CFT/TFT Lecture note

[2/26] 0. Whence CFT&TFT?

Conformal field theory is a speical kind of quangum field theory with a scaling symmetry.

$\rightarrow$ The system is invariant under scale transformation $$\vec{x}\rightarrow \lambda\vec{x}\\ t\rightarrow \lambda t$$

$\bullet$ QFT = quantum mechanics with "$\infty$'ly" many degrees of freedom that fills the space

When the size of lattice spacing is very small, we can treat dynamical degrees of freedom at each point to be effectively continuous $$\phi(n_1\hat{e}_1+n_2\hat{e_2}),\ \frac{1}{a^2}\sum_{n_1,n_2}\rightarrow \phi(x,y),\ \frac{1}{L^2}\int dx\, dy$$ Such a continuum limit is descrived by a QFT.

$\bullet$ In high-energy particle physics, quantum fields describe elementary particles such as quarks, leptons, photons, gauge bosons, ...

$\bullet$ In condensed matter physics, fields can be local spin degrees of freedom of anything that can be associated with space.

$\bullet$ In HEP, quantum fields are conisdered to be elementary.

$\bullet$ In CMP, quantum fields are low-energy effective description. ($E\ll \frac{1}{a}$)

Whence CFT?

Confirmal symmetry $\sim$ "scale invariance"

$\rightarrow$ no characteristic scale

Critical phenomena

eg) Ising model $$H=\sum_{\langle ij\rangle}\sigma_i\sigma_j$$ $$\langle\sigma_i\sigma_j\rangle -\langle \sigma_i\rangle\langle\sigma_j\rangle \sim \exp \left( -\frac{|i-j|}{\xi}\right)$$ For generic $T$, the system have no scale invariance, $(a,\xi)$ introduce scales.

For $T=T_c$, $\xi\rightarrow \infty$. (much larger that $a$).

"critical point" $\Rightarrow$ scale symmetry emergent.

$\bullet$ Ising model is just one among as $\infty$ models that can provide approx. desc. of cpx systems with local interacts "universality class".

$\Rightarrow$ CFT: study of such universality class

Q) Can we calssify "all" such class?

A) Only partically in $1+1d$.

Quantum critical system ($T=0$)

gapless excitation $\Rightarrow$ \langle \sigma\sigma\rangle\sim \frac{1}{|x|^2}$

- edge modes in quantum Hall system

- Knodo effect

Deep inelastic scattering

$\bullet$ scale free natrue of QCD $\leftrightarrow$ "asymptotic freedom"

(scale-invariance at short distance (UV))

String theory

$\bullet$ world-sheet theory has scale invariance

Quantum gravity

Ads/CFT correspondence

[2/28] Whence TFT?

Topological field theory = QFT that does not depend on the choice of spacetime metric $g_{\mu\nu}$

$rightarrow$ It does depend on topology.

$$M \rightarrow Z[M,g] = \int \mathcal{D}\phi \, e^{iS[M,g,\phi]}$$  

TQFT does depend on $M$ but does not depend on $g$. $\Rightarrow$ 'topological' 'topology of $M$'

Physically, $\hat{H}=0$, no local excitations.

all local excitations are massive. ($m\rightarrow \infty$)

$\bullet$ On a space with boundary, gapless edge modes can appear. (QHE)

For any physical system, one would like to know low-$E$ excitations "gapped phase"

$\bullet$ It may look trivial, but on a topologically non-trivial space (locally trivial), non-local observables (eg: line operators) can be non-trivial.

Another case is "gapless phase" which has continuous level of energy.

$\bullet$ trivial means free theory, non-trivial means non-trivial CFT.

1. From particles to fields to particles again

Let particles are on lattice with spring, and their position is $x_n=na$, and $q_n$ be displacement away from equilibrium. $$H=\sum_{n=1}^N\left( \frac{p_n^2}{2m}+\frac{1}{2}K(q_n-q_{n-1})^2\right)+\cdots$$ (Boundary condition: $q_{n+N}=q_n$) Individual fluctuation is linear sum of normal modes.

Translatino invariance: $T:x\mapsto x+a$ $$\hat{q}_k=\frac{1}{\sqrt{N}}\sum_{n=1}^Ne^{ikx_n}q_n\\ \hat{p}_k=\frac{1}{\sqrt{N}}e^{ikx_n}p_n$$ with boundary condition, $$q_n=\frac{1}{\sqrt{N}}\sum_k e^{-ikx_n}\tilde{q}_k$$ $k$ be quantized with $e^{ikNa}=1$, with $k=\frac{2\pi m}{Na}$, $m=a,\cdots, N-1$

Brilliomn zone is $$-\frac{\pi}{a}\le k<\frac{\pi}{a}\quad k\sim k+\frac{2\pi}{a}.$$

System in a box $\Rightarrow$ $k$-space discrete ($q_{n+N}=q_n$) "IR regulator"

System on a lattice $\Rightarrow$ $k$-space finite ($a$) "UV regulator"

$$H=\sum \frac{p_n^2}{2m}+\frac{1}{2}K(q_{n+1}-q_n)^2$$ put $$p_n=\frac{1}{?}\sum e^{-ikx}\tilde{p}_k$$ then $$H=\sum_k\left( \frac{\tilde{p}_k\tilde{p}_{-k}}{2m}+\frac{1}{2}m\omega_k^2\tilde{q}_k\tilde{q}_{-k}\right)$$ with $$\omega_k\equiv 2\sqrt{\frac{k}{m}}\sin \frac{|k|a}{2}$$  

$$i\partial_tq_n=[q_n,H]=\frac{p_n}{m}$$ $$i\partial p_n=[p_n,H]=\sum_n \frac{1}{2}k[p+n,(q_n-q_{n-1})^2=k(q_{n+1}-2q_n-q_{n-1})$$ $$\Rightarrow m\ddot{q}_n=-k(2q_n-q_{n+1}-q_{n-1})$$

then $$m\ddot{tilde{q}}_k=-k(2-2\cos ka)\tilde{q}_k$$ put $$\tilde{q}_k=\sum_\omega e^{-i\omega t}q_{k,\omega}$$ then $$\sum_\omega -m\omega^2 q_{k,\omega}+k(2-2\cos ka)q_{k,\omega}=0$$ $$0=(\omega^2-\omega_k^2)q_{k,\omega}\simeq (w-v_s^2k^2)q_{v,\omega}\quad (k\ll \frac{1}{a})$$ with $$(\partial_t^2-v_s^2\partial_x^2)q(x,t)=0$$ $\Rightarrow$ $q$ is a wave ($v_q\equiv \left. \frac{\partial \omega}{\partial k}\right|_{k=0}=a\sqrt{\frac{k}{m}}$

QM: $$[q_n,p_{n'}]=i\delta_{nn'}$$ $$\Rightarrow [\tilde{q}_k,\tilde{p}_{k'}]=\sum_{nn'}u_{kn}u_{k'n'} [q_n,p_n]$$ ($u_{k,n}=e^{ikx_n}/\sqrt{N}$) $$=\sum_n u_{kn}u_{k'n'}i=i\delta_{k-k'}$$

Finally, $$\hat{q}_k=\sqrt{\frac{\hbar}{2m\omega_k}}(a_k+a_{-k}^\dagger)$$ $$\hat{p}_k=\frac{1}{i}\sqrt{\frac{\hbar m\omega_k}{2}}(a_k-a_{-k}^\dagger)$$ $$[a_k,a^\dagger_{k'}]=\delta_{kk'}$$

$$\Rightarrow H=\sum_{k\ne 0}\hbar \omega_k \left( a^\dagger_{k}a_k+\frac{1}{2}\right) +\frac{p_0^2}{2m}$$ sum of decoupled SHO's + center of mass motion

The ground state $|0\rangle$ $$a_k|0\rangle=0 \quad \forall k,\quad p_0|0\rangle=0$$

First excited state: $$a^\dagger_k|0\rangle$$ (one phonon with momentum $\hbar k$: $|k\rangle$)

Second excited state: $$a^\dagger_k a^\dagger_{k'}|0\rangle$$ $$p=\hbar (k+k')\quad E=\hbar (\omega_k+\omega_{k'})\quad N_k=a^\dagger_ka_k$$

Fork space $\mathcal{H}$: 1 particle Hilbert space

$\mathcal{F}=\otimes \mathcal{H}$ / symmetrize or anti-symmetrize

In above example, $a_k$s are automatically symmetry (spin-statistics for boson).

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Chain of oscillators $$H=\sum_k \hbar \omega_k (a^\dagger_k a_k+\frac{1}{2}+\frac{p_0^2}{2m}$$ with $\omega_k=\sqrt{\frac{2k}{m}}\sim \frac{|k|a}{2}$.

$$\mathcal{H}=\mbox{span}\{ a^\dagger_{k_1}a^\dagger_{k_2}\cdots|0\rangle\}$$

Fork space $\sim \otimes \mathcal{H}_{1-ptc}/$(symmetrization)

In QFT, we naturally get symmetrization with $a_k$s.

