[Francesco CFT] 4. Global Conformal Invariance
This article is one of CFT/TFT.
4.1 The Conformal Group
A conformal transformation of the coordinates is an invertible mapping $x\rightarrow x'$, which leaves the metric tensor invariant up to a scale: $$\begin{align}g'_{\mu\nu}(x')=\Lambda(x)g_{\mu\nu}(x)\end{align}$$ In other words, a conformal transformation is locally equivalent to a (pseudo) rotation and a dilation. The set of conformal transformations manifestly forms a group.
Verify Conformal transformation
Definition (1) on an infinitesimal transformation $x^\mu\rightarrow {x'}^\mu=x^\mu+\epsilon^\mu(x)$. The metric, at first order in $\epsilon$, changes as follows $$g'_{\mu\nu}=\frac{\partial x^\alpha}{\partial {x'}^\mu}\frac{\partial x^\beta}{\partial {x'}^\nu}g_{\alpha\beta} =(\delta^\alpha_\mu-\partial_\mu\epsilon^\alpha)(\delta^\beta_\nu-\partial_\nu\epsilon^\beta)g_{\alpha\beta} =g_{\mu\nu}-(\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu)$$ $$\begin{align}g_{\mu\nu}\rightarrow g_{\mu\nu}-(\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu)\end{align}$$ The requirement that the transformation be conformal implies that $$\begin{align}\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu=f(x)g_{\mu\nu}\end{align}$$ The factor $f(x)$ is determined by taking the trace on both sides: $$\begin{align}f(x)=\frac{2}{d}\partial_\rho\epsilon^\rho\end{align}$$ For simplicity, we assume that the conformal transformation is an infinitesimal deformation of the standard Cartesian metric $g_{\mu\nu}=\eta_{\mu\nu}$, where $\eta_{\mu\nu}=\mbox{diag}(1,1,\cdots,1)$. (If the reader insists on living in Minkowski space, the treatment is identical, except for the explicit form of $\eta_{\mu\nu}$.) By applying an extra derivative $\partial_\rho$ on Eq. (3), $$\partial_\rho\partial_\mu\epsilon_\nu+\partial_\rho\partial_\nu\epsilon_\mu=\partial_\rho f(x)g_{\mu\nu}$$ permuting the indices $$ \partial_\rho\partial_\mu\epsilon_\nu+\partial_\rho\partial_\nu\epsilon_\mu=\partial_\rho f(x)g_{\mu\nu}\\ \partial_\nu\partial_\rho\epsilon_\mu+\partial_\nu\partial_\mu\epsilon_\rho=\partial_\nu f(x)g_{\rho\mu}\\ \partial_\mu\partial_\nu\epsilon_\rho+\partial_\mu\partial_\rho\epsilon_\nu=\partial_\mu f(x)g_{\nu\rho}$$ and taking a linear combination (+use $\partial$ is commute), we arrive at $$\begin{align}2\partial_\mu\partial_\nu\epsilon_\rho=\eta_{\mu\rho}\partial_\nu f+\eta_{\nu\rho}\partial_\mu f-\eta_{\mu\nu}\partial_\rho f\end{align}$$ Upon contracting with $\eta^{\mu\nu}$, this becomes $$\begin{align}2\partial^2\epsilon_\mu=(2-d)\partial_\mu f\end{align}$$ Applying $\partial_\nu$ on this expression and $\partial^2$ on Eq. (3), $$(2-d)\partial_\nu\partial_\mu f=2\partial_\nu\partial^2\epsilon_\mu=2\partial^2\partial_\nu\epsilon_\mu=\partial^2(\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu)=g_{\mu\nu}\partial^2f$$ we find $$\begin{align}(2-d)\partial_\mu\partial_\nu f=\eta_{\mu\nu}\partial^2 f\end{align}$$ Finally, contracting with $\eta^{\mu\nu}$, we end up with $$\begin{align}(d-1)\partial^2f=0\end{align}$$ From Eqs. (3)-(8), we can derive the explicit form of conformal transformations in $d$ dimensions.
Specify each conformal transformation
For the moment, we concentrate on the case $d\ge 3$. Equations (8) and (7) imply that $\partial_\mu\partial_\nu f=0$ (i.e., that the function $f$ is at most linear in the coordinates): $$\begin{align}f(x)=A+B_\mu x^\mu\quad (A,B_\mu \mbox{ constant})\end{align}$$ If we substitute this expression into Eq. (5), we see that $\partial_\mu\partial_\nu\epsilon_\rho$ is constant, which means that $\epsilon_\mu$ is at most quadratic in the coordinates. We therefore write the general expression $$\begin{align}\epsilon_\mu=a_\mu+b_{\mu\nu}x^\nu+c_{\mu\nu\rho}x^\nu x^\rho\, \quad c_{\mu\nu\rho}=c_{\mu\rho\nu}\end{align}$$ Since the constraints (3)-(5) hold for all $x$, we may treat each power of the coordinate separately. It follows that the constant term $a_\mu$ is free of constraints. This term amounts to an infinitesimal translation. Substitution of the linear term into (3)(+(4)) $$\partial_\mu(b_{\nu\rho}x^\rho)+\partial_\nu(b_{\mu\rho}x^\rho)=\frac{2}{d}\partial_\lambda(g^{\lambda\theta}b_{\theta\kappa}x^\kappa)g_{\mu\nu}$$ yields $$\begin{align}b_{\mu\nu}+b_{\nu\mu}=\frac{2}{d}b^\lambda_\lambda\eta_{\mu\nu}\end{align}$$ which implies that $b_{\mu\nu}$ is the sum of an antisymmetric part and a pure trace: $$\begin{align}b_{\mu\nu}=\alpha\eta_{\mu\nu}+m_{\mu\nu}\quad m_{\mu\nu}=-m_{\nu\mu}\end{align}$$ The pure trace represents an infinitesimal scale transformation, whereas the anti-symmetric part is an infinitesimal rigid rotation. Substitution of the quadratic term of (10) into Eq. (5) (+(4))$$2\partial_\mu\partial_\nu(c_{\rho\delta\kappa}x^\delta x^\kappa)=\eta_{\mu\rho}\partial_\nu(\frac{2}{d}\partial_\gamma g^{\gamma\sigma}c_{\sigma\delta'\kappa'}x^{\delta'}x^{\kappa'})-\eta_{\mu\nu}\partial_\rho(\frac{2}{d}\partial_{\gamma'} g^{\gamma'\sigma'}c_{\sigma'\delta''\kappa''}x^{\delta''}x^{\kappa''})$$ yields $$\begin{align}c_{\mu\nu\rho}=\eta_{\mu\rho}b_\nu+\eta_{\mu\nu}b_\rho-\eta_{\nu\rho}b_\mu\quad \mbox{where}\ b_\mu\equiv \frac{1}{d}c^\sigma_{\sigma\mu}\end{align}$$ and the corresponding infinitesimal transformation is $${x'}^\mu=x^\mu+\epsilon^\mu=x^\mu+b_\nu x^\nu x^\mu+b_\rho x^\rho x^\mu-b^\mu \eta_{\nu\rho}x^\nu x^\rho$$ $$\begin{align}{x'}^\mu=x^\mu+2(x\cdot b)x^\mu-b^\mu x^2\end{align}$$ which bears the name of special conformal transformation (SCT).
