 Ze Chen Quantum Field Theory (II) Quantization of Fields

## Quantum Field Theory (III)

Feynman Diagrams

あの日見た1PIの伝播関数は僕たちはまだ知らない

#### Foundations of Field Theory

##### Heisenberg Picture and Interaction Picture
• Heisenberg picture: $\phi(t,\vb*{x}) = e^{iH(t-t_0)} \phi(t_0,\vb*{x}) e^{-iH(t-t_0)}.$
• Interaction picture: $\phi_I(t,\vb*{x}) = e^{iH_0(t-t_0)} \phi(t_0, \vb*{x}) e^{-iH_0(t-t_0)}.$
##### Scalar Field
• Expansion (Schrödinger picture): $\phi(t_0, \vb*{x}) = \int \frac{\dd{^3 p}}{(2\pi)^3} \frac{1}{\sqrt{2E_{\vb*{p}}}} \qty(a_{\vb*{p}} e^{i\vb*{p}\cdot \vb*{x}} + a^\dagger_{\vb*{p}} e^{-i\vb*{p}\cdot \vb*{x}}).$
• Expansion (Interaction picture): $\phi_I(t,\vb*{x}) = \left. \int \frac{\dd{^3 p}}{(2\pi)^3} \frac{1}{\sqrt{2E_{\vb*{p}}}} \qty(a_{\vb*{p}} e^{-ip\cdot x} + a^\dagger_{\vb*{p}} e^{ip\cdot x}) \right\vert_{x^0 = t-t_0}.$
##### Dirac Field
• Expansion: \begin{align*} \psi(x) &= \int \frac{\dd{^3 p}}{(2\pi)^3} \frac{1}{\sqrt{2E_{\vb*{p}}}} \sum_s \qty(a^s_{\vb*{p}} u^s(p) e^{-ip\cdot x} + b^{s\dagger}_{\vb*{p}} v^s(p) e^{ip\cdot x}), \\ \overline{\psi}(x) &= \int \frac{\dd{^3 p}}{(2\pi)^3} \frac{1}{\sqrt{2E_{\vb*{p}}}} \sum_s \qty(b^s_{\vb*{p}} \overline{v}^s(p) e^{-ip\cdot x} + a^{s\dagger}_{\vb*{p}} \overline{u}^s(p) e^{ip\cdot x}). \end{align*}{}
##### Electromagnetic Field
• Free electromagnetic field (under the Lorenz gauge): $\partial^2 A_\mu = 0.$
• Each component of $$A$$ obeys the Klein-Gordon equation with $$m=0$$.
• Expansion: $A_\mu(x) = \int \frac{\dd{^3} p}{(2\pi)^3} \frac{1}{\sqrt{2E_{\vb*{p}}}} \sum_{r=0}^3 \qty(a_{\vb*{p}}^r\epsilon^r_\mu{(p)}e^{-ip\cdot x} + a^{r\dagger}_{\vb*{p}} \epsilon^{r*}_{\mu}(p) e^{ip\cdot x}).$
• In the integration we have $$p^2=0$$, and
• $$\epsilon_\mu(p)$$ is a four-vector.
##### Propagator
• $$\ket{\Omega}$$ is usually different from $$\ket{0}$$.
• Note the extra minus sign for Dirac fields.
• Feynman propagator
• Scalar field (definition): $D_F = \bra{0}T\phi(x)\phi(y) \ket{0}.$
• Scalar field (in momentum space): $D_F(x-y) = \int \frac{\dd{^4 p}}{(2\pi)^4} \frac{i}{p^2 - m^2 + i\epsilon} e^{-ip\cdot (x-y)}.$
• Dirac field (definition): $S_F(x-y) = \bra{0}T\psi(x)\overline{\psi}(y)\ket{0}.$
• Dirac field (in momentum space): $S_F(x-y) = \int \frac{\dd{^4 p}}{(2\pi)^4} \frac{i(\unicode{x2215}\kern-.5em {p} + m)}{p^2 - m^2 + i\epsilon} e^{-ip\cdot (x-y)}.$