$|k|\ll \frac{1}{a}$ $\Rightarrow \omega_k^2\simeq k^2V_s^2$ photon $E=pc$ is same as $\omega=kc$.

$(\partial_t^2-v_s^2\nabla^2)\phi=0$ $\phi=\sum A_\omega e^{i\omega t}$ $\omega^2=v_s^2k^2$

And this can be Lorentz invariant $\partial_\mu \partial^\mu \equiv \partial_t^2-\nabla^2$

Mossbawer effect

Scalar field theory (in 1+1d)

Path-integral formulation $$Z=\int dq_1\, dq_2\, \cdots dq_N\, e^{iS[q_1,\cdots,q_N]}$$ when $S[q]=\int dt [\frac{1}{2}\sum m\dot{q}_n^2-V(q)]$ and $V(q)=\frac{1}{2}k\sum^N_{n=1}(q_{n+1}-q_n)^2$. 

Continuum limit or long-wavelength limit $a\rightarrow 0, N\rightarrow \infty$

then $$(q_{n+1}-q_n)^2\rightarrow a^2(\partial_x q)^2\quad a\sum_n f(x_n)\rightarrow \int dx\, f(x)$$

$Z=\int [Dq]e^{iS[q(x)]}$

$$S[g]=\int dt\int dx(\frac{1}{2}(\partial_tq)^2-\mu V_s^2(\partial_xq)^2-rq^2-sq^4\cdots)=\int dx\, dt\, \mathcal{L}(q,\partial_tq,\partial_xq)$$ Action functional is spacetime integral of Lagrangian density.

(when $k\ll \frac{1}{a}$, higher-derivatives neglected)

$[x,p]=\hbar$ became $[q,\pi]$. momentum conj. for $q(x,t)$ $$\pi(x,t)=\frac{\partial\mathcal{L}}{\partial \dot{q}}=\mu\dot{q}$$ Hamiltonian became $$H=\sum_n p_n\dot{q}_n-L=\int dx(\frac{\pi(x)^2}{2\mu}+\mu V_s^2(\partial_xq)^2+\cdots)$$  

In $\infty$-vol limit, $$\frac{1}{L^d}\sum_k\rightarrow \int \frac{d^dk}{(2\pi)^d}$$ $$L^d\delta_{kk'}\rightarrow (2\pi)^d\delta^{(d)}(k-k')$$ Continuum scalar field theory in (d+1)-dim. $$S[\phi]=\int d^dx\, dt\, (\frac{1}{2}\dot{\phi}^2-\frac{1}{2}v_s^2(\nabla\phi)^2-V(\phi))$$ EOM comes from "Least(Stationary) Action Principle" $$0=S[\phi+\delta\phi]-S[\phi]\Leftrightarrow \frac{\delta S}{\delta \phi}=0$$ This became $$\delta S=\int d^dx\, dt\, (-\ddot{\phi}+v_s\nabla^2\phi-V'(\phi))\delta\phi=0$$ for example, $V(\phi)=\frac{1}{2}m^2\phi^2$ and $V_s=1$, this become $(-\partial_t+\nabla^2-m^2)\phi=0$

Conjugate momentum is $$\pi(x)=\frac{\partial\mathcal{L}}{\partial \dot{\phi}}=\dot{\phi}\\ \Rightarrow H=\int d^dx(\frac{1}{2}\pi(x)^2+\frac{1}{2}V_s^2(\nabla \phi)^2+\frac{1}{2}m^2\phi^2)\ge 0$$ $$\phi(x)=\int \frac{d^dk}{(2\pi)^d}e^{i\vec{k}\cdot\vec{x}}\phi_\vec{k}$$ $$\pi(x)=\int \frac{d^dk}{(2\pi)^d}e^{-i\vec{k}\cdot\vec{x}}\pi_\vec{k}$$ $$\Rightarrow H=\int \frac{d^dk}{(2\pi)^d}(\frac{1}{2}\pi_\vec{k}\pi_{-\vec{k}}+\frac{1}{2}(V_s^2\vec{k}^2+m^2)\phi_k\phi_{-k})$$ Dispersion relation is $$\omega_\vec{k}=m^2+V_s^2\vec{k}^2$$ Now we can represent as creation and annihilation operator $$\phi_k\equiv \frac{1}{\sqrt{2\omega_k}}(a_k+a^\dagger_{-k})\\ \pi_k\equiv \frac{1}{i}\sqrt{\frac{\omega_k}{2}}(a_k-a^\dagger_{-k})$$ and $$[\phi(x),\pi(y)]=i\delta^{(d)}(x-y)\Rightarrow [a_k,a^\dagger_{k'}]=(2\pi)^d\delta^{(d)}(k-k')$$ $$\Rightarrow H=\int \frac{d^dk}{(2\pi)^d}\omega_k(a^\dagger_k a_k+\frac{1}{2}L^d)$$

$$\pi(x)=\int \frac{d^dk}{(2\pi)^d}\frac{1}{\sqrt{2\omega_k}}(e^{i\vec{k}\cdot\vec{x}}a_\vec{k}+e^{-i\vec{k}\cdot\vec{x}}a^\dagger_\vec{k})$$ $$\pi(x)=\int \frac{d^dk}{(2\pi)^d}\sqrt{\frac{\omega_k}{2}}(e^{i\vec{k}\cdot\vec{x}}a_\vec{k}-e^{-i\vec{k}\cdot\vec{x}}a_{-\vec{k}}^\dagger)$$ For time difference, $\phi(x,t)=e^{iHt}\phi(x)e^{-iHt}$. 

[3/6]

Symmetries in field theory

$$\mathcal{L}=\mathcal{L}(\phi,\partial_\mu\phi)$$ (which means locality) $$S[\phi]=\int d^{d+1}x\, \mathcal{L}(\phi,\partial_\mu\phi)$$

eg) $\mathcal{L}_{KG}=\frac{1}{2}(\partial_\mu\phi)(\partial^\mu\phi)-\frac{1}{2}m^2\phi^2(-V(\phi))$ real massive scalar field.

eg) $\mathcal{L}_{EM}=-\frac{1}{4e^2}F_{\mu\nu}F^{\mu\nu}$ where $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ and $A_\mu=(\phi,\vec{A})$

dimensional analysis 

$$\hbar=c=1$$ "natural unit" $$\mbox{mass}\sim\mbox{energy}\sim\mbox{length}^{-1}\sim\mbox{time}^{-1}$$ $$\Rightarrow [p_\mu]=[m]=[\partial_\mu]=+1$$ $$[X^\mu]=-1$$ $$[S]=0\quad \Rightarrow [\mathcal{L}]=d+1=D$$

Equation of motion ($\Leftarrow$ least action principle)

  $$\frac{\delta S}{\delta \phi(x)}=0=\int d^Dx\left[ \frac{\partial\mathcal{L}}{\partial \phi}\delta \phi+\frac{\partial\mathcal{L}}{\partial (\partial_\mu\phi)}\delta(\partial_\mu\phi)\right]=\partial_\mu \left[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\delta\phi\right]-\left[ \partial_\mu\left( \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)\right]\delta\phi$$ $$=\int d^Dx\left[ \frac{\partial\mathcal{L}}{\partial\phi}-\partial_\mu\left( \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right)\right]\delta\phi+\int (\mbox{total derivative})=0$$  

Noether's theorem

Suppose $S$ is invariant under a continuous transformation $$\phi\rightarrow \phi'(x)+\epsilon\Delta\phi(x)$$ (continuous) symmetry $$\delta S=S[\phi+\epsilon(x^\mu)\Delta\phi]-S[\phi]=\int d^Dx(\partial_\mu\epsilon)j^\mu=^{IBP}-\int d^Dx\, \epsilon(\partial_\mu j^\mu)=0$$ by EOM is satisfied. so, $\partial_\mu j^\mu=0$. Which means conserved currrent.

$$\partial_0j^0-\vec{\nabla}\cdot\vec{j}=0$$ $$Q_R\equiv \int_R d^dx\, j^0$$ then $\partial_t Q_R=\int d^dx\, \partial_0j^0=-\int_Rd^dx\, \nabla\cdot \vec{j}=0$Q$ is conserved. 

