Lecture 5

Chasse_neige

Massless Amplitudes

(Bootstrap)

$$M=M(p_1,\cdots ,p_n)e_1\cdots e_l{e}_{l+1}^{\ast }\cdots e_n^{\ast }$$

but we have $p_i^2=0$ and $e$s are non Lorentz covariant.

Spinor Helicity Formalism

We use the fact that the Lorentz group is, at least locally, a product of two $\text{SO}(3)$ groups: $\text{SO}(3,1)=\text{SO}(3{)}_L\otimes \text{SO}(3{)}_R$. So, we use spinors of two $\text{SO}(3)$'s to construct everything. More explicitly, the generators of two $\text{SO}(3)$ can be written in terms of Lorentz generators as

$$SO(3,1)=SO(3{)}_L\otimes SO(3{)}_R$$$$J_L^i=\frac{J_i-i{K}_i}{2}{J}_R^i=\frac{J_i+i{K}_i}{2}$$

SO(3): spinor ${\psi }_a(a=1,2)$ - $c$ number

$${\psi }_a^{\prime }=D_{ab}{\psi }_a$$$$D_{ab}(\gamma ,\beta ,\alpha )=e^{-i\gamma \frac{{\sigma }_3}{2}}{e}^{-i\gamma \frac{{\sigma }_2}{2}}{e}^{-i\gamma \frac{{\sigma }_3}{2}}$$

Two types of spinors:

$$\{\begin{array}{l}(\frac{1}{2},0)\text{left spinor},{\psi }_a,a=1,2,\cdots \\ (0,\frac{1}{2})\text{right spinor},{\tilde{\psi }}_{\dot{a}},\dot{a}=1,2,\cdots \end{array}$$

A index raised/lowered by ${ϵ}_{ab}$, ${ϵ}^{ab}$, ${ϵ}_{\dot{a}\dot{b}}$, ${ϵ}^{\dot{a}\dot{b}}$.

$$p^{\mu }:(\frac{1}{2},\frac{1}{2})\to p_{a\dot{a}}$$$$p_{a\dot{a}}\equiv p_{\mu }({\sigma }^{\mu }{)}_{a\dot{a}}\text{by}{\sigma }^{\mu }\equiv (1,{\sigma }_i)$$$$({\overline{\sigma }}^{\mu }{)}^{\dot{a}a}={ϵ}^{\dot{a}\dot{b}}{ϵ}^{ab}({\sigma }^{\mu }{)}_{b\dot{b}}=(1,-{\sigma }_i)$$$$p^{\dot{a}a}=p_{\mu }({\overline{\sigma }}^{\mu }{)}^{\dot{a}a}$$$$({\sigma }^{\mu }{)}_{a\dot{a}}({\overline{\sigma }}^{\nu }{)}^{\dot{a}a}=-2{\eta }^{\mu \nu }\Rightarrow p_{a\dot{a}}{p}^{\dot{a}a}=-2p_{\mu }{p}^{\mu }$$$$p^{\mu }{p}_{\mu }=0\Rightarrow det{p}_{a\dot{a}}=0$$$$p_{a\dot{a}}=-{\psi }_a{\tilde{\psi }}_{\dot{a}}\Rightarrow p^{\dot{a}a}=-{\tilde{\psi }}^{\dot{a}}{\psi }^a$$$$|p⟩\equiv {\psi }_a⟨p|\equiv {\tilde{\psi }}^{\dot{a}}[p|\equiv {\tilde{\psi }}_{\dot{a}}$$$$\Rightarrow p_{a\dot{a}}=-|p⟩[p|p^{\dot{a}a}=-|p]⟨p|$$

Inner product

$$⟨p_1{p}_2⟩\equiv {\psi }_1^a{\psi }_{2a}[p_1{p}_2]\equiv {\tilde{\psi }}_{1\dot{a}}{\tilde{\psi }}_2^{\dot{a}}$$

are antisymmetric

$$⟨pq⟩[pq]=2p_{\mu }{q}^{\mu }$$$$p_{a\dot{a}}=\left(\begin{array}{cc}-p^0+p^3& p^1-i{p}^2\\ p^1+i{p}^2& -p^0-p^3\end{array}\right)$$$$|p⟩=\frac{1}{\sqrt{p^0+p^3}}\left(\begin{array}{c}-p^1+i{p}^2\\ p^0+p^3\end{array}\right)$$$$|p]=\frac{1}{\sqrt{p^0+p^3}}\left(\begin{array}{c}{p}^0+p^3\\ p^1+i{p}^2\end{array}\right)$$