• Two-point correlation function, or two-point Green's function: $\bra{\Omega}T\phi(x)\phi(y)\ket{\Omega},$ where $$\ket{\Omega}$$ is the ground state of the interacting theory.
• Källén-Lehmann spectral representation: analytic structure of two-point correlation, $\color{red} \bra{\Omega} T\phi(x)\phi(y) \ket{\Omega} = \int_0^\infty \frac{\dd{M^2}}{2\pi} \rho(M^2) D_F(x-y, M^2).$
• Spectral density: $\color{red} \rho(M^2) = \sum_\lambda (2\pi) \delta(M^2 - m_\lambda^2) \abs{\bra{\Omega} \phi(0) \ket{\lambda_0}}^2.$
• $\rho(M^2) = 2\pi\delta(M^2 - m^2) \cdot Z + (\text{nothing else until } M^2 \gtrapprox (2m)^2).$
• The leading contribution is from the single-particle state.
• There are poles from single-particle bounded states near $$M^2 = (2m)^2$$.
• $$Z$$ is referred to as the field-strength renormalization.
• Physical mass: $$m$$ is the exact mass of a single particle, i.e. the exact energy eigenvalue at rest, different from the value of the mass parameter in the Lagrangian.
• Only the physical mass is directly observable.
• Bare mass: $$m_0$$ in the Lagrangian.
• $$\ket{\lambda_{0}}$$: an eigenstate of $$H$$ annihilated by $$\vb*{P}$$.
• Källén-Lehmann spectral representation for Scalar fields: $\int \dd{^4 x} e^{ip\cdot x} \bra{\Omega} T\phi(x) \phi(0) \ket{\Omega} = \frac{iZ}{p^2 - m^2 + i\epsilon} + \int_{\sim 4m^2}^\infty \frac{\dd{M^2}}{2\pi} \rho(M^2) \frac{i}{p^2 - M^2 + i\epsilon}.$
• $$Z_2$$: the probability for the quantum field to create or annihilate an exact one-particle eigenstate of $$H$$, $\sqrt{Z_2} = \bra{\Omega} \phi(0) \ket{p}.$
• Källén-Lehmann spectral representation for Dirac fields: $\int \dd{^4 x} e^{ip\cdot x} \bra{\Omega} T\psi(x)\overline{\psi}(0) \ket{\Omega} = \frac{i Z_2 (\unicode{x2215}\kern-.5em {p} + m)}{p^2 - m^2 + i\epsilon} + \cdots.$
• $$Z_2$$: the probability for the quantum field to create or annihilate an exact one-particle eigenstate of $$H$$, $\sqrt{Z_2} u^s(p) = \bra{\Omega}\psi(0)\ket{p,s}.$
##### Path Integral
• Scalar field: \begin{align*} \color{orange} Z_0[J] &\color{orange}= \int \mathcal{D}\varphi\, e^{i\int \dd{^4 x} [\mathcal{L_0} + J\varphi]} \\ &\color{orange}= \exp\qty[\frac{1}{2}\iint \dd{^4 x} \dd{^4 x'} J(x) S_F(x-x') J(x')]. \end{align*}
• \begin{align*} \color{darkcyan} Z[J] &\color{darkcyan}= e^{i\int \dd{^4} \mathcal{L}_1(\frac{1}{i}\frac{\delta}{\delta J(x)})} Z_0[J]. \end{align*}
• $$\mathcal{L}_1$$ is a perturbation, $\mathcal{L} = \mathcal{L}_0 + \mathcal{L}_1.$
##### Formulation of Scattering Problem