The finite transformations corresponding to the above are the following: $$\begin{align}\begin{cases}\mbox{(translation)}& {x'}^\mu=x^\mu+a^\mu\\ \mbox{(dilation)}& {x'}^\mu=\alpha x^\mu\\ \mbox{(rigid rotation)}& {x'}^\mu=M^\mu_\nu x^\nu\\ \mbox{(SCT)}& {x'}^\mu=\frac{x^\mu-b^\mu x^2}{1-2b\cdot x+b^2x^2}\end{cases}\end{align}$$ The first three of the above "exponentiations" are fairly familiar, whereas the last one is not. We shall not demonstrate its validity here, but it is trivial to verify that its infinitesimal version is indeed (14), and straightforward to show that it is indeed conformal, with a scale factor $\Lambda(x)$ given by $$\begin{align}\Lambda(x)=(1-2b\cdot x+b^2x^2)^2\end{align}$$ The SCT can also be expressed as $$\begin{align}\frac{{x'}^\mu}{{x'}^2}=\frac{x^\mu}{x^2}-b^\mu\end{align}$$ Manifestly, the SCT is nothing but a translation, preceded and followed by an inversion $x^\mu\rightarrow x^\mu/x^2$.
Generator Form
We recall the definition of the generator of an infinitesimal transformation. General infinitesimal transformation with $\{\omega_a\}$ is $${x'}^\mu=x^\mu+\omega_a\frac{\delta x^\mu}{\delta \omega_a}\\ {\Phi'}(x')=\Phi(x)+\omega_a\frac{\delta\mathcal{F}}{\delta \omega_a}(x)$$ We defined the generator $G_a$ of a symmetry transformation by the following expression for the infinitesimal transformation at a same point: $$\delta_\omega\Phi(x)\equiv \Phi'(x)-\Phi(x)\equiv -i\omega_a G_a\Phi(x)$$ Put above infinitesimal transformation, $$\Phi'(x')=\Phi(x)+\omega_a\frac{\delta\mathcal{F}}{\delta \omega_a}(x)=\Phi(x')-\omega_a\frac{\delta x^\mu}{\delta \omega_a}\partial_\mu \Phi(x')+\omega_a\frac{\delta \mathcal{F}}{\delta \omega_a}(x')$$ Then the explicit expression for the generator is therefore $$iG_a\Phi=\frac{\delta x^\mu}{\delta \omega_a}\partial_\mu\Phi-\frac{\delta\mathcal{F}}{\delta \omega_a}$$
If we suppose for the moment that the fields are unaffected by the transformation (i.e., $\mathcal{F}(\Phi)=\Phi$).
1. Translation by a vector $\omega^\mu$, one has $$\delta x^\mu/\delta \omega^\nu=\delta^\mu_\nu$$ and $\delta\mathcal{F}/\delta \omega^\nu=0$. Then generator is $$P_\nu=-i\partial_\nu.$$
2. Lorentz transformation $${x'}^\mu=x^\mu+\omega^\mu_\nu x^\nu=x^\mu+\omega_{\rho\nu}\eta^{\rho\mu}x^\nu$$ variation of the coordinate under an infinitesimal Lorentz transformation as $$\frac{\delta x^\mu}{\delta \omega{\rho\nu}}=\frac{1}{2}(\eta^{\rho\mu}x^\nu-\eta^{\nu\mu}x^\rho)$$ Then $$\frac{1}{2}i\omega_{\rho\nu}L^{\rho\nu}\Phi=\frac{1}{2}\omega_{\rho\nu}(x^\nu\partial^\rho-x^\rho \partial^\nu)\Phi$$ And $$L^{\rho\nu}=i(x^\rho\partial^\nu-x^\nu\partial^\rho)$$
3. Dilation $${x'}^\mu=\alpha x^\mu$$ has $\delta x^\mu/\delta \alpha=x^\mu$ then $$D=-ix^\mu\partial_\mu$$
4. SCT $${x'}^\mu=x^\mu+2(x\cdot b)x^\mu-b^\mu x^2$$ then $$\frac{\delta x^\mu}{\delta b^\nu}=2x_\nu x^\mu-x^2$$ then $$K_\mu=-i(2x_\mu x^\nu\partial_\nu-x^2\partial_\mu)$$
In conclusion, $$\begin{align}\begin{cases}\mbox{(translation)}& P_\mu=-i\partial_\mu\\ \mbox{(dilation)}& D=-ix^\mu \partial_\mu\\ \mbox{(rigid rotation)}& L_{\mu\nu}=i(x_\mu\partial_\nu-x_\nu\partial_\mu)\\ \mbox{(SCT)}& K_\mu=-i(2x_\mu x^\nu\partial_\nu-x^2\partial_\mu)\end{cases}\end{align}$$
Commutation Relation
These generators obey the following commutation rules, which in fact define the conformal algebra. $$[D,P_\mu]=-x^\nu\partial_\nu \partial_\mu+\partial_\mu x^\nu \partial_\nu=(\partial_\mu x^\nu)\partial_\nu=\delta_\mu^\nu\partial_\nu=iP_\mu$$
