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The EM tensor of the $bc$-CFT is $$ T(z) = \colon \partial b c \colon - \lambda \partial \colon b c \colon $$ After expanding in a mode expansion, we find $$ T(z) = \sum_{m} \frac{1}{z^{m+2}} \sum_n \left( \lambda m - n \right) \colon b_n c_{m-n} \colon \implies L_m = \sum_n \left( \lambda m - n \right) \colon b_n c_{m-n} \colon ~~...(1) $$ Now, one would like to rewrite this expression using creation-annihilation (CA) normal ordering ${}_\circ^\circ~~{}_\circ^\circ$ instead of conformal-normal order $\colon~~\colon$. A simple calculation reveals $$ L_m = \sum_n \left( \lambda m - n \right) {}_\circ^\circ b_n c_{m-n}{}_\circ^\circ + \frac{\lambda \left( 1 - \lambda \right)}{2} \delta_{m,0} $$ The above expression was derived by requiring that $L_m$ satisfy the Virasoro algebra.

I am now looking to provide an alternative derivation of the above expression. Here's my procedure. We start with the result of exercise 2.13(a) in Polchinski (which I have already verified) $$ \colon b(z) c(w) \colon - {}_\circ^\circ b(z) c(w) {}_\circ^\circ = \frac{(z/w)^{1 - \lambda} - 1 }{z - w} $$ Now, taking a limit of $w \to z$, we find $$ \colon b(z) c(z) \colon = {}_\circ^\circ b(z) c(z) {}_\circ^\circ + \frac{1-\lambda}{z} $$ from where we can derive the difference between the CA normal ordering and conformal ordering of the mode coefficients $$ \colon b_m c_n \colon = {}_\circ^\circ b_m c_n {}_\circ^\circ + \left( 1 - \lambda \right) \delta_{m+n,0} $$ Now, let us start from the expression $(1)$ and plug in the above. We find \begin{equation} \begin{split} L_m &= \sum_n \left( \lambda m - n \right) \colon b_n c_{m-n} \colon \\ &= \sum_n \left( \lambda m - n \right) \left[ {}_\circ^\circ b_n c_{m-n} {}_\circ^\circ + \left( 1 - \lambda \right) \delta_{m,0} \right] \\ &= \sum_n \left( \lambda m - n \right) {}_\circ^\circ b_n c_{m-n} {}_\circ^\circ + \delta_{m,0} \sum_n n \left( \lambda - 1 \right) \\ \end{split} \end{equation} Comparing with the expression mentioned above, we are required to have

$$ \sum_{n=-\infty}^\infty n \left( \lambda - 1 \right) = \frac{\lambda \left( 1 - \lambda \right)}{2} $$ I am not able to show this. Anyone has any idea?

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  • $\begingroup$ I think the conclusion $\colon b_m c_n \colon = {}_\circ^\circ b_m c_n {}_\circ^\circ + \left( 1 - \lambda \right) \delta_{m+n,0}$ is incorrect. Any comments? $\endgroup$
    – Prahar
    Dec 30, 2013 at 21:44
  • $\begingroup$ Not sure it helps, but you have [$\sum\limits_{n=1}^{+\infty} (n-\lambda)]_R - [\sum\limits_{n=1}^{+\infty} n]_R = \dfrac{\lambda(1-\lambda)}{2}$, for $ 0 \leq \lambda \leq 1$ $\endgroup$
    – Trimok
    Dec 31, 2013 at 13:06
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    $\begingroup$ @Trimok - I realized that one cannot describe the relation between CA normal ordering and conformal normal ordering in terms of the mode coefficients. The sum on the right is in fact zero. $\endgroup$
    – Prahar
    Dec 31, 2013 at 14:59
  • $\begingroup$ My skills being limited, I would be very interested in a more detailed answer/comment. $\endgroup$
    – Trimok
    Dec 31, 2013 at 18:43
  • $\begingroup$ I wanted to answer by calculating $\sum_M^\infty n = M(1-M)/2-1/12$ for you - a generalization of the usual sum of positive integers - but I decided that it wouldn't be helpful because you're confused about too many other things, too. For example, the strategy should be to rewrite the sums as "normal-ordered products" plus "c-number" and the normal-ordered products must be summed over positive $n$ only, not all $n$. The sums over $1$ over the whole axis is $0$ while the sum over the linear one in $n$ is $2\times (-1/12)$, a term you are also neglecting. $\endgroup$ Jan 1, 2014 at 5:56

1 Answer 1

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Indeed the line

$$\colon b_m c_n \colon = {}_\circ^\circ b_m c_n {}_\circ^\circ + \left( 1 - \lambda \right) \delta_{m+n,0}$$ is incorrect. There is however a relation between the modes, which we will try to derive below.

