Conformal field theory does not have... conformal symmetry? This post is about 1+1d. It is often said that conformal field theory has an infinite-dimensional symmetry generated by the Virasoro algebra:
$$ [L_n,L_m] =  (n-m) L_{n+m} + \frac{c}{12} n (n^2-1) \delta_{n+m,0}. $$
(Similarly for the anti-holomorphic branch with generators $\bar L_n$.)
But (at least in radial quantization) the Hamiltonian is $H = L_0 + \bar L_0$. This obviously does not commute with the above generators, since
$[L_n,L_0] = nL_n$.
In other words, it seems the Virasoro algebra functions as a 'spectrum-generating algebra' (since $L_n$ maps eigenspaces of $H$ to eigenspaces of $H$), rather than as a symmetry? Am I misunderstanding something?
 A: The Virasoro algebra is a true symmetry of the theory, in the sense that the action of a conformal field theory is conformally invariant if it exists, and in the sense that the algebra elements map solutions to the equations of motion (quantumly: eigenstates of the Hamiltonian) to solutions of the equations of motion. 
However, the generators indeed do not commute with the Hamiltonian because they correspond to time-dependent transformations. $[Q,H] = 0$ is only the condition for a symmetry if the symmetry does not transform the time coordinate - the statement for a time-dependent classical symmetry generator is $[Q,H] + \partial_t Q = 0$.
Note that the classical infinitesimal symmetry the $L_n$ correspond to is $z\mapsto z + \epsilon z^{n+1}$, and since $z$ is a mixture of time and space coordinates, the generator $L_n = z^{n+1}\partial_z$ is explicitly time-dependent and you cannot expect the quantum generators to commute with the Hamiltonian. 
Exactly the same is true in a much less confusing theory: The Lorentz boost generators, whose classical expression $t\partial_{x^i} - x^i \partial_t$ is also explicitly time-dependent, do not commute with the zeroth component of momentum - the Hamiltonian - either!
