Consider a Coulomb gas CFT described by a scalar $X$ with background charge $Q$. The modes of the scalar are given by $\alpha_n$ $$ i \partial X = \sum_{n \in \mathbb Z} \frac{\alpha_n}{z^{n+1}}. $$ Then a state of the CFT is given by (I am not normalizing the states) $$ |\psi\rangle = \prod_{n > 0} \alpha_{-n}^{N_n} |p\rangle $$ where $|p\rangle$ is the vacuum with momentum $p$ obtained by acting with the zero-mode and $N_n$ is the number operator for level $n$. Let's call this the oscillator basis. The Virasoro operators are $$ L_n = \sum_{n \in \mathbb Z} \frac{{:\mathrel{\alpha_n \alpha_{-n}}:}}{z^{n+2}}. $$

I have read and hear several times the statement that there is an equivalence between the oscillator basis and Virasoro basis, the latter being made of states $$ |\psi'\rangle = \prod_{n > 0} L_{-n}^{N_n} |p\rangle. $$ In general I found this in the case where one has null states (for example when studying the ground ring and the so-called discrete states in 2d gravity – but I don't know much about this). For example if the combination $$ \big( L_{-2} - \frac{3}{2} L_{-1}^2 \big) |0\rangle $$ is null, then the corresponding state $$ \big( \alpha_{-2} - \frac{3}{2} \alpha_{-1}^2 \big) |0\rangle $$ vanishes identically.

I can understand that it is possible since we have $$ [L_0, L_{-n}] = n L_{-n}, \qquad [L_0, \alpha_{-n}] = n \alpha_{-n} $$ which shows that both $L_{-n}$ and $\alpha_{-n}$ increases the $L_0$ eigenvalue by the same amount.

But more generally I did not find much details in the literature and I was wondering if someone could explain more or points towards useful references?

  • $\begingroup$ $\uparrow$ Read where? $\endgroup$ – Qmechanic Nov 2 '16 at 18:30
  • $\begingroup$ This is simply due to the energy-momentum tensor in mode expansion (Virasoro mode) can be constructed or expressed in mode expansion of the field (oscillator mode). For free boson, it is $L_n=\frac{1}{2}\sum_{r\in\mathbb{Z}}:\alpha_{n-r}\alpha_r:$. $\endgroup$ – Tom Gao Nov 2 '16 at 20:03
  • $\begingroup$ @Qmechanic: Before posting the question I looked again after the references but I did not find them so it's why I did not give more details. But when I will find them I will add them. $\endgroup$ – Harold Nov 4 '16 at 10:57

The following blog post might answer your question:


In your question, you are missing the precise (quadratic) relation between Virasoro and free scalar modes. Using that relation, you can rewrite your Virasoro null state in terms of free scalar modes. The result should be identically zero.

For example, consider the state $L_{-1}|0\rangle$, where $|0\rangle$ is a primary state of dimension zero. This is a null state because it is killed by $L_{n>0}$. Now, in terms of free scalar modes, we have $L_{-1}|0\rangle = -2\alpha_0\alpha_{-1}|0\rangle = 0 $, using $\alpha_0|p\rangle = p|p\rangle$.


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