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Aspects of Conformal Field Theory

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there is a conjecture along those lines but has not been proven.

in 2 dimensions, the only scalar is the Ricci. thereofore, there
is no ambiguity since any violation of scaling has to proportional to
Ricci. However, in higher dimensions, there are other
possibilities. For example, I could take two Weyl tensor and make a
scalar. In principle, this could show up in conformal anomaly as
well. However, it seems that (as far as I could remember) that in
known examples, the term proportional to Ricci always carries the
information of number of degrees of freedom of the conformal field
theory. It cannot be proven this is always the case. However, it is
the conjecture that this is always the case.  


> number of degrees of freedom of 是指中心荷吗?


right. Polchinski's book is very clear on this point. In bosonic
string, central charge is the number of dimensions, which is the
number of world sheet fields which is the number of degree of freedom  


> What do you think about the Fermi bc ghost field?
> This guy has a quadratic form of central charge(depend on lambda,
> which is the spin) .
> I do not very understand the physics meaning of Fermi bc ghost field,
> could you explain it?

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I am not an expert on conformal field theory either.

As far as I can see, bc CFT is another example of CFT. It is useful
when one BRST quantize the bosonic worldsheet CFT in string theory. In
that case, it plays the role of the FP ghost in gauge theory to make
the theory unitary.

the physical meaning of ghosts is that they are not physical states
(since physical states would have to conserve probability and so
on). however, they are useful tools in quantizing gauge theory.  


> 另外CFT是不是可以看作是共形变换下invariant的QFT,能不能作这样理解,也就
> 是说看作一种有限制的QFT。
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well. it is a special QFT in which: there is no renormalization group
running of the coupling constant, there is no S-matrix, Green's
functions are largely fixed by conformal symmetry....  


> 二维 CFT 的 2-point 及
> 3-point correlation function 可以被 conformal symmetry 确定到只差一个
> 常数,4-point correlation function 可以被确定到只差一个任意函数,等。
> (an answer supplied by Changhai)


> 如何理解没有S矩阵呢?

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there is no way to separate particles, i.e., there is not asymptotic
states which we can call one particle.  


二零零五年六月十五日