Annihilation operators in a vertex algebra

Suppose there exists $v,w \in V(d)$ such that $v_{(d)}w \neq 0$. And now consider $(Tv)_{(d)}w = -2d \,v_{(d)}w \neq 0$. Notice also that by skew-symmetry your condition being true for $d>n$ implies the same condition for $n < d$.

The answer seems to be yes for quasi-primary $v$ if $V$ has a suitable invariant bilinear form. Then one can identify $v_{(n)} w$ with its pairing with the vacuum, and obtains it as the appropriate coefficient of $$(\mathbf{1}, Y(v, x)w) = (-x^{-2})^d (Y(v, x^{-1})\mathbf{1}, w) = (-x^{-2})^d (e^{x^{-1} L(-1)} v, w) = 0$$ since the weight of $v$ is greater than that of $w$.

Of course, this argument does not work if $v$ is not quasi-primary since then $v$ must be replaced with $e^{x L(1)} v$ after the first equality in the calculation.

Note that it isn’t necessary to assume the bilinear form is nondegenerate, just nondegenerate on the one-dimensional vacuum space.