Conformal Field Theory

  1. 2月前

    Phantom_Ghost

    1楼 10月17日 数学版主, 物理版主, 优秀回答者
    2月前Phantom_Ghost 重新编辑

    Disclaimer

    The content is attributed to the lecture course TVI/TMP-MD4 Conformal Field Theory from LMU during WS-18/19. In this post, the lecture note and the excercises with solutions will be presented. Any repost and copy for business commercial purposes are forbiddened. All rights are reserved.

    images.png

  2. Phantom_Ghost

    2楼 10月17日 数学版主, 物理版主, 优秀回答者
    2月前Phantom_Ghost 重新编辑

    1. Foreword

    Conformal field theory, abbreviated as CFT...

    What is it ?
    ---- translation, rotation, dilation.....symmetry transformation that keep certain (confromal) property;
           fields or statistical systems that have such symmetry.

    Why ?
    ---- 1) critical phenomena, second phase transition...
           2) constructive QFT at work in 2D

    \[\]

    Content:

    1. Foreword
    2. Conformal transformation
    3. Critical phenomenon
    4. Scale invariance vs. conformal invariance
    5. Introduction to CFT
    6. Primary field
    7. Radial quantization
    8. Conformal bootstrap
    9. CFT in 2D
    10. Virasoro algebra, Verma module
    11. Minimal models
    12. Boundary CFT
    13. Kac-Moody algebra, Superconformal symmetry

    \[\]

    References can be found in the link of the course.

  3. Phantom_Ghost

    3楼 10月18日 数学版主, 物理版主, 优秀回答者
    2月前Phantom_Ghost 重新编辑

    2. Conformal transformation

    Continous map between manifolds: $\varphi: M\to N$

    1.jpg

    Pushforward: $d\varphi:T_pM\to T_{\varphi(p)}N$ (also denoted as $\varphi_*$)

    Diffeomorphism: $\varphi$ bijective and differentiable s.t. its inverse map $\varphi^{-1}$ is also differentiable.

    Assume now manifolds are equipped with a metric, i.e. $(M,\tilde{g})$, $(N,g)$, then
    \begin{align*}
    (\varphi^*g)_p(x,y)&=g_{\varphi(p)}(d\varphi(x),d\varphi(y))\,,\,\forall x,y\in T_pM\\
    \tilde{g}_{\alpha\beta}&=(\varphi^* g)_{\alpha\beta}=g_{\mu\nu}({\varphi(p)})\frac{\partial\varphi^\mu(p)}{\partial x^\alpha}\frac{\partial\varphi^\nu(p)}{\partial x^\beta}
    \end{align*}
    ($\varphi^*:T_{\varphi(p)}^*N\to T_p^* M$ is the pullback of $\varphi$)

    Definition (conformal diffeomorphism):
    Specially let $M=\mathbb{R}^n$, $\varphi: U\to V$ where $U,V\subset M$ s.t. $(\varphi^* g)_p(x,y)=\Omega^2(p)g_p(x,y)$$^{(!)}$ $\forall x,y\in T_pM=\mathbb{R}^n$ ($\Omega>0$ is called scale factor).
    $(!)$:Here in flat sapce $\mathbb{R}^n$, metric is $\delta$ hence $g_p=g_{\varphi(p)}=\delta$

    Example: Isometries of $\mathbb{R}^n$ are conformal diffeomorphism with $\Omega=1$.

    2.jpg

    Infinitesimal version:
    Choose a coordinate $\{x^\mu\}$, $x^\mu\to x^\mu+\varepsilon f^\mu(x)$, $f^\mu$ is a infinitesimal conformal diffeomorphism, iff $\boxed{f_{\mu;\nu}+f_{\nu;\mu}=(\Omega^2-1)g_{\mu\nu}}$ namely the "conformal Killing eq." holds ($f_{\mu;\nu}$ is covariant derivative of $f_\mu$; when in flat spacetime, i.e. $g_{\mu\nu}=\eta_{\mu\nu}$ or $\delta_{\mu\nu}$, it then becomes $f_{\mu,\nu}=\partial_\nu f_\mu$).

    Contract both side with $g^{\mu\nu}$, one gets $f^\mu_{\;,\mu}+f^\nu_{\;,\nu}=(\Omega^2-1)g^{\mu\nu}g_{\mu\nu}$ $\Rightarrow$ $\frac{2}{n}f^\alpha_{\;,\alpha}=\Omega^2-1$ $\Rightarrow$ $\boxed{f_{\mu,\nu}+f_{\nu,\mu}=\frac{2}{n}f^\alpha_{\;,\alpha}\,g_{\mu\nu}}$.