$$[D,K_\mu]=-x^\nu\partial_\nu(2x_\mu x^\rho\partial_\rho-x^2\partial_\mu)+(2x_\mu x^\rho\partial_\rho-x^2\partial_\mu)x^\nu\partial_\nu\\ =+2x^\nu x_\mu x^\rho\partial_\nu\partial_\rho-(\partial_\nu x_\mu x^\rho) 2x^\nu\partial_\rho-2x_\mu x^\rho x^\nu \partial_\rho\partial_\nu+(\partial_\mu x^\nu)2x_\mu x^\rho \partial_\nu\\ -x^\nu x^2\partial_\nu\partial_\mu+(\partial_\nu x^2)x^\nu \partial_\mu+x^2x^\nu\partial_\mu\partial_\nu-(\partial_\mu x^\nu)x^2\partial_\nu\\ =2x_\mu x^\rho\partial_\rho+x_\nu x^\nu \partial_\mu=-iK_\mu$$
$$[K_\mu,P_\nu]=-(2x_\mu x^\rho\partial_\rho-x^2\partial_\mu)\partial_\nu+\partial_\nu(2x_\mu x^\rho\partial_\rho-x^2\partial_\mu)\\ =2(\partial_\nu x_\mu x^\rho)\partial_\rho-(\partial_\nu x^2)\partial_\mu=2\eta_{\mu\nu}x^\rho\partial_\rho+2x_\mu\partial_\nu-2x_\nu\partial_\mu=2i(\eta_{\mu\nu}D-L_{\mu\nu})$$
$$[K_\rho,L_{\mu\nu}]=(2x_\rho x^\sigma\partial_\sigma-x^2\partial_\rho)(x_\mu\partial_\nu-x_\nu\partial_\mu)-(x_\mu\partial_\nu-x_\nu\partial_\mu)(2x_\rho x^\sigma\partial_\sigma-x^2\partial_\rho)\\ =-2x_\rho x^\sigma x_\mu \partial_\sigma\partial_\nu+(\partial_\sigma x_\mu)2x_\rho x^\sigma\partial_\nu+x_\mu 2x_\rho x^\sigma\partial_\nu\partial_\sigma-(\partial_\nu x_\rho x^\sigma)x_\mu\partial_\sigma\\ +x^2x_\mu\partial_\rho\partial_\nu-(\partial_\rho x_\mu)x^2\partial_\nu-x_\mu x^2\partial_\nu\partial_\rho+(\partial_\nu x^2)x_\mu\partial_\rho\\ +2x_\rho x^\sigma x_\nu \partial_\sigma\partial_\mu-(\partial_\sigma x_\nu)2x_\rho x^\sigma\partial_\mu-x_\nu 2x_\rho x^\sigma\partial_\mu\partial_\sigma+(\partial_\mu x_\rho x^\sigma)x_\nu\partial_\sigma\\ -x^2x_\nu\partial_\rho\partial_\mu+(\partial_\rho x_\nu)x^2\partial_\mu+x_\nu x^2\partial_\mu\partial_\rho-(\partial_\mu x^2)x_\nu\partial_\rho\\ = \eta_{\rho\mu}(2x_\nu x^\sigma\partial_\sigma-x^2\partial_\nu)-\eta_{\rho\nu}(2x_\mu x^\sigma\partial_\sigma-x^2\partial_\mu)$$
Finally, $$\begin{align}\begin{matrix}[D,P_\mu]&=& iP_\mu\\ [D,K_\mu]&=& -iK_\mu\\ [K_\mu,P_\nu]&=& 2i(\eta_{\mu\nu}D-L_{\mu\nu})\\ [K_\rho,L_{\mu\nu}] &=& i(\eta_{\rho\mu}K_\nu-\eta_{\rho\nu}K_\mu)\\ [P_\rho,L_{\mu\nu}]&=&i(\eta_{\rho\mu}P_\nu-\eta_{\rho\nu}P_\mu)\\ [L_{\mu\nu},L_{\rho\sigma}]&=&i(\eta_{\nu\rho}L_{\mu\sigma}+\eta_{\mu\sigma}L_{\nu\rho}-\eta_{\mu\rho}L_{\nu\sigma}-\eta_{\nu\sigma}L_{\mu\rho})\end{matrix}\end{align}$$ In order to put the above commutation rules into a simpler form, we define the following generators: $$\begin{align}\begin{matrix}J_{\mu\nu}=L_{\mu\nu},& J_{-1,\mu}=\frac{1}{2}(P_\mu-K_\mu)\\ J_{-1,0}=D,& J_{0,\mu}=\frac{1}{2}(P_\mu+K_\mu)\end{matrix}\end{align}$$ where $J_{ab}=-J_{ba}$ and $a,b\in \{-1,0,1,\cdots,d\}$. These new generators obey the $SO(d+1,1)$ commutation relations:
$$[J_{-1,0},J_{-1,\mu}]=\frac{1}{2}iP_\mu+i\frac{1}{2}K_\mu=iJ_{0,\mu}$$
$$[J_{-1,\mu},J_{0,\mu}]=i(\eta_{\mu\mu}D-L_{\mu\mu})=iJ_{-1,0}$$
$$[J_{0,\mu},J_{-1,0}]=\frac{1}{2}iP_\mu-i\frac{1}{2}K_\mu=iJ_{-1,\mu}$$
$$[J_{-1,0},L_{\mu\nu}]=0$$
$$[J_{-1,\rho},L_{\mu\nu}]=i\eta_{\rho\mu}J_{-1,\mu}-i\eta_{\rho\nu}J_{-1,\mu}$$
$$[J_{0,\rho},L_{\mu\nu}]=i\eta_{\rho\mu}J_{0,\mu}-i\eta_{\rho\nu}J_{0,\mu}$$
$$\begin{align}[J_{ab},J_{cd}]=i(\eta_{ad}J_{bc}+\eta_{bc}J_{ad}-\eta_{ac}J_{bd}-\eta_{bd}J_{ac})\end{align}$$ where the diagonal metric $\eta_{ab}$ is $\mbox{diag}( -1, 1, 1, \cdots,1)$ if space-time is Euclidian (otherwise an additional component, say $\eta_{dd}$, is negative). This shows the isomorphism between the conformal group in d dimensions and the group $SO(d + 1, 1)$, with $\frac{1}{2}(d + 2)(d + 1)$ parameters.