Polarization $e_{\mu }^{\pm }(p)$: $e_{\mu }^{\pm }$ has a L-index $\Rightarrow e_{\mu }^{\pm }=|⟩[|\times \text{scalar}$

has helicity $\pm 1$. $\Rightarrow |p{]}^2,\frac{|p]}{|p⟩},\frac{1}{|p{⟩}^2}$ ( $|p⟩\to e^{-i\delta }|p⟩$ leads to $⟨p|\to e^{-i\delta }⟨p|$; $|p]\to e^{i\delta }|p]$ leads to $[p|\to e^{i\delta }[p|$.)

$[e_{\mu }^{\pm }=0]$ but $[|p⟩]=\frac{1}{2}$

“Ref momentum” q $q^2=0,q\cdot p\ne 0$

$$e^{+}\propto \frac{|q⟩[p|}{⟨pq⟩}$$

satisfies all conditions.

$$e^{-}=\frac{|q]⟨p|}{[pq]}$$

Also define spin-s

$$\begin{array}{c}{e}^{+s}={\left(\frac{|q⟩[p|}{⟨pq⟩}\right)}^s\\ e^{-s}={\left(\frac{|q]⟨p|}{[pq]}\right)}^s\end{array}$$

Consider the polarization tensors under LGT

$$\begin{array}{c}|p⟩\to e^{-i\delta }|p⟩\\ |q⟩\to \alpha |q⟩+\beta |p⟩\end{array}$$$$e^{+}\to e^{2i\delta }(\frac{|q⟩[p|}{⟨pq⟩}+\frac{\alpha }{\beta }\frac{|p⟩[p|}{⟨pq⟩})$$

3-point Massless Amplitudes

Couting the DoF of the system: $4\times 3-3-10=-1$, so this process is non-physical.

$$p_1\cdot p_2=\cdots =0$$

In the spinor helicity formalism

$$⟨p_1{p}_2⟩[p_1{p}_2]=0$$

Analytic continuation: $p_i^{\mu }\in \mathbb{C}$, then the on-shell conditions will become

$$p_1\cdot p_2=0\Rightarrow \{\begin{array}{l}⟨12⟩=0,[12]\ne 0\\ ⟨12⟩\ne 0,[12]=0\end{array}$$

Convention: for all outgoing particles, let $p^{\mu }=-p^{\mu }$, which will lead to

$$\sum _a{p}_a^{\mu }=0,e^{\pm }\to e^{\mp }$$

So we can get the most general 3pt amplitudes of species $a,b,c$, momentum $p_1,p_2,p_3$ and helicity $h_1,h_2,h_3$

$$M(1_a,2_b,3_c)=M_{abc}({\psi }_1,{\psi }_2,{\psi }_3,{\tilde{\psi }}_{\dot{1}},{\tilde{\psi }}_{\dot{2}},{\tilde{\psi }}_{\dot{3}})$$

We can use the principles to constrain the form of $M$

1. Poincare inv. + LGT

   $$p_i\cdot p_j=0\Rightarrow ⟨ij⟩[ij]=0(i\ne j)$$

   let $⟨12⟩=0$, we can derive

   $$\sum _i|i⟩[i|=0\Rightarrow ⟨13⟩=⟨23⟩=0$$

   Similarly, we can get if $[12]=0$, then

   $$\sum _i|i⟩[i|=0\Rightarrow [13]=[23]=0$$

   This will constrain $M$ in the form of

   $$M(1,2,3)=M_{abc}^L({\psi }_1,{\psi }_2,{\psi }_3)\text{or}{M}_{abc}^R({\tilde{\psi }}_{\dot{1}},{\tilde{\psi }}_{\dot{2}},{\tilde{\psi }}_{\dot{3}})$$

   LGT: Rotate particle i around ${\overrightarrow{p}}_i$

   $$z_i\equiv e^{-i{\theta }_i\overrightarrow{J}\cdot {\hat{p}}_i}$$

   Under this rotation, the spinors will act as

   $$\begin{array}{c}{\psi }_i\to z_i^{-\frac{1}{2}}{\psi }_i\\ {\tilde{\psi }}_i\to z_i^{\frac{1}{2}}{\tilde{\psi }}_i\end{array}$$