Noether's method

$$\epsilon\rightarrow \epsilon(x^\mu)$$

symmetry: $$\mathcal{L}(\phi',\partial\phi')=\mathcal{L}(\phi,\partial\phi)+\epsilon(\partial_\mu\Theta^\mu)$$ Taylor expand, $$\mathcal{L}(\phi',\partial\phi')=\mathcal{L}(\phi,\partial\phi)+\epsilon\left(\frac{\partial \mathcal{L}}{\partial\phi}\Delta \phi+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\partial_\mu\Delta\phi\right)=\epsilon\left( \frac{\partial\mathcal{L}}{\partial\phi}-\partial_\mu\left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}\right)\right)\Delta\phi+\epsilon\partial_\mu\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi\right]$$ first term vanish with EOM, $$\partial_\mu j^\mu=0$$ with $$j^\mu=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\Delta\phi-\Theta^\mu$$  

Example) $\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)(\partial^\mu\phi)$ have "shift symmetry" $\phi\rightarrow \phi'=\phi+\epsilon$ $$S[\phi+\epsilon(x)]-S[\phi]=\int d^Dx\frac{1}{2}(\partial_\mu(\phi+\epsilon))^2-\frac{1}{2}(\partial_\mu\phi)^2=\int (\partial_\mu\phi)(\partial^\mu\epsilon)=-\int\epsilon\partial_\mu\partial^\mu\phi$$ so we conclude $$j^\mu=\partial^\mu\phi$$ and $\partial^\mu j_\mu=0$ satisfy from EOM.

With $Q=\int j^0$, $\delta\phi\equiv \phi'-\phi=\epsilon$ $$=i\epsilon[Q,\phi]=i\epsilon[\int d^dy\partial_t \phi(y),\phi(x)]=i\epsilon \int d^dx[\pi(y),\phi(x)]=\epsilon$$ when use $\dot{\phi}=\frac{\partial\mathcal{L}}{\partial\phi}=\pi$, $[\pi,\phi]=-i\delta$.

Example) Complex scalar field $\phi\equiv =\phi_1+i\phi_2$ $$\mathcal{L}=\partial_\mu\phi\partial^\mu\phi^*-m^2\phi^*\phi-V(|\phi|^2)$$ $$\phi\rightarrow e^{i\alpha}\phi\\ \phi^*\rightarrow e^{-i\alpha}\phi$$ $e^{i\alpha}\in U(1)$, expand with $\phi\rightarrow \phi'=\phi+i\alpha\phi$ $$\delta S=S[\phi+i\alpha\phi]-S[\phi]=\int d^Dx\alpha\partial_\mu(i\phi^*\partial^\mu\phi-i\phi\partial^\mu\phi^*)$$ which derive current.

With quantization. $$j^\mu(x)=\int \frac{d^dk}{(2\pi)^d}(a^\dagger_ka_k-b^\dagger_kb_k)=N_p-N_\bar{p}$$

Spacetime translation

$$x^\mu\rightarrow x'^\mu=x^\mu-a^\mu$$ $$\Rightarrow \phi(x)\rightarrow \phi(x')=\phi(x)+a^\nu\partial_\nu\phi+O(a^2)$$ Current became for each direction $\nu$, $$\partial_\mu T^\mu_\nu=0$$ which called stress-energy tensor.

Symmetry: $$\mathcal{L}(\phi(x),\partial\phi(x))\rightarrow \mathcal{L}(\phi(x+a),\partial_\mu\phi(x+a))=\mathcal{L}+a^\mu\frac{d\mathcal{L}}{dx^\mu}=\mathcal{L}+a^\nu \frac{d}{dx^\mu}(\delta^\mu_\nu\mathcal{L})$$ $$\Rightarrow T^\mu_\nu=\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}(\partial_\nu\phi)-\mathcal{L}\delta^\mu_\nu$$ $T^0_0$ is Hamiltonian, which is $\frac{\partial\mathcal{L}}{\partial\dot{\phi}}\dot{\phi}-\mathcal{L}=\pi\dot{\phi}-\mathcal{L}=\mathcal{H}$. $P_i$ is momentum, $\int T^0_id^dx=\int \left(\frac{\partial\mathcal{L}}{\partial \dot{\phi}}(\partial_i\phi)\right)d^dx=\int \pi\partial_i\phi d^dx$ which is momentum carried by fields

Scale transformation

$$x^\mu\rightarrow \lambda x^\mu$$ $$\phi(x^\mu)\rightarrow \lambda^{d_\phi}\phi(\lambda x^\mu)$$ which is called dilatation, $d_\phi$ is scaling dim of $\phi$.

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RG flows & Scale invariance

Slogan: Every theory is an effective theory

$\bullet$ we need to specify the relevant scale.

$\bullet$ QFT defined UV-cutoff $\Lambda$ 

Renormalization group flow $\Rightarrow$ a way to parametrize our ignorance of physics beyound $\:ambda$ in terms of low energy dof. 

$\bullet$ Single scalar in (Euclidean) $d$-dim $$\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)^2+\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$ $$Z=\int [\mathcal{D}\phi]_\Lambda e^{-S}$$ $$[\mathcal{D}\phi]_\lambda=\prod_{|\vec{k}|<\lambda} d\phi(\vec{k})$$ $$\phi(x)=\int \frac{d^dk}{(2\pi)^d}e^{ik\cdot x}\phi(k)$$  

We would like to know how the description in terms of low $E$ dof depending on $\Lambda$.

Pick $0<b<1$ $$\hat{\phi}(k)=\begin{cases}\phi(k)&\mbox{for } b\Lambda\le k<\Lambda\\ 0& \mbox{for }|k|<b\Lambda\end{cases}$$ $$\phi_{new}(k)=\begin{cases}\phi(k)& \mbox{for }k<b\Lambda\\ 0 & \mbox{otherwise}\end{cases}$$ and $\phi(k)=\hat{\phi}(k)+\phi_{new}(k)$. $$Z=\int [\mathcal{D}\phi]_\Lambda e^{-\int d^dx\, \mathcal{L}(\phi,\partial\phi)}=\int [\mathcal{D}\phi]_{b\Lambda}[\mathcal{D}\hat{\phi}]_\Lambda e^{-\int d^dx\, \mathcal{L}(\phi+\hat{\phi})}=\int [\mathcal{D}\phi]_{b\Lambda} e^{-\int d^dx\, \mathcal{L}(\phi)}\int [\mathcal{D}\hat{\phi}]_\Lambda e^{-\int d^dx(\frac{1}{2}(\partial\hat{\phi})^2+\frac{1}{2}m^2\hat{\phi}^2+\frac{\lambda}{6}\phi^3\hat{\phi}+\cdots)}$$  

$\bullet$ Assume $m$, $\lambda$ small, & expand order by order

$\bullet$ Integrate out $\hat{\phi}$, term by term.

$\rightarrow$ Recast everything as an exponential for the low $E$ dof.

$\Rightarrow$ $Z=\int [\mathcal{D}\phi]_{b\Lambda}e^{-\int d^dx\, \mathcal{L}_{eff}(\phi)}$ $$\mathcal{L}_{eff}=\frac{1}{2}(\partial_\mu \phi)^2+\frac{m^2}{2}\phi^2+\frac{\lambda}{4!}\phi^4+\frac{1}{2}\mu^2\phi^2+\frac{\xi}{4!}\phi^4+\Delta C(\partial_\mu\phi)^2+\Delta D(\phi^6)+\cdots$$

$\bullet$ RG flow

$$Z=\int [\mathcal{D}\phi]_{b\Lambda}e^{-\int \mathcal{L}_{eff}}$$ rescale distance $$x'=xb\quad k'=k/b\quad |k'|<\Lambda$$ $$\int d^dx\, \mathcal{L}_{eff}=\int d^dx'\, b^{-d}(\frac{1}{2}(1+\Delta z)b^2(\partial_\mu\phi)^2+\frac{1}{2}(m^2+\Delta m^2)\phi^2+\cdots)$$ & Rescale $\phi\Rightarrow \phi'=(b^{2-d}(1+\Delta Z))^{1/2}\phi$ $$\int d^dx\, \mathcal{L}_{eff}=\int d^dx(\frac{1}{2}(\partial_\mu \phi)^2+\frac{1}{2}m'^2\phi^2+\frac{\lambda'}{4!}\phi^4+\cdots)$$ when $$m'^2=(m^2+\Delta m^2)(1+\Delta Z)^{-1}b^{-2}$$ $$\lambda'=(\lambda+\Delta\lambda)(1+\Delta Z)^{-2}b^{d-4}$$ $$c'=(c+\Delta c)(1+\Delta Z)^{-3}b^d$$ $$\Rightarrow Z=\int [\mathcal{D}\phi]_\Lambda e^{-S'_{eff}}$$  

$\bullet$ $b: S\rightarrow S'_{eff}$ The process of integrating out high-E doe =transform of $\mathcal{L}$. 

$\bullet$ $b=1-\delta$ $\delta\ll 1$ $\Rightarrow$ "RG flow"

Scale-invariant theory

eg) Gaussian fixed point : $$\mathcal{L}_0=\frac{1}{2}(\partial_\mu\phi)^2$$ $\rightarrow$ does not RG flow "fixed point": Scale invariant theory

Consider $\mathcal{L}$ close to $\mathcal{L}_0$. $$m'^2=m^2b^{-2}$$ $$\lambda'=\lambda b^{d-4}$$ $$c'=cb^d$$ when $b=1-\delta$, $\delta\ll 1$.