We end this section by constructing conformal invariants, that is, functions $\Gamma(x_i)$ of $N$, points $x_i$ that are left unchanged under all types of conformal transformations. Translation and rotation invariance imply that $\Gamma$ can depend only on the distances $|x_i-x_j|$ between pairs of distinct points. Scale invariance implies that only ratios of such distances. such as $$\frac{|x_i-x_j|}{|x_k-x_l|}$$ will appear in $\Gamma$. Finally, under a special conformal transformation, the distance separating two points $x_i$ and $x_j$ becomes $$\begin{align}|x'_i-x'_j|=\frac{|x_i-x_j|}{(1-2b\cdot x_i+b^2x_i^2)^{1/2}(1-2b\cdot x_j+b^2x^2_j)^{1/2}}\end{align}$$ It is therefore impossible to construct an invariant $\Gamma$ with only 2 or 3 points. The simplest possibilities are the following functions of four points: $$\begin{align}\frac{|x_1-x_2||x_3-x_4|}{|x_1-x_3||x_2-x_4|}\quad \frac{|x_1-x_2||x_3-x_4|}{|x_2-x_3||x_1-x_4|}\end{align}$$ Such expressions are called anharmonic ratios or cross-ratios. With $N$ distinct points, $N(N - 3)/2$ independent anharmonic ratios may be constructed.
4.2 Conformal Invariance in Classical Field Theory
4.2.1 Representations of the Conformal Group in $d$ Dimensions
We first show how classical fields are affected by conformal transformation. Given an infinitesimal conformal transformation parametrized by $\omega_g$, we seek a matrix representation $T_g$ such that a multicomponent field $\Phi(x)$ transforms as $$\begin{align}\Phi'(x')=(1-i\omega_gT_g)\Phi(x)\end{align}$$ The generator $T_g$ must be added to the space-time part given in (18) to obtain the full generator of the symmetry. (like above generator, $\mathcal{F}$.) In order to find out the allowed form of these generators, we start from subgroup of the Poincare group that leaves the point $x=0$ invariant, that is, the Lorentz group. We then introduce a matrix representation $S_{\mu\nu}$ to defined the action of infinitesimal Lorentz transformations on the field $\Phi(0)$: $$\begin{align}L_{\mu\nu}\Phi(0)=S_{\mu\nu}\Phi(0)\end{align}$$ $S_{\mu\nu}$ is the spin operator associated with the field $\Phi$. Next, by use of the commutation relations of the Poincare group, we transflate the generator $L_{\mu\nu}$ to a nonzero value of $x$: $$\begin{align}e^{ix^\rho P_\rho}L_{\mu\nu}e^{-ix^\rho P_\rho}=S_{\mu\nu}-x_\mu P_\nu +x_\nu P_\mu\end{align}$$ The above translation is explicitly calculated by use of the Hausdorff formula: $$\begin{align}e^{-A}Be^A=B+[B,A]+\frac{1}{2!}[[B,A],A]+\frac{1}{3!}[[[B,A],A],A]+\cdots\end{align}$$ This allows us to write the action of the generators: $$\begin{align}\begin{matrix}P_\mu\Phi(x)&=& -i\partial_\mu \Phi(x)\\ L_{\mu\nu}\Phi(x)&=& i(x_\mu \partial_\nu -x_\nu\partial_\mu)\Phi(x)+S_{\mu\nu}\Phi(x)\end{matrix}\end{align}$$
We proceed in the same way for the full conformal group. The subgroup that leaves the origin $x = 0$ invariant, so remove translation generator. We then denote by $S_{\mu\nu}$, $\tilde{\Delta}$, and $\kappa_\mu$ the respective values of the generators $L_{\mu\nu}$, $D$, and $K_\mu$ at $x=0$. These must form a matrix representation of the reduced algebra $$\begin{align}\begin{matrix}[\tilde{\Delta},S_{\mu\nu}]&=& 0\\ [\tilde{\Delta},\kappa_\mu]&=& -i\kappa_\mu\\ [\kappa,\kappa_\mu] &=&0\\ [\kappa_\rho,S_{\mu\nu}] &=& i(\eta_{\rho\mu}\kappa_\nu-\eta_{\rho\nu}\kappa_\mu)\\ [S_{\mu\nu},S_{\rho\sigma}]&=& i(\eta_{\nu\rho}S_{\mu\sigma}+\eta_{\mu\sigma}S_{\nu\rho}-\eta_{\mu\rho}S_{\nu\sigma}-\eta_{\nu\sigma}S_{\mu\rho})\end{matrix}\end{align}$$ The commutations (19) then allow us to translate the generators, using the Hausdorff formula (27): $$\begin{align}\begin{matrix} e^{ix^\rho P_\rho}De^{-ix^\rho P_\rho}&=&D+x^\nu P_\nu\\ e^{ix^\rho P_\rho}K_\mu e^{-ix^\rho P_\rho}&=& K_\mu+2x_\mu D-2x^\nu L_{\mu\nu}+2x_\mu(x^\nu P_\nu)-x^2P_\mu\end{matrix}\end{align}$$ form which we arrive finally at the following extra transformation rules: $$\begin{align}\begin{matrix}D\, \Phi(x) &=& (-ix^\nu\partial_\nu+\tilde{\Delta})\Phi(x)\\ K_\mu \Phi(x) &=& \left\{ \kappa_\mu+2x_\mu \tilde{\Delta}-x^\nu S_{\mu\nu}-2ix_\mu x^\nu \partial_\nu+ix^2\partial_\mu\right\} \Phi(x)\end{matrix}\end{align}$$
If we demand that the field $\Phi(x)$ belong to an irreducible representation of the Lorentz group, then, by Schur's lemma, any matrix that commutes with all the generators $S_{\mu\nu}$ must be a multiple of the identity. Consequently, the matrix $\tilde{\Delta}$ is a multiple of the identity and the algebra (29) forces all the matrices $\kappa_\mu$ to vanish. $\tilde{\Delta}$ is then simply a number, manifestly equal to $-i\Delta$, where $\Delta$. is the scaling dimension of the field $\Phi$, as defined in Eq. (121). That the eigenvalue of $\tilde{\Delta}$ is not real simply reflects the non-Hermiticity of the generator $\tilde{\Delta}$ (i.e., representations of the dilation group on classical fields are not unitary).
