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?

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    $\begingroup$ Hi Ruben, I think this is physically reasonable, since the $L_n$ are shape deformations of the circle (eg. L_2 is an elliptical deformation), and so they will change the energy density. Only spacetime symmetries which preserve the spatial slice can act on the Hilbert space. So boosts don't act on the Hilbert space either! $\endgroup$ Commented Jun 21, 2018 at 19:46
  • $\begingroup$ @RyanThorngren Hey Ryan! Thanks for the comment. What do you mean by 'doesn't act on the Hilbert space'? After all, all $L_n$ have a well-defined action on the Hilbert space, right? (Indeed, we use them to construct our spectrum.) $\endgroup$ Commented Jun 22, 2018 at 11:32
  • $\begingroup$ @RyanThorngren, wait what? Do you mean that you have different Hilbert spaces for different slices, and boosts take you between those? If so I think one should say that Lorenz invariance means that there is a separate isomorphism between all these spaces, and by composition you can define an action of boost generators in a fixed Hilbert space. $\endgroup$ Commented Jun 23, 2018 at 8:24

1 Answer 1


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!

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    $\begingroup$ +1 well written! I inserted a short discussion (from a general viewpoint not referring to Virasoro algebra) on this point in the lecture notes of one of my courses of the last semester www.science.unitn.it/%7Emoretti/MFQM.pdf from p. 238 on. $\endgroup$ Commented Jun 21, 2018 at 20:20
  • $\begingroup$ Nice answer, thank you! Makes a lot of sense. This should also give an answer to the question of whether or not the ground state of a CFT spontaneously breaks the conformal symmetry (this was claimed by Maldacena and Stanford: physics.stackexchange.com/questions/302026/… ). Your argument implies: the system doesn't spontaneously break it, rather, any choice of time-slice can be said to hide the symmetry. $\endgroup$ Commented Jun 22, 2018 at 11:37
  • $\begingroup$ @RubenVerresen Indeed. The standard notion of "spontaneous symmetry breaking" only makes sense for time-independent symmetries. $\endgroup$
    – ACuriousMind
    Commented Jun 22, 2018 at 19:57
  • $\begingroup$ @RubenVerresen Ground state of a CFT is not invariant under Virasoro symmetry. Therefore, the Virasoro symmetry is spontaneously broken. If vacuum was invariant, one would necessarily have $c=0$ and furthermore the theory would be topological. $\endgroup$ Commented Jun 23, 2018 at 8:08
  • $\begingroup$ Also, this answer is overly classical. In quantum theory, any unitary operator (or, infinitesimally, hermitian operator) can be called a symmetry. It is ok for symmetries to not commute with each other (so that you have a non-abelian symmetry group). Of course there are tons of uninteresting unitary operators; the criterion of whether they are interesting is simplicity of their action on local operators and other natural objects (such simplicity allows you to use them to constrain stuff). $\endgroup$ Commented Jun 23, 2018 at 8:14

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