3 categories of couplings

$\bullet$ $(\ )\times b^{#<0}$ $\Rightarrow$ grows under RG flow $\Rightarrow$ away from the FP "relevant" ex) $\phi^2$

$\bullet$ $(\ )\times b^{#>0}$ $\Rightarrow$ decays along RG. $\Rightarrow$ "irrelevant" ex) $\phi^4$ for $d\ge 4$

$\bullet$ $(\ )\times b^0$ $\Rightarrow$ "marginal" 

For $m$, $\lambda$, when $d\ge 4$, $m$ bigger and bigger, $\lambda$ smaller and smaller. We assume there is fixed point in high $m$.

For $d<4$, $\phi^4$ is relevant, $m$ and $\lambda$ grow. We assume another fixed point in high $\lambda$. This called Wilson-Fisher Fixed Point, which is non-trivial CFT.

[3/13]

Conformal transformations

$\mathcal{M}$ is spacetime, $g_{\mu\nu}$ is metric on $\mathcal{M}$ then $$ds^2=g_{\mu\nu}(x)dx^\mu dx^\nu$$  

Def: Conformal transformation of coordinates is an invertible map $x\rightarrow x'(x)$ which leaves the metric invariant up to a local rescaling $$g'_{\mu\nu}(x')=\Lambda(x)g_{\mu\nu}(x)$$

Remarks 

i) $\Lambda(x)=1$ $\Rightarrow$ isometry

eg) If $\mathcal{M}=\mathbb{R}^{3,1}$, $g_{\mu\nu}=\eta_{\mu\nu}$ isometries = Poincare group

ii) $\Lambda(x)=$const. "scale transformation" or dilatations $$x^\mu\rightarrow \lambda x^\mu\quad \lambda\in\mathbb{R}>0$$

For an infinitesimal coord. tr. $$x^\mu\rightarrow x'^\mu=x^\mu+\epsilon^\mu(x)$$ $$g_{\mu\nu}\rightarrow g'_{\mu\nu}=\left( \frac{\partial x^\alpha}{\partial x'^\mu}\right)\left(\frac{\partial x^\beta}{\partial x'^\nu}\right)g_{\alpha\beta}=g_{\mu\nu}-(\partial_\mu \epsilon_\nu+\partial_\nu\epsilon_\mu)+O(\epsilon^2)$$ Let $$\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu=f(x)g_{\mu\nu}$$ $$2\partial^\mu\epsilon_\mu=f(x)\eta^{\mu\nu}\eta_{\mu\nu}$$ (we set Euclidean $g_{\mu\nu}=\eta_{\mu\nu}$ and $\eta^{\mu\nu}\eta_{\mu\nu}=d$) $$\Rightarrow f(x)=\frac{2(\partial\cdot \epsilon)}{d}$$ $$\partial_\rho\partial_\mu\epsilon_\nu+\partial_\rho\partial_\nu\epsilon_\mu=(\partial_\rho f)\eta_{\mu\nu}$$ we can get $$2\partial_\mu\partial_\nu\epsilon_\rho=\eta_{\mu\rho}\partial_\nu f+\eta_{\nu\rho}\partial_\mu f-\eta_{\mu\nu}\partial_\rho f$$ $$2\partial^2\epsilon_\mu=(2-d)\partial_\mu f$$ $$(2-d)\partial_\mu \partial_\mu f=2\partial^\rho\partial_\rho(\partial_\nu \epsilon_\mu)=\eta_{\mu\nu}\partial^2 f$$ finally, $$(1-d)\partial^2 f=0$$

Assume $d>2$, let   $$\partial_\mu\partial_\nu f=0$$ and $$f=A+B_\mu x^\mu$$ $$\epsilon_\mu =a_\mu+b_{\mu\nu}X^\nu+C_{\mu\nu\rho}x^\nu x^\rho$$ we put this in $\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu=f(x)g_{\mu\nu}$ then we can find

- $a_\mu$ no constraints $\rightarrow$ translation ($\dim=d$) 

- $b_{\mu\nu}=\alpha\eta_{\mu\nu}+m_{\mu\nu}$ when $\alpha$ is scale tr. and $m_{\mu\nu}=-m_{\nu\mu}$ Lorentz tr. 

- $C_{\mu\nu\rho}=\eta_{\mu\rho}b_\nu+\eta_{\mu\nu}b_\rho-\eta_{\nu\rho}-\eta_{\nu\rho}b_\mu$ when $b_\mu$ vector 

conclusion, $$x^\mu\rightarrow x^\mu +a^\nu\quad\dim=d$$ $$x^\mu\rightarrow (1+\alpha) x^\mu \quad\dim=1$$ $$x^\mu\rightarrow M^\mu_\nu x^\nu\quad\dim=\frac{1}{2}d(d-1)$$ $$x^\mu\rightarrow x^\mu+2(x-\cdot b)x^\mu-b^\mu x^2\quad\dim=d$$ last term called special conformal transformation. In finite version(not infinitesimal) $$x^\mu=\frac{x^\mu-b^\mu x^2}{1-2b\cdot x+b^2x^2}$$  

# of generators = $\dim$ of conformal group = $\frac{1}{2}(d+1)(d+2)$

If we define inversion, $$I:x^\mu\rightarrow x'^\mu=\frac{x^\mu}{x^2}$$ then $I^2=1$ then $$SCT=IT_bI$$  

Conformal Group

$$x\rightarrow x'=x'(x)$$ $$\Phi(x)=\Phi'(x')$$ $$x'^\mu=x^\mu+\omega)a\frac{\delta x^\mu}{\delta \omega^a}\quad |\omega_a|\ll 1$$ $$\delta_\omega\Phi\equiv \Phi'(x)-\Phi(x)=-i\omega_a G_a\Phi$$ $$\Rightarrow iG_a\Phi=\frac{\delta x^\mu}{\delta \omega_a}\partial_\mu\Phi$$  

eg) Translation $\frac{\delta x^\mu}{\delta x^\nu}=\delta^\mu_\nu$ $\Rightarrow$ Generator $P_\mu=-i\partial_\mu$

eg) Lorentz tr. $x'^\mu=x^\mu+\omega^\mu_\nu x^\nu=x^\mu+\omega_{\rho\sigma}\eta^{\rho\mu}x^\rho$ ($\omega_{\rho\sigma}=-\omega_{\sigma\rho}$

then $$\frac{\delta x^\mu}{\delta \omega_{\rho\sigma}}=\frac{1}{2}(\eta^{\rho\mu}x^\sigma-\eta^{\rho\mu}x^\rho)$$ then $$L_{\mu\nu}=i(X_\mu\partial_\nu-X_\nu\partial_\mu)$$ other things are $$D=-ix^\mu\partial_\mu$$ $$SCT=K_\mu=-i(2x^\mu x^\nu \partial_\nu-x^2\partial_\mu)$$  

Conformal algebra

Generators $D,P_\mu,K_\mu,L_{\mu\nu}$ satisfies commutation relation $$[D,L_{\mu\nu}]=[P_\mu,P_\mu]=[K_\mu,K_\nu]=0$$ $$[D,P_\mu]=iP_\mu\\ [D,K_\mu]=-iK_\mu\\ [K_\mu,P_\nu]=2i(\eta_{\mu\nu}D-L_{\mu\nu})\\ [L_{\mu\nu},P_\rho]=-i(\eta_{\mu\rho}P_\nu-\eta_{\nu\rho}P_\mu)\\ [L_{\mu\nu},K_\rho]=-i(\eta_{\mu\rho}K_\nu-\eta_{\nu\rho}K_\mu)\\ [L_{\mu\nu},L_{\rho\sigma}]=i(L_{\mu\rho}\eta_{\nu\sigma}+(\mbox{perm}))$$

Defining $$J_{\mu,\nu}=L_{\mu\nu}\\ J_{-1,\mu}=\frac{1}{2}(P_\mu-K_\mu)\\ J_{-1,0}=D\\ J_{0,\mu}=\frac{1}{2}(P_\mu+K_\mu)\\ a,b=-1,0,1,\cdots,d$$ $$J_{a,b}=-J_{b,a}$$ $J_{a,b}\Rightarrow$generates the Lorentz group in $(d+1,1)$ $$SO(d+1,1)\quad (\mbox{Euclidean})\\ SO(d,2)\quad (\mbox{Minkowski})$$  

Is scale invariance (Inv. in $D$) gives conformal invariance(Inv. in $D,K_\mu$)? With Poincare symmetry, yes.

[Homework 1]

[3/18]

Conformal Algebra

Conformal Algebra $\supset$ Scale symm. gp $\simeq$ Poincare group $\ltimes$ $\{D\}$

$\bullet$ Any unitary, Poincare inv, QFT w/ a scale symmetry $\rightarrow$ Conformal symmetry

(proven in $d=2,3,4$ but there is conterexample otherwise.)