In principle, we can derive from the above the change in under a finite conformal transformation. However, we shall give the result only for spinless fields ($S_{\mu\nu}=0$). Under a conformal transformation $x\rightarrow x'$, a spinless field $\phi(x)$ transforms as $$\begin{align}\phi(x)\rightarrow \phi'(x')=\left| \frac{\partial x'}{\partial x}\right|^{-\Delta/d}\phi(x)\end{align}$$ where $|\partial x'/\partial x|$ is the Jacobian of the conformal transformation of the coordinates, related to the scale factor $\Lambda(x)$ of Eq. (1) by $$\begin{align}\left| \frac{\partial x'}{\partial x}\right| =\Lambda(x)^{-d/2}\end{align}$$ A field transforming like the above is called "quasi-primary."
4.2.2 The Energy-Momentum Tensor
Under an arbitrary transformation of the coordinates $x^\mu\rightarrow x^\mu+\epsilon^\mu$, the action changes as follows: $$\begin{align}\delta S=\int d^dx\, T^{\mu\nu}\partial_\mu\epsilon_\nu=\frac{1}{2}\int d^dx\, T^{\mu\nu}(\partial_\mu\epsilon_\nu+\partial_\nu\epsilon\mu)\end{align}$$ where $T^{\mu\nu}$ is the energy-momentum tensor, assumed to be symmetric. This is valid even if the equations of motion are not satisfied. The definition (3) of an infinitesimal conformal transformation implies that the corresponding variation of the action is $$\begin{align}\delta S=\frac{1}{d}\int d^dx\, T^\mu_\mu \partial_\rho\epsilon^\rho\end{align}$$ The tracelessness of the energy-momentum tensor then implies the invariance of the action under conformal transformations. The converse is not true, since $\partial_\rho\epsilon^\rho$ is not an arbitrary function.
We first consider a generic field theory with scale invariance in dimension $d>2$. The conserved current associated with the infinitesimal dilation $$\begin{align}{x'}^\mu=(1+\alpha)x^\mu\quad \mathcal{F}(\Phi)=(1-\alpha\Delta)\Phi\end{align}$$ is, Noether current is $$\begin{align}j^\mu_D=-\mathcal{L}x^\mu+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi)}x^\nu\partial_\nu\Phi+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi)}\Delta\Phi\nonumber\\ =T^\mu_{c\nu}x^\nu+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi)}\Delta\Phi\end{align}$$ where $T^{\mu\nu}_c$ is the canonical energy-momentum tensor. Since by hypothesis this current is conserved, we have $$\begin{align}\partial_\mu j^\mu_D=T^\mu_{c\mu}+\Delta\partial\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi)}\Phi\right)=0\end{align}$$
We now define the virial of the field $\Phi$: $$\begin{align}V^\mu=\frac{\delta\mathcal{L}}{\delta(\partial_\rho\Phi)}(\eta^{\mu\rho}\Delta+iS^{\mu\rho})\Phi\end{align}$$ where $S^{\mu\rho}$ is the spin operator of the field $\Phi$. We also assume that the virial is the divergence of another tensor $\sigma^{\alpha\mu}$: $$\begin{align}V^\mu=\partial_\alpha\sigma^{\alpha\mu}\end{align}$$ This last condition is obeyed in a large class of physical theories. Then we define $$\begin{align}\sigma^{\mu\nu}_+=\frac{1}{2}(\sigma^{\mu\nu}+\sigma^{\nu\mu})\nonumber\\ X^{\lambda\rho\mu\nu}=\frac{2}{d-2}\left\{\eta^{\lambda\rho}\sigma^{\mu\nu}_+-\eta^{\lambda\mu}\sigma^{\rho\nu}_+-\eta^{\lambda\mu}\sigma^{\nu\rho}_++\eta^{\mu\nu}\sigma^{\lambda\rho}_+\\ +\frac{1}{d-1}(\eta^{\lambda\rho}\eta^{\mu\nu}-\eta^{\lambda\mu}\eta^{\rho\nu})\sigma^\alpha_{+\alpha}\right\}\end{align}$$ and we consider the following modified energy-momentum tensor: $$\begin{align}T^{\mu\nu}=T^{\mu\nu}_c+\partial_\rho B^{\rho\mu\nu}+\frac{1}{2}\partial_\lambda\partial_\rho X^{\lambda\rho\mu\nu}\end{align}$$ The first two terms of the above expression constitute the Belinfante tensor. The last term is an addition that will make $T^{\mu\nu}$ traceless. Because of the symmetry properties of $X^{\lambda\rho\mu\nu}$, this additional term does not spoil the conservation law: $$\begin{align}\partial_\mu\partial_\lambda\partial_\rho X^{\lambda\rho\mu\nu}=0\end{align}$$ Indeed, the addition would not be conserved if $X^{\lambda\rho\mu\nu}$ had a part completely symmetric in the first three indices, but this is not the case. This new term does not spoil the symmetry of the Belinfante tensor either, since the part of $X^{\lambda\rho\mu\nu}$ antisymmetric in $\mu$, $\nu$ is $$X^{\lambda\rho\mu\nu}-X^{\lambda\rho\nu\mu}=\frac{2}{(d-2)(d-1)}\sigma^\alpha_{+\alpha}(\eta^{\lambda\mu}\eta^{\rho\nu}-\eta^{\lambda\nu}\eta^{\rho\mu})$$ and it vanishes upon contraction with $\partial_\lambda\partial_\rho$. Finally, the trace of the new term is $$\begin{align}\frac{1}{2}\partial_\lambda\partial_\rho X^{\lambda\rho\mu}_\mu=\partial_\lambda\partial_\rho\sigma^{\lambda\rho}_+=\partial_\mu V^\mu\end{align}$$ Since $$\partial_\rho B^{\rho\mu}_\mu=\frac{1}{2}i\partial_\rho\left(\frac{\delta\mathcal{L}}{\delta(\partial^\mu\Phi)}S^{\mu\rho}\Phi\right)$$ it follows from (38) and (39) that $$\begin{align}T^\mu_\mu=\partial_\mu j^\mu_D\end{align}$$ and therefore scale invariance implies that the modified energy-momentum tensor (42) is traceless, provided, of course, that the virial satisfies condition (40). This relation also means that the dilation current may be generally written as $$\begin{align}j^\mu_D=T^\mu_\nu x^\nu\end{align}$$
4.3 Conformal Invariance in Quantum Field Theory
4.3.1 Correlation Functions
Consider the two-point function $$\begin{align}\langle\phi_1(x_1)\phi_2(x_2)\rangle=\frac{1}{Z}\int [d\Phi]\phi_1(x_1)\phi_2(x_2)\exp -S[\Phi] \end{align}$$ where $\phi_1$ and $\phi_2$ are quasi-primary fields (not necessarily distinct). $\Phi$ denotes the set of all functionally independent fields in the theory (to which $\Phi_1$ and $\Phi_2$ may belong), and $S[\Phi]$ is the action, which we assume to be conformally invariant.
The assumed conformal invariance of the action and of the functional integration measure leads to the following transformation of the correlation function, for spinless fields for simplicity: $$\begin{align}\langle\phi_1(x_1)\phi_2(x_2)\rangle=\left|\frac{\partial x'}{\partial x}\right|^{\Delta_1/d}_{x=x_1}\left|\frac{\partial x'}{\partial x}\right|^{\Delta_2/d}_{x=x_2}\langle\phi_1(x'_1)\phi_2(x'_2)\rangle\end{align}$$ If we speccialize to a scale transformation $x\rightarrow \lambda x$ we obtain $$\begin{align}\langle\phi_1(x_1)\phi_2(x_2)\rangle=\lambda^{\Delta_1+\Delta_2}\langle\phi_1(\lambda x_1)\phi_2(\lambda x_2)\rangle\end{align}$$ Rotation and translation invariance require that $$\begin{align}\langle\phi_1(x_1)\phi_2(x_2)\rangle=f(|x_1-x_2|)\end{align}$$ where $f(x)=\lambda^{\Delta_1+\Delta_2}f(\lambda x)$ by virtue of (49). In other words, $$\begin{align}\langle\phi_1(x_1)\phi_2(x_2)\rangle=\frac{C_{12}}{|x_1-x_2|^{\Delta_1+\Delta_2}}\end{align}$$ where $C_{12}$ is a constant coefficient. It remains to use the invariance under special conformal transformations. We recall that, for such a transformation, $$\begin{align}\left| \frac{\partial x'}{\partial x}\right|=\frac{1}{(1-2b\cdot x+b^2x^2)^d}\end{align}$$ Given the transformation (22) for the distance $|x_1-x_2|$, the covariance of the correlation function (51) implies $$\begin{align}\frac{C_{12}}{|x_1-x_2|^{\Delta_1+\Delta_2}}=\frac{C_{12}}{\gamma^{\Delta_1}_1\gamma^{\Delta_2}_2}\frac{(\gamma_1\gamma_2)^{(\Delta_1+\Delta_2)/2}}{|x_1-x_2|^{\Delta_1+\Delta_2}}\end{align}$$ with $$\begin{align}\gamma_i=(1-2b\cdot x_i+b^2x_i^2)\end{align}$$ This constraint is identically satisfied only if $\Delta_1=\Delta_2$. In other words, two quasi-primary fields are correlated only if they have the same scaling dimension: $$\begin{align}\end{align}$$
$$\begin{align}\langle\phi_1(x_1)\phi_2(x_2)\rangle=\begin{cases}\frac{C_{12}}{|x_1-x_2|^{2\Delta_1}}& \mbox{if }\Delta_1=\Delta_2\\ 0& \mbox{if }\Delta_1\ne\Delta_2\end{cases}\end{align}$$
