phi^2(x):\equiv\lim_{y\to x}\left[\phi (y)\phi (x)-D(y-x) \right]. \end{align}

Setting the VEV to zero is an arbitrary choice and one could equally well define a finite quadratic operator which differs from textbook normal ordering by bi-solutions to the K-G e.o.m. which remain bounded as $y\to x$. i.e., choose instead to define

\begin{align} :\widetilde{\phi^2} (x):\equiv \lim_{y\to x} \left[ \phi(y)\phi(x)-D(y-x)+G(y,x) \rangle \right], \end{align} where $G(y,x)$ is a solution to the K-G e.o.m. which remains finite in the limit $y\to x $.

The ambiguity in defining the renormalized quadratic operator for the free theory is then just proportional to the identity operator, $I$.

Now, consider defining a renormalized cubic operator, $:\phi^3:$. In analogy to the quadratic definition, I expect something like

\begin{align} :\phi^3(x):\equiv \lim_{y,z\to x} \left[\phi(z)\phi(y)\phi(x)-D(z-y)\phi(x)-D(y-x)\phi(z)-D(z-x)\phi(y)\right].\\ \end{align}

And limits can also be taken in different orders: e.g., taking $y\to x$ before $z\to x$,

\begin{align} :\phi^3(x):=\lim_{z\to x} \left[ \phi(z):\phi^2(x):-2D(z-x)\phi(x)\right]. \end{align}

If one doesn't insist on the VEV of the cubic operator being zero, can one choose $:\phi^3:$ independent of the choice for $:\phi^2:$ or is the cubic operator fixed by the definition of the quadratic operator? And so on for higher powers?



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