Action on operators /fields

Any symmetry should be realized as operators acting on $\mathcal{H}$ (Schrodinger) or on local operators (Heisenberg picture)

$$\phi_\alpha(x)=e^{-ip\cdot x}\phi_\alpha(0)e^{ip\cdot x}$$ $$\partial_\mu \phi_\alpha(x)=e^{-ip\cdot x}(-iP_\mu \phi_\alpha(0)+\phi_\alpha(0)(iP_\mu)e^{-p\cdot x}=-i[P_\mu,\phi_\alpha(x)]$$

$\bullet$ $[P_\mu,\phi_\alpha(x)]=i\partial_\mu\phi_\alpha(x)$

other generators? $L_{\mu\nu},D,K_\mu$ $\rightarrow$ leave origin inv.

$\bullet$ Consider $\Phi_\alpha(0)$. satisfying $$[D,\Phi_\alpha(0)]=i\Delta \Phi_\alpha(0)$$ $$[L_{\mu\nu},\Phi_\alpha(0)]=i(S_{\mu\nu})_\alpha^\beta \Phi_\beta(0)$$ ($\Phi_\alpha$ is a irreducible representation of $SO(d)$) $$[K_\mu,\Phi_\alpha(0)]=0$$

$\Delta$: Scaling dim. of $\Phi_\alpha$

$S_{\mu\nu}$: irred. rep of Lorentz gp that acts on $\Phi_\alpha$

then $\Phi_\alpha(x)$ primary operator/field

then $\Phi_\alpha(x)=e^{-ip\cdot x}\Phi_\alpha(0)e^{ip\cdot x}$

$$[D,\phi_\alpha(x)]=De^{-ip\cdot x}\phi_\alpha(0)e^{ip\cdot x}-e^{-ip\cdot x}\phi_\alpha(0)e^{ip\cdot x}D=e^{-ip\cdot x}(e^{ip\cdot x}De^{-ip\cdot x}\phi_\alpha(0)-\phi_\alpha(0)e^{ip\cdot x}De^{-ip\cdot x})e^{ip\cdot x}=e^{-ip\cdot x}[\hat{D},\phi_\alpha(0)]e^{ip\cdot x}$$ $$\hat{D}=e^{ip\cdot x}De^{-ip\cdot x}$$

$$\hat{D}=(1+ixp-\frac{(xp)^2}{2}+\cdots)D(1-ixp-\frac{(xp)^2}{2}+\cdots)=D+ix^\mu [x_\mu,D]-\frac{1}{2}x^\mu x^\nu [x_\mu,[x_\nu,D]]=D+x^\mu p_\mu$$

$$[D,\Phi_\alpha(x)]=i(\Delta+x^\mu \partial_\mu)\Phi_\alpha(x)$$

$$[L_{\mu\nu},\Phi_\alpha(x)]=-i(x_\mu\partial_\nu-x_\nu\partial_\mu)\Phi_\alpha(x)+i(S_{\mu\nu})_\alpha^\beta \Phi_\beta(x).$$

$$[K_\mu,\phi_\alpha(x)]=2ix^\mu \Delta \Phi_\alpha(x)+i(2x^\mu x^\nu \partial_\nu -x^2\partial_\mu)\Phi_\alpha(x)+2ix^\rho(S_{\rho\mu})_\alpha^\beta \Phi_\beta(x)$$

eg) $\varphi(x)$: scalar primary w/ scaling dim $\Delta$

$$\varphi(x)\rightarrow \varphi(x')=\left| \frac{\partial x'}{\partial x}\right|^{-\Delta/d}\varphi(x)$$

$$\left| \frac{\partial x'}{\partial x}\right| =\Lambda(x)^{-d/2}(\mbox{Jacobian})$$

$$\varphi(x')=\Lambda(x)^{\Delta/2}\phi(x)$$

In QM we want to know eigenstates (values of $H=P^0$)

In CFT, we focus on "eigenstates(primary)" of $D$, & eigenvalues $\Delta$

$$[D,P_\mu]=iP_\mu$$

$$[D,K_\mu]=-iK_\mu$$

Suppose $\mathcal{O}_\Delta$ is a primary $$D\cdot \mathcal{O}\equiv [D,\mathcal{O}_\Delta]=i\Delta \mathcal{O}_\Delta$$ $$D\cdot P_\mu\mathcal{O}=([D,P_\mu]+P_\mu D)\cdot \mathcal{O}=i(\Delta+1)P_\mu \mathcal{O}$$

$P_\mu \rightarrow$ raising op.

$K_\mu \rightarrow$ lower op.

"Ground state" $\rightarrow$ annihilated by $K_\mu$ $$K_\mu \mathcal{O}_\Delta=0$$ primary states/op.

$$\Rightarrow P_\mu \mathcal{O}_\Delta \rightarrow D=\Delta+1\\ P_\mu P_\nu \mathcal{O}_\Delta \rightarrow \Delta+2$$

For a given primary $\mathcal{O}_\Delta$ $\rightarrow$ can form "descendants" $$P_{\mu_1}\cdots P_{\mu_r}\mathcal{O}_\Delta \sim \partial_{\mu_1}\cdots\partial_{\mu_r}\mathcal{O}_\Delta(x)$$  

Dilatation op. can be thought of as a Hamiltonian via certain map

In $\mathbb{R}^d$, $$ds^2=dr^2+r^2d\Omega_{d-1}=r^2\left[\frac{dr^2}{r^2}+d\Omega_{d-1}\right]$$ let $t=\log r$, then $\frac{\dr^2}{r^2}=dt^2$. $$ds^2=dt^2+d\Omega_{d-1}:\mathbb{R}\times S^{d-1}$$

$\Rightarrow$ By conformal symmetry, CFT on $\mathbb{R}^d$ $\Leftrightarrow$ CFT on $\mathbb{R}_t\times S^{d-1}$

(that picture)

State-operator correspondence

$$\mathbb{R}^d\quad \mathbb{R}\times S^{d-1}\\ D\leftrightarrow H\\ \mathcal{O}_\Delta(x)\leftrightarrow |\Delta,j\rangle \in \mathcal{H},\ K^\dagger=P$$

 

[3/20]

Consequences of Conformal Symmetry

Noether's theorem: To $\forall$ continuous symmetry, $\exists$ a current $j_\mu$, $$\partial^\mu j_\mu=0$$ $$x^\mu\rightarrow x'^\mu=x^\mu+\omega_a\frac{\delta x^\mu}{\delta \omega_a}$$ $$\phi\rightarrow \phi'(x')=\phi(x)+\omega_a\frac{\delta F}{\delta \omega_a}$$ Symmetry: $\delta S=0$ when $\omega_a\rightarrow \omega_a(x)$ $$\delta S=\int d^dx\, j^\mu_a(\partial_\mu \omega^a)=-\int d^dx (\partial_\mu j^\mu_a)\omega^a=0$$ by EOM, so $\partial_\mu j^\mu_a=0$.

$$j^\mu_a=\left( \frac{\mathcal{L}}{\partial(\partial_\mu\phi)}\partial_\mu \phi-\delta^\mu_\nu \mathcal{L}\right)\times \frac{\delta x^\nu}{\delta \omega^a}-\frac{\partial \mathcal{L}}{\partial (\partial_\mu\phi)}\frac{\delta F}{\delta \omega^a}$$

$$Q_a=\int d^{d-1}_{x^0=const}x\, j^0_a$$

$$j^\mu_a\rightarrow j^\mu_{new,a}=j^\mu_a+\partial_\nu \beta^{\mu\nu}_a$$ ($\beta^{\mu\nu}=-\beta^{\nu\mu}$) we call this "improvement"

$$\partial_\mu j^\mu_{new,a}=0$$

Stress-energy tensor

=Noether current for the translation symmetry $x^\mu\rightarow x^\mu +\epsilon^\mu$

$$\frac{\delta x^\mu}{\delta x^\nu}=\delta^\mu_\nu,\quad \frac{\delta F}{\delta \epsilon}=0$$

$$T^{\mu\nu}_c=\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}\partial^\nu \phi-\eta^{\mu\nu}\mathcal{L}$$

$$\partial_\mu T^{\mu\nu}_c=0$$ $$p^\nu=\int d^{d-1}x\, T^{0\nu}$$ $$H=p^0=\int d^{d-1}T^{00}=\int d^{d-1}x\left( \frac{\partial\mathcal{L}}{\partial\phi}\dot{\phi}-\mathcal{L}\right)$$ which is Hamiltonian.

For a Poincare invariant theory, we can "improve" $$T^{\mu\nu}_c\rightarrow T^{\mu\mu}$$

such that $$T^{\mu\nu}=T^{\nu\mu}$$

$$x^\mu\rightarrow x^\mu=x^\mu+\epsilon^\mu(x)$$

$$\delta S=-\int d^dx T^{\mu\nu}(\partial_\mu \epsilon_\nu)=-\frac{1}{2}\int d^dx T^{\mu\nu}(\partial_\mu \epsilon_\nu+\partial_\nu\epsilon_\mu)$$

$T^{\mu\nu}$ generates the deformations under general coordinates transformations. 

(이거 아노말리 렉노에서 나온 $\mathacal{J}$ 랑 같은말인가?)