We solve the problem where

• the incident particles occupy a volume of length $$l_B$$ with number density $$\rho_B$$ and velocity (uniformly) $$v$$;
• the target particles occupy a volume of length $$l_A$$ with number density $$\rho_A$$; and
• the area common to the two branches is $$A$$.

We define a few quantities below.

• Cross section: $\color{orange}\sigma = \frac{\text{Number of scattering events}}{\rho_A l_A \rho_B l_B A},$
• Equivalently, \begin{align*} &\text{Number of events} \\ &= \sigma l_A l_B \int \dd{^2 x} \rho_A(x) \rho_B(x) \\ &= \frac{\sigma N_A N_B}{A}. \end{align*}
• Differential cross section: $\color{orange}\frac{\dd{\sigma}}{\dd{^3 p_1}\cdots \dd{^3 p_n}}$
• $$p_1,\cdots,p_n$$ denote the momenta of final states;
• integration over any small $$\dd{^3 p_1}\cdots \dd{^3 p_n}$$, gives the cross section for scattering into the final-state momentum space.
• Decay rate: $\color{orange}\Gamma = \frac{\text{Number of decays per unit time}}{\text{Number of } A \text{ particles present}}.$
• Breit-Wigner formula: near the reasonance energy, the scattering amplitude is given by \begin{align*} & f(E) \propto \frac{1}{E - E_0 + i\Gamma/2}\\ &\xlongequal{\text{relativistic}} \frac{1}{p^2 - m^2 + im\Gamma} \\ &= \frac{1}{2E_{\vb*{p}}(p^0 - E_{\vb*{p}} + i(m/E_{\vb*{p}})\Gamma/2)}, \end{align*}
• and the cross section by $\sigma \propto \frac{1}{(E-E_0)^2 + \Gamma^2/4}.$
##### Cross Section and S-Matrix

We use Heisenberg picture in this subsection.

We reduce the cross section to the $$\mathcal{M}$$ matrix.

• $$B$$ denotes the incident particles.
• Normalization: $\ket{\vb*{k}} = \sqrt{2E_{\vb*{k}}}a_{\vb*{k}}^\dagger\ket{0}.$
• $$\bra{\phi}\ket{\phi} = 1$$ if $\int \frac{\dd{^3 k}}{(2\pi)^3} \abs{\phi(\vb*{k})}^2 = 1.$
• Expansion of in states: plane wave to in state, $\ket{\phi_A \phi_B}_{\text{in}} = \int \frac{\dd{^3 k_A}}{(2\pi)^3} \int \frac{\dd{^3 k_B}}{(2\pi)^3} \frac{\phi_A(\vb*{k}_A) \phi_B(\vb*{k}_B) e^{-i\vb*{b}\cdot \vb*{k}_B}}{\sqrt{(2E_A) (2E_B)}}\ket{\vb*{k}_A \vb*{k}_B}_{\text{in}}.$
• Constructed in the far past.
• $$\phi_{B}(\vb*{k}_B)$$ is the state of the incident particle, with $$\vb*{b} = 0$$.
• Expansion of out states: plane wave to out state, $_{\text{out}}\bra{\phi_1\phi_2\cdots} = \qty(\prod_f \int \frac{\dd{^3 p_f}}{(2\pi)^3} \frac{\phi_f (\vb*{p}_f)}{\sqrt{2E_f}}) {_{\text{out}}\bra{\vb*{p}_1 \vb*{p}_2 \cdots}}.$
• Constructed in the far future.
• $$S$$-matrix: \begin{align*} &\color{orange}\phantom{{}={}} _{\text{out}}\bra{\vb*{p}_1\vb*{p}_2\cdots}\ket{\vb*{k}_A \vb*{k}_B}_{\text{in}} \\ &\color{orange}= \bra{\vb*{p}_1\vb*{p}_2\cdots} e^{-iH(2T)} \ket{\vb*{k}_A \vb*{k}_B} \\ &\color{orange}= \bra{\vb*{p}_1\vb*{p}_2\cdots} S \ket{\vb*{k}_A \vb*{k}_B}. \end{align*}
• $$T$$-matrix: $\color{orange} S = \mathbb{1} + iT$
• Invariant matrix element $$\mathcal{M}$$: brakets between in and out plane waves, $\color{orange} \bra{\vb*{p}_1\vb*{p}_2\cdots} iT \ket{\vb*{k}_A \vb*{k}_B} = (2\pi)^4 \delta^{(4)}\qty(k_A + k_B - \sum p_f)\cdot i\mathcal{M}\qty(k_A,k_B \rightarrow p_f),$
• The $$\delta^{(4)}$$ is imposed by the conservation of momentum.
• The probability of getting a certain out state (depending on the impact parameter): brakets of in and out states to probability, $\mathcal{P}(AB\rightarrow 1,2,\cdots,n) = \qty(\prod_f \frac{\dd{^3 p_f}}{(2\pi)^3}\frac{1}{2E_f})\abs{{_{\text{out}}\bra{\vb*{p}_1\cdots \vb*{p}_n}\ket{\phi_A \phi_B}_{\text{in}} }}^2.$
• Cross section: probability to cross section, summing over all impact parameters, $\sigma = \int \dd{^2 b} \mathcal{P}(\vb*{b}).$
• Putting together: \begin{align*} \dd{\sigma} &= \qty(\prod_f \frac{\dd{^3 p_f}}{(2\pi)^3}\frac{1}{2E_f}) \frac{\abs{\mathcal{M}(p_A,p_B\rightarrow \qty{p_f})}^2}{2E_A 2E_B \abs{v_A - v_B}} \int \frac{\dd{^3 k_A}}{(2\pi)^3} \int \frac{\dd{^3 p_B}}{(2\pi)^3} \\ &{\phantom{{}={}}} \times \abs{\phi_A \qty(\vb*{k}_A)}^2 \abs{\phi_B\qty(\vb*{k}_B)}^2 (2\pi)^4 \delta^{(4)}\qty(k_A + k_B - \sum p_f). \end{align*}{}
• Conclusion: $$\mathcal{M}$$ matrix to cross section, \begin{align*} \color{red}\dd{\sigma} &= \color{red}\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}) \\ &{\phantom{{}={}}} \color{red}\times \abs{\mathcal{M}(p_A,p_B\rightarrow \qty{p_f})}^2 (2\pi)^4 \delta^{(4)}\qty(p_A+p_B - \sum p_f). \end{align*}{}
• We demanded the wave packet be smooth (not so sharp).