    Enumerate the solutions:
    \begin{array}{|l|c|l|}
    \hline \text{transform type} & x^\mu+f^\mu & \;\;\;\;\; \# \\
    \hline \text{Isometry}\;(f_{\mu,\nu}=0\,,\,\Omega^2=1) & x^\mu\to x^\mu+a^\mu & n \\
            & x^\mu\to x^\mu+\Lambda^\mu_{\;\nu}x^\nu & \frac{1}{2}n(n-1) \\
    \hline \text{Dilation}\;(\Omega^2\neq 1) & x^\mu\to x^\mu+\lambda x^\mu \;(\lambda\leq 0) & 1 \\
    \hline \text{Special conf. transform} & x^\mu\to x^\mu+2(c\cdot x)x^\mu-c^2 x^2\;(c^\mu\;\text{constant}) & n\\
    \hline
    \end{array}
    Altogether $\#=\frac{1}{2}(n+1)(n+2)$.

    Case $n=2$: Let $g_{\mu\nu}=\delta_{\mu\nu}$, $\varphi:\mathbb{R}^2\to\mathbb{R}^2$ ($\varphi^1=u$, $\varphi^2=v$) ; $\tilde{g}_{\alpha\beta}=g_{\mu\nu}\frac{\partial\varphi^\mu}{\partial_\alpha}\frac{\partial\varphi^\nu}{\partial x^\beta}=\Omega^2 g_{\mu\nu}$.

    $\Rightarrow$
    \begin{align*}
    g_{11}(\frac{\partial \varphi^1}{\partial x^1})^2+g_{22}(\frac{\partial \varphi^2}{\partial x^1})^2&=\Omega^2{g}_{11}\\
    g_{11}(\frac{\partial \varphi^1}{\partial x^2})^2+g_{22}(\frac{\partial \varphi^2}{\partial x^2})^2&=\Omega^2{g}_{22}\\
    g_{11}\frac{\partial \varphi^1}{\partial x^1}\frac{\partial \varphi^1}{\partial x^2}+g_{22}\frac{\partial \varphi^2}{\partial x^1}\frac{\partial \varphi^2}{\partial x^2}&=\Omega^2{g}_{12}=0
    \end{align*}
    $\Rightarrow$
    \begin{align*}
    &(\frac{\partial u}{\partial{x^1}})^2+(\frac{\partial v}{\partial{x^1}})^2=(\frac{\partial u}{\partial{x^2}})^2+(\frac{\partial v}{\partial{x^2}})^2=\Omega^2\\
    &\frac{\partial u}{\partial{x^1}}\frac{\partial u}{\partial{x^2}}+\frac{\partial v}{\partial{x^1}}\frac{\partial v}{\partial{x^2}}=0
    \end{align*}
    $\Rightarrow$
    \begin{align*}
    \left\{
    \begin{array}{rl}
    &\frac{\partial u}{\partial{x^1}}=\frac{\partial v}{\partial{x^2}}\\
    &\frac{\partial u}{\partial{x^2}}=-\frac{\partial v}{\partial{x^1}}
    \end{array}
    \right. \;\;\;(\text{Riemann-Cauchy eq.})
    \end{align*}
    This implies a conformal diffeomorphism is a holomorphic function and vice versa.

    The derivation above shows that it gives rise a linear complex structure to real plane such that $\mathbb{R}^2\cong\mathbb{C}$ ($z=x^1+i x^2$): As for a diffeomorphism, we know it is a smooth function, then consequently analytic (holomorphic) over $\mathbb{C}$. Therefore from another way around, conformal transformations are carried out by holomorphic functions, which are natural objects over $\mathbb{C}$, bring our works into the complex plane.

    Interpretation:
    $\tilde{g}_{\mu\nu}=g_{\alpha\beta}\frac{\partial\varphi^\alpha}{\partial x^\mu}\frac{\partial\varphi^\beta}{\partial x^\nu}=(M^T g M)_{\alpha\beta}=\Omega(x)^2g_{\mu\nu}$, $\frac{\partial\varphi^\alpha}{\partial x^\mu}=M^\alpha_{\;\mu}$ ; if $\Omega=1$ $\Rightarrow$ Lorentz transformation.

    Conformal diffeomorphism can be thought as a composition of rotation with a rescaling locally, preserving the angle.

    3.jpg

    Local property (generators)
    For each mapping $\leftrightarrow$ generator
    \begin{array}{|l|c|}
    \hline \text{transformation} & \text{generator} \\
    \hline \text{translation} & P_\mu=-i\partial_\mu \\
    \hline \text{rotation} & M_{\mu\nu}=i(x_\mu\partial_\nu-x_\nu\partial_\mu) \\
    \hline \text{dilation} & D=-i x^\mu\partial_\mu \\
    \hline \text{special conf. transform.} & K_\mu=-i(2x_\mu x^\nu\partial_\nu-(x\cdot x)\partial_\mu) \\
    \hline
    \end{array}

