TriTertButoxy said:
Oh yes. Thank you for spending the time on this, samalkhaiat. I really appreciate that.
The conformal group of the two-dimensional space consists of all substitutions of the forms
z \rightarrow w(z), \ \ \bar{z} \rightarrow \bar{w}(\bar{z}), \ \ (1)
where w and \bar{w} are arbitrary analytical functions.
For most purposes, it is convenient to treat z and \bar{z} not as complex conjugated but as two independent complex variables, i.e., to deal with the complex space \mathbb{C}^{2}. In this case it is clear from eq(1) that the conformal group G is a direct product
G = \Gamma \otimes \bar{\Gamma},
where \Gamma ( \bar{\Gamma}) is a group of the analytical substitutions of the variable z ( \bar{z}).
So, rank (h , \bar{h}) conformal tensor (primary field) is defined by the following transformation law;
<br />
T( z , \bar{z}) \rightarrow \tilde{T}( w , \bar{w}) = ( \frac{dw}{dz})^{-h} ( \frac{d \bar{w}}{d \bar{z}})^{-\bar{h}}\ T( z , \bar{z})<br />
For our purpose, we only need to consider the (0 , 0) conformal scalar
\tilde{\Phi}(w) = \Phi (z) , \ \ (2)
and the ( h , 0) conformal tensor;
\tilde{V}(w) = \left( \frac{dw}{dz} \right)^{-h} \ V(z). \ \ (3)
To study the Lie algebra of the group \Gamma, we consider an element w \in \Gamma close to the identity;
w(z) = z e^{i \epsilon (z) }\approx z + i z \epsilon (z).
Expanding both sides of eq(2) to the 1st order in \epsilon (z), we find
<br />
\delta \Phi = \tilde{\Phi}(z) - \Phi (z) = - i z \epsilon (z) \frac{d}{dz} \Phi (z). \ \ (4)<br />
Now, we make a Laurent expansion of \epsilon (z);
\epsilon (z) = \sum_{n \in \mathbb {Z}} \epsilon_{n} z^{n}. \ \ (5)
Thus, eq(4) becomes
\delta \Phi = - i \sum_{n} \epsilon_{n} z^{n + 1} \frac{d}{dz}\Phi(z), \ \ (6)
and, we are led to introduce the operators;
\ell_{n} = - z^{n + 1} \frac{d}{dz}, \ \ n \in \mathbb{Z}. \ \ (7)
These generate the (infinite-dimensional) Lie algebra (the Witt algebra);
[\ell_{n} , \ell_{m}] = (n - m) \ell_{n + m}. \ \ (8)
The Witt algebra associated with the infinite-dimensional group \Gamma coincides with the algebra of vector fields on a circle; that is the algebra of diffeomorphisms of S^{1}. Indeed, it is easy to see that the realization of the generators \ell_{n} acting on functions on the unit circle z = \exp ( i \theta ), is given by
\ell_{n} = i e^{i n \theta } \frac{d}{d \theta } \ \ (9)
A representation ( L_{n}) of the algebra (8) will be unitary if the hermiticity condition
L_{n}^{\dagger} = L_{- n} \ \ (10)
holds.
The field \Phi (z) will then transform as
\tilde{\Phi}(z) = U \Phi (z) U^{\dagger}. \ \ (11)
If we write
U \approx 1 - i \sum_{n} \epsilon_{n} L_{n},
U^{\dagger} = U^{-1} \approx 1 + i\sum_{n} \epsilon_{n}L_{n},
then eq(11) becomes;
\delta \Phi = [-i\sum_{n} \epsilon_{n}L_{n} , \Phi (z)] \ \ (12)
Comparing this with eq(6), we find
[L_{n} , \Phi (z) ] = - \ell_{n}\Phi (z) . \ \ (13)
Now, let us go back to eq(3) and do the same thing, i.e.,
1) expand both sides:
<br />
\tilde V(z) + i\sum_{n} \epsilon_{n} z^{n + 1} \frac{dV}{dz} = \left( 1 - ih \sum_{n} (n + 1) z^{n}\right) V(z) <br />
and write it as
<br />
\delta V(z) = -i \sum_{n} \epsilon_{n}\left( z^{n + 1}\frac{dV}{dz} + h ( n + 1 ) z^{n}V(z) \right)<br />
2) compare the above equation with the infinitesimal unitary transformation;
\delta V(z) = [ -i \sum_{n} \epsilon_{n} L_{n} , V(z) ]
This way you get your first equation
[L_{n}, V(z) ] = - \ell_{n}V(z) + h (n + 1) z^{n} V(z) \ \ (14)
When h = 1, this lead to a very useful result in CFT, that is
[ L_{n} , \oint \ V(z) \ dz ] = 0
****
Ok, let’s go to string theory and consider a local operator V( \tau , \sigma ) at the \sigma = 0 end of an open string; V( \tau , 0) \equiv V( \tau ).
Due to reparametrization invariance, the “conformal” weight of V( \tau ) is defined by
<br />
\bar{V}( \bar{\tau}) = \left( \frac {d \bar{ \tau }}{ d \tau}\right)^{-h} V( \tau ), \ \ (15)<br />
where
\tau \rightarrow \bar{ \tau}( \tau ),
is an arbitrary change of variables. The infinitesimal change
\bar{\tau} = \tau + \delta \tau \ \ (16)
is generated by [eq(9)]
\ell_{n} = i e^{i n \tau} \frac{d}{d \tau },
with
\delta \tau = \sum_{n} a_{n} e^{i n \tau } \ \ (17)
Now, if you expand (15) using (16) and (17), you find
\delta V(\tau) = -i \sum_{n} a_{n}e^{i n \tau }\{ -i \frac{dV}{d \tau} + nhV \}.
In terms of the generators L_{n}, we find
<br />
[L_{n} , V( \tau )] = e^{i n \tau } \{ -i \frac{dV}{d \tau } + nh V( \tau ) \} \ \ (18)<br />
From this it follows that
[L_{0}, V( \tau )] = -i \frac{dV}{d \tau }
as it should because L_{0} is the string Hamiltonian.
Finally, if you wants, you can introduce the variable z = \exp ( i \tau ) into eq(18) and arrive at your second equation;
[L_{n} , V(z) ] = -\ell_{n} V(z) + nhz^{n} V(z)
regards
sam
