Lecture 4

Chasse_neige

On-Shell Factorization

A model of light bulbs

Assume that there exists a local operator $O(x)$ having a non-zero element between vacuum & IPS

$$⟨\overrightarrow{P}|O(0)|\mathrm{\Omega }⟩=1$$$$\Rightarrow ⟨\overrightarrow{P}|O(x)|\mathrm{\Omega }⟩=e^{-i\overrightarrow{p}\cdot \overrightarrow{x}},p^0=\sqrt{p^2+m^2}$$

Modify Hamiltonian $H\to H+g\int {\mathrm{d}}^{}\overrightarrow{x}j(t,\overrightarrow{x})O(t,\overrightarrow{x})$, where $|g{|}^2$ means the “lumnosity”

$$S_{\varphi \psi }[j]=⟨{\mathrm{\Phi }}_{-}|T\{e^{-ig\int {\mathrm{d}}^{}t{\mathrm{d}}^{}\overrightarrow{x}jO}\}|{\mathrm{\Psi }}_{+}⟩$$

$O$ is the Heisenberg picture operator, it evolves by $H$ (not $H+g\int jO$)

If $g$ is small, we can expand $S_{\varphi \psi }[j]$ in $g$, where the most general object is

$$G_{\varphi \psi }(x_1,\cdots ,x_n)=⟨{\mathrm{\Phi }}_{-}|T\{O(x_1)\cdots O(x_n)\}|{\mathrm{\Psi }}_{+}⟩$$

Let $y_i=x_i-x_1$ ($y_1=0$), then

$$⟨{\mathrm{\Phi }}_{-}|T\{O(x_1)\cdots O(x_n)\}|{\mathrm{\Psi }}_{+}⟩=e^{i(p_{\varphi }-p_{\psi })\cdot x_1}\times ⟨{\mathrm{\Phi }}_{-}|T\{O(0)\cdots O(y_n)\}|{\mathrm{\Psi }}_{+}⟩$$

On-shell Factorization

In momentum space the n-point Green function is

$$G(k_1,\cdots ,k_n)\tilde{\delta }(\sum _i{k}_i)\equiv \int {{\mathrm{d}}^{4}x}_a{e}^{-i{k}_a\cdot x_a}⟨\mathrm{\Omega }|O_1(x_1)\cdots O_n(x_n)|\mathrm{\Omega }⟩$$

scalar particles only. $G$ is a function of $K_{12}^2,\cdots$

$$K_{1\cdots n}^{\mu }=k_1^{\mu }+\cdots +k_n^{\mu }$$

We want $G$‘s behavior on the $s_l=-K_{1\cdots l}^2$ plane by the condtion $K_{1\cdots l}^0<0$.

$G(s_l)$ develops a simple pole at $s^2=m^2$ when approaching it from $K_{1\cdots n}^0<0$ , if there exists 1 PS of A, $|\overrightarrow{P}⟩$, with mass $m$ having non vanishing M-matrix element

$$G_E\tilde{\delta }(K_{1\cdots l}+p)\equiv \int ⟨\overrightarrow{P}|T\{O_1(x_1)\cdots O_l(x_l)\}|\mathrm{\Omega }⟩$$$$G_L\tilde{\delta }(K_{l+1\cdots n}-p)\equiv \int ⟨\mathrm{\Omega }|T\{O_{l+1}(x_{l+1})\cdots O_n(x_n)\}|\overrightarrow{P}⟩$$

And we have

$$\underset{S_l\to m^2,K_{1\cdots l}^0<0}{lim}G(S_l)=G_L\times \frac{-i}{-S_l+m^2-iϵ}\times G_E$$

In reality the 2 $\to$ 2 scattering never divs, for is A is stable, it requires $m_A<2m_{\varphi }$, and if we cutoff by $\mathrm{\Gamma }$, the pole will shift by

$$m^2\to {(m-\frac{i}{\mathrm{\Gamma }})}^2$$

Proof. (sketch)

The pole is from on-shell propagation of a 1 PS $\propto \theta (\text{min}\{x_{l+1}^0,\cdots ,x_n^0\}-\text{max}\{x_1^0,\cdots ,x_l^0\})$

$$\theta (z)=-\frac{1}{2\pi i}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{d}}^{}\omega }{\omega +iϵ}{e}^{-i\omega z}$$$$G(k_1,\cdots ,k_n)\tilde{\delta }(\sum _i{k}_i)\equiv \int {{\mathrm{d}}^{4}x}_a{e}^{-i{k}_a\cdot x_a}\frac{-1}{2\pi i}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{d}}^{}\omega }{\omega +iϵ}{e}^{-i\omega [min-max]}⟨\mathrm{\Omega }|O_1(x_1)\cdots O_n(x_n)|\mathrm{\Omega }⟩$$$$⟨\mathrm{\Omega }|O_1\cdots O_n|\mathrm{\Omega }⟩=⟨\mathrm{\Omega }|T\{O_{l+1}\cdots O_n\}T\{O_1\cdots O_l\}|\mathrm{\Omega }⟩$$

