Ze Chen
 Quantum Field Theory (IV) Renormalization

## Quantum Field Theory (V)

Applications of Feynman Diagrams

1PIを剃る。そして断面積を求める。

#### Foundations of Field Theory

##### Scattering
• Mandelstam variables: for $$1+2\rightarrow 3+4$$
• $s = (p_1 + p_2)^2 = (p_3 + p_4)^2.$
• $t = (p_1 - p_3)^2 = (p_4 - p_2)^2.$
• $u = (p_1 - p_4)^2 = (p_3 - p_2)^2.$
• $s+t+u = m_1^2 + m_2^2 + m_3^2 + m_4^2.$
$$s$$-channel $$t$$-channel $$u$$-channel
$$\displaystyle \mathcal{M}\propto \frac{1}{s-m_\phi^2}{}$$ $$\displaystyle \mathcal{M}\propto \frac{1}{t-m_\phi^2}{}$$ $$\displaystyle \mathcal{M}\propto \frac{1}{u-m_\phi^2}{}$$
Images from Mandelstam variables.

#### Tricks

##### Spin Sums

\begin{align*} \sum_s u^s(p) \overline{u}^s(p) &= \unicode{x2215}\kern-.5em {p} + m, \\ \sum_s v^s(p) \overline{v}^s(p) &= \unicode{x2215}\kern-.5em {p} - m. \end{align*}

##### Photon Polarization Sums
• Ward Identity: if during a calculation of the amplitude for some QED process involving an external photon with momentum $$k$$, we encounter a invariant matrix of the form $\mathcal{M}(k) = \mathcal{M}^\mu(k)\epsilon^*_\mu(k),$ then we could safely assume that $\color{red} k_\mu \mathcal{M}^\mu(k) = 0.$
• Ward-Takahashi Identity: Let $$\mathcal{M}_0$$ be the diagram without the insertion of photon $$k$$. We have $\sum_{\substack{\mathrm{insertion} \\ \mathrm{points}}} k_\mu \mathcal{M}^\mu(k;p_1\cdots p_n;q_1\cdots q_n) = e\sum_i \qty[{ \mathcal{M}_0(p_1\cdots p_n; q_1\cdots (q_1 - k) \cdots) - \mathcal{M}_0(p_1\cdots (p_i + k) \cdots; q_1\cdots q_n) }].$ This does not contribute to the final $$S$$-matrix.
• Let $S(p) = \frac{i}{\unicode{x2215}\kern-.5em {p} - m - \Sigma(p)}.$ Applying the Ward-Takahashi identity to the case where $$\mathcal{M}_0$$ is a two-point correlation, $-i k_\mu \Gamma^\mu(p+k, p) = S^{-1}(p+k) - S^{-1}(p).$
• Substitution: $\sum_{\text{polarizations}} \epsilon^*_\mu \epsilon_\nu \mathcal{M}^\mu(k) \mathcal{M}^{\nu*}(k) = -g_{\mu\nu} \mathcal{M}^\mu(k) \mathcal{M}^{\nu*}(k).$
• Time-like and longitudinal photons could be omitted.
##### Trace Technology
• Traces of products: \begin{align*} \tr(\mathbb{1}) &= 4, \\ \tr(\text{odd number of } \gamma) &= 0, \\ \tr{\gamma^\mu \gamma^\nu} &= 4g^{\mu\nu}, \\ \tr(\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma) &= 4(g^{\mu\nu} g^{\rho\sigma} - g^{\mu\rho}g^{\nu\sigma} + g^{\mu\sigma}g^{\nu\rho}), \\ \tr(\gamma^5) &= 0, \\ \tr(\gamma^\mu \gamma^\nu \gamma^5) &= 0, \\ \tr{\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma \gamma^5} &= -4i \epsilon^{\mu\nu\rho\sigma}. \end{align*}{}
• Order be reversed: $\tr(\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma \cdots) = \tr(\cdots \gamma^\sigma \gamma^\rho \gamma^\nu \gamma^\mu ).$
• Contraction: \begin{align*} \gamma^\mu \gamma_\mu &= 4, \\ \gamma^\mu \gamma^\nu \gamma_\mu &= -2\gamma^\nu, \\ \gamma^\mu \gamma^\nu \gamma^\rho \gamma_\mu &= 4g^{\nu\rho}, \\ \gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma \gamma_\mu &= -2\gamma^\sigma \gamma^\rho \gamma^\nu. \end{align*}{}
##### Crossing Symmetry
• Crossing symmetry: $\mathcal{M}(\phi(p) + \cdots \rightarrow \cdots) = \mathcal{M}(\cdots \rightarrow \cdots + \overline{\phi}(k)).$
• A particle of momentum $$p$$ on the LHS moved to an anti-particle of momentum $$k=-p$$ on the RHS.
• Only in the sense of analytic continuation.
• There is an overall minus sign and should be manually removed.

#### General Procedures

##### How do We Obtain Cross Sections?
1. Draw the diagrams for the desired process.
2. Use the Feynman rules to write down the amplitude $$\mathcal{M}$$.
3. Square the amplitude and average or sum over spins, using the spin sum relations.
4. Evaluate traces using the trace identities.
5. Specialize to a particular frame of reference, and draw a picture of the kinematic variables in that frame. Express all 4-momentum vectors in terms of a suitably chosen set of variables, e.g. $$E$$ and $$\theta$$.
6. Plug the resulting expression for $$\abs{\mathcal{M}}^2$$ into the cross-section formula $\dd{\sigma} = \frac{1}{2E_A 2E_B \abs{v_a - v_B}} \qty(\prod_f \frac{\dd{^3 p_f}}{(2\pi)^3} \frac{1}{2E_f})\times \abs{\mathcal{M}(p_A,p_B \rightarrow \qty{p_f})}^2 (2\pi)^4 \delta^{(4)}(p_A + p_B - \sum p_f).$ Then we integrate over phase-space variables to obtain a cross section in the desire form.
##### Scattering of a Specific Helicity?

Projection onto left-helicity of $$u$$ is onto right-helicity of $$v$$, and vice versa.

• Projection of helicity:
• Onto left-handed (for $$u$$, right-handed for $$v$$): $\frac{1+\gamma^5}{2} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}.$
• Onto right-handed (for $$u$$, left-handed for $$v$$): $\frac{1-\gamma^5}{2} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}.$
##### Feynman Parameters: Loop Diagrams
1. Draw the diagrams and write down the amplitude.
2. Sqeezing the denominator: \begin{align*} \frac{1}{AB^n} &= \int_0^1 \dd{x} \dd{y} \delta(x+y-1) \frac{ny^{n-1}}{[xA + yB]^{n+1}}; \\ \frac{1}{A_1 \cdots A_n} &= \int_0^1 \dd{x_1} \cdots \dd{x_n} \delta(\sum x_i - 1) \frac{(n-1)!}{[x_1A_1 + \cdots + x_n A_n]^n}; \\ \frac{1}{A_1^{m_1} \cdots A_n^{m_n}} &= \int_0^1 \dd{x_1} \cdots \dd{x_n} \delta(\sum x_i - 1) \frac{\prod x_i^{m_i - 1}}{\qty[\sum x_i A_i]^{\sum m_i}} \frac{\Gamma(m_1 + \cdots + m_n)}{\Gamma(m_1) \cdots \Gamma(m_n)}. \end{align*}
3. Completing the square in the new denominator by shifting to a new loop momentum $$l$$. Therefore, the denominator is of the form $$D^m$$ where $$D$$ is an even function of $$l$$, $D = l^2 - \Delta + i\epsilon.$
4. Eliminate some terms in the numerator by symmetry, e.g. \begin{align*} \int \frac{\dd{^4 l}}{(2\pi)^4} \frac{l^\mu}{D^m} &= 0; \\ \int \frac{\dd{^4 l}}{(2\pi)^4} \frac{l^\mu l^\nu}{D^m} &= \int \frac{\dd{^4 l}}{(2\pi)^4} \frac{g^{\mu\nu} l^2/4}{D^m}; \\ \int \frac{\dd{^d l}}{(2\pi)^d} \frac{l^\mu l^\nu}{D^m} &= \int \frac{\dd{^4 l}}{(2\pi)^4} \frac{g^{\mu\nu} l^2/d}{D^m}. \end{align*} Note that contraction identities should be modified in the case of dimensional regularization to \begin{align*} \gamma^\mu \gamma^\nu \gamma_\mu &= -(2-\epsilon) \gamma^\nu; \\ \gamma^\mu \gamma^\nu \gamma^\rho \gamma_\mu &= 4g^{\nu\rho} - \epsilon \gamma^\nu \gamma^\rho; \\ \gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma \gamma_\mu &= -2\gamma^\sigma \gamma^\rho \gamma^\nu + \epsilon \gamma^\nu \gamma^\rho \gamma^\sigma. \end{align*}{} In this way, we rewrite the integral in a isotropic form.
• We may also write the numerator in the form $\gamma^\mu \cdot A + (p'^\mu + p^\mu) \cdot B + q^\mu \cdot C.$
1. Regularize the integral using, for example, Pauli-Villars prescription, or dimensional regularization.
• How do we choose the regulator?