When doing integration to get the total cross section or decay rate for a final state of $$n$$ particles, one should either restrict the integration to inequivalent configurations or divide the result by $$n!$$.

• Two-body scattering:
• In the center-of-mass frame, $$\vb*{p_1} = -\vb*{p}_2$$.
• $$E_{\mathrm{cm}}$$ denote the total energy.
• Cross section of two-particles: $\color{red} \qty(\dv{\sigma}{\Omega})_{\text{CM}} = \frac{1}{2E_A 2E_B \abs{v_A - v_B}} \frac{\abs{\vb*{p}_1}}{(2\pi)^2 4E_{\text{cm}}}\abs{\mathcal{M}(p_A,p_B \rightarrow p_1,p_2)}^2.$
• Identical mass: $\color{red} \qty(\dv{\sigma}{\Omega})_{\text{CM}} = \frac{\abs{\mathcal{M}}^2}{64\pi^2 E^2_{\text{cm}}}.$
• Decay rate: $\color{red} \dd{\Gamma} = \frac{1}{2m_A}\qty(\prod_f \frac{\dd{^3 p_f}}{(2\pi)^3} \frac{1}{2E_f}) \abs{\mathcal{M}(m_A \rightarrow \qty{p_f})}^2 (2\pi)^4 \delta^{(4)}\qty(p_A - \sum p_f).$
###### Remark on Decay Rate
• $$M^2(p^2)$$ denotes the sum of all 1PI insertions into the $$\phi^4$$ propagator. Let $$m$$ be defined by $m^2 - m_0^2 - \Re M^2(m^2) = 0.$
• The two-point correlation is found to be $\frac{i Z}{p^2 - m^2 - iZ \Im M^2(p^2)}.$
• Decay rate identified to be $\Gamma = -\frac{Z}{m} \Im M^2(m^2).$
• cf. Breit-Wigner formula $\sigma \propto \abs{\frac{1}{p^2 - m^2 + im\Gamma}}^2.$
##### Born Approximation

Born approximation: $\bra{p'}iT\ket{p} = -i\tilde{V}\qty(\vb*{q})(2\pi)\delta \qty(E_{\vb*{p}'} - E_{\vb*{p}})$ where $$\vb*{q} = \vb*{p}' - \vb*{p}$$.