    Generators & Algebra
    translational & rotaional (sub)group $ISO(n)$ (or in Minkovski spacetime $ISO(n-1,1)$)
    \begin{align*}
    \left.
    \begin{array}{rl}
    [M_{\mu\nu},M_{\alpha\beta}]&=\cdots\\
    [M_{\mu\nu},P_\alpha]&=0
    \end{array}
    \right\} \text{Poincar}\acute{\text{e}}\;\text{algebra}
    \end{align*}
    Dilation
    \begin{align*}
    [D,P_\mu]&=-iP_\mu\;\;\;(\text{rescale of a translation is still translation})\\
    [D,K_\mu]&=iK_\mu
    \end{align*}
    Special conformal transformation
    \begin{align*}
    [P_\mu,K_\nu]=2i(\delta_{\mu\nu}D-M_{\mu\nu})
    \end{align*}

    - Conformal algebra is isomorphic to $\mathfrak{so}(n+1,1)$, i.e. the isometry group in Minkovski spacetime $\mathbb{R}^{n+1,1}$
    \begin{align*}
    &(\mathbb{R}^{n+2},\eta_{AB})\;,\;
    \eta_{AB}=
    \left(
    \begin{array}{cc}
    \mathbb{I}_n & \\
      & -\mathbb{I}_1 \\
    \end{array}
    \right)\;\;\text{for}\;\;A=1,\dots, N+2\\
    &x^A=M^A_{\;B}x^B\;,\;M^A_{\;B}\in SO(n+1,1)
    \end{align*}
    - $SO(n+1,1)\overset{\sim}{\longrightarrow}SO(n,2)$ (two "time" direction, not physical)
    From $\mathbb{R}^{n+2}\to \mathbb{R}^{n,2}$:

    • $\mathbb{R}^{n+2}$, light-cone is preserved by conformal transformation.
    • surface of light cone: $X^A X_A=X^2=1$ (1 dim. less)
    • Lorentz transformation on the surface

    $\rightarrow$ go to light cone surface (a) initially $n+2$ dim. (b) fixed by the surface of light cone :
    (1) $X^1,\dots,X^n;X^+,X^-$
    (2) For $X^+=x^{n+2}+x^{n+1}$ , $X^-=x^{n+2}-x^{n+1}$
    (3) $ds^2=dX^A dX_A=dX^\mu dX_\mu-dX^+ dX^-$
    (4) For $X^+=f(X^\mu=x^\mu)$, $X^+=f(x^\mu)$, $X^-=\frac{x^\mu x_\mu}{X^+}$ $\Rightarrow$ $dX^+=\frac{\partial f}{\partial x^\mu}dx^\mu$ (choosing a section).
    \begin{align*}
    ds^2|_\text{section}&=dx^\mu dx_\mu-dX^+ dX^-=dX^2|_\text{section}\\
    &\overset{X^a\to\lambda X^A}{=}\lambda^2(x)dX^2+\underset{dX^A=0\;(\text{constant surf.})}{\underbrace{d\lambda X^A dX^A}}+\underset{X^AX_A=0}{\underbrace{(d\lambda)^2X^AX_A}}
    \end{align*}

    mapping the Lorentz transformation back to the section by rescaling $\lambda(x)$, which depends on the chosen section.

    $SO(n,2)$ Isometry group of $\text{AdS}_{n+1}$, $x^A x^B\eta_{AB}=-1$, $\eta_{AB}=
    \left(
    \begin{array}{cc}
    \mathbb{I}_n & \\
      & -\mathbb{I}_2 \\
    \end{array}
    \right)$.

    Revisit $n=2$ case: $z=x^1+ix^2$, $z\mapsto f(z)$, $f$ locally holomorphic, i.e. conformal diffeomorphism.

    If $f(z)=a+b z+c z^2+\cdots$

    • $f(z)=z+a$ translation
    • $f(z)=bz=z+(b-1)z\;,\;|b|=1$ rotation

    Globalization:
    We want $f$ to be globally holomorphic.
    Riemann sphere $S^2\cong\mathbb{C}\cup\{\infty\}$ (conformal compatification)

    images.png

    The solution now yields rational function over Riemann sphere $f(z)=\frac{P(z)}{Q(z)}=\frac{az+b}{cz+d}$, $a,b,c,d\in\mathbb{C}$ (only single zero, maps pole to pole).

    Infinitesimal version (linearization):
    \begin{align*}
    f(z)&=\frac{z+\varepsilon az+\varepsilon b}{\varepsilon c z+1+\varepsilon d}\\
    &\approx z+\varepsilon b+\varepsilon az-\varepsilon c z^2-\varepsilon d z\\
    &=\underset{\text{translation}\;(\ell_{-1})}{\underbrace{z+\varepsilon b}}+\underset{(c\cdot x)x\;(\ell_0)}{\underbrace{z(\varepsilon a-\varepsilon d)}}-\underset{c x^2\;(\ell_1)}{\underbrace{z^2\varepsilon c}}
    \end{align*}

    Generators: $\ell_{-1}=-\partial_z$, $\ell_0=-z\partial_z$, $\ell_1=-z^2\partial_z$
    Algebra: $[\ell_n,\ell_m]=(n-m)\ell_{n+m}$ ($n,m=-1,0,1$)

    these generate the Lie group $SL(2,\mathbb{C})/\mathbb{Z}_2$ (called the Möbius group), and $SL(2,\mathbb{C})=SL(2,\mathbb{R})_{\mathbb{C}}=SL(2,\mathbb{R})\times SL(2,\mathbb{R})$.