The question of which regulator to use has no a priori answer in quantum field theory. Often the choice has no effect on the predictions of the theory. When two regulators gives different answers for observable quantities, it is generally because some symmetry (such as the Ward idensity) is being violated by one (or both) of them. In this case we take the symmetry to be fundamental anddemand that it be preserved by the regulator.

2. Apply Wick rotation:
• For the integration in $$l^0$$, the poles lies at $l^0 = \pm\qty(-\sqrt{\vb*{l}^2 + \Delta} + i\epsilon).$ Therefore, we are entitled to rotate the path by $$\pi/2$$.
• The integration is rewritten as $\int \frac{\dd{^4 l}}{(2\pi)^4} \frac{1}{[l^2 - \Delta]^m} = \frac{i(-1)^m}{(2\pi)^4} \int \dd{\Omega_4} \int_0^\infty \dd{l_E} \frac{l_E^3}{[l_E^2 + \Delta]^m}.$
• Examples: \begin{align*} \int \frac{\dd{^4 l}}{(2\pi)^4} \frac{1}{[l^2 - \Delta]^m} &= \frac{i(-1)^m}{(4\pi)^2} \frac{1}{(m-1)(m-2)} \frac{1}{\Delta^{m-2}}, \\ \int \frac{\dd{^4 l}}{(2\pi)^4} \frac{l^2}{[l^2 - \Delta]^m} &= \frac{i(-1)^{m-1}}{(4\pi)^2} \frac{2}{(m-1)(m-2)(m-3)} \frac{1}{\Delta^{m-3}}. \end{align*}
##### Physical Discontinuity
1. Cut through the diagrams in all possible ways such that the cut propagators can simultaneously be put on shell.
2. For each cut, replace $\frac{1}{p^2 - m^2 + i\epsilon} \rightarrow -2\pi i \delta(p^2 - m^2)$ in each cut propagator, and then perform the loop integrals.
3. Sum the contributions of all possible cuts.
##### Dimensional Regularization
• Area of a $$d$$-dimensional unit sphere: $\int \dd{\Omega_d} = \frac{2\pi^{d/2}}{\Gamma(d/2)}.$
• Propagators: \begin{align*} \int \frac{\dd{^d l_E}}{(2\pi)^d} \frac{1}{(l^2_E + \Delta)^n} &= \frac{1}{(4\pi)^{d/2}} \frac{\Gamma(n - d/2)}{\Gamma(n)} \qty(\frac{1}{\Delta})^{n-d/2}; \\ \int \frac{\dd{^d l_E}}{(2\pi)^d} \frac{l^2_E}{(l^2_E + \Delta)^n} &= \frac{1}{(4\pi)^{d/2}} \frac{d}{2} \frac{\Gamma(n-d/2-1)}{\Gamma(n)}\qty(\frac{1}{\Delta})^{n-d/2-1}. \end{align*}

#### Diagrams

##### 1PI Diagrams
• 1PI diagram:
• one-particle irreducible diagram,
• cannot be split into two by removeing a single line.

#### φ⁴ Theory: Loop

##### One-Loop
1. Cutting the diagram: Working in the COMF, with $$k = (k^0, \vb{0})$$, and now we evaluate the integral $i\delta \mathcal{M} = \frac{\lambda^2}{2} \int \frac{\dd{^4 q}}{(2\pi)^4} \frac{1}{(k/2 - q)^2 - m^2 + i\epsilon} \frac{1}{(k/2 + q)^2 - m^2 + i\epsilon}.$
2. Completing the path below, and replacing $\frac{1}{(k/2 + q)^2 - m^2 + i\epsilon} \rightarrow -2\pi i \delta((k/2 + q)^2 - m^2)$ to pick up the residue at $q^0 = -\frac{1}{2}k^0 + E_{\vb*{q}},$ we find $i\delta M = -2\pi i \frac{\lambda^2}{2} \frac{4\pi}{(2\pi)^4} \int_m^\infty \dd{E_{\vb*{q}}} E_{\vb*{q}} \abs{\vb*{q}} \frac{1}{2E_{\vb*{q}}} \frac{1}{k^0 (k^0 - 2E_{\vb*{q}})}.$ The discontinuity is given by replacing the pole with a delta function, i.e. another replacement $\frac{1}{(k/2 - q)^2 - m^2 + i\epsilon} \rightarrow -2\pi i \delta((k/2 - q)^2 - m^2).$ With the result to the first order for the $$\phi^4$$-theory, i.e. $$\mathcal{M}(k) = -\lambda$$, we verified to $$\lambda^2$$ that (cf. the optical theorem) \begin{align*} \operatorname{Disk} \mathcal{M}(k) &= 2i \Im \mathcal{M}(k) \\ &= \frac{i}{2} \int \frac{\dd{^3 p_1}}{(2\pi)^3} \frac{1}{2E_1} \frac{\dd{^3 p_2}}{(2\pi)^3} \frac{1}{2E_2} \abs{\mathcal{M}(k)}^2 (2\pi)^4 \delta^{(4)}(p_1 + p_2 - k). \end{align*}{}
3. We don't have other contributions yet.

#### Quantum Electrodynamics: Tree Level

##### Electron–Positron Annihilation: to Muon
###### Unpolarized Cross Section
1. Feynman diagram:
2. Amplitude: \begin{align*} i\mathcal{M} &= \overline{v}^{s'}(p')(-ie\gamma^\mu) u^s(p) \qty(\frac{-ig_{\mu\nu}}{q^2})\overline{u}^r(k)(-ie\gamma^\nu) v^{r'}(k') \\ &= \frac{ie^2}{q^2}\qty({\overline{v}(p')\gamma^\mu u(p)})\qty({\overline{u}(k)\gamma_\mu v(k')}). \end{align*}{}
3. Amplitude squared and averaged: \begin{align*} \frac{1}{4}\sum_{s,s',r,r'} \abs{\mathcal{M}(s,s' \rightarrow r,r')}^2 &= \frac{e^4}{4q^4} \tr\qty[(\unicode{x2215}\kern-.5em {p}' - m_e)\gamma^\mu (\unicode{x2215}\kern-.5em {p}+m_e)\gamma^\nu] \tr\qty[(\unicode{x2215}\kern-.5em {k} + m_\mu)\gamma_\mu (\unicode{x2215}\kern-.5em {k}' - m_\mu)\gamma_\nu]. \end{align*}{}
4. Traces evaluated: $\frac{1}{4}\sum_{s,s',r,r'} \abs{\mathcal{M}(s,s' \rightarrow r,r')}^2 = \frac{8e^4}{q^4}\qty[(p\cdot k)(p'\cdot k')+(p\cdot k')(p'\cdot k) + m_\mu^2 (p\cdot p')].$
5. New kinematic variables: \begin{align*} p &= (E,E\hat{\vb*{z}}), \\ p' &= (E, -E\hat{\vb*{z}}), \\ k &= (E,\vb*{k}), \\ k' &= (E,-\vb*{k}). \end{align*}{}
• Dot products: $\begin{gather*} q^2 = (p+p')^2 = 4E^2, \\ p\cdot p' = 2E^2, \\ p\cdot k = p'\cdot k' = E^2 - E\abs{\vb*{k}} \cos\theta, \\ p\cdot k' = p'\cdot k = E^2 + E\abs{\vb*{k}} \cos\theta. \end{gather*}{}$
• Amplitude simplified: $\frac{1}{4}\sum_{s,s',r,r'} \abs{\mathcal{M}(s,s' \rightarrow r,r')}^2 = e^4 \qty[\qty(1+\frac{m_\mu^2}{E^2}) + \qty(1-\frac{m_\mu^2}{E^2})\cos^2\theta].$
• With Mandelstam variables: in massless limit, $\frac{1}{4}\sum_{s,s',r,r'} \abs{\mathcal{M}(s,s' \rightarrow r,r')}^2 = \frac{8e^4}{s^2}\qty[\qty(\frac{t}{2})^2 + \qty(\frac{u}{2})^2].$
6. Differential cross section: \begin{align*} \dv{\sigma}{\Omega} &= \frac{1}{2E_{\mathrm{cm}}^2} \frac{\abs{\vb*{k}}}{16\pi^2 E_{\mathrm{cm}}} \cdot \frac{1}{4}\sum_{s,s',r,r'} \abs{\mathcal{M}(s,s' \rightarrow r,r')}^2 \\ &= \frac{\alpha^2}{4E_{\mathrm{cm}}^2}\sqrt{1-\frac{m_\mu^2}{E^2}}\qty[\qty(1+\frac{m_\mu^2}{E^2}) + \qty(1-\frac{m_\mu^2}{E^2})\cos^2\theta]. \end{align*}{} Total cross section: $\sigma = \frac{4\pi \alpha^2}{3E_{\mathrm{cm}}^2} \sqrt{1-\frac{m_\mu^2}{E^2}}\qty(1+\frac{1}{2}\frac{m_\mu^2}{E^2}).$
###### Helicity Structure