• Assuming nonrelativistic normalization: $\bra{p'}\ket{p} = \delta(p'-p).$
##### LSZ Reduction Formula

LSZ Reduction Formula \color{red} \begin{align*} & \prod_1^n \int \dd{^4 x_i} e^{ip_i\cdot x_i} \prod_1^m \int \dd{^4 y_j} e^{-ik_i \cdot y_j} \bra{\Omega} T\qty{\phi(x_1) \cdots \phi(x_n) \phi(y_1) \cdots \phi(y_m)} \ket{\Omega} \\ & \underset{\substack{p_i^0 \rightarrow E_{\vb*{p}_i} \\ k_j^0 \rightarrow E_{\vb*{p}_j}}}{\sim} \qty( \prod_1^n \frac{\sqrt{Z} i}{p_i^2 - m^2 + i\epsilon} )\qty( \prod_1^m \frac{\sqrt{Z} i}{k_j^2 - m^2 + i\epsilon} ) \bra{\vb*{p}_1 \cdots \vb*{p}_n} S \ket{\vb*{k_1} \cdots \vb*{k}_m}. \end{align*}

##### The Optical Theorem
• From the unitarity of $$S$$: \begin{align*} &\phantom{{}={}}-i[\mathcal{M}(k_1k_2 \rightarrow p_1p_2) - \mathcal{M}^*(p_1p_2 \rightarrow k_1k_2)] \\ &= \sum_n \qty(\prod_{i=1}^n \int \frac{\dd{^3 q_i}}{(2\pi)^3} \frac{1}{2E_i}) \mathcal{M}^*(p_1p_2 \rightarrow \qty{q_i}) \mathcal{M}(k_1k_2 \rightarrow \qty{q_i}) \times (2\pi)^4 \delta^{(4)}(k_1 + k_2 - \sum_i q_i). \end{align*}{}
• Abbreviated as $-i[\mathcal{M}(a\rightarrow b) - \mathcal{M}^*(b\rightarrow a)] = \sum_f \int \dd{\Pi_f} \mathcal{M}^*(b\rightarrow f) \mathcal{M}(a\rightarrow f).$
• With the factors in the formula for cross section, we obtain the optical theorem, $\color{red} \Im \mathcal{M}(k_1, k_2 \rightarrow k_1, k_2) = 2 E_{\mathrm{cm}} p_{\mathrm{cm}} \sigma_{\mathrm{tot}}(k_1,k_2 \rightarrow \mathrm{anything}).$
• Feynman diagrams yields imaginary part only when the virtual particles in the diagram go on-shell.
• Discontinuity: above the energy threshold, $\operatorname{Disc} \mathcal{M}(s) = 2i \Im \mathcal{M}(s+i\epsilon).$

#### Perturbation Expansion of Correlation Functions

The perturbed Hamiltonian has the form $H = H_0 + H_{\text{int}} = H_0 + H_{\text{Klein-Gordon}}.$

In the $$\phi^4$$ theory $H = H_0 + H_{\text{int}} = H_{\text{Klein-Gordon}} + \int \dd{^3 x} \frac{\lambda}{4!} \phi^4(x).$