    Comparing to the Lorentz group in 4D: $SO(3,1)=SL(2,\mathbb{R})\times SL(2,\mathbb{R})$, which shows the compatiblility.

    From above we see that the conformal diffeomorphisms are local holomorphic functions, however these functions may not be able to analytically extended to the entire complex plane. Therefore one generally considers the family of meromorphic functions in complex plane and the infinitesimal version of them (generators) form a infinite dimensional Lie algebra called Witt algebra: $[\ell_n,\ell_m]=(n-m)\ell_{n+m}\;(n,m\in\mathbb{Z})$.

    Conformal transformation:
    Def. : A Weyl transformation composes with a conformal diffeomorphism.

    Weyl transformation: $(p,g_p)\mapsto(p,\tilde{\Omega}^2(p)g_p)$

    Now choose $\tilde{\Omega}^2=\frac{1}{\Omega^2}$ $\Rightarrow$ Conformal transformation: $g_p\overset{\text{Conf. Diff.}\;\varphi}{\mapsto}\Omega^2 g_{\varphi(p)}\overset{\text{Weyl transf.}}{\mapsto}g_{\varphi(p)}$.

  4. Phantom_Ghost

    4楼 10月18日 数学版主, 物理版主, 优秀回答者
    2月前Phantom_Ghost 重新编辑

    Exercise sheet 1

    0001.jpg

    0002.jpg

    \[\]

    Solution
    \[\]\[\]\[\]
    Ex.1

    Ex.2

    Ex.3

  5. Phantom_Ghost

    5楼 10月24日 数学版主, 物理版主, 优秀回答者
    2月前Phantom_Ghost 重新编辑

    3. Critical Phenomena

    Review: Curie phase transition (magnetization)

    img19.gif

    second phase transition

    Experiment:
    Magnetization $\langle M_z\rangle\propto|T-T_c|^\beta$, $\beta\approx 0.33$

    Magnetic susceptibility $\chi=\frac{\partial\langle M_z\rangle}{\partial B_z}\Big|_{B_z=0}\propto |T-T_c|^{\gamma}$, $\gamma\approx 1.25$

    4.jpg

    $\beta,\gamma$ (roughly) independent of the ferromagnetic materials $\longrightarrow$ universality!

    Why?

    • In critical point, correlation length is very large (divergence), much larger then characteristic quantities of the detail system (lattice constant $a$, coupling constant $J$, etc.) $\xi\gg a$
    • Independent on scale, i.e. scale invarience

    Example (Ising model) (classical statistical system)
    Lattice $x\in\Lambda\subset\mathbb{Z}^n$ ($x$ is lattice site)

    dBCLt.png

    Hamiltonian
    \begin{align*}
    H[\{\sigma_x\}]=-J\sum_{\langle i j\rangle}\sigma_{x_i}\sigma_{x_j}
    \end{align*}
    where $\langle i j\rangle=\{x_i,x_j\in\Lambda||x_i-x_j|=a\}$ and $\sigma_x\in\{\pm 1\}$

    Summing over distribution $\{\sigma_x\}$ with weight $e^{-\beta H[\{\sigma_x\}]}$, we define the partition function
    \begin{align*}
    Z(\beta)=\sum_{\{\sigma_x\}}e^{-\beta H[\{\sigma_x\}]}
    \end{align*}
    (two point) correlation function
    \begin{align*}
    \langle\sigma_x\sigma_y\rangle&=\frac{1}{Z}\sum_{\{\sigma_{x'}\}}\sigma_x\sigma_y e^{-\beta H[\{\sigma_{x'}\}]}\\
    &\overset{\overset{\Lambda\to\mathbb{Z}^n}{|x-y|\to\infty}}{=}
     \left\{
    \begin{array}{rl}
    e^{-|x-y|/\xi}\;\;\;,\;\;\;\beta< \beta_c\\
    \text{constant}\;\;\;,\;\;\;\beta> \beta_c\\
    \frac{\text{const.}}{|x-y|^{n-2+\eta}}\;\;\;,\;\;\;\beta= \beta_c
    \end{array}
    \right.
    \end{align*}
    where
    \begin{align*}
    \eta=
     \left\{
    \begin{array}{rl}
    \frac{1}{4}\;\;\;&,\;\;\;n=2\;(\text{exact solution, Onsager})\\
    \text{constant}&,\;\;\;n=3\;(\text{numeric simulation, e.g. MC algorithm})\\
    0\;\;\;&,\;\;\;n\geq 4
    \end{array}
    \right.
    \end{align*}
    $\eta$ is a kind of anomalous dimension revelant to IR (long distance) behavior, since the fluctuation in low dimensional space has very strong impact, in higher dimension it gets wearker impact (e.g. 1D Ising model, fluctuation are so strong that there is no phase transition along all tempreture)