A similar analysis may be performed on three-point functions. Covariance under rotations, translations, and dilations forces a generic three-point function to have the following form: $$\begin{align}\langle\phi_1(x_1)\phi_2(x_2)\phi_3(x_3)\rangle=\frac{C^{(abc)}_{123}}{x_{12}^ax_{23}^b x_{13}^c}\end{align}$$ where $x_{ij}=|x_i-x_j|$ and with $a,b,c$ such that $$\begin{align}a+b+c=\Delta_1+\Delta_2+\Delta_3\end{align}$$ Actually, a sum (over $a, b, c$) of such terms is also acceptable, as long as the above equality is satisfied. Under special conformal transformations Eq. (57) becomes $$\frac{C^{(abc)}_{123}}{\gamma_1^{\Delta_1}\gamma_2^{\Delta_2}\gamma^{\Delta_3}_3}\frac{(\gamma_1\gamma_2)^{a/2}(\gamma_2\gamma_3)^{b/2}(\gamma_1\gamma_3)^{c/2}}{x_{12}^ax_{23}^bx_{13}^c}$$ For this expression to be of the same form as Eq. (57), all the factors involving the transformation parameter $b^\mu$ must disappear, which leads to the following set of constraints: $$\begin{align}a+c=2\Delta_1\quad a+b=2\Delta_2\quad b+c=2\Delta_3\end{align}$$ The solution to these constraints is unique: $$\begin{align}\begin{matrix}a &=& \Delta_1+\Delta_2-\Delta_3\\ b&=& \Delta_2+\Delta_3-\Delta_1\\ c &=& \Delta_3+\Delta_1-\Delta_2\end{matrix}\end{align}$$ Therefore. the correlator of three quasi-primary fields is made of a single term of the form (57). namely $$\begin{align}\langle\phi_1(x_1)\phi_2(x_2)\phi_3(x_3)\rangle=\frac{C_{123}}{x^{\Delta_1+\Delta_2-\Delta_3}_{12}x^{\Delta_2+\Delta_3-\Delta_1}_{23}x^{\Delta_3+\Delta_1-\Delta_2}_{13}}\end{align}$$
Four-point function is $$\begin{align}\langle\phi_1(x_1)\cdots\phi_4(x_4)\rangle=f\left(\frac{x_{12}x_{34}}{x_{13}x_{24}},\frac{x_{12}x_{34}}{x_{23}x_{14}}\right)\prod^4_{i<j}x_{ij}^{\Delta/3-\Delta_i-\Delta_j}\end{align}$$ where we have defined $\Delta=\sum^4_{i=1}\Delta_i$.
4.3.2 Ward Identities
We shall now write the Ward identities implied by conformal invariance. according to the general identity (2.157). The Ward identity associated with translation invariance appears in Eq. (2.183) and we reproduce it here: $$\begin{align}\partial_\mu\langle T^\mu_\nu X\rangle=-\sum_i\delta(x-x_i)\frac{\partial}{\partial x^\nu_i}\langle X\rangle\end{align}$$ This identity holds even after a modification of the energy-momentum tensor. as in Eq. (42). Recall that $X$ stands for a product of n local fields. at coordinates $i=1,\cdots,n$.
We consider now the Ward identity associated with Lorentz (or rotation) invariance. Once the energy-momentum has been made symmetric. the associated current $j^{\mu\nu\rho}$ has the form given in Eq. (2.172): $$\begin{align}j^{\mu\nu\rho}=T^{\mu\nu}x^\rho-T^{\mu\rho}x^\nu\end{align}$$ The generator of Lorentz transformations is given by Eq. (2.134). Consequently, the Ward identity is $$\begin{align}\partial_\mu\langle(T^{\mu\nu}x^\rho-T^{\mu\rho}x^\nu)X\rangle=\sum_i\delta(x-x_i)\left[(x^\nu_i\partial^\rho_i-x^\rho_i\partial^\nu_i)\langle X\rangle-iS^{\nu\rho}_i\langle X\rangle\right]\end{align}$$ where $S^{\nu\rho}_i$ is the spin generator appropriate for the $i$-th field of the set $X$. The derivative on the l.h.s. of the above equation may act either on the energy-momentum tensor or on the coordinates. Using the first Ward identity (63), we reduce the above to $$\begin{align}\langle(T^{\rho\nu}-T^{\nu\rho})X\rangle=-i\sum_i \delta(x-x_i)S^{\nu\rho}_i\langle X\rangle\end{align}$$ which is the Ward identity associated with Lorentz (rotation) invariance. It states that the energy-momentum tensor is symmetric within correlation functions, except at the position of the other fields of the correlator.
Finally, we consider the Ward identity associated with scale invariance. We shall assume that the dilation current $j^\mu_D$ may be written as in Eq. (46), which supposes that the energy-momentum tensor has been suitably modified (if needed) to be traceless. So far we have not shown how this can be done generally in two dimensions, although we hold that it can be done. In the next chapter we shall provide an alternate derivation of the Ward identity, which circumvents this problem. Since the generator of dilations is $D=-ix^\nu\partial_\nu-i\Delta$ for a field of scaling dimension $\Delta$, the Ward identity is $$\begin{align}\partial_\mu\langle T^\mu_\nu x^\nu X\rangle=-\sum_i\delta(x-x_i)\left\{ x^\nu_i\frac{\partial}{\partial x^\nu_i}\langle X\rangle+\Delta_i\langle X\rangle\right\}\end{align}$$ Here again the derivative $\partial_\mu$, may act on $T^\mu_\nu$ and on the coordinate. Using Eq. (63), this identity reduces to $$\begin{align}\langle T^\mu_\mu X\rangle=-\sum_i \delta(x-x_i)\Delta_i\langle X\rangle\end{align}$$ Eqs. (63), (66), and (68) are the three Ward identities associated with conformal invariance.
