I'm reading Zavialov's book "Renormalized Quantum Field Theory" and there is a statement there (in section 1.2 of Chapter I), that I was interested in finding out how to prove. The statement is as follows:
If $[A(x), j_{\{\lambda\}}(y)]=0$ for $(x-y)^2<0$ then $A(x)$ is a local polynomial.
The relevant definitions are:
$$A = \sum_n \int A_n(x_1,\ldots, x_n) :\phi(x_1)\cdots\phi(x_n): dx_1 \cdots dx_n$$
$$j_{\{\lambda\}}(x)=:\phi_{(\lambda_1)}(x)\cdots \phi_{(\lambda_n)}(x):$$
$$\phi_{(\lambda_1)}(x) = \left(\frac{\partial}{\partial x}\right)^{(\lambda_i)}\phi(x) = \left(\frac{\partial}{\partial x_0}\right)^{\lambda_{i0}}\left(\frac{\partial}{\partial x_1}\right)^{\lambda_{i1}}\left(\frac{\partial}{\partial x_2}\right)^{\lambda_{i2}}\left(\frac{\partial}{\partial x_3}\right)^{\lambda_{i3}}\phi(x)$$
and a local polynomial is just a linear combination of $j_{\{\lambda\}}(x)$.
Can anyone help me prove this fact?
These are the main thoughts I came up trying to prove the statement:
The first thing is that the book does not explicitly define $A(x)$, but I think it is probably defined by the natural
$$A(x) = \sum_n \int A_n(x,x_1,\ldots, x_n) :\phi(x_1)\cdots\phi(x_n): dx_1 \cdots dx_n$$
From here, the first thing I thought was that the computation of $[A(x), j(y)]$ reduces to know the value of the commutators:
$$[:\phi(x_1)\cdots \phi(x_n):, j(y)]$$
And since the derivatives can be extracted from the commutator, we will need to compute:
$$[:\phi(x_1)\cdots \phi(x_n):, :\phi(y_1)\cdots \phi(y_m):]$$
Now, using the Wick theorem we can get rid of the normal ordering, and the relation $[A,BC]=B[A,C]+[A,B]C$, we should be able to reduce everything to the commutator $[A(x), \phi(y)]$. So, either I'm neglecting something important or the assumption $[A,j]=0 \ \forall \ j \ $ seems to be too strong, needing only $[A,\phi]=0$, no?
Then we can write the commutators
$$[\phi(x_1)\cdots \phi(x_n), \phi(y)] = \sum_{i=1}^n [\phi(x_i), \phi(y)] \phi(x_1)\cdots\phi(x_{i-1})\phi(x_{i+1})\cdots\phi(x_n)$$
My idea then is to try proving that this sum cannot vanish unless each particular term vanishes. We know that $[\phi(x_i), \phi(y)]$ vanishes when $(x_i-y)^2<0$ so the condition $[A,\phi]=0$ would imply that the functions $A_n(x,x_1,\ldots, x_n)$ must vanish whenever $(x_i-y)^2\geq 0$.
Since this must be true for all $y$ such that $(x-y)^2<0 \ $ this implies that $A_n(x,x_1,\ldots, x_n)$ must be zero whenever $x_i\neq x$ and therefore all the functions $A_n(x,x_1,\ldots, x_n)$ would be linear combinations of $\delta(x-x_i)$ and their derivatives. Which would prove that $A(x)$ is a local polynomial.
I am unsure whether this is the correct approach or how to fill in the details. The part that I think I'm more distant to prove (with some rigour) is proving that indeed the sum cannot be zero unless each individual term vanishes, which seems plausible to me. But I have no clue.
If anyone knows the proof or wants to share any thoughts about it, I'd like to hear you.