    Thermal quantities
    \begin{align*}
    M&=\lim_{B\to 0}\frac{1}{N}\sum_{y\in\Lambda}\langle\sigma_y\rangle_B
    =\lim_{B\to 0}\frac{1}{N}\sum_{y\in\Lambda}\frac{1}{Z}\sum_{\{\sigma_x\}}\sigma_y e^{-\beta(H[\{\sigma_x\}]+B\sum_{z\in\Lambda}\sigma_z)}\\
    \chi&=\frac{\partial M(B)}{\partial B}\Big|_{B\to 0}=
    \left\{
    \begin{array}{rl}
    \infty\;\;\;&\;\;\;N\to\infty\;,\;\beta\to\beta_c\\
    \text{finite}&\;\;\;N\to\infty\;,\;\beta<\beta_c
    \end{array}
    \right.
    \end{align*}
    $N=|\Lambda|=\#\;\text{of lattice sites (size of the lattice)}$.
    \[\]

    Renormalization group

    Block spin: replace $\Lambda_a$ by $\Lambda_{2a}$

    (a) $\sigma_i\to\sum_{i=1}^4\langle\sigma_i\rangle=\sigma^{(2)}_i$
    (b) $H\to -J^{(2)}\sum_{|x-y|=2a}\sigma^{(2)}_x\sigma^{(2)}_y-B^{(2)}\sum_x\sigma^{(2)}_x$ (renormalizable theory)
    (c) Expand $\xi\to\frac{\xi}{2}$
    (d) Free energy $F=-\frac{1}{\beta V}\ln Z$ $\Rightarrow$ $2^{n+1}F=-\frac{1}{\beta V}\ln Z$

    The idea: $J\to J^{(2)}\to J^{(4)}\to\cdots J^{(L)}\cdots$
     

    314997_1_En_4_Fig2_HTML.gif

    RG flow: $g=\beta J$, $g^{(L)}=u_L(g,B)$, $B^{(L)}=v_L(g,B)$ renormalized coupling constants (functions that keep physics unchanged)

    Def.: critical coupling $g^{*(L)}=g^*$, $B^{*(L)}=B^*$ (critical point is fixed point)

    At the critical point, we as well replace the lattice by a continuum model. Then $L$ is a continuous variable. What happens if it goes to the neighborhood of the critical point?
    \begin{align*}
    L\frac{dg^{(L)}}{dL}&=\lim_{\delta\to 0}\frac{g^{((1+\delta)L)}-g^{(L)}}{\delta}=\mu(g^{(L)})\\
    &=(g^{(L)}-g^*)\underset{y}{\underbrace{\frac{\partial\mu}{\partial g^{(L)}}\Big|_{g^{(L)}=g^*}}}+\cdots
    \end{align*}
    $\Rightarrow$ linearize flow: $L\frac{dg^{(L)}}{dL}=(g^{(L)}-g^*)y$ tell us if a critical point is attractive or not.
    Correlation length $\xi(g)=L\xi(g^{(L)})$ is a scale invariant function of $g$ (depends on the microscopic model), acts on it with $L\frac{d}{dL}$ and get $\xi(g^{(L)})+L\frac{dg^{(L)}}{dL}\frac{\partial\xi}{\partial g^{(L)}}=0$ $\Rightarrow$ $\xi(g^{(L)})+(g^{(L)}-g^*)y\frac{\partial\xi}{\partial g^{(L)}}=0$, the solution $\xi(g^{(L)})\propto (g^{(L)}-g^*)^{-\frac{1}{y}}$.

    $g^*=\frac{J}{k_B T_c}$, $g^{(L)}=\frac{J}{k_BT}$, $t=\frac{T_c-T}{T_c}$ (reduced temperature), $T\leq T_c$ $\Rightarrow$ $\boxed{\xi(t)\propto t^{-\frac{1}{y}}}$.

    3-Figure1-1.png

    Find the right (renormalizable) continuous field theory: compute correlation functions

    Def. scaling limit
    \begin{align*}
    \langle\sigma_{x_1}\cdots\sigma_{x_N}\rangle^*=\lim_{\lambda\to\infty}\lambda^{(n-2+\eta)\frac{N}{2}}\langle\sigma_{\lambda\cdot x_1}\cdots\sigma_{\lambda\cdot x_N}\rangle
    \end{align*}
    $\langle\sigma_{\kappa\cdot x_1}\cdots\sigma_{\kappa\cdot x_N}\rangle^*=\kappa^{-(n-2+\eta)\frac{N}{2}}\langle\sigma_{x_1}\cdots\sigma_{x_N}\rangle^*$ ,$\kappa\in\mathbb{R}_+$ $\Rightarrow$ replace the lattice model with continuum field theory $\langle\sigma_{x_1}\cdots\sigma_{x_N}\rangle^*=\langle\phi(x_1)\cdots\phi(x_N)\rangle$

