Will $L_n$ commute with $f(z)I$? and Ward identity for 0 conformal weight This might be a stupid question.
In some situation I have to calculate a form like this.
$$
L_n z^N |\psi\rangle \overset{?}{=} z^N L_n|\psi\rangle
$$
In the appendix $B$ of BPZ https://www.sciencedirect.com/science/article/abs/pii/055032138490052X


From B.4 B.6 to B.7, the author seems to have assumed $[L_n, z^N]=0$
I got confused because I somehow think it shouldn't commute since we have for example $L_{-1}=\partial_z$.
In my understanding general holomorphic functions are not primary fields and the only primary field of weight $(0,0)$ is constant, so we can not directly apply the Ward identity for primary field.
Is the infinitesimal transformation of $f(z)I$ also generated by $T(z)$?
This problem has haunted me for a while.
 A: The answer is Yes, they commute. I have this confusion because $L_n$ is represented as differential operator when acted on primary fields and I saw no reason to expect that $[L_n, f(z)I]=0$. After several hours of investigation into literatures, I found a way to justify it. The key word is vertex operator algebra. VOA is a good tool for axiomatize CFT and helps me to clarify the confusion in state-operator correspondence.
We start with a complex vector space $V$, which could be physically regarded the state space (not all state physical,i.e. not necessarily Hilbert).

*

*There is a unique vector $|0\rangle$， which is called vacuum state.


*There is a linear map $Y: V\to End(V)[[z^{\pm 1}]]$, which we call state-operator correspondence, where $End(V)[[z^{\pm 1}]]$ is the formal Laurent series of the endomorphism(matrics) over $V$. For a general vector $|\Psi\rangle\in V$
$$
Y(|\Psi\rangle, z) =\sum_{n\in\mathbb{Z}} Y(|\Psi\rangle)_n z^{-n-1}
$$


*$Y(|0\rangle,z) = id_V$


*$\exists |\omega\rangle \in V$ s.t.
$$
Y(|\omega\rangle, z) = \sum_n L_n z^{-n-2}:= T(z)
$$


*some other axioms...
The axiom may or may not require $L_n$ to satisfy Virasoro algebra, but if it does, it describes a CFT.
We will not go further in the mathematical definition. In physics literature the coordinate dependent operators $\Psi(z)$ is just $Y(|\Psi\rangle, z)$.
The commutator $[L_n,Y(|\Psi\rangle, z)]= \sum_j [L_n, Y(|\Psi\rangle)_j] z^{-j-1} $ is now indeed commutator of matrices expaned with formal parameter $z$.
When the field is primary with weight $\Delta$ it means the commutator satisfies the following algebra,
$$
[L_n,Y(|\Delta\rangle)_j] = ((j-n) + \Delta (n+1)) Y(|\Delta\rangle)_{j-n}  
$$
so that we can compactly write
$$
[L_n,Y(|\Psi\rangle, z)] = ( z^{n+1}\partial_z + \Delta(n+1) z^n) Y(|\Psi\rangle, z)
$$
So in this sense, differential operator is indeed only a representation and  clearly we have $[L_n, f(z) I ]=\sum_j [L_n,  I]* f_j *z^j=0$.