4.3.3 Tracelessness of $T_{\mu\nu}$ in Two Dimensions
We consider the two-point function of the energy-momentum tensor (called the Schwinger function): $$\begin{align}S_{\mu\nu\rho\sigma}(x)=\langle T_{\mu\nu}(x)T_{\rho\sigma}(0)\rangle\end{align}$$ Since by assumption the theory is translation and rotation invariant, $T_{\mu\nu}$ is conserved and symmetric (or can be made symmetric). The symmetry of $T_{\mu\nu}$ implies that $$\begin{align}S_{\mu\nu\rho\sigma}=S_{\nu\mu\rho\sigma}=S_{\mu\nu\sigma\rho}=S_{\nu\mu\sigma\rho}\end{align}$$ Translation invariance implies that $$\begin{align}S_{\mu\nu\rho\sigma}(x)=\langle T_{\mu\nu}(0)T_{\rho\sigma}(-x)\rangle=T_{\rho\sigma}(-x)T_{\mu\nu}(0)\rangle=S_{\rho\sigma\mu\nu}(-x)\end{align}$$ If the theory is invariant under parity, we conclude that $$\begin{align}S_{\mu\nu\rho\sigma}(x)=S_{\rho\sigma\mu\nu}(x)\end{align}$$ Finally, scale invariance implies that $T_{\mu\nu}$ transforms covariantly under scale transformations, with scaling dimension 2 since it is a density. This means that $$\begin{align}S_{\mu\nu\rho\sigma}(\lambda x)=\lambda^{-4}S_{\mu\nu\rho\sigma}(x)\end{align}$$
All these constraints restrict the most general form that $S_{\mu\nu\rho\sigma}$ can take: $$\begin{align}S_{\mu\nu\rho\sigma}(x)=(x^2)^{-4}\left\{A_1g_{\mu\nu}g-{\rho\sigma}(x^2)^2+A_2(g_{\mu\rho}g_{\nu\sigma}+g_{\mu\sigma}g_{\nu\rho})(x^2)^2\nonumber\\ +A_3(g_{\mu\nu}x_\rho x_\sigma+g_{\rho\sigma}x_\mu x_\nu)x^2+A_4 x_\mu x_\nu x_\rho x_\sigma \right\}\end{align}$$ The constants $A_1$ to $A_4$ are not all arbitrary. Indeed, the conservation law $\partial^\mu T_{\mu\nu}=0$ obviously extends to the Schwinger function. Taking the derivative, we find $$\begin{align}\partial^\mu S_{\mu\nu\rho\sigma}(x)=-(x^2)^{-4}\left\{3(A_4+2A_3)x_\nu x_\rho x_\sigma+(4A_1+3A_3)g_{\rho\sigma}x_\nu x^2+(4A_2-A_3)(g_{\rho\nu}x_\sigma+g_{\nu\sigma}x_\rho)x^2\right\}\end{align}$$ This vanishes everywhere only if each combination of coefficients in parentheses vanishes. This leaves only one arbitrary constant: $$\begin{align}A_1=3A\quad A_2=-A\quad A_3=-4A\quad A_4=8A\end{align}$$ Upon inserting these values into Eq. (74), we find $$\begin{align}S_{\mu\nu\rho\sigma}(x)=\frac{A}{(x^2)^4}\left\{(3g_{\mu\nu}g_{\rho\sigma}-g_{\mu\rho}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho})(x^2)^2\nonumber\\ -4x^2(g_{\mu\nu}x_\rho x_\sigma+g_{\rho\sigma}x_\mu x_\nu)+8x_\mu x_\nu x_\rho x_\sigma \right\}\end{align}$$ It is then straightforward to show that the trace $$\begin{align}S^{\mu\sigma}_{\mu\sigma}(x)=\langle T^\mu_\mu(x)T^\sigma_\sigma(0)\rangle\end{align}$$ vanishes everywhere. In particular $\langle T^\mu_\mu(0)^2\rangle$, which implies that the operator $T^\mu_\mu$ has zero expectation value and zero standard deviation in the ground state. In fact, the general result is the Ward identity (68), which states that $T^\mu_\mu(x)$ vanishes within correlation functions, except when $x$ coincides with the position of another field present in the correlator.
[Homework 2 (PH754 2024S)]
1. Consider the canonical energy-momentum tensor for the free boson in $d > 2$. Find an improvement term that makes it classically traceless without spoiling classical conservation.
2. Consider two-dimensional Liouville theory whose Lagrangian density is given as $$\mathcal{L}=\frac{1}{2}\partial_\mu\phi\partial^\mu\phi+\frac{1}{2}m^2e^\phi.$$ Write down the canonical energy-momentum tensor and verify that it is conserved. Add a term so that it is also traceless without spoiling the conservation.
3. Prove the following property under special conformal transformations $$|x'_i-x'_j|=\frac{|x_i-x_j|}{\gamma_i^{1/2}\gamma_j^{1/2}}$$ where $\gamma_i=1-2b\cdot x_i+b^2x_i^2$.
4. Consider the inversion tensor $I_{\mu\nu}(x)\equiv \eta_{\mu\nu}-2\frac{x_\mu x_\nu}{x^2}$. Show that $$I_{\mu\alpha}(x)I^{\alpha\beta}(x-y)I_{\beta\nu}(y)=I_{\mu\nu}(x'-y'),$$ where $(x')^\mu=x^\mu/x^2$, $(y')^\mu=y^\mu/y^2$.
5. Consider the OPE $\phi_1(x)\phi_2(0)$ and the contribution to this OPE from a single primary $\phi_\Delta(0)$ plus all its tower of descdents: $$\phi_1(x)\phi(0)|0\rangle=\frac{const}{|x|^{\Delta_1+\Delta_2-\Delta}}(\phi_\Delta(0)+\alpha x^\mu\partial_\mu\phi_\Delta(0)+\cdots)|0\rangle$$ Wehaveshowninthe lecture how to fix $\alpha$ by acting $K_\mu$ on both sides. In the lecture, we have outlined another procedure of deriving the expansion by computing the 3-point function $\langle \phi_1(x)\phi_2(0)\phi_\Delta(z)\rangle$. Derive the above expression using this up to quadratic order in $x$.