  6. Phantom_Ghost

    6楼 11月12日 数学版主, 物理版主, 优秀回答者
    2月前Phantom_Ghost 重新编辑

    Exercise sheet 2

    ps2-1.jpg

    ps2-2.jpg

  7. Phantom_Ghost

    7楼 11月12日 数学版主, 物理版主, 优秀回答者
    2月前Phantom_Ghost 重新编辑

    Simplest continuum (field) theory

    Euclidean field (for statistical model)
    Action: $S[\phi(x)]=\int d^nx\left(\frac{1}{2}\partial_\mu\phi\partial^\mu\phi+\frac{1}{2}m^2\phi^2\right)$

    Correlation function: $\langle\phi(x_1)\cdots\phi(x_N) \rangle=\frac{1}{Z}\int\mathcal{D}[\phi(x)]\phi(x_1)\cdots\phi(x_N)e^{-S[\phi]}$, $Z=\int\mathcal{D}[\phi] e^{-S[\phi]}$.

    $\phi(x)=\sum_n c_n\phi_n(x)$, $(-\Delta+m^2)\phi_n(x)=\lambda_n\phi_n(x)$; $\int\mathcal{D}[\phi]=\int\prod_n dc_n$.

    $Z[J]=e^{-W[J]}:=\int\mathcal{D}[\phi]e^{-S[\phi]-\int d^nx J(x)\phi(x)}$, $\langle\phi(x_1)\cdots\phi(x_N) \rangle=\left.\frac{\delta^{N}W[J]}{\delta J(x_1)\cdots\delta J(x_M)}\right|_{J=0}$.

    Integrate out $\phi$:
    \begin{align*}
    e^{-W[J]}&=\int\mathcal{D}[\phi]e^{-\int d^nx ( \frac{1}{2}\phi(-\Delta+m^2)\phi+J\phi )}\\
    &\propto\frac{1}{\text{det}(\Delta+m^2)}e^{\int dx^n\int d^ny\;J(x)(-\Delta+m^2)^{-1}J(y)}\\
    &=\frac{1}{\text{det}(\Delta+m^2)}e^{\int dx^n\int d^ny\;J(x)G(x,y)J(y)}
    \end{align*}
    $(-\Delta+m^2)G(x,y)=\delta^{(n)}(x-y)$. Then the two-point correlation function yields the Green function $\langle\phi(x)\phi(y)\rangle=\left.\frac{\delta^2W[J]}{\delta J(x)\delta J(y)}\right|_{J=0}=G(x,y)$.
    \[
    G(x,y)=
    \left\{
    \begin{array}{rl}
    \frac{e^{-m|x-y|}}{|x-y|^{n-2}}\;\;\;,\;\;\;|x-y|\to\infty\;(n\neq 2)\\ \frac{1}{|x-y|^{n-2}}\;\;\;,\;\;\;|x-y|\to\infty\;(n\neq 2)\\
    \ln|x-y|\;\;\;,\;\;\;|x-y|\to 0\;(n=2)\\
    e^{-m|x-y|}\;\;\;,\;\;\;|x-y|\to\infty\;(n=2)
    \end{array}
    \right.
    \]
    Note:
    \begin{align*}
    \frac{\delta^2W[J]}{\delta J(x)\delta J(y)}&=-\frac{\delta}{\delta J(x)}\frac{\delta\ln Z[J]}{\delta J(y)}=\frac{\delta}{\delta J(x)}\left(\frac{1}{Z[J]}\frac{\delta Z[J]}{\delta J(y)}\right)\\
    &\xrightarrow{J\to 0}\frac{1}{Z}\int\mathcal{D}[\phi]\phi(x)\phi(y)e^{-S[\phi]}-\left(\frac{1}{Z}\int\mathcal{D}[\phi]\phi(x)e^{-S[\phi]}\right)\left(\frac{1}{Z}\int\mathcal{D}[\phi]\phi(y)e^{-S[\phi]}\right)\\
    &=\langle\phi(x)\phi(y)\rangle-\underset{=0}{\underbrace{\langle\phi(x)\rangle}}\underset{=0}{\underbrace{\langle\phi(y)\rangle}}=\langle\phi(x)\phi(y)\rangle_\text{connected}\end{align*}

    We can see that $G(x,y)$ and $\langle\sigma_x\sigma_y\rangle^*$ for statistical lattice model at critical point coincide iff. $m=0$.

    \[\]

    Ginzburg-Landau theory

    $S[\phi]=\frac{1}{2}\int d^nx[(\partial\phi)^2+m^2\phi^2+V(\phi)]$, $m^2=\frac{T-T_c}{T_c}$, $V(\phi)=\lambda\phi^4$. $\phi$ is order parameter.