$\bullet$ Translation $\partial_\mu T^{\mu\nu}=0$

Poincare $T^{\mu\nu}=T^{\nu\mu}$

For a conf. tr. $$\partial_\mu \epsilon_\nu+\partial_\nu \epsilon_\mu=f(x)\eta_{\mu\nu}$$ $$\delta S=-\frac{1}{2}\int d^dx \,T^\mu_\mu f(x)$$ Scale transformations is $f(x)=\alpha=$const. $$\delta S\sim \alpha\int d^dx T^\mu_\mu=0$$ $T^\mu_\mu$ is a total derivative, $$T^\mu_\mu=\partial_\nu T^\nu$$ for some $J$.

If the theory is invariant under SCT, then $$T^\mu_\mu=0$$

$\Leftrightarrow$ Conformally invariant.

ex) free boson. $$S[\varphi]=\frac{1}{2}\int d^dx\, (\partial_\mu\varphi)(\partial^\mu \varphi)$$

$$T^{\mu\nu}_c=(\partial^\mu \varphi)(\partial^\nu \varphi)-\frac{1}{2}\eta^{\mu\nu}(\partial_\rho\varphi)(\partial^\rho\varphi)$$ $$T^{\mu\nu}_c=T^{\mu\nu}_c$$

$$T^\mu_\mu=(\partial^\mu\varphi)(\partial_\mu\varphi)-\frac{d}{2}(\partial_\rho\varphi)(\partial^\rho\varphi)=\frac{2-d}{2}(\partial^\mu\varphi)(\partial_\mu\varphi)=0$$ for $d=2$ conformal. $$=#\partiall(\varphi\partial\varphi)$$ for $d\ne 2$

Can add an improvement term s.t. $$\tilde{T}^\mu_\mu=0$$

$$T^{\mu\nu}=\frac{2}{\sqrt{g}}\left. \frac{\delta S}{\delta g^{\mu\nu}}\right|_{g^{\mu\nu}=\eta^{\mu\nu}}$$ ($g=\det g_{\mu\nu}$) $$S[\varphi,g_{\mu\nu}]=\frac{1}{2}\int d^dx\sqrt{g}g^{\mu\nu}\partial_\mu \varphi\partial_\nu\varphi$$ $$g_{\mu\nu}\rightarrow g_{\mu\nu}+\delta g_{\mu\nu}$$

Implication for correctors

Correlation functions $$\langle \phi(x_1)\phi(x_2)\cdots \phi(x_n)\rangle=\frac{1}{Z}\int [\mathcal{D}\phi]e^{-S[\phi]}\phi(x_1)\cdots \phi(x_n)$$  

$\bullet$ Suppose the action is inv. under a symmetry $S[\phi]\rightarrow S[F(\phi)]$ 

$\bullet$ $[\mathcal{D}\phi]\rightarrow [\mathcal{D}\phi']=[\mathcal{D}\phi]$ for a scale-symmetry $\mathcal{D}\phi'\ne \mathcal{D}\phi$ $\exists$ uv cut-off $\Lambda$.

at RG-fixed point, scale symmetry becomes exact!

$$\langle \phi(x_1')\phi(x_2')\cdots\phi(x_n')\rangle =\langle \phi'(x_1')\phi'(x_2')\cdots\phi'(x_n')\rangle=\langle F(\phi(x_1))F(\phi(x_2))\cdots F(\phi(x_n))\rangle$$ $F(\phi(x_1))=\phi'(x_1)$

eg) Translation $x'=x+a$, $F(\phi)=\phi$.

$$\langle \phi(x_1+a)\cdots\phi(x_n+a)\rangle=\langle \phi(x_1)\cdots\phi(x_n)\rangle$$

eg) Rotation of scalars $x'^\mu=\Lambda^\mu_\nu x^\nu$ $\phi\rightarrow \phi$ $$\langle \phi(\Lambda x_1)\cdots\phi(\Lambda x_n)\rangle=\langle \phi(x_1)\cdots\phi(x_n)\rangle$$

$\bullet$ In a CFT $\rightarrow$ all operators are primary($\phi$) of descendant ($\partial_\mu\partial_\nu\cdots\phi$. 

For primary scalars $$\langle \phi(x_1')\cdots\phi(x_n')\rangle=\left|\frac{\partial x_1'}{\partial x_1}\right|^{-\Delta^1/d}\cdots \left| \frac{\partial x_n'}{\partial x_n}\right|^{-\Delta^n/d}\times \langle \phi(x_1)\cdots\phi(x_n)\rangle$$  

$$x\rightarrow \lambda x\\ \phi(\lambda x)=\lambda^{-1}\phi(x)$$ (for $d=4$, free masless scalar.)

2-point function 

($\phi$: primary of scale dim. $\Delta_i$, which is $K_\mu\phi(0)=0,\ D\phi(0)=\Delta\phi(x)$) 

Translation: $$\langle \phi(x_1)\phi(x_2)\rangle =f(|x_1-x_2|)$$  

Scale tranformation $x \rightarrow \lambda x$ $$\langle \phi(x_1)\phi(x_2)\rangle=\lambda^{\Delta_1+\Delta_2}\langle \phi(\lambda x_1)\phi(\lambda x_2)\rangle$$ $$f(x)=\lambda^{\Delta_1+\Delta_2}f(\lambda x)$$

$$\langle \phi(x_1)\phi(x_2)\rangle=\frac{C_{12}}{|x_1-x_2|^{\Delta_1+\Delta_2}}$$

SCT $\vec{x}'_\mu=\frac{\vec{x}_\mu-\vec{b}_\mu x^2}{1-2(b\cdot x)+b^2x^2}$ $|x'_1=x'_2|=\frac{|x_1-x_2|}{\gamma^{1/2}_1\gamma^{1/2}_2}$,  $\gamma_i\equiv 1-2b\cdot x_i+b^2x_i^2$: $$\langle \phi(x_1)\phi(x_2)\rangle =\left| \frac{\partial x_1'}{\partial x_1}\right|^{\Delta_1/d}\left| \frac{\partial x_2'}{\partial x_2}\right|^{\Delta/2}\langle \phi(x_1')\phi(x_2')\rangle=\frac{1}{\gamma_1^{\Delta_1}}\frac{1}{\gamma_2^{\Delta_2}}\langle \phi(x_1')\phi(x_2')\rangle$$ $$\frac{c_{12}}{|x_1-x_2|^{\Delta_1+\Delta_2}}=\frac{1}{\gamma_1^{\Delta_1}\gamma_2^{\Delta_2}}\frac{c_{12}}{|x_1'-x_2'|^{\Delta_1+\Delta_2}}=\frac{\gamma_1\gamma_2)^{(\Delta_1+\Delta_2)/2}}{\gamma_1^{\Delta_1}\gamma_2^{\Delta_2}}\frac{c_{12}}{|x_1-x_2|^{\Delta_1+\Delta_2}}$$

$$\langle \phi(x_1)\phi(x_2)\rangle =\begin{cases}\frac{1}{|x_1-x_2|{2\Delta_1}}& (\Delta_1=\Delta_2)\\ 0& (\Delta_1\ne \Delta_2)\end{cases}$$ $c_{12}$ can be normalized to 1.

3-point function

$$\langle \phi(x_1)\phi(x_2)\phi(x_3)\rangle=\frac{c_{123}}{|x_{12}|^a|x_{23}|^b|x_{31}|^c}$$ when $x_{ij}=x_i-x_j$.

scale-invariance $\Rightarrow a+b+c=\Delta_1+\Delta_2+\Delta_3$

SCT $$\Rightarrow a=\Delta_1+\Delta_2-\Delta_3,\ b=\Delta_2+\Delta_3-\Delta_1,\ c=\Delta_3+\Delta_1-\Delta_2$$ and $c_{123}\rightarrow$ not fixed by symmetry! contains dynamics of CFT. 

and all the n-point function determined by 3-point function.

[3/25]

CFT data is made from spectrum of primary operators $(\Delta,l)$ and OPE coefficients.

Spin-1 field 2 point function

$$\langle J_\mu(x)J_\nu(y)\rangle=\frac{\alpha_{\mu\nu}(x-y)}{|x-y|^{2\Delta}}$$

where $\alpha_{\mu\nu}$: symmetric $\mu\leftrightarrow \nu$.

$$\alpha_{\mu\nu}(x)=\eta_{\mu\nu}+\alpha\frac{X_\mu X_\nu}{X^2}$$  

$$J'(x')=\frac{\partial x^\nu}{\partial x'^\mu}J_\nu(x)$$

$$\langle_\mu(x')J_\nu(y')\rangle=\langle J'_\mu(x')J'_\nu(y')\rangle$$ $$\rightarrow \alpha=2\, \alpha_{\mu\nu}(x)=I_{\mu\nu}(x)=\eta_{\mu\nu}-2\frac{x_\mu x_\nu}{x^2}$$ where $I_{\mu\nu}$ is invariant tensor. ($I_{\mu\nu}I^{\nu\rho}=\delta^\rho_\mu$) 

Conserved currents

If $J_\mu$ is a conserved current $$\partial^\mu J_\mu=0$$ $$\Rightarrow \langle \partial^\mu_x  J_\mu(x)J_\nu(y)\rangle=0$$ $$\partial^\mu_x\left( \frac{I_{\mu\nu}(x-y)}{|x-y|^{2\Delta}}\right)=0\quad\Rightarrow \Delta=d-1$$ generally, $\Delta\ge d-1$, so $\partial^\mu J_\mu=0\Leftrightarrow \Delta=d-1$

For a scalar primary $$\Delta_\phi\ge \frac{d-2}{2}$$ then $$\Delta=\frac{d-2}{2}\Leftrightarrow \partial^2\phi=0$$

then $\phi$ is a free scalar.