##### Obtaining the Field Operator%%% Obtaining $$\phi$$
• $$\phi$$ under Schrödinger picture: $\color{darkcyan}\phi(t,\vb*{x}) = U^\dagger(t,t_0)\phi_I(t,\vb*{x})U(t,t_0)$
• Time evolution operator under the interaction picture: $\color{orange}U(t,t_0) = e^{iH_0(t-t_0)}e^{-iH(t-t_0)}$
• Equation of motion: $i\pdv{}{t}U(t,t_0) = H_I(t) U(t,t_0).$
• Interaction Hamiltonian: $\color{orange}H_I(t) = e^{iH_0(t-t_0)} (H_{\mathrm{int}}) e^{-iH_0 (t-t_0)}.$
• Dyson series: \begin{align*} U(t,t_0) &= \mathbb{1} + (-i) \int_{t_0}^t \dd{t_1} H_I(t_1) + (-i)^2 \int_{t_0}^{t} \dd{t_1} \int_{t_0}^{t_1} \dd{t_2} H_I(t_1) H_I(t_2) + (-i)^3 \int_{t_0}^{t} \dd{t_1} \int_{t_0}^{t_1} \dd{t_2} \int_{t_0}^{t_2} \dd{t_3} H_I(t_1) H_I(t_2) H_I(t_3) + \dots \\ &= 1 + (-i) \int_{t_0}^t \dd{t_1} H_I(t_1) + \frac{(-i)^2}{2!} \int_{t_0}^t \dd{t_1} \dd{t_2} T\qty{H_I(t_1)H_I(t_2)} + \cdots \\ \color{red}U(t,t') &\color{red}= T\qty{\exp\qty[-i\int_{t'}^t\dd{t''}H_I(t'')]}. \end{align*}
• $$U(t,t')$$ is only defined for $$t>t'$$.
• For $$t_1 \ge t_2 \ge t_3$$,
• $U(t_1,t_2) U(t_2,t_3) = U(t_1,t_3),$
• $U(t_1,t_3)[U(t_2,t_3)]^\dagger = U(t_1,t_2).$
##### Obtaining the Vacuum State%%% Obtaining $$\ket{\Omega}{}$$
• $\ket{\Omega} = e^{-iHT} \ket{0}.$
• $\color{red} T \rightarrow \infty(1-i\epsilon).$
• $E_0 = \bra{\Omega}H\ket{\Omega}.$
• $H_0\ket{0} = 0.$
• $U(t,t') = e^{iH_0(t-t_0)} e^{-iH(t-t')}e^{-iH_0(t'-t_0)}.$
• $$\Omega$$ from vacuum:
• $\ket{\Omega} = \qty(e^{-iE_0(t_0 - (-T))}\bra{\Omega}\ket{0})^{-1} U(t_0,-T)\ket{0}.$
• $\bra{\Omega} = \bra{0}U(T,t_0) \qty(e^{-iE_0(T-t_0)}\bra{0}\ket{\Omega})^{-1}.$
• Normalization: $\bra{\Omega}\ket{\Omega} = 1.$
Derivation

With $e^{-iHT}\ket{0} = e^{-iE_0 T} \ket{\Omega}\bra{\Omega}\ket{0} + \sum_{n\neq 0} e^{-iE_n T}\ket{n}\bra{n} \ket{0},$ we find that $\ket{\Omega} = \qty(e^{-iE_0 T}\bra{\Omega}\ket{0})^{-1} e^{-iHT}\ket{0},$ which simplifies to $\ket{\Omega} = \qty(e^{-iE_0(t_0 - (-T))}\bra{\Omega}\ket{0}) U(t_0, -T)\ket{0}.$

What are the Consequences of the Imaginary Part of $$T$$?

After applying the momentum space Feynman rules, we are facing with $\int_{-T}^T \dd{z^0} \int \dd{^3 z} e^{-i(p_1 + p_2 + p_3 - p_4)\cdot z}.$ To prevent the integral from blowing up, we demand $p^0 \propto (1+i\epsilon),$ which is equivalent to the Feynman prescription of propagator.

##### Obtaining the Correlation Function (Canonical Quantization)
• Two-point correlation function: $\color{red} \bra{\Omega}T\qty{\phi(x)\phi(y)}\ket{\Omega}= \dfrac{\bra{0}T\qty{\phi_I(x)\phi_I(y)\exp\qty[-i\int_{-T}^{T}\dd{t}H_I(t)]}\ket{0}}{\bra{0}T\qty{\exp\qty[-i\int_{-T}^T \dd{t} H_I(t)]}\ket{0}}.$ For each factor inserted to the LHS, we inserted an identical one on the RHS.
##### Obtaining the Correlation Function (Path Integral)
• Path integral formulation: $\color{red} \bra{\Omega} T\phi_H(x_1) \phi_H(x_2) \ket{\Omega} = \dfrac{\int \mathcal{D}\phi\, \phi(x_1) \phi(x_2) \exp\qty[i\int_{-T}^T \dd{^4 x}\mathcal{L}]}{\int \mathcal{D}\phi\, \exp\qty[i\int_{-T}^T \dd{^4 x}\mathcal{L}]}.$
• From the generating functional: $\color{red} \bra{\Omega} T\phi(x_1)\phi(x_2) \ket{\Omega} = \frac{1}{Z_0}\left.\qty(-i\frac{\delta}{\delta J(x_1)})\qty(-i\frac{\delta}{\delta J(x_2)}) Z[J]\right|_{J=0}.$