    images.png

    \begin{align*}
    \langle\phi(x)\phi(y)\rangle&=\frac{1}{Z}\int\mathcal{D}[\phi]\phi(x)\phi(y) e^{-\int d^nx[\frac{1}{2}(\partial\phi)^2+\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4]}\\
    &\approx\frac{1}{Z}\int\mathcal{D}[\phi]\phi(x)\phi(y)\left(1+\frac{\lambda}{4!}\int d^nx\phi^4\right)e^{-\int d^nx[\frac{1}{2}(\partial\phi)^2+\frac{1}{2}m^2\phi^2]}+O(\lambda^2)
    \end{align*}
    4.jpg

    \begin{align*}
    G(x,y)&=G(|x-y|)=\int\frac{d^nk}{(2\pi)^n}e^{ik\cdot(x-y)}\Delta(k)\\
    \Delta(k)&=\underset{=\frac{1}{k^2+m^2}}{\underbrace{\Delta_0(k)}}-\frac{\lambda}{2}\Delta_0(k)\underset{=\Sigma(0)}{\underbrace{\left(\int d^nq\frac{1}{q^2+m^2}\right)}}\Delta_0(k)+O(k)
    \end{align*}
    self-energy: $\Sigma(0)\overset{\text{dim. regularization}}{=}\frac{(m^2)^{\frac{n-2}{2}}\Gamma(\frac{2-n}{2})}{(\sqrt{4\pi})^n}$

    Classical approximation
     extremizes the potential in Lagrangian to find the critical point (only classical motion invovled)

    images (1).png

    critical exponent: $\chi=\left.\frac{\partial M}{\partial B}\right|_{B=0}=\frac{1}{(T-T_c)^\gamma}$ $(T\geq T_c)$.
    \begin{align*}
    \chi&=\int d^nx C(x)\propto\xi^2\;,\;\gamma\approx 1.25\\
    C(x-y)&=\langle\sigma_x\sigma_y\rangle^*=\frac{e^{-|x-y|/\xi}}{|x-y|^{n-2}}
    \end{align*}
    $\lambda=0$: $C(x-y)=G(x,y)$ $\Rightarrow$ $\chi\propto\frac{1}{m^2}\propto\frac{1}{T-T_c}$ $\Rightarrow$ $\gamma=1$.

  8. Phantom_Ghost

    8楼 11月12日 数学版主, 物理版主, 优秀回答者

    Exercise sheet 3

    ps3-1.jpg

  9. 2周前

    Phantom_Ghost

    9楼 12月30日 数学版主, 物理版主, 优秀回答者
    2周前Phantom_Ghost 重新编辑

    5. Introduction to CFT

    Noether current with charges
    Canonical approach: Poincar$\acute{\text{e}}$ transformation+dilatation

    $\tilde{x}^\mu=x^\mu+\xi^\mu(x)$

    $J^\mu=\int_{\Sigma_t}(\xi^\nu T^\mu_{\;\nu}-\frac{1}{4}\xi^{\alpha}_{\;,\alpha}\underset{\text{virial current}}{\underbrace{V^\mu}})$ , $T^0_{\;\mu}=P_\mu\sim-i\hbar\partial_\mu$

    $\phi(x)=\lambda^\delta\tilde{\phi}(\tilde{x})$ , $\delta\phi(x)=[J_0,\phi(x)]=-i(\xi^\mu\partial_\mu\phi-\frac{1}{n}\xi^\alpha_{\;\alpha}\Delta\phi)$ ($\frac{1}{n}\xi^\alpha_{\;\alpha}=\lambda$ for $\tilde{x}^\mu=x^\mu+\lambda x^\mu$ dilatation)

    Conceptually, conformal transformation is the composition of conformal diffeomorphism and Weyl transformation.

    Conformal diffeomorphism: $\tilde{x}^\mu=x^\mu+\xi^\mu(x)$ , $(1+\theta)g^{\mu\nu}=\tilde{g}^{\mu\nu}+\xi_{\mu,\nu}+\xi_{\nu,\mu}$ , $\tilde{\phi}(x)=\phi(x)-\xi^\mu\partial_\mu\phi$

    Weyl transformation: $\tilde{x}=x$ , $\tilde{g}^{\mu\nu}=(1-\theta)g^{\mu\nu}$ , $\tilde{\phi}(x)=(1+\Delta\sigma)\phi(x)$

    $\text{Weyl trans.}\circ\text{conformal Diff.}$: $\tilde{x}^\mu=x^\mu+\xi^\mu$ , $\tilde{g}^{\mu\nu}=g^{\mu\nu}$ , $\tilde{\phi}(x)=\phi(x)-\xi^\mu\partial_\mu\phi(x)+\Delta\sigma\phi(x)$ ($\sigma=\xi^\alpha_{\;,\alpha}$)

    Explicit example:
    $T_{\mu\nu}=\partial_\mu\phi\partial_\nu\phi$ , $T^\mu_{\;\mu}=\partial^\mu\partial_\mu(\phi^2)$

    $\Rightarrow$ $V_\mu=\partial_\mu(\varphi^2)$ , $V_0=2(\partial_0\phi)\phi=2\pi_\phi\phi$ , $[\pi_\phi\phi(x),\phi(y)]=-\delta(x-y)\phi(x)$.

