# CFT: equivalence of oscillator and Virasoro basis

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?

• $\uparrow$ Read where? – Qmechanic Nov 2 '16 at 18:30
• 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 $L_n=\frac{1}{2}\sum_{r\in\mathbb{Z}}:\alpha_{n-r}\alpha_r:$. – Tom Gao Nov 2 '16 at 20:03
• @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. – Harold Nov 4 '16 at 10:57

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 $\alpha_0|p\rangle = p|p\rangle$.