Relativistic limit: assuming massless. $e^-_R e^+_L \rightarrow \mu^-_R \mu^+_L.$

1. Feynman diagram: in the previous example.
2. Amplitude: we should apply projections onto specific helicities to $$u$$, $$v$$, etc. We apply it only to one of the incident particles and one of the outcoming particles, since $\overline{v}\gamma^\mu\qty(\frac{1+\gamma^5}{2})u = \overline{v}\gamma^\mu\qty(\frac{1+\gamma^5}{2})^2 u = {v}^\dagger \qty(\frac{1+\gamma^5}{2}) \gamma^0 \gamma^\mu\qty(\frac{1+\gamma^5}{2})u.$ We see the opposite helicities automatically applies to $$u$$ and $$v$$.
3. We hypocritically sum over all spins again, only to exploit the spin sum identities, although we know that some helicities are effectively thrown out.
4. Spin sum carried out. Trace Evaluated. \begin{align*} \sum_{s,s'} \abs{\overline{v}(p')\gamma^\mu \qty(\frac{1+\gamma^5}{2})u(p)}^2 &= \tr\qty[\unicode{x2215}\kern-.5em {p}'\gamma^\mu \unicode{x2215}\kern-.5em {p} \gamma^\nu \qty(\frac{1+\gamma^5}{2})] \\ &= 2(p'^\mu p^\nu + p'^\nu p^\mu - g^{\mu\nu}p\cdot p' - i\epsilon^{\alpha\mu\beta\nu} p'_\alpha p_\beta). \\ \sum_{r,r'} \abs{\overline{u}(k)\gamma^\mu \qty(\frac{1+\gamma^5}{2})v(k')}^2 &= 2(k_\mu k'_\nu - k_\nu k'_\mu - g_{\mu\nu} k\cdot k' - i\epsilon_{\rho\mu\sigma\nu} k^\rho k'^\sigma). \end{align*}{}
5. Using new kinematic variables: $\abs{\mathcal{M}}^2 = \qty(1+\cos\theta)^2.$
6. Differential cross section: \begin{align*} \dv{\sigma}{\Omega}\qty(e^-_R e^+_L \rightarrow \mu^-_R \mu^+_L) = \dv{\sigma}{\Omega}\qty(e^-_L e^+_R \rightarrow \mu^-_L \mu^+_R) = \frac{\alpha^2}{4E_{\mathrm{cm}}^2}\qty(1-\cos\theta)^2, \\ \dv{\sigma}{\Omega}\qty(e^-_R e^+_L \rightarrow \mu^-_L \mu^+_R) = \dv{\sigma}{\Omega}\qty(e^-_L e^+_R \rightarrow \mu^-_R \mu^+_L) = \frac{\alpha^2}{4E_{\mathrm{cm}}^2}\qty(1-\cos\theta)^2. \end{align*}

Brute-force approach: electrons and muons in relativistic limit,

• Amplitude $\mathcal{M} = \frac{e^2}{q^2}\qty(\overline{v}\qty{p'}\gamma^\mu u\qty{p})\qty(\overline{u}\qty(k)\gamma_\mu v\qty(k')).$
• $$u$$ and $$v$$ in relativistic limit: \begin{align*} u(p) &= \sqrt{2E}\cdot \frac{1}{2} \begin{pmatrix} (1-\hat{p}\cdot \vb*{\sigma}) \xi \\ (1+\hat{p}\cdot \vb*{\sigma}) \xi \end{pmatrix}, \\ v(p) &= \sqrt{2E}\cdot \frac{1}{2} \begin{pmatrix} (1-\hat{p}\cdot \vb*{\sigma}) \xi \\ -(1+\hat{p}\cdot \vb*{\sigma}) \xi \end{pmatrix}. \end{align*}{}
• Electron half by brute-force: $\overline{v}(p') \gamma^\mu u(p) = -2E(0,1,i,0).$
• Muon half by rotation: $\overline{u}(k) \gamma^\mu v(k') = -2E(0,\cos\theta,-i,-\sin\theta).$
• Amplitude $\mathcal{M}\qty(e^-_R e^+_L \rightarrow \mu^-_R \mu^+_L) = -e^2(1+\cos\theta).$

Brute-force approach: electrons in relativistic limit, muons in nonrelativistic limit,

• Amplitude: in the previous example.
• $$p$$ in relativistic limit: in the previous example. $$k$$ in nonrelativistic limit: $u(k) = \sqrt{m}\begin{pmatrix} \xi \\ \xi \end{pmatrix},\quad v(k') = \sqrt{m} \begin{pmatrix} \xi' \\ -\xi' \end{pmatrix}.$
• Electron half: in the previous example.
• Muon half: $\overline{u}(k) \gamma^\mu v(k') = m \begin{pmatrix} \xi^\dagger & \xi^\dagger \end{pmatrix}\begin{pmatrix} \overline{\sigma}^\mu & 0 \\ 0 & \sigma^\mu \end{pmatrix}\begin{pmatrix} \xi' \\ -\xi' \end{pmatrix} = \begin{cases} 0, & \text{if } \mu = 0, \\ -2m\xi^\dagger \sigma^i \xi', & \text{if } \mu = i. \end{cases}$
• Amplitude: $\mathcal{M}(e_R^- e_L^+ \rightarrow \mu^+ \mu^-) = -2e^2 \xi^\dagger \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \xi'.$
• Muon spins summed over: all four particles have spin $$\uparrow$$, \begin{align*} \mathcal{M}^2 &= 4e^4, \\ \dv{\sigma}{\Omega}(e_R^- e_L^+ \rightarrow \mu^+ \mu^-) &= \frac{\alpha^2}{E_{\mathrm{cm}}^2} \frac{\abs{\vb*{k}}}{E}. \end{align*}
##### Electron-Position Annihilation: to Bound States
• Amplitude of general spin configurations: $i\mathcal{M}(\text{something} \rightarrow \vb*{k}, \vb*{k}') = \xi^\dagger [\Gamma\qty(\vb*{k})] \xi = \tr(\xi' \xi^\dagger \Gamma(\vb*{k})) = \tr\qty(\frac{\vb*{n}^*\cdot \vb*{\sigma}}{\sqrt{2}}\Gamma\qty(\vb*{k})).$
• Substitution: $\xi' \xi^\dagger \rightarrow \frac{1}{\sqrt{2}}\vb*{n}^*\cdot \vb*{\sigma}.$
• $\uparrow\uparrow \quad\mathbin{\leftrightarrow}\quad \vb*{n} = \frac{\hat{\vb*{x}} + i\hat{\vb*{y}}}{\sqrt{2}},$
• $\downarrow\downarrow \quad\mathbin{\leftrightarrow}\quad \vb*{n} = \frac{\hat{\vb*{x}} - i\hat{\vb*{y}}}{\sqrt{2}},$
• $\frac{\uparrow\downarrow + \downarrow\uparrow}{\sqrt{2}} \quad\mathbin{\leftrightarrow}\quad \vb*{n} = \hat{\vb*{z}}.$
• Scattering into bound states: see bound states for the expansion of a bound state. \begin{align*} \mathcal{M}(\uparrow \uparrow \rightarrow B) &= \sqrt{2M} \int \frac{\dd{^3 k}}{(2\pi)^3} \tilde{\psi}^*(\vb*{k}) \frac{1}{\sqrt{2m}} \frac{1}{\sqrt{2m}} \mathcal{M}(\uparrow \uparrow \rightarrow \vb*{k}\uparrow, -\vb*{k}\uparrow) \\ &= \sqrt{\frac{2}{M}}(-2e^2)\psi^*(0). \end{align*}{}
• For general spin configurations, $i\mathcal{M}(\text{something}\rightarrow B) = \sqrt{\frac{2}{M}} \int \frac{\dd{^3 k}}{(2\pi)^3} \tilde{\psi}^*(\vb*{k}) \tr\qty(\frac{\vb*{n}^* \cdot \vb*{\sigma}}{\sqrt{2}}\Gamma\qty(\vb*{k})).$
• For the current case: $\mathcal{M}(e_R^- e_L^+ \rightarrow B) = \sqrt{\frac{2}{M}}(-2e^2)(\vb*{n}^* \cdot \vb*{\epsilon}_+) \psi^*(0).$
• Polarization states of incidental electrons: $\vb*{\epsilon}_+ = \frac{\hat{\vb*{x}} + i\hat{\vb*{y}}}{\sqrt{2}}.$
• Polarization of outcoming muons: $$\vb*{n}^*$$.
• Total cross section: averaged over all possible direction of $$\vb*{n}$$, \begin{align*} \sigma(e^+ e^- \rightarrow B) &= \frac{1}{2} \frac{1}{2m} \frac{1}{2m} \int \frac{\dd{^3 K}}{(2\pi)^3} \frac{1}{2E_{\vb*{K}}}(2\pi)^4 \delta^{(4)}(p+p'-K)\cdot \frac{2}{M}(4e^4)\frac{1}{2}\abs{\psi(0)}^2 \\ &= 64\pi^3 \alpha^2 \frac{\abs{\psi(0)}^2}{M^3}\delta(E^2_{\mathrm{cm}} - M^2). \end{align*}{}
##### Bound States Decay: to Electron-Position Pairs
• Decay of bound states: \begin{align*} \Gamma(B \rightarrow e^+ e^-) &= \frac{1}{2M} \int \dd{\Pi_2} \abs{\mathcal{M}}^2 \\ &= \frac{1}{2M}\int \qty(\frac{1}{8\pi} \frac{\dd{\cos\theta}}{2}) \frac{8e^4}{M}\abs{\psi(0)}^2\qty(\abs{\vb*{n}\cdot \vb*{\epsilon}}^2 + \abs{\vb*{n}\cdot \vb*{\epsilon}^*}^2). \end{align*}
• Averaged over all three directions of $$\vb*{n}$$: $\Gamma(B\rightarrow e^+ e^-) = \frac{16\pi \alpha^2}{3} \frac{\abs{\psi(0)}^2}{M^2}.$
##### Electron-Muon Scattering
1. Feynman diagram:
2. Amplitude: $i\mathcal{M} = \frac{ie^2}{q^2} \overline{u}(p'_1)\gamma^\mu u(p_1) \overline{u}(p'_2) \gamma_\mu u(p_2).$
3. Amplitude squared and averaged: $\frac{1}{4}\sum_{\text{spins}} \abs{\mathcal{M}}^2 = \frac{e^4}{4q^4}\tr\qty[(\unicode{x2215}\kern-.5em {p}'_1 + m_e)\gamma^\mu (\unicode{x2215}\kern-.5em {p}_1 + m_e)\gamma^\nu]\tr\qty[(\unicode{x2215}\kern-.5em {p}'_2 + m_\mu)\gamma_\mu (\unicode{x2215}\kern-.5em {p}_2 + m_\mu)\gamma_\nu].$
4. Traces evaluated $\frac{1}{4}\sum_{\text{spins}} \abs{\mathcal{M}}^2 = \frac{8e^4}{q^4}\qty[(p_1 \cdot p'_2)(p'_1 \cdot p_2) + (p_1\cdot p_2)(p'_1\cdot p'_2) - m_\mu^2 (p_1\cdot p'_1)].$
5. New kinematic variables: \begin{align*} p_1 &= (k,k\hat{\vb*{z}}), \\ p_2 &= (E, -k\hat{\vb*{z}}), \\ p'_1 &= (k,\vb*{k}), \\ p'_2 &= (E,-\vb*{k}). \end{align*}{}
• Dot products: $\begin{gather*} p_1 \cdot p_2 = p'_1 \cdot p'_2 = k(E+k), \\ p'_1 \cdot p_2 = p_1 \cdot p'_2 = k(E+k\cos\theta), \\ p_1 \cdot p'_1 = k^2(1-\cos\theta), \\ q^2 = -2p_1\cdot p'_1 = -2k^2(1-\cos\theta). \end{gather*}$
• Amplitude simplified: $\frac{1}{4}\sum_{\text{spins}} \abs{\mathcal{M}}^2 = \frac{2e^4}{k^2(1-\cos\theta)^2}\qty(\qty(E+k)^2 + \qty(E+k\cos\theta)^2 - m_\mu^2\qty(1-\cos\theta)).$
• With Mandelstam variables: in massless limit, $\frac{1}{4}\sum_{\text{spins}} \abs{\mathcal{M}}^2 = \frac{8e^4}{s^2}\qty[\qty(\frac{s}{2})^2 + \qty(\frac{u}{2})^2].$
6. Differential cross section: $\qty(\dv{\sigma}{\Omega})_{\mathrm{CM}} = \frac{\alpha^2}{2k^2(E+k)^2(1-\cos\theta)^2} \qty(\qty(E+k)^2 + \qty(E+k\cos\theta)^2 - m_\mu^2\qty(1-\cos\theta)).$
• Note the $$\theta^{-4}$$ singularity as $$\theta$$ approaches $$0$$.
##### Compton Scattering
1. Feynman diagrams:

$$+$$

1. Amplitude: \begin{align*} i\mathcal{M} &= \overline{u}(p')(-ie\gamma^\mu)\epsilon^*_\mu(k') \frac{i( \unicode{x2215}\kern-.5em {p} + \unicode{x2215}\kern-.5em {k} + m )}{(p+k)^2 - m^2}(-ie\gamma^\nu)\epsilon_\nu(k)u(p) + \overline{u}(p')(-ie\gamma^\nu)\epsilon_\nu(k) \frac{i( \unicode{x2215}\kern-.5em {p} - \unicode{x2215}\kern-.5em {k}' + m )}{(p-k')^2 - m^2}(-ie\gamma^\mu)\epsilon^*_\mu(k')u(p) \\ &= -ie^2 \epsilon^*_\mu(k')\epsilon_\nu(k)\overline{u}(p')\qty[\frac{\gamma^\mu \unicode{x2215}\kern-.5em {k} \gamma^\nu + 2\gamma^\mu p^\nu}{2p\cdot k} + \frac{-\gamma^\nu \unicode{x2215}\kern-.5em {k}' \gamma^\mu + 2\gamma^\nu p^\mu}{-2p\cdot k'}]u(p). \end{align*}{}
2. Amplitude squared and averaged: with the substitution applied, \begin{align*} \frac{1}{4}\sum_{\text{spins}}\abs{\mathcal{M}}^2 &= \frac{e^4}{4}g_{\mu\rho} g_{\nu\sigma} \cdot \tr\qty{ (\unicode{x2215}\kern-.5em {p}' + m)\qty[\frac{\gamma^\mu \unicode{x2215}\kern-.5em {k} \gamma^\nu + 2\gamma^\mu p^\nu}{2p\cdot k} + \frac{-\gamma^\nu \unicode{x2215}\kern-.5em {k}' \gamma^\mu + 2\gamma^\nu p^\mu}{-2p\cdot k'}] (\unicode{x2215}\kern-.5em {p} + m)\qty[\frac{\gamma^\sigma \unicode{x2215}\kern-.5em {k} \gamma^\rho + 2\gamma^\rho p^\sigma}{2p\cdot k} + \frac{-\gamma^\rho \unicode{x2215}\kern-.5em {k}' \gamma^\sigma + 2\gamma^\sigma p^\rho}{-2p\cdot k'}] } \\ &= \frac{e^4}{4}\qty{\frac{\vb{I}}{(2p\cdot k)^2} + \frac{\vb{II}}{(2p\cdot k)(2p\cdot k')} + \frac{\vb{III}}{(2p\cdot k')(2p\cdot k)} + \frac{\vb{IV}}{(2p\cdot k')^2}}. \end{align*}{}
3. Trace evaluated: $\frac{1}{4}\sum_{\text{spins}}\abs{\mathcal{M}}^2 = 2e^4 \qty[\frac{p\cdot k'}{p\cdot k} + \frac{p\cdot k}{p\cdot k'} + 2m^2\qty(\frac{1}{p\cdot k} - \frac{1}{p\cdot k'}) + m^4 \qty(\frac{1}{p\cdot k} - \frac{1}{p\cdot k'})^2].$
4. New kinematic variables: in the lab frame, \begin{align*} k &= (\omega, \omega \hat{\vb*{z}}), \\ p &= (m, 0), \\ k' &= (\omega', \omega' \sin\theta, 0, \omega' \cos\theta), \\ p' &= (E', \vb*{p}'). \end{align*}
• Dot products: $p\cdot k = m\omega,\quad p\cdot k' = m\omega'.$
• Compton's formula: $\omega' = \frac{\omega}{\displaystyle 1 + \frac{\omega}{m}(1-\cos\theta)}.$
5. Phase space integral: \begin{align*} \int \dd{\Pi_2} &= \int \frac{\dd{^3 k}}{(2\pi)^3} \frac{1}{2\omega'} \frac{\dd{^3 p'}}{(2\pi)^3} \frac{1}{2E'} (2\pi)^4 \delta^{(4)}(k'+p'-k-p) \\ &= \frac{1}{8\pi} \int \dd{\cos\theta} \frac{(\omega')^2}{\omega m}. \end{align*}{} Differential cross section: \begin{align*} \dv{\sigma}{\cos\theta} &= \frac{1}{2\omega} \frac{1}{2m} \cdot \frac{1}{8\pi} \frac{(\omega')^2}{\omega m}\cdot \qty(\frac{1}{4} \sum_{\text{spins}} \abs{\mathcal{M}}^2) \\ &= \frac{\pi \alpha^2}{m^2}\qty(\frac{\omega'}{\omega})^2 \qty[\frac{\omega'}{\omega} + \frac{\omega}{\omega'} - \sin^2\theta]. \end{align*}{}

High-energy behavior:

1. New kinematic variables: in the center-of-mass frame, \begin{align*} k &= (\omega, \omega \hat{\vb*{z}}), \\ p &= (E, -\omega \hat{\vb*{z}}), \\ k' &= (\omega, \omega \sin\theta, 0, \omega\cos\theta), \\ p' &= (E', -\omega \sin\theta, 0, -\omega\cos\theta). \end{align*}{}
• Amplitude simplified: for $$E\gg m$$ and $$\theta\approx \pi$$, $\frac{1}{4}\sum_{\text{spins}}\abs{\mathcal{M}}^2 = 2e^4\cdot \frac{E + \omega}{E + \omega \cos\theta}.$
2. Differential cross section: \begin{align*} \dv{\sigma}{\cos\theta} &= \frac{1}{2}\cdot \frac{1}{2E} \cdot \frac{1}{2\omega} \cdot \frac{\omega}{(2\pi)4(E+\omega)} \cdot \frac{2e^4 (E+\omega)}{E+\omega\cos\theta} \\ &\approx \frac{2\pi \alpha^2}{2m^2 + s(1+\cos\theta)}. \end{align*}{} Total cross section $\sigma \approx \frac{2\pi\alpha^2}{s}\log\qty(\frac{s}{m^2}).$
###### Helicity Structure
• $$u$$-channel diagram:
• Invariant matrix: $i\mathcal{M} = ie^2 \epsilon_\mu(k)\epsilon^*_\nu(k') \overline{u}(p')\gamma^\mu \frac{\unicode{x2215}\kern-.5em {p} - \unicode{x2215}\kern-.5em {k}' + m}{(p-k')^2 - m^2}\gamma^\nu u(p).$
• $\chi = \pi - \theta.$
• Region: $\frac{m}{\omega} \ll \chi \ll 1.$ Contribution from $(\unicode{x2215}\kern-.5em {p} - \unicode{x2215}\kern-.5em {k}').$
• Initial electron right-handed: final electron right-handed, initial photon right-handed, final photon right-handed, \begin{align*} \mathcal{M}(e^-_R \gamma_R \rightarrow e^-_R \gamma_R) &= -e^2 \epsilon_\mu(k) \epsilon^*_\nu(k')u^\dagger_R(p')\sigma^\mu \frac{\overline{\sigma}\cdot (p-k')}{-(\omega^2 \chi^2 + m^2)}\sigma^\nu u_R(p) &\approx \frac{4e^2\chi}{\chi^2 + m^2/\omega^2}. \end{align*}{}
• Equal contribution from left-handed initial electron.
• Region: direct backward scattering, $\chi = 0.$ Contribution from $$m$$.
• $\mathcal{M}(e^-_R \gamma_L \rightarrow e^-_L \gamma_R) = \frac{4e^2 m/\omega}{\chi^2 + m^2/\omega^2}.$
• Equal contribution from left-handed initial electron.
• Differential cross section: \begin{align*} \dv{\sigma}{\cos\theta} &= \frac{1}{2} \frac{1}{2E} \frac{1}{2\omega} \frac{\omega}{(2\pi)4(E+\omega)}\qty[\mathcal{M}(e^-_R \gamma_R \rightarrow e^-_R \gamma_R) + \mathcal{M}(e^-_R \gamma_L \rightarrow e^-_L \gamma_R)] \\ &= \frac{4\pi\alpha^2}{s(\chi^2 + 4m^2/s)}. \end{align*}{}
##### Electron-Position Annihilation: to Photons
1. Feynman diagrams:

$$+$$

1. Making replacements from the Compton scattering: $p \rightarrow p_1, \quad p' \rightarrow -p_2,\quad k\rightarrow -k_1,\quad k'\rightarrow -k_2.$
2. Amplitude squared and averaged, and trace evaluated: $\frac{1}{4}\sum_{\text{spins}}\abs{\mathcal{M}}^2 = -2e^4\qty[\frac{p_1 \cdot k_2}{p_1 \cdot k_1} + \frac{p_1 \cdot k_1}{p_1\cdot k_2} + 2m^2\qty(\frac{1}{p_1\cdot k_1} + \frac{1}{p_1 \cdot k_2}) - m^4 \qty(\frac{1}{p_1 \cdot k_1} + \frac{1}{p_1 \cdot k_2})^2].$
• Overall minus sign from the crossing relation and should be removed.
3. New kinematic variables: \begin{align*} p_1 &= (E,p\hat{\vb*{z}}), \\ p_2 &= (E,-p\hat{\vb*{z}}), \\ k_1 &= (E,E\sin\theta, 0, E\cos\theta), \\ k_2 &= (E, -E\sin\theta, 0, -E\cos\theta). \end{align*}{}
4. Differential cross section: $\dv{\sigma}{\cos\theta} = \frac{2\pi\alpha^2}{s}\qty(\frac{E}{p})\qty[\frac{E^2 + p^2 \cos^2\theta}{m^2 + p^2\sin^2\theta} + \frac{2m^2}{m^2 + p^2\sin^2\theta} - \frac{2m^4}{(m^2 + p^2\sin^2\theta)^2}].$

#### Quantum Electrodynamics: Loop

##### Bremsstrahlung: Classical Computation
• Classical process: an electron get a kicked at $$t=0$$ and $$\vb*{x} = 0$$, and undergoes a momentum change $p \rightarrow p'.$
• The energy radiated is given by \begin{align*} \mathrm{Energy} &= \int \frac{\dd{^3 k}}{(2\pi)^3} \sum_{\lambda=1,2} \frac{e^2}{2}\abs{\vb*{\epsilon}_\lambda(\vb*{k})\cdot \qty(\frac{\vb*{p}'}{k\cdot p'} - \frac{\vb*{p}}{k\cdot p})}^2 \\ &= \frac{e^2}{(2\pi)^2} \int \dd{k} \mathcal{I}(\vb*{v}, \vb*{v}') \\ &= \frac{\alpha}{\pi} \cdot k_{\mathrm{max}} \cdot \mathcal{I}(\vb*{v}, \vb*{v}'), \end{align*} where
• we have carefully picked a frame such that $$p^0 = p'^0$$, and defined $p^\mu = E(1,\vb*{v}), \quad p'^\mu = E(1,\vb*{v}');$
• differential intensity $\mathcal{I}(\vb*{v}, \vb*{v}') = \int \frac{\dd{\Omega_{\hat{\vb*{k}}}}}{4\pi} \qty({ \frac{2(1-\vb*{v}\cdot \vb*{v}')}{(1 - \hat{\vb*{k}}\cdot \vb*{v})(1 - \hat{\vb*{k}} \cdot \vb*{v}')} - \frac{m^2/E^2}{(1-\hat{\vb*{k}}\cdot \vb*{v}')^2} - \frac{m^2/E^2}{(1-\hat{\vb*{k}}\cdot \vb*{v})^2} });$
• cutoff at $$k_{\mathrm{max}}$$.
• In extreme relativistic limit: $$E\gg m$$,
• Differential intensity $\mathcal{I}(\vb*{v}, \vb*{v}') \approx 2 \log\qty(\frac{-q^2}{m^2}),$ where $q^2 = (p - p')^2.$
• Radiated energy $\mathrm{Energy} = \frac{2\alpha}{\pi} \int_0^{k_{\mathrm{max}}} \dd{k} \log\qty(\frac{-q^2}{m^2}).$
• Number of photons radiated $\mathrm{Number\ of\ photons} = \frac{\alpha}{\pi} \int_0^{k_{\mathrm{max}}} \dd{k} \frac{1}{k} \mathcal{I}(\vb*{v}, \vb*{v}').$
##### Bremsstrahlung: Quantum Computation
• Quantum process: one photon radiated during the scattering of an electron, before and after the interaction with the external field.

• Notations and assumptions:
• $$\mathcal{M}_0$$ the part of the amplitude that comes from the electrons's interaction with the external field.
• Soft radiation: $\abs{\vb*{k}} \ll \abs{\vb*{p}' - \vb*{p}}.$
• Approximation: $\mathcal{M}_0(p', p-k) \approx \mathcal{M}_0(p'+k, p) \approx \mathcal{M}_0(p', p).$
• $$\unicode{x2215}\kern-.5em {k}$$ ignored in the numerator.
• Amplitude: $i\mathcal{M} \approx \underbrace{\overline{u}(p')[\mathcal{M}_0(p',p)]u(p)}_{\text{elastic scattering}}\cdot \underbrace{\qty[{ e\qty({ \frac{p'\cdot \epsilon^*}{p'\cdot k} - \frac{p\cdot \epsilon^*}{p\cdot k} }) }]}_{\text{emission of photon}}.$
• Differential cross section: $\dd{\sigma(p\rightarrow p' + \gamma)} = \dd{\sigma(p\rightarrow p')} \cdot \int \dd{(\mathrm{prob})},$ where $\dd{(\mathrm{prob})} = \frac{\dd{^3 k}}{(2\pi)^3} \sum_\lambda \frac{e^2}{2k} \abs{\vb*{\epsilon}_\lambda \cdot \qty({ \frac{\vb*{p}'}{p'\cdot k} - \frac{\vb*{p}}{p\cdot k} })}^2$ gives the probability of emission of a single photon. This should be re-interpreted as the number of photon radiated.
• Total probability: photon mass $$\mu$$, $\mathrm{Total\ probability} \approx \frac{\alpha}{\pi} \int_{\mu}^{\sim \abs{\vb*{q}}} \dd{k} \frac{1}{k} \mathcal{I}(\vb*{v}, \vb*{v}').$ Therefore, $\dd{\sigma}(p\rightarrow \gamma(k)) \approx \dd{\sigma}(p\rightarrow p')\cdot \frac{\alpha}{\pi} \log\qty(\frac{-q^2}{\mu^2})\log\qty(\frac{-q^2}{m^2}).$
##### Electron Vertex Function: Formal Structure