   So the M will perform as

   $$M_{abc}^L({\psi }_1,{\psi }_2,{\psi }_3)\to z_1^{-h_1}{z}_2^{-h_2}{z}_3^{-h_3}{M}_{abc}^L({\psi }_1,{\psi }_2,{\psi }_3)$$$$M_{abc}^R({\tilde{\psi }}_{\dot{1}},{\tilde{\psi }}_{\dot{2}},{\tilde{\psi }}_{\dot{3}})\to z_1^{h_1}{z}_2^{h_2}{z}_3^{h_3}{M}_{abc}^R({\tilde{\psi }}_{\dot{1}},{\tilde{\psi }}_{\dot{2}},{\tilde{\psi }}_{\dot{3}})$$

   Ansats: $M_{abc}^L={\lambda }_{abc}^L⟨12{⟩}^{n_1}⟨23{⟩}^{n_2}⟨31{⟩}^{n_3}$, solve the results of the LGTs, the amplitude can be represented as

   $${\lambda }_{abc}^L⟨12{⟩}^{h_3-h_1-h_2}⟨23{⟩}^{h_1-h_2-h_3}⟨31{⟩}^{h_2-h_1-h_3}$$

   Similarly, the right hand amplitude will have the form

   $${\lambda }_{abc}^R[12{]}^{-h_3+h_1+h_2}[23{]}^{-h_1+h_2+h_3}[31{]}^{-h_2+h_1+h_3}$$

3. Locality

   $$[⟨12{⟩}^{h_3-h_1-h_2}⟨23{⟩}^{h_1-h_2-h_3}⟨31{⟩}^{h_2-h_1-h_3}]=-(h_1+h_2+h_3)$$$$[[12{]}^{-h_3+h_1+h_2}[23{]}^{-h_1+h_2+h_3}[31{]}^{-h_2+h_1+h_3}]=h_1+h_2+h_3$$

   because the negative dimensions requires something like $\frac{1}{p}$, which is not allowed (singularity here must come from on-shell factorization), and dim-0 also has similar problems

   $$h_1+h_2+h_3=0\Rightarrow \{\begin{array}{l}{h}_1=h_2=h_3=0\\ h_i>0,h_j<0(\text{cross})\end{array}$$

   We can choose the left or right hand form of the amplitude according to whether $\sum _i{h}_i>0$.

Examples

3 particles have the same spin-s (parity even)

$$M(1_a^{+},2_b^{+},3_c^{+})={\lambda }_{abc}([12][23][31]{)}^s$$$$M(1_a^{-},2_b^{-},3_c^{-})={\lambda }_{abc}(⟨12⟩⟨23⟩⟨31⟩{)}^s$$$$M(1_a^{+},2_b^{+},3_c^{-})={\kappa }_{abc}{\left(\frac{[12{]}^3}{[23][31]}\right)}^s$$$$M(1_a^{-},2_b^{-},3_c^{+})={\kappa }_{abc}{\left(\frac{⟨12{⟩}^3}{⟨23⟩⟨31⟩}\right)}^s$$

Spin-statistics (generalized)

$$M(1_a^{+},2_b^{-})\leftrightarrow M(2_b^{-},1_a^{+})$$

For odd/even $s$, the coupling $\lambda /\kappa$ are antisymmetric/symmetric under $i\leftrightarrow j$ exchange.

Massless odd-spin particles cannot have cubic self-interactions with less 3 species.

In particular, a photon cannot have cubic self-interaction (non-perturbatively).

Spin-1 $N_F\ge 3$

$$M(1_a^{+},2_b^{+},3_c^{+})={\lambda }_{abc}[12][23][31]$$$$M(1_a^{+},2_b^{+},3_c^{-})={\kappa }_{abc}\frac{[12{]}^3}{[23][31]}$$

Where

$$\begin{array}{c}[\lambda ]=-2\\ [\kappa ]=0\end{array}$$

We call the most important coupling in the macroscopic world minimal coupling ($\kappa$ here).

Spin-2

$$M(1^{+},2^{+},3^{+})=\lambda ([12][23][31]{)}^2$$$$M(1^{+},2^{+},3^{-})=\kappa {\left(\frac{[12{]}^3}{[23][31]}\right)}^2$$

Where

$$\begin{array}{c}[\lambda ]=-5\\ [\kappa ]=-1\end{array}$$

Minimal coupling for gravitation is $[\kappa ]=-1$.