    Include special conformal transformation: Noether current from $T_{\mu\nu}$ and $V_\mu$
    $\partial^\mu(x^\nu T_{\mu\nu}+\xi^\alpha_{\;,\alpha}V_\mu)$ , if $V_\mu=\partial_\nu L^{\mu\nu}$ ($L^{\mu\nu}$ local field)

    $\delta x^\mu=\xi^\mu=2(c\cdot x)x^\mu-x^2 c^4$

    $J^\mu=\xi^\nu T^\mu_{\;\nu}-2(c\cdot x)V^\mu+2c_\nu L^{\nu\mu}$

    Example: free scalar field in $\mathbb{R}^n$
    $T^\alpha_{\;\alpha}=\partial_\mu\partial^\mu(\phi^2)$ , $L^{\mu\nu}=g^\mu\nu\phi^2$

    \[\]

    Question: Can scale invariance imply conformal invariance ?

    $n=2$ + unitarity (+ Lorentz invariance + conservation) show that $T^\mu_{\;\mu}=0$

    $\tilde{T}_{\mu\nu}(p)=\int d^2x e^{ip\cdot x}T_{\mu\nu}(x)$

    $\langle 0|\tilde{T}_{\mu\nu}(p)\tilde{T}_{\alpha\beta}(-p)|0\rangle=(p_\mu p_\nu-\eta_{\mu\nu p^2})(p_\alpha p_\beta-\eta_{\alpha\beta}p^2)\frac{f(p^2)}{p^2}+(\nu\leftrightarrow\alpha)\frac{g(p^2)}{p^2}$

    $\langle 0|T^{\mu}_{\;\mu}(p)T^\alpha_{\;\alpha}(-p)|0\rangle=(f+2g)p^2$

    $\langle 0|T^{\mu}_{\;\mu}(x)T^\alpha_{\;\alpha}(y)|0\rangle=(f+2g)\square\delta^{(2)}(x-y)$

    This can be removed by certain improvement term such as $S[\phi]=\int[\partial_\alpha\phi\partial^\alpha\phi+V(\phi)]+\underset{\text{geo. local counter term}}{\underbrace{\int\sqrt{g}R}}$

    $\Rightarrow$ $\langle 0|(T_\text{imp})^{\mu}_{\;\mu}(x)(T_\text{imp})^\alpha_{\;\alpha}(y)|0\rangle=0$ , $\sum_n\langle 0|(T_\text{imp})^{\mu}_{\;\mu}(x)|n\rangle\langle n|(T_\text{imp})^\alpha_{\;\alpha}(y)|0\rangle\geq 0$ (insert complete set of states)

    Counter example: (violation of unitarity)
    $S[\phi]=\int(\phi^2\partial_\alpha\partial^\alpha\phi+\phi\square\phi)$

    $T_{\mu\nu}=\phi^2(\partial_\mu\phi\partial_\nu\phi-\frac{1}{2}\eta_{\mu\nu}\partial_\alpha\phi\partial^\alpha\phi)$ , $T^\mu_{\;\mu}=\phi^2\partial_\alpha\phi\partial^\alpha\phi=\partial_\mu V^\mu$ , $V_\mu=\phi^2\partial_\mu(\phi^2)$

    $\Rightarrow$ $\partial^\mu V_\mu=\partial^\mu\phi^2\partial_\mu\phi^2+\phi^2\partial^\mu\partial_\mu\phi^2$

    e.o.m. $\frac{\delta}{\delta\phi}\partial_\mu(\phi^2\partial_\alpha\phi)=0$ , $\frac{\delta}{\delta\phi}\phi\partial_\alpha\phi\partial^\alpha=0$

    Maxwell theory ($n=3$)
    $S[A_\mu]=\int\,d^3x\,F_{\mu\nu}F^{\mu\nu}$ , $T^{\mu}_{\;\mu}=\partial_\mu V^\mu\rightarrow A_\nu F^{\mu\nu}$ , $\int F_{\mu\nu}F^{\mu\nu}=\int \partial_\alpha\phi\partial^\alpha\phi$

    $\varepsilon^{\mu\nu\lambda}F_{\mu\nu}=B^\lambda$ , $\partial_{[\mu}F_{\lambda\rho]}=0$ $\Rightarrow$ $B_\lambda=\partial_\lambda\phi$ , $\partial^\mu F_{\mu\nu}=0$ $\Rightarrow$ $\square\phi=0$

  10. Phantom_Ghost

    10楼 12月30日 数学版主, 物理版主, 优秀回答者

    Exercise sheet 4

    ps4-1.jpg

 

后才能发言