$$=$$ $$+$$ $$+$$ $$+ \cdots$$

• Sum of vertex diagrams: $\begin{gather*} i\mathcal{M} (2\pi) \delta(p'^0 - p^0) = -ie \overline{u}(p') {\color{red} \gamma^\mu} u(p) \cdot \tilde{A}^{\mathrm{cl}}_\mu(p' - p) \\ \Downarrow \text{Vectex corrections} \\ i\mathcal{M} (2\pi) \delta(p'^0 - p^0) = -ie \overline{u}(p') {\color{red} \Gamma^\mu(p', p)} u(p) \cdot \tilde{A}^{\mathrm{cl}}_\mu(p' - p). \end{gather*}$
• Assuming perturbation of the form $\Delta H_{\mathrm{int}} = \int \dd{^3 x} eA^{\mathrm{cl}}_\mu \overline{\psi} \gamma^\mu \psi.$
• Form of $$\Gamma^\mu$$: $\color{red} \Gamma^\mu(p',p) = \gamma^\mu F_1(q^2) + \frac{i\sigma^{\mu\nu} q_\nu}{2m} F_2(q^2).$
• To the lowest order, $F_1 = 1, \quad F_2 = 0.$
###### Correction to Electron Charge

The amplitude of electron scattering from an electric field is given by $i\mathcal{M} = -ie F_1(0)\tilde{\phi}(\vb*{q}) \cdot 2m \xi'^\dagger \xi$ under non-relativistic limit for slowly varying field, which indicate a potential given by $V(\vb*{x}) = eF_1(0) \phi(\vb*{x}).$

###### Correction to Electron Magnetic Moment

The amplitude of electron scattering in an magnetic field is given by \begin{align*} i\mathcal{M} &= +ie\qty[{ \overline{u}(p') \qty({ \gamma^i F_1 + \frac{i\sigma^{i\nu} q_\nu}{2m} F_2 }) u(p) }] \tilde{A}^i_{\mathrm{cl}}(\vb*{q}) &= -i(2m) \cdot e \xi'^\dagger \qty({ \frac{-1}{2m} \sigma^k \qty[F_1(0) + F_2(0)] }) \xi \tilde{B}^k(\vb*{q}). \end{align*} This is equivalent to the potential $V(\vb*{x}) = -\langle \vb*{\mu} \rangle \cdot \vb*{B}(\vb*{x}),$ where \begin{align*} \langle \vb*{\mu} \rangle &= \frac{e}{m} \qty[F_1(0) + F_2(0)] \xi'^\dagger \frac{\vb*{\sigma}}{2} \xi; \\ \vb*{\mu} &= g\qty(\frac{e}{2m}) \vb*{S}; \\ g &= 2\qty[F_1(0) + F_2(0)] = 2 + 2F_2(0). \end{align*}

##### Electron Vertex Function: Evaluation

1. Tree-level correction plus loop: $\Gamma^\mu = \gamma^\mu + \delta \Gamma^\mu$ where $\overline{u}(p') \delta \Gamma^\mu(p',p) u(p) = 2ie^2 \int \frac{\dd{^4 k}}{(2\pi)^4} \frac{\overline{u}(p')\qty[ \unicode{x2215}\kern-.5em {k} \gamma^\mu \unicode{x2215}\kern-.5em {k} + m^2\gamma^\mu - 2m(k+k') ]u(p)}{((k-p)^2 + i\epsilon)(k'^2 - m^2 + i\epsilon)(k^2 - m^2 + i\epsilon)}.$
2. Feynman parameters: $\frac{1}{((k-p)^2 + i\epsilon)(k'^2 - m^2 + i\epsilon)(k^2 - m^2 + i\epsilon)} = \int_0^1 \dd{x} \dd{y} \dd{z} \delta(x+y+z-1) \frac{2}{D^3}.$
3. Completing the square: $D = l^2 - \Delta + i\epsilon$ and $\Delta = -xyq^2 + (1-z)^2 m^2.$
4. Simplifying the numerator: $\overline{u}(p') \delta \Gamma^\mu(p',p) u(p) = 2ie^2 \int \frac{\dd{^4 l}}{(2\pi)^4} \int_0^1 \dd{x} \dd{y} \dd{z} \delta(x+y+z-1) \frac{2}{D^3} \overline{u}(p')\qty[\mathrm{Numerator}] u(p)$ where $[\mathrm{Numerator}] = \gamma^\mu \cdot \qty({-\frac{1}{2} l^2 + (1-x)(1-y) q^2 + (1-4z+z^2)m^2 }) + \frac{i\sigma^{\mu\nu}q_\nu}{2m}(2m^2 z(1-z)).$
5. Regularizing the integral: introduce a large $$\Lambda$$ and replace the photon propagator $\frac{1}{(k-p)^2 + i\epsilon} \rightarrow \frac{1}{(k-p)^2 + i\epsilon} - \frac{1}{(k-p)^2 - \Lambda^2 + i\epsilon}.$ For the $$\Lambda$$ term, the denominator is altered by $\Delta \rightarrow \Delta_\Lambda = -xyq^2 + (1-z)^2 m^2 + z\Lambda^2.$ This replacement introduces a correction $\int \frac{\dd{^4 l}}{(2\pi)^4} \qty({ \frac{l^2}{[l^2 - \Delta]^3} - \frac{l^2}{[l^2 - \Delta_\Lambda]^3} }) = \frac{i}{(4\pi)^2} \log \qty(\frac{\Delta_\Lambda}{\Delta}) \approx \frac{i}{(4\pi)^2} \log \qty(\frac{z\Lambda^2}{\Delta}).$ We are counting on this correction to fix the divergence of the $$l^2$$-term in the numerator.
6. Applying Wick rotation, we obtain $\overline{u}(p') \delta \Gamma^\mu(p',p) u(p) = \frac{\alpha}{2\pi} \int_0^1 \dd{x}\dd{y}\dd{z} \delta(x+y+z-1) \overline{u}(p') \qty(\mathrm{Mess}) u(p)$ where $(\mathrm{Mess}) = \gamma^\mu \qty[{ \log \frac{z\Lambda^2}{\Delta} + \frac{1}{\Delta}\qty({ (1-x)(1-y)q^2 + (1-4z+z^2)m^2 }) }] + \frac{i\sigma^{\mu\nu} q_\nu}{2m}\qty[{ \frac{1}{\Delta} 2m^2 z(1-z) }].$

A few notes are in order:

• The correction to $$F_1$$ is nonzero at $$q = 0$$. Therefore, we manually set $\delta F_1(q^2) \rightarrow \delta F_1(q^2) - \delta F_1(0),$ which is justified later.
• $$\Delta$$ in the denominator introduces a divergence. We set the photon mass to a nonzero $$\mu$$, and therefore $\Delta \rightarrow \Delta + z\mu^2.$
• We are now able to obtain $F_2(q^2=0) = \frac{\alpha}{2\pi},$ and therefore the anomalous magnetic moment of electron $a_e = \frac{g-2}{2} = \frac{\alpha}{2\pi}.$
##### Electron Vertex Function: Electron Self-Energy Correction
• With the electron self-energy correction, the exact vertex function should be $Z_2 \Gamma^\mu(p',p) = \gamma^\mu F_1(q^2) + \frac{i\sigma^{\mu\nu} q_\nu}{2m} F_2(q^2).$
• $$F_2$$ is still of order $$\alpha$$.
• $$F_1(q^2)$$ receives a correction from $$Z_2$$, $F_1(q^2) = 1 + \delta F_1(q^2) + \delta Z_2 = 1 + [\delta F_1(q^2) - \delta F_1(0)]$ since $\delta Z_2 = -\delta F_1(0).$
##### Electron Vertex Function: Infrared Divergence
• Infrared divergence: only in $$F_1$$ near $$z=1$$ and $$x=y=0$$. Retaining only the terms divergent when $$\mu\rightarrow 0$$, and set $$x=y=0$$ and $$z=1$$ in the denominator, with the substitution $y = (1-z) \xi,\quad w = (1-z)$ we find \begin{align*}F_1(q^2) &= \frac{\alpha}{4\pi} \int_0^1 \dd{\xi} \qty[{ \frac{-2m^2 + q^2}{m^2 - q^2\xi(1-\xi)}\log\qty({ \frac{m^2 - q^2 \xi(1-\xi)}{\mu^2} }) + 2\log\qty(\frac{m^2}{\mu^2}) }]\\ &= 1 - \frac{\alpha}{2\pi} f_{\mathrm{IR}}(q^2)\log\qty({ \frac{-q^2\ \mathrm{or}\ m^2}{\mu^2} }) + \mathcal{O}(\alpha^2), \end{align*} where $f_{\mathrm{IR}}(q^2) = \int_0^1 \dd{\xi} \qty(\frac{m^2 - q^2/2}{m^2 - q^2 \xi(1-\xi)}) - 1.$
• Easily verified that $\color{red} \mathcal{I}(\vb*{v}, \vb*{v}') = 2f_{\mathrm{IR}}(q^2).$
• Divergence behavior of $$F_1$$: $F_1(-q^2 \rightarrow \infty) = 1 - \frac{\alpha}{2\pi} \log \qty(\frac{-q^2}{m^2}) \log \qty(\frac{-q^2}{\mu^2}) + \mathcal{O}(\alpha^2).$
• Since infrared divergence only affects $$F_1$$, the correction to the amplitude of electron scattering is given by replacing $e \rightarrow e\cdot F_1(q^2).$
###### Dawn: It Cancels
• Correction from the vertex factor: $\dv{\sigma}{\Omega}(p\rightarrow p') = \qty(\dv{\sigma}{\Omega})_0\qty[{ 1 - \frac{\alpha}{\pi}\log\qty(\frac{-q^2}{m^2})\log\qty(\frac{-q^2}{\mu^2}) + \mathcal{O}(\alpha^2) }];$
• Correction from Bremsstrahlung: $\dv{\sigma}{\Omega}(p\rightarrow p' + \gamma) = \qty(\dv{\sigma}{\Omega})_0\qty[{+\frac{\alpha}{\pi}\log\qty(\frac{-q^2}{m^2})\log\qty(\frac{-q^2}{\mu^2}) + \mathcal{O}(\alpha^2) }].$
###### Measurement
• Events are identified as $p \rightarrow p'$ if the Bremsstrahlung $$\gamma$$ is lower than a certain threshold. From the formula for the total probability we find \begin{align*} \qty(\dv{\sigma}{\Omega})_{\mathrm{measured}} &= \dv{\sigma}{\Omega}(p\rightarrow p') + \dv{\sigma}{\Omega}(p\rightarrow p' + \gamma(k < E_l)) \\ &= \qty(\dv{\sigma}{\Omega})_0\qty[{ 1 - \frac{\alpha}{\pi} f_{\mathrm{IR}}(q^2) \log\qty(\frac{-q^2\ \mathrm{or}\ m^2}{E_l^2}) + \mathcal{O}(\alpha^2) }]. \end{align*}
• In the $$-q^2 \gg m^2$$ limit we find $\qty(\dv{\sigma}{\Omega})_{\mathrm{measured}} = \qty(\dv{\sigma}{\Omega})_0\qty[{ 1 - \frac{\alpha}{\pi} \log\qty(\frac{-q^2}{m^2}) \log\qty(\frac{-q^2}{E_l^2}) + \mathcal{O}(\alpha^2) }].$
##### To Wrap Up: Beyond One Photon
• To get a singular denominator in an electron propagator, we need the momentum to be near the mass shell. Therefore, we consider only the diagrams in which an arbitrary hard process is modified by the addition of soft real and virtual photons on the electron legs.
• Emission at the outgoing leg: killing $$\unicode{x2215}\kern-.5em {k}$$ in the numerator and $$\mathcal{O}(k^2)$$ terms in the denominator,

and taking into account all permutations, the amplitude is given by \begin{align*} & \phantom{{}={}} \sum_{\mathrm{permutations\ on\ }(1,\cdots,n)} \overline{u}(p') \qty(e\frac{p'^{\mu_1}}{p'\cdot k_1}) \qty(e\frac{p'^{\mu_2}}{p'\cdot (k_1 + k_2)}) \cdots \qty(e\frac{p'^{\mu_2}}{p'\cdot (k_1 + \cdots + k_n)}) \cdots \\ &= \overline{u}(p')\qty(e\frac{p'^{\mu_1}}{p'\cdot k_1})\cdots \qty(e\frac{p'^{\mu_n}}{p'\cdot k_n}). \end{align*}
• Emission at the incoming leg: killing $$\unicode{x2215}\kern-.5em {k}$$ in the numerator and $$\mathcal{O}(k^2)$$ terms in the denominator,

the amplitude is obtained similarly.
• All possible combinations: $$p_i$$ may be on the incoming or outgoing leg,

the amplitude is given by $\overline{u}(p') i\mathcal{M}_{\mathrm{hard}} u(p) \cdot e\qty({ \frac{p'^{\mu_1}}{p'\cdot k_1}-\frac{p^{\mu_1}}{p\cdot k_1} })\cdots e\qty({ \frac{p'^{\mu_n}}{p'\cdot k_n}-\frac{p^{\mu_n}}{p\cdot k_n} }).$
• Virtual photons:
1. Pick two photon momenta $$k_i$$ and $$k_j$$, set $$k_j = -k_i = k$$;
2. Multiply by the photon propagator;
3. Integrate over $$k$$. This introduces a factor $\vb{X} = \frac{e^2}{2} \int \frac{\dd{^4 k}}{(2\pi)^4} \frac{-i}{k^2 + i\epsilon} \qty({ \frac{p'}{p'\cdot k} - \frac{p}{p\cdot k} }) \cdot \qty({ \frac{p'}{-p'\cdot k} - \frac{p}{-p\cdot k} }) = -\frac{\alpha}{2\pi} f_{\mathrm{IR}}(q^2) \log\qty(\frac{-q^2}{\mu^2})$ in the amplitude.
• Real photons: we apply Ward's identity here,
1. Multiply by its polarization vector;
2. Sum over polarizations;
3. Integrate the squared matrix element over the photon's phase space. This introduces a factor $\vb{Y} = \int \frac{\dd{^3 k}}{(2\pi)^3} \frac{1}{2k} e^2(-g_{\mu\nu}) \qty({ \frac{p'^\mu}{p'\cdot k}-\frac{p^{\mu}}{p\cdot k} }) \qty({ \frac{p'^\nu}{p'\cdot k}-\frac{p^{\nu}}{p\cdot k} }) = \frac{\alpha}{\pi} f_{\mathrm{IR}}(q^2) \log\qty(\frac{E^2_l}{\mu^2})$ in the cross section.
• To wrap up: \begin{align*} & \phantom{{}={}} \qty(\dv{\sigma}{\Omega})_{\mathrm{measured}}\qty(\vb*{p} \rightarrow \vb*{p}' + \mathrm{any\ number\ of\ photons\ with\ }k<E_l) \\ &= \qty(\dv{\sigma}{\Omega})_0 \times \qty(\sum_{m=0}^\infty \frac{\vb{X}^m}{m!})^2 \sum_{n=0}^\infty \frac{\vb{Y}^n}{n!} \\ &= \qty(\dv{\sigma}{\Omega})_0 \times \exp\qty[-\frac{\alpha}{\pi} f_{\mathrm{IR}}(q^2) \log \qty(\frac{-q^2}{E_l^2})]. \end{align*}
##### Electron Self-Energy