Due to an unfortunate collision of notation, $$Z_0$$ may stands for both the generating functional $$Z_0[J_A,J_B,\cdots]$$ or the constant $Z_0 = Z_0[0,0,\cdots].$

1. Write down the perturbation $$\mathcal{L}_1$$, which may contain various kinds of fields.
• It's important that the fields occur in product form, e.g. $\mathcal{L}_1 = g\phi_A \phi_B,$ so that we may write the perturbation as products of variations of the generating functional.
2. Obtain the unperturbed generating functional $$Z_0[J_A, J_B, \cdots]$$. This should admit the form \begin{align*} Z_0[J_A,J_B,\cdots] &= \qty(\int \mathcal{D}\varphi_A \int \mathcal{D} \varphi_B \cdots) \exp\qty[i\int \dd{^4 x} \qty[\mathcal{L}_A + J_A \phi_A]] \exp\qty[i\int \dd{^4 x} \qty[\mathcal{L}_B + J_B \phi_B]] \\ &= Z_0 \cdot \exp\qty[-\frac{1}{2}\iint \dd{^4 x} \dd{^4 x'} J_A(x) D^A_F(x-x') J_A(x')] \exp\qty[-\frac{1}{2}\iint \dd{^4 x \dd{^4 x'}} J_B(x) D^B_F(x-x') J_B(x')]\cdots. \end{align*}
3. To obtain the perturbed generating functional, we
1. replace each item in the perturbation as a functional derivative with respect to the corresponding source: $g\phi_A \phi_B \rightarrow g\qty(-i\frac{\delta}{\delta J_A})\qty(-i\frac{\delta}{\delta J_B}).$
• Extra care should be taken for the possible minus sign for Dirac fields.
2. then exponentiate it and let it act on the unperturbed generating functional: \begin{align*} Z[J_A, J_B,\cdots] &= Z_0 \cdot \exp\qty[\int \dd{^4 x} ig\qty(-i\frac{\delta}{\delta J_A(x)})\qty(-i\frac{\delta}{\delta J_B(x)})]Z_0[J_A,J_B,\cdots] \\ &= Z_0 e^{iW[J_A, J_B, \cdots]}. \end{align*}{}
4. To obtain the correlation amplitude, we apply functional derivatives to the perturbed generating functional: \begin{align*} \color{red} \bra{\Omega} T\phi_A(x_1) \phi_A(x_2) \ket{\Omega} &\color{red} = \frac{Z_0}{Z[0,0,\cdots]} \left.\qty(-i \frac{\delta}{\delta J_A(x_1)})\qty(-i \frac{\delta}{\delta J_A(x_2)}) \cdot \exp\qty[\int \dd{^4 x} ig\qty(-i\frac{\delta}{\delta J_A(x)})\qty(-i\frac{\delta}{\delta J_B(x)})]Z_0[J_A,J_B,\cdots]\right\vert_{J_A = J_B = \cdots = 0} \\ &\color{red} = \left.\qty(-i \frac{\delta}{\delta J_A(x_1)})\qty(-i \frac{\delta}{\delta J_A(x_2)}) \cdot i W[J_A, J_B, \cdots]\right\vert_{J_A = J_B = \cdots = 0}. \end{align*}{}
##### Obtaining the T-Matrix
• T-matrix: \begin{align*} &\bra{\vb*{p}_1 \cdots \vb*{p}_n} iT \ket{\vb*{p}_A \vb*{p}_B} \\ &= \qty({_0\bra{\vb*{p}_1 \cdots \vb*{p}_1}T\qty(\exp\qty[-i\int_{-T}^T \dd{t} H_I(t)])\ket{\vb*{p}_A \vb*{p}_B}_0})_{\substack{\text{connected,}\\\text{amputated}}}. \end{align*}

#### Wick's Theorem: Correlation Amplitude to Feynman Propagators

In contractions, $$\phi$$ is always under the interaction picture even without the subscript $$I$$.