We can insert a complete basis into the two time-evolving operators, and only the 1 PS in the basis lead to poles.

$$⟨\mathrm{\Omega }|O_1\cdots O_n|\mathrm{\Omega }⟩=\int \frac{{\mathrm{d}}^3\overrightarrow{p}}{(2\pi {)}^3}\frac{1}{2E_p}⟨\mathrm{\Omega }|T\{O_{l+1}\cdots O_n\}|\overrightarrow{P}⟩⟨\overrightarrow{P}|T\{O_1\cdots O_l\}|\mathrm{\Omega }⟩+\cdots$$$$⟨\mathrm{\Omega }|T\{O_{l+1}(x_{l+1})\cdots O_n(x_n)\}|\mathrm{\Omega }⟩\to e^{i\overrightarrow{p}\cdot {\overrightarrow{x}}_{l+1}}⟨\mathrm{\Omega }|T\{O_{l+1}(0)\cdots O_n(y_n)\}|\mathrm{\Omega }⟩(y_i=x_i-x_{l+1})$$

Integrate other variables, we can get a form similar to

$$\frac{1}{\omega +iϵ}\to \frac{1}{{\overrightarrow{K}}_{1\cdots l}^0+\sqrt{K_{1\cdots l}^2+m^2}+iϵ}$$

For non-zero spin

$$\underset{S_l\to m^2}{lim}G(S_l)=\sum _h{G}_L(h)\frac{-i}{-S_l+m^2-iϵ}{G}_E(-h)$$

LSZ Reduction

$$G_{\mathrm{\Omega }\mathrm{\Omega }}\tilde{\delta }={\int }_x{e}^{-i\sum k\cdot x}⟨\mathrm{\Omega }|T\{O_1(x_1)\cdots O_n(x_n)\}|\mathrm{\Omega }⟩$$

Put all external momenta on shell

$$k_i^0=\mp \sqrt{{\overrightarrow{k}}_i^2+m_i^2}\text{is}\{\begin{array}{l}-:i=1,\cdots ,l\\ +:i=l+1,\cdots ,n\end{array}$$

Let $k_1^2\to -m_1^2$ and $k_1^0<0$

$$G_L\tilde{\delta }=\int {\mathrm{d}}^{4}x{e}^{-ik\cdot x}⟨\mathrm{\Omega }|O_1(x)|\overrightarrow{P}⟩=(2\pi {)}^4{\delta }^4(k-p)$$$$G_L=1$$$$\underset{k_1^2\to -m_1^2,k_1^0<0}{lim}{G}_{\mathrm{\Omega }\mathrm{\Omega }}(k_1,\cdots ,k_n)=\frac{-i}{k_1^2+m_1^2-iϵ}\times G_{{\overrightarrow{k}}_1\mathrm{\Omega }}(k_2,\cdots ,k_n)$$$$\underset{x_1^0\to \mathrm{\infty }}{lim}G(x_1,\cdots ,x_n)\to e^{i{E}_p{x}_1^0-i{\overrightarrow{p}}_1\cdot {\overrightarrow{x}}_1}\tilde{G}(x_2,\cdots ,x_n)$$

and notice that $\int {{\mathrm{d}}^{4}x}_1{e}^{-i{k}_1\cdot x_1}G$ is divergent when $k_1^{\mu }\to -p^{\mu }$

Similarly, we put all $p_i^2\to -m_i^2$, and this will lead to

$$G_{\mathrm{\Omega }\mathrm{\Omega }}\to \prod _{i=1}^l\frac{-i}{k_i^2+m_i^2-iϵ}$$

So we can get the LSZ reduction formula

$$\underset{k_a^2\to -m_a^2}{lim}{G}_{\mathrm{\Omega }\mathrm{\Omega }}(k_1,\cdots k_n)=\prod _{a=1}^n\frac{-i}{k_a^2+m_a^2-iϵ}⟨\{{\overrightarrow{p}}_{l+1}\cdots {\overrightarrow{p}}_n{\}}_{-}|\{{\overrightarrow{p}}_1\cdots {\overrightarrow{p}}_n{\}}_{+}⟩$$

Weinberg Soft Theorem

Try to find the relationship between the scattering amplitude of $n$ particles and $n+1$ particles, where the extra particle is soft, i.e. $k\to 0$. We call the amplitude $M_{\varphi ,\psi }$ and $M_{\varphi \gamma ,\psi }$

Minimal Coupling

Consider a massless spin-1 particle (photon), the Green function of a scalar particle $\varphi$ and a photon $\gamma$ is

$$G(p,p^{\prime },q)=g{e}_{\pm \mu }^{\ast }(q)p^{\mu }F(p^2,p^{\prime 2},q^2)$$

We assume that all the particles are on-shell, so the form factor $F$ is a constant, and we can set $F=1$. The amplitude is

$$G=g{e}_{\pm }^{\ast }\cdot p(G\to G+\alpha gq\cdot p)$$

Soft Theorem

Assume that there are $L$ particles in the initial state and $N$ particles in the final state, and we want to find the relationship between $M_{\varphi \gamma ,\psi }$ and $M_{\varphi ,\psi }$.