$$+$$

1. Free propagator $\frac{i(\unicode{x2215}\kern-.5em {p} + m_0)}{p^2 - m_0^2 + i\epsilon}$ plus one-loop correction $\frac{i(\unicode{x2215}\kern-.5em {p} + m_0)}{p^2 - m_0^2}[-i\Sigma_2(p)]\frac{i(\unicode{x2215}\kern-.5em {p} + m_0)}{p^2 - m_0^2 + i\epsilon}$ where $-i\Sigma_2(p) = (-ie)^2 \int \frac{\dd{^4 k}}{(2\pi)^4} \gamma^\mu \frac{i(\unicode{x2215}\kern-.5em {k} + m_0)}{k^2 - m_0^2 + i\epsilon} \gamma_\mu \frac{-i}{(p-k)^2 - \mu^2 + i\epsilon}.$
2. Feynman parameters: $\frac{1}{k_2 - m_0^2 + i\epsilon} \frac{1}{(p-k)^2 - \mu^2 + i\epsilon} = \int_0^1 \dd{x} \frac{1}{D^2}.$
3. Completing the square: $D = l^2 - \Delta + i\epsilon$ and $\Delta = -x(1-x) p^2 + x\mu^2 + (1-x) m_0^2.$
4. Simplifying the numerator: $-i\Sigma_2(p) = -e^2 \int_0^1 \dd{x} \int \frac{\dd{^4 l}}{(2\pi)^4} \frac{-2 x\unicode{x2215}\kern-.5em {p} + 4m_0}{D^2}.$
5. Regularizing the integral: introduce a large $$\Lambda$$ and replace the photon propagator $\frac{1}{(p-k)^2 + i\epsilon} \rightarrow \frac{1}{(p-k)^2 + i\epsilon} - \frac{1}{(k-p)^2 - \Lambda^2 + i\epsilon}.$ For the $$\Lambda$$ term, the denominator is altered by $\Delta \rightarrow \Delta_\Lambda = -x(1-x)p^2 + x\mu^2 + (1-x)m_0^2.$ This replacement introduces a correction which gives $\frac{i}{(4\pi)^2} \int_0^\infty \dd{l_E^2} \qty({ \frac{l_E^2}{[l_E^2 + \Delta]^2} - \frac{l_E^2}{[l_E^2 + \Delta_\Lambda]^2} }) = \frac{i}{(4\pi)^2} \log\qty(\frac{\Delta_\Lambda}{\Delta}).$
6. Applying Wick rotation, we obtain (with $$\Lambda \rightarrow \infty$$) $\Sigma_2(p) = \frac{\alpha}{2\pi} \int_0^1 \dd{x} (2m_0 - x\unicode{x2215}\kern-.5em {p}) \log \qty(\frac{x\Lambda^2}{(1-x) m_0^2 + x\mu^2 - x(1-x)p^2}).$
• Analytic Behavior of $$\Sigma_2$$: branch cut occurs when $p^2 = (m_0 + \mu)^2.$
• 1PI diagrams: denoted by $$-i\Sigma(p)$$, which equals to $$-i\Sigma_2(p)$$ to the first order of $$\alpha$$.
• Two-point correlation with correction \begin{align*} \int \dd{^4 x} \bra{\Omega} T\psi(x)\overline{\psi}(0)\ket{\Omega} e^{-ip\cdot x} &= \frac{i(\unicode{x2215}\kern-.5em {p} + m_0)}{p^2 - m_0^2} + \frac{i(\unicode{x2215}\kern-.5em {p} + m_0)}{p^2 - m_0^2} (-i\Sigma) \frac{i(\unicode{x2215}\kern-.5em {p} + m_0)}{p^2 - m_0^2} + \cdots \\ &= \frac{i}{\unicode{x2215}\kern-.5em {p} - m_0 - \Sigma(\unicode{x2215}\kern-.5em {p})}. \end{align*}
• Position of pole determined by $\qty[\unicode{x2215}\kern-.5em {p} - m_0 - \Sigma(\unicode{x2215}\kern-.5em {p})]\vert_{\unicode{x2215}\kern-.5em {p} = m} = 0.$
• $$Z_2$$ determined by $Z_2^{-1} = 1 - \left.\dv{\Sigma}{\unicode{x2215}\kern-.5em {p}}\right|_{\unicode{x2215}\kern-.5em {p} = m}.$
• To order $$\alpha$$, $\delta m = m - m_0 \approx \Sigma_2(\unicode{x2215}\kern-.5em {p} = m_0) \xrightarrow{\Lambda \rightarrow \infty} \frac{3\alpha}{4\pi} m_0 \log\qty(\frac{\Lambda^2}{m_0^2}).$
• Not hard to show that $\delta F_1(0) + \delta Z_2 = 0.$
##### Renormalization of Charge

1. One-loop correction \begin{align*}i\Pi^{\mu\nu}_2(q) &= -(-ie)^2 \int \frac{\dd{^4 k}}{(2\pi)^4} \tr\qty[{ \gamma^\mu \frac{i(\unicode{x2215}\kern-.5em {k} + m)}{k^2 - m^2} \gamma^\nu \frac{i(\unicode{x2215}\kern-.5em {k} + \unicode{x2215}\kern-.5em {q} + m)}{(k+q)^2 - m^2} }] \\ &= -4e^2 \int \frac{\dd{^4 k}}{(2\pi)^4} \frac{k^\mu (k+q)^\nu + k^\nu (k+q)^\mu - g^{\mu\nu}(k\cdot (k+q) - m^2)}{(k^2 - m^2)((k+q)^2 - m^2)}. \end{align*}{}
2. Feynman parameters: $\frac{1}{(k^2 - m^2)((k+q)^2 - m^2)} = \int_0^1 \dd{x} \frac{1}{D^2}.$
3. Completing the square: $D = l^2 - \Delta + i\epsilon$ and $\Delta = m^2 - x(1-x) q^2.$
4. Simplifying the numerator: $\mathrm{Numerator} = 2l^\mu l^\nu - g^{\mu\nu}l^2 - 2x(1-x) q^\mu q^\nu + g^{\mu\nu} \qty(m^2 + x(1-x) q^2) + (\mathrm{terms\ linear\ in\ }l).$
5. Applying Wick rotation, $i\Pi^{\mu\nu}_2(q) = -4ie^2 \int_0^1 \dd{x} \int \frac{\dd{^4 l_E}}{(2\pi)^4} \frac{-(1/2)g^{\mu\nu} l_E^2 + g^{\mu\nu} l_E^2 - 2x(1-x)q^\mu q^\nu + g^{\mu\nu}(m^2 + x(1-x) q^2)}{(l_E^2 + \Delta)^2}.$
6. Regularizing the integral: using dimensional regularization. \begin{align*} i\Pi_2^{\mu\nu}(q) &= -4ie^2 \int_0^1 \dd{x} \frac{1}{(4\pi)^{d/2}} \frac{\Gamma(2-d/2)}{\Delta^{2-d/2}} \times \qty[{ g^{\mu\nu}(-m^2 + x(1-x) q^2) + g^{\mu\nu}(m^2 + x(1-x)q^2) - 2x(1-x)q^\mu q^\nu }] \\ &= (q^2 g^{\mu\nu} - q^\mu q^\nu) \cdot i\Pi_2(q^2) \end{align*} where \begin{align*} \Pi_2(q^2) &= \frac{-8e^2}{(4\pi)^{d/2}} \int_0^1 \dd{x} x(1-x) \frac{\Gamma(2-d/2)}{\Delta^{2-d/2}} \\ &\xrightarrow{d\rightarrow 4} -\frac{2\alpha}{\pi} \int_0^1 \dd{x} x(1-x) \qty(\frac{2}{\epsilon} - \log \Delta - \gamma + \log(4\pi)). \end{align*}
• Ward identity violated without regularization.
• The observed difference is given by $\hat{\Pi}_2(q^2) = \Pi_2(q^2) - \Pi_2(0) = -\frac{2\alpha}{\pi} \int_0^1 \dd{x} x(1-x) \log\qty(\frac{m^2}{m^2 - x(1-x) q^2}).$
• Branch cut of $$\Pi_2$$ lies at $q^2 > 4m^2.$
• Discontinuity by $\Im[\hat{\Pi}_2(q^2 \pm i\epsilon)] = \mp \frac{\alpha}{3} \sqrt{1-\frac{4m^2}{q^2}}\qty(1+\frac{2m^2}{q^2}).$
• Let $$i\Pi^{\mu\nu}(q)$$ denote the 1PI diagram of photon. With the Ward identity we identify $\color{red} \Pi^{\mu\nu}(q) = (q^2 g^{\mu\nu} - q^\mu q^\nu) \Pi(q^2).$
• Exact two-point function: \begin{align*} &\phantom{{}={}} \frac{-ig_{\mu\nu}}{q^2} + \frac{-ig_{\mu\nu}}{q^2}\qty[i(q^2 g^{\rho\sigma} - q^\rho q^\sigma)\Pi(q^2)]\frac{-ig_{\sigma\nu}}{q^2} + \cdots \\ &= \frac{-i}{q^2(1-\Pi(q^2))}\qty(g_{\mu\nu} - \frac{q_\mu q_\nu}{q^2}) + \frac{-i}{q^2}\qty(\frac{q_\mu q_\nu}{q^2}) \\ &\doteq \frac{-i g_{\mu\nu}}{q^2(1-\Pi(q^2))}. \end{align*}
• The last equality holds because of Ward identity, i.e. terms proportional to $$q_\mu$$ sums to zero.
• The residue of the $$q^2 = 0$$ pole: $\frac{1}{1-\Pi(0)} = Z_3.$
• Renormalization of charge: $e = \sqrt{Z_3} e_0.$
• $$Z_3 = 1$$ to the lowest order.
• For $$q^2 \neq 0$$, replace $\alpha_0 \rightarrow \alpha_{\mathrm{eff}}(q^2) = \frac{e_0^2/4\pi}{1-\Pi(q^2)} \approx \frac{\alpha}{1-[\Pi_2(q^2) - \Pi_2(0)]}.$

Uehling potential: recalculate the tree level $e^- + e^- \rightarrow e^- + e^-$ we find, for $$r\gg 1/m$$, \begin{align*} V(\vb*{x}) &= \int \frac{\dd{^3 q}}{(2\pi)^3} e^{i\vb*{q} \cdot \vb*{x}} \frac{-e^2}{\abs{\vb*{q}}^2 [1 - \hat{\Pi}_2(-\abs{\vb*{q}}^2)]} \\ &\approx -\frac{\alpha}{r}\qty(1 + \frac{\alpha}{4\sqrt{\pi}} \frac{e^{-2mr}}{(mr)^{3/2}}). \end{align*} However, at small distance, or $$-q^2 \gg m^2$$, $\alpha_{\mathrm{eff}}(q^2) = \frac{\alpha}{\displaystyle 1-\frac{\alpha}{3\pi} \log\qty(\frac{-q^2}{Am^2})}$ where $$A = \exp(5/3).$$

#### Miscellaneous

##### Asymptotic Series
 Index Index

2021/4/4 5:36:25

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