# Conformal field theory: vertex operator

TriTertButoxy
I am getting apparently conflicting statements about the conformal transformation law of the vertex operator appearing in and 2D QFT (such as in bosonic string theory). For example, according to http://en.wikipedia.org/wiki/Conformal_field_theory" [Broken] (eqn 64 on page 15), the transformation law is

$$[L_n,V(z)] = z^n\left(z\frac{d}{dz}+(n+1)h\right)V(z),$$​

where $L_n$ are the Virasoro generators and h is the conformal weight of V(z). But, according to Green, Schwarz, Witten's string theory book (Vol 1, eqn 7.1.3, p 356), the transformation law is

$$[L_n,V(z)] = z^n\left(z\frac{d}{dz}+nh\right)V(z),$$​

which has n instead of (n+1). These two equations describe pretty different transformation properties, and I don't know who to believe. The first one makes more sense, but when I actually try to derive the transformation law, I find the get the second one.

How do I reconcile this difference?

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I am getting apparently conflicting statements about the conformal transformation law of the vertex operator appearing in and 2D QFT (such as in bosonic string theory). For example, according to http://en.wikipedia.org/wiki/Conformal_field_theory" [Broken] (eqn 64 on page 15), the transformation law is

$$[L_n,V(z)] = z^n\left(z\frac{d}{dz}+(n+1)h\right)V(z),$$​

where $L_n$ are the Virasoro generators and h is the conformal weight of V(z). But, according to Green, Schwarz, Witten's string theory book (Vol 1, eqn 7.1.3, p 356), the transformation law is

$$[L_n,V(z)] = z^n\left(z\frac{d}{dz}+nh\right)V(z),$$​

which has n instead of (n+1). These two equations describe pretty different transformation properties, and I don't know who to believe. The first one makes more sense, but when I actually try to derive the transformation law, I find the get the second one.

How do I reconcile this difference?

By checking how the two formulas behave under subsequent calculations. This is the usual way one finds out which version of sloppily stated formulas makes sense. Almost without exception, only one of them is consistent. (The method breaks down only if there are too many wrongly written formulas.)

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TriTertButoxy
Sorry, that was not very helpful. I have gone through subsequent calculations, and it is not very clear to me which of the two is correct. I am hoping for someone with experience with CFTs to shed light on this issue.

Gaberdiel's eqs.(61-63) seem pretty convincing evidence that eq.(64) is correct. But I don't have GSW to see what they did.

TriTertButoxy
Actually, most string theory books (including GSW, Kaku, ) use the second form (n instead of n+1). But then they make statements about the finite transformation properties of the vertex function that are inconsistent with that form.

Sorry, that was not very helpful.

The point is that if two authorities disagree, it is not enough to get the advice of other, lower authorities, but you need to have criteria that make you yourself the authority to decide.

I have gone through subsequent calculations, and it is not very clear to me which of the two is correct. I am hoping for someone with experience with CFTs to shed light on this issue.

One of the two formulas defines a representation of the Virasoro algebra on functions of V, the other doesn't. You can find out which one is correct by checking whether both sides of the Jacobi identity
$$[L_m,[L_n,V(x)]] - [L_n,[L_m,V(x)]] = [[L_m,L_n],V(x)]$$
are identical.

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TriTertButoxy
Hey, that's a useful trick! I did the calculation for both cases, and found that they both satisfy the Jacobi identity.

In particular I checked it for
$$[L_n,V(z)] = z^n\left(z\frac{d}{dz}+(n+s)h\right)V(z),$$​
for arbitrary s, and found that the identity is satisfied for all s. So unfortunately, that doesn't decide which one is correct :(

but that identity looked like a good idea.

Hey, that's a useful trick! I did the calculation for both cases, and found that they both satisfy the Jacobi identity.

In particular I checked it for
$$[L_n,V(z)] = z^n\left(z\frac{d}{dz}+(n+s)h\right)V(z),$$​
for arbitrary s, and found that the identity is satisfied for all s. So unfortunately, that doesn't decide which one is correct :(

but that identity looked like a good idea.

Ah, that explains why different sources use different formulas: Both are correct!

Indeed, both describe the same algebra, but in different notations. You can go from the representation with s=0 to the one with arbitrary s by replacing V(z) by e^{shz}V(z).

But of course, one needs to keep the same normalization of V(z) throughout the calculations.

These two equations describe pretty different transformation properties

True.

and I don't know who to believe.

Both!

The first one makes more sense, but when I actually try to derive the transformation law, I find the get the second one.

How do I reconcile this difference?

I will make some notes and come back to you soon.

sam

TriTertButoxy
Oh yes. Thank you for spending the time on this, samalkhaiat. I really appreciate that.

[text]z \rightarrow w(z), \ \ \bar{z} \rightarrow \bar{w}(\bar{z}), \ \ (1)[/tex]

Could you please edit this and a few other formulas so that they print correctly?

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;

$$T( z , \bar{z}) \rightarrow \tilde{T}( w , \bar{w}) = ( \frac{dw}{dz})^{-h} ( \frac{d \bar{w}}{d \bar{z}})^{-\bar{h}}\ T( z , \bar{z})$$

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

$$\delta \Phi = \tilde{\Phi}(z) - \Phi (z) = - i z \epsilon (z) \frac{d}{dz} \Phi (z). \ \ (4)$$

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:

$$\tilde V(z) + i\sum_{n} \epsilon_{n} z^{n + 1} \frac{dV}{dz} = \left( 1 - ih \sum_{n} (n + 1) z^{n}\right) V(z)$$

and write it as

$$\delta V(z) = -i \sum_{n} \epsilon_{n}\left( z^{n + 1}\frac{dV}{dz} + h ( n + 1 ) z^{n}V(z) \right)$$

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

$$\bar{V}( \bar{\tau}) = \left( \frac {d \bar{ \tau }}{ d \tau}\right)^{-h} V( \tau ), \ \ (15)$$

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

$$[L_{n} , V( \tau )] = e^{i n \tau } \{ -i \frac{dV}{d \tau } + nh V( \tau ) \} \ \ (18)$$

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

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