[Physics] general relationship between the conformal weight of a field and its (classical) scaling dimension

conformal-field-theoryquantum-field-theoryscaling

A field $\phi(z)$ has the conformal weight $h$, if it transforms under $z\rightarrow z_1(z)$ as

$$
\phi(z) = \tilde{\phi}(z_1)\left(\frac{dz_1}{dz}\right)^h
$$

The (classical) scaling dimension can be obtained for each field by appearing in the Lagrangian by making use of the constraint that has to be dimensionless, resulting for example in

$$
[\phi] = [A^{\mu}] = 1
$$

for a scalar and a gauge field or

$$
[\Psi_D] = [\Psi_M] = [\chi] = [\eta] = \frac{3}{2}
$$

for Dirac, Majorana, and Weyl spinors.

Are these two concepts of scaling dimension and conformal weight somehow related?

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From "Perturbative quantum field theory" Edward Witten (page 446 in volume 1 of "Quantum fields and strings : A course for mathematicians"):

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String Theory – Understanding Dimensions of CFT Operators

Just to be sure, MAGOO is the AdS/CFT Bible

http://arxiv.org/abs/hep-th/9905111

whose authors are usually listed alphabetically. Concerning your questions,

Conformal dimension or scaling dimension or mass dimension are meant to be the same thing in the context of conformal field theories in more than 2 dimensions. In 2 dimensions, one may distinguish the left-moving (holomorphic) and right-moving (antiholomorphic) dimensions or their sum - but the separate chiral "dimensions" are usually called "weights", anyway. The dimension pretty much counts the exponent of "kilogram" in the unit of the corresponding operator - except that in quantum mechanics, the exponent is often fractional. The dimensions are usually positive - they get mapped to the energy which is positive as well. Operators have positive dimensions of mass. If your cryptic "-ve" meant that there is a minus sign, then there is no minus sign. If you prefer to use units of length, then their powers are negative, but you should switch to masses as the base whose powers count the dimension.
a) The field $\phi$, as opposed to $\phi_0$, is dimensionless because it is an actual field in the AdS space (the bulk), unlike $\phi_0$ that is defined on the boundary. They mean that it is dimensionless under the conformal transformations of the boundary - and that's true because the conformal transformations are realized as simple isometries in the bulk, so they can't rescale the AdS field $\phi$ - at most, they move it to another point. Also, on that page, one deals with particular finite modified boundary conditions for $\phi$ at the boundary of the AdS space so it must be finite even at $z=0$.
b) In equation (3.15), the left hand side is dimensionless. The right hand side has a power of $\epsilon$ whose dimension is $length^{d-\Delta}$ because $\epsilon$ has units of length (on boundary), and $\phi_0$ whose dimension must be $length^{\Delta-d}$ as a consequence, to get a dimensionless product. In equation (3.13), the exponent on the left hand side has $d^4x$ whose dimension is $length^d$ - note that $d=4$ for $AdS_5$; $\phi_0$ whose dimension is $length^{\Delta-d}$ as I said in the previous sentence - note that the $d$ term in the exponent cancels; and $O$ whose dimension must therefore be $length^{-\Delta}$ which means $mass^{\Delta}$. As I said, the standard base for counting dimensions is mass, so we also say that $O$ has dimension $\Delta$. So I suspect that your sign error is simply caused by the point 1) - namely by your incorrect assumption that dimensions refer to the powers of length.

When we say "dimension" without extra specifications, we mean the power of the mass, not length. It's a healthy convention because the operators end up having non-negative dimensions. You may want to remember the dimensions of basic operators in 4 dimensions. The identity operator is always dimensionless (and $x$-independent): the dimension is 0. Bosonic scalar fields $\phi_0$ and potentials $A_\mu$ have dimension 1, their field strength has dimension $2$, fermions like the Dirac $\psi$ have dimension 3/2, and all terms in the Lagrangian density have dimension $4$. These are classical dimensions; each derivative adds $1$ to the dimension. Products of these operators have dimensions that are sums of the dimensions of the factors - except that quantum mechanics also adds "anomalous dimensions" to this classical form of the dimension, so that the total dimensions may have fractional terms that are proportional to powers of $g$ etc.

Quantum Field Theory – Dirac Spinors Under Parity Transformation: Understanding Weyl Spinors in Dirac Spinors

You are looking for a unitary representation of parity on spinors. That it should be unitary can be seen from the fact, that parity commutes with the Hamiltonian. Compare this to time-reversal and charge conjugation, which anticommute with $P^0$ and hence need be antiunitary and antilinear. They involve complex conjugation.

As demonstrated parity transforms a $(\frac{1}{2},0)$ into a $(0,\frac{1}{2})$ representation. Hence it cannot act on any such representation alone in a meaningful way. The Dirac-spinors in the Weyl-basis on the other hand contain a left- and right-handed component $$ \Psi = \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} $$

As a linear operator on those spinors - a matrix in a chosen basis - it mixes up the spinor components. After what has been said before, left- and right-handed components should transform into each other. The only matrix one can write down that does this is $\gamma^0$. There could in principle be a phase factor. In a theory with global $U(1)$-symmetry this may be set to one however.

Edit: Statements like $\chi_L \rightarrow P\chi_L=\chi_R $ for a Weyl-Spinor $\chi_L$ are not sensible. The Weyl-spinors are reps. of $\mathrm{Spin(1,3)}$, whereas $P\in \mathrm{Pin(1,3)}$. One cannot expect that some representation is also a representation of a larger group. Dirac-spinors on the other hand are precisely irreps. of $\mathrm{Spin(1,3)}$ including parity, which cannot act in any other sensible way than by exchanging the chiral components.

Think of what representation means. It's a homomorphism from a group to the invertible linear maps on a vector space. $$ \rho: G \rightarrow GL(V)$$ Particularly, for any $g\in G$ and $v\in V$, $\rho(g)v\in V$. Now set $V$ to be the space of say left-handed Weyl-spinors and $g=P\in\mathrm{Pin(1,3)}$ the parity operation. As you have shown above, the image of a potential $\rho(P)$ is not a left-handed Weyl-spinor, hence is not represented.

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Related Solutions

String Theory – Understanding Dimensions of CFT Operators

Quantum Field Theory – Dirac Spinors Under Parity Transformation: Understanding Weyl Spinors in Dirac Spinors

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