EM tensor 2 point function

$$\langle T_{\mu\nu}(x)T_{\rho\sigma}(y)\rangle=\frac{c}{|x-y|^{2\Delta}}\left(\frac{1}{2}I_{\mu\sigma}(x-y)I_{\nu\rho}(x-y)+I_{\mu\rho}(x-y)I_{\nu\sigma}(x-y)-\frac{1}{d}\eta_{\mu\nu}\eta_{\rho\sigma}\right)$$ $c$ is unambiguous "central charge" or "weyl anomaly". This measure DoF.

$$\partial_\mu T^{\mu\nu}=0\quad \Leftrightarrow \Delta=d$$

4-point function

$$\langle \phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\rangle=\frac{I(\mu,\nu)}{\prod_{i<j}|x_{ij}|^{\delta_{ij}}}$$ where $\sum_{j\ne i}\delta_{ij}=\Delta_i$ 

corss-ratios $$u=\frac{x_{12}^2x_{34}^2}{x_{13}^2x_{24}^2},\ v=\frac{x_{14}^2x_{23}^2}{x_{13}^2x_{24}^2}$$

Radial Quantization

In a QFT, spacetime is foliation of time sheets $\Sigma_{t_i}$, Hilbert space defined each slice $\mathcal{H}_{\Sigma_i}$. Let incoming state $O_1,O_2,O_3$, denote as $|\psi_{in}\rangle$ and out state $O'_1,O'_2,O'_3$ denote as $\langle \psi_{out}|$ then Correlators is $$\langle O'\cdots O\rangle\sim \langle \psi_{out}|u|\psi_{in}\rangle,\quad u=e^{-iP_0\Delta t}$$ when Hamiltonian $P_0$ commute with any quantum number.

In CFT, sheets become spheres $\Sigma$. (Euclidian space) In CFT, we foliate the space by $S^{d-1}$ with the origin at the center.

In Hilbert space $\mathcal{H}_\Sigma$, in state $|\psi_{in}\rangle$ means operators $O_1,O_2,O_3$ are in $\Sigma$, out state $\langle \psi_{out}|$ means operators $O'_1,O'_2,O'_3$ are out $\Sigma$. Then correlator is $$\langle \psi_{out}|u|\psi_{in}\rangle,\quad u=e^{iD\Delta \tau},\ \tau=\log r$$ all states on $\mathcal{H}_\Sigma$ are classified in terms of $(\Delta,l)$. $$D|\Delta,l\rangle=i\Delta |\Delta,l\rangle\\ L_{\mu\nu}|\Delta,l\rangle_\alpha=i(S_{\mu\nu})^\alpha_\beta|\Delta,l\rangle_\beta$$ when $S_{\mu\nu}$ is generators of angular momentum operator $L_{\mu\nu}$. also which is equal to primary operator.

[4/1]

State operator correspondence

$$\psi\in \mathcal{H}(S^{d-1})$$  

$\leftrightarrow$ inserting op. inside $S^{d-1}$

ex) $|0\rangle$ vacuum state ($\Delta=0\ l=0$)  $\leftrightarrow$ no insertion (annihilated by $\forall$ generators $P_\mu,\ K_\mu,L_{\mu\nu},D$)

ex) Insert a primary op. at the origin

$$\phi_\Delta(0)\ \leftrightarrow \ |\Delta\rangle=\phi_\Delta(0)|0\rangle$$

$$D|\Delta\rangle=D\phi_\Delta(0)|0\rangle\\ =[D,\phi_\Delta(0)]|0\rangle\\ =i\Delta\phi_\Delta(0)|0\rangle\\ =i\Delta |\Delta\rangle$$ $|\Delta\rangle$ is an eigentate of $D$.

WTS: $|\Delta\rangle$ is a primary state. ($K_\mu|\Delta\rangle=0$)

$$K|\Delta\rangle=K\phi_\Delta(0)|0\rangle=[K,\phi_\Delta(0)]|0\rangle=0$$

$|\Delta,z\rangle=\phi_\Delta(z)|0\rangle$ Not a primary.

Because $$|psi\rangle=\phi_\Delta(x)|0\rangle=e^{-ixp}\phi_\Delta(0)e^{ixp}|0\rangle\\ =e^{-ixp}\phi_\Delta(0)|0\rangle=(1-ixp+\cdots)\phi_\Delta(0)|0\rangle\\ =|\Delta\rangle-ix\cdot p|\Delta\rangle+\cdots$$

$$D(P_\mu|\Delta\rangle)=([D,P_\mu]+P_\mu D)|\Delta\rangle=i(\Delta+1)P_\mu|\Delta\rangle$$

Descendant state of $|\Delta\rangle$

$P_\mu|\Delta\rangle$ is a Not primary state. ($K_\nu P_\mu|\Delta\rangle\ne 0$)

Because $$K_\nu P_\mu|\Delta\rangle=([K_\nu,P_\mu]+P_\mu K_\nu)|\Delta\rangle=[K_\nu,P_\mu]|\Delta\rangle\ne 0$$

$P_\mu$ on $|\Delta\rangle$ $\rightarrow$ scaling dim $=\Delta+1$

$P_\mu P_\nu$ on $|\Delta\rangle$ $\rightarrow$ $\Delta+2$

Looks like $P$ is raising op, $K$ is lowering op. and $|\Delta\rangle$ is ground state.

Operator Product Expansion

$$O_1(x)O_2(y)=\sum (x-y)^n\partial_n O(x)\sim \sum_j C^{(x,y)}_{Rj}O_{Rj}(x)$$

In CFT, $|\psi\rangle=\phi_1(x)\phi_2(0)|0\rangle$ $$|psi\rangle=\sum_n C_n(x)|E_n\rangle$$ where $|E_n\rangle$ is eigenstate of $D$.

$|E_n\rangle$ is a linear conbination of primaries & descendants $$\phi_1(x)\phi_2(0)|0\rangle=\sum C_\Delta(x,\partial)\phi_\Delta(0)|0\rangle$$ for primary operators $\phi_\Delta$. This is more constraint than generic QFT. 

$$\phi_1(x)\phi_2(0)|0\rangle=\frac{const}{|x|^k}(\phi_\Delta(0)+\cdots)|0\rangle$$

$$D\phi_1(x)\phi_2(0)|0\rangle=i(\Delta_1+x^\mu\partial_\mu)\phi_1(x)\phi_2(0)|0\rangle$$

$$[D\phi_1(x)]\phi_2(0)|0\rangle+\phi_1(x)D\phi_2(0)=i(\Delta_1+x^\mu\partial_\mu)\phi_1(x)\phi_2(0)|0\rangle+\phi_1(x)(i\Delta_2)\phi_2(0)|0\rangle=i(\Delta_1-\Delta_2-k)\frac{const}{|x|^k}(\cdots)$$

$$D(RHS)=i\Delta \frac{const}{|x|^k}(\phi_\Delta(0)+\cdots)|0\rangle\\ k=\Delta_1-\Delta_2+\Delta$$ In conclusion, we can fix primary part (other parts are descendents) by OPE $$\phi_1(x)\phi_2(0)=\frac{consst}{|x|^{\Delta_1-\Delta_2+\Delta}}(\phi_\Delta(0)+\alpha x^\mu \partial_\mu \phi_\Delta+\cdots)$$

We can fix first descendent term from $$[K_\mu,\phi_\Delta(x)]=2ix_\mu\Delta \phi_\Delta(x)+i(2x_\mu x^\nu\partial_\nu-x^2\partial_\mu)\phi_\Delta(x)$$ $$[K_\mu,\phi_1(x)]\phi_2(0)=ix_\mu(\Delta_1-\Delta_2+\Delta)\frac{const}{|x|^{\Delta_1+\Delta_2-\Delta}}\times (\phi_\Delta(0)+\cdots)|0\rangle$$

$$K_\mu(RHS)=\cdots$$ $$\alpha=\frac{\Delta_1-\Delta_2+\Delta}{2\Delta}$$

Another way to computing is $$C_\Delta(x,\partial)$$ by 3-point function $$\langle\phi_1(x)\phi_2(x)\phi_\Delta(z)\rangle=\sum_{\Delta'\in prim}C_{R\Delta'}C_{\Delta'}(x,\partial)\langle \phi_{\Delta'}(y)\phi_\Delta(z)\rangle$$ $\Delta=\Delta'$ then $$=C_{12\Delta}C_\Delta(x,\partial)\frac{1}{|x-y|^{2\Delta}}$$ $$=\frac{C_{12\Delta}}{x^{\Delta_1+\Delta_2-\Delta}z^{\Delta_2+\Delta-\Delta_1}-|x-z|^{\Delta_1+\Delta-\Delta_2}}$$

