Complete characterization of local monomials

[Gaussian97]

TL;DR Summary

If $[A(x), j_{\{\lambda\}}(y)]=0$ for $(x-y)^2<0$ then $A(x)$ is a local polynomial.

Hi, I'm reading Zavialov's book on QFT and there's a statement there I was interested in finding how to prove it. 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:
$$A = \sum_n \int A_n(x_1,\ldots, x_n) :\phi(x_1)\cdots\phi(x_n): dx_1 \cdots dx_n$$
$$j_{\{\lambda\}}(x)=:\phi_{(\lambda_1)}(x)\cdots \phi_{(\lambda_n)}(x):$$
$$\phi_{(\lambda_1)}(x) = \left(\frac{\partial}{\partial x}\right)^{(\lambda_i)}\phi(x)$$
and a local polynomial is just a linear combination of $j_{\{\lambda\}}(x)$.

Does anyone have a proof of this fact?

This 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 commutator $[A(x), j(y)]$ 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 at the end we 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 using the relation $[A,BC]=B[A,C]+[A,B]C$, we should really 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 innecesary, just needing $[A,\phi]=0$, 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 trying to prove 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 don't know if anyone knows if this is the right approach. The part that I think I'm more distant to actually prove (with some rigour) is to prove that indeed the sum cannot be zero unless each individual term, 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.

 

[eljose79]

Thank you for your interest in the statement from Zavialov's book on QFT. I can offer some insights and suggestions on how to prove this fact.

Firstly, I agree with your approach of using the Wick theorem to get rid of the normal ordering and reducing the commutator to $[A(x), \phi(y)]$. This is a common technique in QFT and can simplify the calculations.

Regarding the assumption $[A,j]=0$, I believe it is necessary in this case. While it is true that $[\phi(x_i), \phi(y)]=0$ when $(x_i-y)^2<0$, this does not necessarily mean that $[\phi(x_1)\cdots \phi(x_n), \phi(y)]=0$ for all values of $y$. In fact, the commutator might not even be well-defined for some values of $y$. Therefore, the assumption $[A,j]=0$ is needed to ensure that the commutator is well-defined and can be used in the proof.

Your idea of proving that the sum cannot vanish unless each particular term vanishes is a good one. However, I would suggest using a more rigorous argument to prove this. One approach could be to use the fact that the commutator $[\phi(x_i), \phi(y)]$ is a c-number (a constant) and therefore can be pulled out of the sum. Then, you can use the properties of the delta function to show that the sum cannot vanish unless each term is zero.

Lastly, your conclusion that $A(x)$ is a local polynomial is correct. This follows from the fact that the functions $A_n(x,x_1,\ldots, x_n)$ must be linear combinations of $\delta(x-x_i)$ and their derivatives, as you stated.

I hope this helps in your proof and I wish you all the best in your studies of QFT.