Option 1

The soft photon is emitted in the initial state

$$M_{\varphi \gamma ,\psi }^{(a)\pm }=M_{\varphi ,\psi }\times \frac{-i}{(p_a-q{)}^2+m^2-iϵ}\times g_a{e}_{\mu }^{\pm \ast }(q)p_a^{\mu }$$

while $q\to 0$

$$M_{\varphi \gamma ,\psi }^{(a)\pm }\to M_{\varphi ,\psi }\times \frac{g_a{e}_{\mu }^{\pm \ast }(q)p_a^{\mu }}{-2p_a\cdot q-iϵ}$$

Option 2

The soft photon is emitted in the final state

$$M_{\varphi \gamma ,\psi }^{(b)\pm }\to M_{\varphi ,\psi }\times \frac{g_b{e}_{\mu }^{\pm \ast }(q)p_b^{\prime \mu }}{2p_b^{\prime }\cdot q-iϵ}$$

Option 3

The soft photon is emitted in the internal line, which is regular when $q\to 0$.

So the total amplitude gives out

$$\underset{q\to 0}{lim}{M}_{\varphi \gamma ,\psi }=M_{\varphi ,\psi }\times (\sum _{a=1}^L\frac{g_a{e}_{\mu }^{\pm \ast }(q)p_a^{\mu }}{-2p_a\cdot q-iϵ}+\sum _{b=1}^N\frac{g_b{e}_{\mu }^{\pm \ast }(q)p_b^{\prime \mu }}{2p_b^{\prime }\cdot q-iϵ})+O(q^0)$$

Implications

Lorents inv. of $M_{\varphi \gamma ,\psi }$ requires that $\sum _a{g}_a=\sum _b{g}_b$, which is the charge conservation law.

Proof: do a LGT to the photon polarization vector

$$e_{\mu }^{\pm \ast }\to e_{\mu }^{\pm \ast }+\alpha q_{\mu }$$$$\Rightarrow \mathrm{\Delta }{M}_{\varphi \gamma ,\psi }\propto -\frac{\alpha }{2}[\sum _{a=1^L}{q}_a-\sum _{b=1}^N{g}_b]=0$$

$g_a$: coupling constant of the $a$-th particle, which is also the charge of the particle. So we have the charge conserved in the scattering process.

The minimal coupling to a massless spin-1 particle demands charge conservation.

Minimal coupling actually refers to a long-range interaction under classical limit.

Spin-2 (Graviton)

Minimal Coupling

Consider a massless spin-2 particle, the Green function of a scalar particle $\varphi$ and a graviton $h$ is

$$G(p,p^{\prime },q)=g{e}_{\pm \mu \nu }^{\ast }(q)p^{\mu }{p}^{\nu }F(p^2,p^{\prime 2},q^2)$$

We assume that all the particles are on-shell, so the form factor $F$ is a constant, and we can set $F=1$. The amplitude is

$$G=g{e}_{\mu \nu }^{\pm \ast }{p}^{\mu }{p}^{\nu }$$

Soft Theorem

Similarly, we can get the difference of the amplitude with and without a soft graviton

$$\underset{q\to 0}{lim}{M}_{\varphi h,\psi }^{\pm }=M_{\varphi \psi }\times [\sum _{a=1}^L\frac{g_a{e}_{\mu \nu }^{\pm \ast }{p}^{\mu }{p}^{\nu }}{-2p_a\cdot q}+\sum _{b=1}^N\frac{g_b{e}_{\mu \nu }^{\pm \ast }{p}^{\prime \mu }{p}^{\prime \nu }}{2p_b^{\prime }\cdot q}]+O(q^0)$$

Implication

Using the L-inv. of the $M_{\varphi h,\psi }^{\pm }$, we can get that all the coupling constants, $g_a$ and $g_b$, are the same, which is the equivalence principle in classical gravity.

The minimal coupling to a massless spin-2 particle is universal.

Spin-s

$$\sum _{a=1}^L{g}_a{p}^{{\mu }_1}\cdots p^{{\mu }_{s-1}}=\sum _{b=1}^N{g}_b{p}_b^{\prime {\mu }_1}\cdots p_b^{\prime {\mu }_s}$$

Massless particles with spin $\ge$ 3 cannot mediate long-range force.