[4/8]

OPE in CFT. $$|\psi\rangle\in \mathcal{H}_{S^{d-1}}$$ $$|\psi\rangle=\phi_1(x)\phi_2(0)|0\rangle$$ $$|\psi\rangle=\sum_n c_n(x)|E_n\rangle$$ $$\phi_1(x)\phi_2(0)   =\sum_{\phi\in prim\ in \phi_1\times \phi_2}C_\Delta(x,\partial)\phi_\Delta(0)$$ "Operator algebra" $C_\Delta$ is fixed by conformal symmetry 

Consider $$\langle \phi_1(x)\phi_2(0)\phi_\Delta(z)\rangle=\sum_{\Delta'\in \phi_1\times \phi_2}C_{12\Delta'}C_{\Delta'}(x,\partial_y)\langle \phi_{\Delta'}(y)\phi_\Delta(z)\rangle|_{y=0}$$ where $\langle \phi_{\Delta'}(y)\phi_\Delta(z)\rangle|_{y=0}=\delta_{\Delta \Delta'}\times \frac{1}{|y-z|^{2\Delta}}$

LHS is $$\frac{C_{12\Delta}}{|x|^{\Delta_1+\Delta_2-\Delta}|z|^{\Delta_2+\Delta-\Delta_1}|x-z|^{\Delta_1+\Delta-\Delta_2}}$$ and RHS is $$\left.\frac{C_\Delta(x,\partial_y)}{|y-z|^{2\Delta}\right|_{y=0}$$

Look at 4-point function

scalar primary with scaling dimension $\Delta$ $$\langle \phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\rangle=\frac{G(u,v)}{|x_{12}^{2\Delta_E}|x_{34}|^{2\Delta E}}$$ where $$u=\frac{x_{12}^2x_{34}^2}{x_{13}^2x_{24}^2}\quad v=\frac{x_{14}^2x_{23}^2}{x_{12}^2x_{34}^2}$$ are invariant of conformal group.

We have to fix $G(u,v)$. $$\phi(x_1)\phi(x_2)=\sum_{\Delta \in \phi\times \phi}c_\Delta C_\Delta (x_{12},\partial_y)\phi_\Delta(y)|_{y=\frac{x_1+x_2}{2}}$$ $$\phi(x_2)\phi(x_4)=\sum_{\Delta \in \phi\times \phi}c_\Delta C_\Delta (x_{34},\partial_z)\phi_\Delta(z)|_{z=\frac{x_3+x_4}{2}}$$ $c_\Delta$: "OPE coeff": $\phi_{\Delta_E}\phi_{\Delta_#}\phi_\Delta$

$$\langle \phi\phi\phi\phi\rangle= \sum_{\Delta\in\phi\times\phi}C_\Delta^2[C_\Delta(x_{12},\partial_y)C_\Delta(x_{34},\partial_z)\langle \phi_\Delta(y)\phi_\Delta(z)\rangle |_{y=\frac{x_1+x_2}{2},z=\frac{x_3+x_4}{2}}]$$ we just need $C_\Delta$. other things are completely fixed by symmetry. $$=\sum_\Delta C_\Delta^2 \frac{G_{\Delta,l}(u,v)}{|x_{12}|^{2\Delta_E}|x_{34}|^{2\Delta_E}}$$ and $G_{\Delta,l}(u,v)$ are completely fixed by symmetry and it called 4-point conformal block. this depends op "external ops" and intermediate ops.

$C^2_\Delta$ and $\Delta\in\phi\times \phi$ are data from 3-point functions.

2-dim CFT

infinitesimal coordinate transformation $$x\rightarrow {x'}^\mu=x^\mu+\epsilon^\mu(x)$$ $$g_{\mu\nu}\rightarrow g_{\mu\nu}'=\frac{\partial x^\alpha}{\partial {x'}^\mu}\frac{\partial x^\beta}{\partial {x'}^\nu}g_{\alpha\beta}=(\delta^\alpha_\mu-\partial_\mu \epsilon^\alpha)(\delta^\beta_\nu-\partial_\nu\epsilon^\beta)\eta_{\alpha\beta}=\eta_{\mu\nu}-(\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu)+O(\epsilon^2)$$

Conformal tr: $$g_{\mu\nu}\rightarrow g_{\mu\nu}'=\Lambda(x)g_{\mu\nu}$$ when $\Lambda(x)\simeq 1-f(x)$ $$g^{\mu\nu}(\partial_\mu\epsilon_\nu+\partial+\nu\epsilon_\mu)=(f(x)g_{\mu\nu})g^{\mu\nu}$$ $$2\partial_\mu\epsilon^\mu=f(x)d$$

$$\partial_\mu\epsilon_\nu+\partial_\nu\epsilon_\mu=\frac{2}{d}(\partial_\rho\epsilon^\rho)\eta_{\mu\nu}$$  

when $d=2$, $$\partial_0\epsilon_0=\partial_1\epsilon_1,\quad \partial_0\epsilon_1=-\partial_1\epsilon_0$$ "Cauchy-Riemann eqn."

$z\equiv x^0+ix^1,\quad \bar{z}=x^0-ix^1$

$\partial\equiv \partial_z=\frac{1}{2}(\partial_0-i\partial_1)$ 

$\bar{\partial}\equiv \partial_{\bar{z}}=\frac{1}{2}(\partial_0+i\partial_1)$

$\epsilon(z)=\epsilon^0+i\epsilon^1$, $\bar{\epsilon}(z)=\epsilon^0-i\epsilon^1$ then $\epsilon(z)$ is arbitrary holomorphic, $\bar{\epsilon}(z)$ is anti-holomorphic.

2d confirmal tr. consider w/ analytic coord tr. $$z\rightarrow z'=f(z)\\ \bar{z}\rightarrow \bar{z}'=\bar{f}(\bar{z})$$

$$\Rightarrow$ $\infty$-dim'l algebraic of conformal tranformation.

Consider $z'=z+\epsilon(z)$ when $$\epsilon(z)=\sum_{n\in\mathbb{Z}} C_nz^{n+1}$$  

$\phi(z,\bar{z))\rightarrow$ spin-less $\Delta=0$

$$\phi'(z',\bar{z}')=\phi(z,\bar{z})$$ $$\delta\phi=\phi'(z,\bar{z})-\phi(z,\bar{z})\\ =\phi(z-\epsilon,\bar{z}-\epsilon)-\phi(z,\bar{z})\\ \sum_n(c_n l_n+\bar{c}_n\bar{l}_n)\phi(z,\bar{z})$$ with $$l_n\equiv -z^{n+1}\partial_z\\ \bar{l}_n\equiv -\bar{z}^{n+1}\partial_\bar{z}$$

Witt algebra: $$[l_m,l_n]=(m-n)l_{m+n}$$ $$[\bar{l}_m,\bar{l}_n]=(m-n)\bar{l}_{m+n}$$ $$[l_m,\bar{l}_n]=0$$ Conf. algebra in 2d = Widd $\oplus$ Witt (Classically)

$\{l_n,\bar{l}_n\}$ generates "local" conformal algebra.

Not globally well-defined on the Riemann sphere. $$S^2=\mathbb{C}\cup \{\infty\}$$

[Homework 2]

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.

Sol)

$$\begin{align}T^{\mu\nu}_c=-\eta^{\mu\nu}\mathcal{L}+\frac{\partial\mathcal{L}}{\partial(\partial_\mu\Phi)}\partial^\nu\Phi\end{align}$$ Free boson is $$\mathcal{L}=\partial_\mu \varphi\partial^\mu\varphi$$ Then canonical energy momentum tensor is $$T^{\mu\nu}_c=-\eta^{\mu\nu}\partial_\rho\varphi\partial^\rho\varphi+\partial_\mu\varphi\partial^\nu\varphi$$ Spin is 0. We can make Belinfante tensor and virial: 

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.

Sol)

Belinfante tensor and virial:

3. Prove the following property under special conformal transformations $$|{x'}_i-{x'}_j|=\frac{|x_i-x_j|}{\gamma^{1/2}_i\gamma^{1/2}_j}$$ where $\gamma_i=1-2b\cdot x_i+b^2x_i^2$.

Sol)

4. Consider the inverse 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 $(x')^\mu=x^\mu/x^2$, $(y')^\mu=y^\mu/y^2$.

Reference

(Song, Jaewon)[PH754] Advanced Particle Physics<Conformal and Topological Field Theory>