Firstly, I believe the intuition for the two modes is important and here's my understanding ( and please let me know if you think otherwise): the CNO (conformal normal ordering) $\colon \colon$ refers to that ordering which puts annihilators of the SL2(R) vacuum on the right, whereas CAO (creation annihilation ordering) ${}_\circ^\circ {}_\circ^\circ$ places lowering modes on the right (up to anticommutations).

Now onto the derivation of the relation between the two modes.

Start with:

$$\colon b(z) c(w) \colon - {}_\circ^\circ b(z) c(w) {}_\circ^\circ = \frac{(z/w)^{1 - \lambda} - 1 }{z - w}$$

Let us note the RHS can be written as

$$\frac{(z/w)^{1 - \lambda} - 1 }{z - w}=-\frac{1}{z}(1+\frac{w}{z}+...+\frac{w^{\lambda-2}}{z^{\lambda-2}})$$

Expanding the LHS in modes, we now have:

$$\colon \sum\limits_{n',m'}\frac{b_{n'} c_{m'}}{z^{n'+\lambda}w^{m'+1-\lambda}} \colon - {}_\circ^\circ \sum\limits_{n',m'}\frac{b_{n'} c_{m'}}{z^{n'+\lambda}w^{m'+1-\lambda}} {}_\circ^\circ = -\frac{1}{z}(1+\frac{w}{z}+...+\frac{w^{\lambda-2}}{z^{\lambda-2}})$$

Let us multiply both sides by $\frac{1}{z^{-n-\lambda+1}w^{-m+\lambda}}$ and integrate around z=0, w=0 (ignoring factors of $\frac{1}{2\pi i})$. Notice the LHS simply gives

$$\colon b_n c_m \colon - {}_\circ^\circ b_n c_m {}_\circ^\circ$$

while the RHS gives

$$-\oint \oint dz dw \frac{1}{z^{-n-\lambda+1}w^{-m+\lambda}}\frac{1}{z}(1+\frac{w}{z}+...+\frac{w^{\lambda-2}}{z^{\lambda-2}})$$

When is the RHS not equal to zero? when $\textbf{both}$ the z $\textbf{and}$ w integrals are non zero.

Now note the following: By the residue theorem, The z integral is non zero only when the integrand is of the form $\frac{1}{z}$, and that may only occur for values of $n$ such that $n=k+1-\lambda$, with $k=0,1,2,...,\lambda-2$, seen by multiplying out the brackets.

However, this fixes also the power of w appearing in the second integral: therefore, the only values of m for which the second integral is nonzero, given the first integral is nonzero, are for $m=-k-1+\lambda$, that is, $m=-n$.

We have thus shown that only for the cases $-m=n=-1,-2,...1-\lambda$ the two modes are $\textbf{not}$ equal, and in those cases we have

$$\colon b_n c_{-n} \colon = {}_\circ^\circ b_n c_{-n} {}_\circ^\circ -1, n=-1,...,1-\lambda$$

and otherwise we have

$$\colon b_n c_m \colon = {}_\circ^\circ b_n c_m {}_\circ^\circ$$

Let us use this corrected relation between the modes to now demonstrate the normal ordering constant of $L_m$:

\begin{equation} \begin{split} L_m &= \sum_n ( \lambda m - n ) \colon b_n c_{m-n} \colon \\ &= (\sum\limits_{n=-\infty}^{-\lambda}+\sum\limits_{n=0}^{\infty}) ( \lambda m - n ) {}_\circ^\circ b_n c_{m-n} {}_\circ^\circ + \sum\limits_{n=1-\lambda}^{-1}( \lambda m - n)({}_\circ^\circ b_n c_{m-n} {}_\circ^\circ -1\delta_{m,0}) \\ &= \sum_n ( \lambda m - n ) {}_\circ^\circ b_n c_{m-n} {}_\circ^\circ + \sum\limits_{n=1-\lambda}^{-1}( - n)(-1\delta_{m,0}) \\ &= \sum_n ( \lambda m - n ) {}_\circ^\circ b_n c_{m-n} {}_\circ^\circ + \frac{\lambda \left( 1 - \lambda \right)}{2} \delta_{m,0}\\ \\ \end{split} \end{equation} as required.

Note:

The above derivation was done to demonstrate carefully that there is a relation between the modes, which you can derive from the original non-mode expanded field definitions $ \colon b(z) c(z) \colon$ & ${}_\circ^\circ b(z) c(z) {}_\circ^\circ $.

However, a quicker derivation of the normal order constant $\frac{\lambda( 1 - \lambda )}{2}$ could be obtained by simply using the definitions of CNO and CAO I gave at the start (figure out which $b_n,c_m$ modes kill the SL2(R) vacuum and that defines CNO for you, CAO is simply puts positive modes to the right and negative modes to the left. Then use the commutation relation between $b_n$ and $c_m$ to relate the two definitions). I'll leave it as an exercise.

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