• $$\phi^+$$ and $$\phi^-$$ denote the annihilation and creation parts of $$\phi$$ respectively, i.e.
• $\color{warning} \phi^+_I(x) = \int \frac{\dd{^3 p}}{(2\pi)^3} \frac{1}{\sqrt{2E_{\vb*{p}}}} a_{\vb*{p}} e^{-ip\cdot x};$
• $\color{warning} \phi^-_I(x) = \int \frac{\dd{^3 p}}{(2\pi)^3} \frac{1}{\sqrt{2E_{\vb*{p}}}} a^\dagger_{\vb*{p}} e^{+ip\cdot x}.$

The time-ordered product of a Dirac field contains a factor of the sign of the permutation, e.g. $T\qty(\psi_1 \psi_2 \psi_3 \psi_4) = (-1)^3 \psi_3 \psi_1 \psi_4 \psi_2$ if $x_3^0 > x_1^0 > x_4^0 > x_2^0.$

• $$N(XYZ)$$ that turns $$XYZ$$ into normal order,
• Scalar field:$\color{warning} N(a_{\vb*{p}}a_{\vb*{k}}^\dagger a_{\vb*{q}}) = a^\dagger_{\vb*{k}}a_{\vb*{p}}a_{\vb*{q}}.$
• Dirac field:$\color{warning}N(a_{\vb*{p}} a_{\vb*{q}} a_{\vb*{r}}^\dagger) = (-1)^2 a_{\vb*{r}}^\dagger a_{\vb*{p}} a_{\vb*{q}} = (-1)^3 a_{\vb*{r}}^\dagger a_{\vb*{q}} a_{\vb*{p}}.$
• Wick contraction: different from the convention in most literatures, here we use color to denote contraction, ${\color{blue}\phi}(x){\color{blue}\phi}(y) \color{orange} = T\qty{\phi(x)\phi(y)} - N\qty{\phi(x)\phi(y)}.$ More generally, ${\color{blue} A}{\color{blue} B} = T\qty{AB} - N\qty{AB}.$
• For a scalar field:${\color{blue}\phi}(x){\color{blue}\phi}(y) \color{orange} = \begin{cases} \qty[\phi^+(x), \phi^-(y)] & \text{for } x^0 > y^0; \\ \qty[\phi^+(y), \phi^-(x)] & \text{for } y^0 > x^0. \\ \end{cases}$
• The contraction yields the Feynman propagator, ${\color{blue}\phi}(x){\color{blue}\phi}(y) = D_F(x-y).$
• Surprisingly, the contraction yields a number, without bra-keting with $$\ket{0}$$.
• For a Dirac field:${\color{blue}\psi}(x) {\color{blue} \overline{\psi}}(y) = \begin{cases} \phantom{-}\qty{\psi^+(x), \overline{\psi}^-(y)}, & \text{for } x^0 > y^0; \\ -\qty{\overline{\psi}^+(y),\psi^-(x)}, & \text{for } x^0 < y^0. \end{cases}$
• The contraction yields the Feynman propagator:${\color{blue}\psi}(x) {\color{blue} \overline{\psi}}(y) = S_F(x-y),$
• or vanishes: ${\color{blue}\psi}{\color{blue}\psi} = {\color{blue}\overline{\psi}}{\color{blue}\overline{\psi}} = 0.$
• The following notation is used, although it seems that contraction should not be put in $$N\qty{\cdots}$$:
• Scalar field: $N\qty{{\color{blue}\phi_1}\phi_2{\color{blue}\phi_3}\phi_4} = D_F(x_1 - x_3)\cdot N\qty{\phi_2\phi_4}.$
• Dirac field: $N({\color{blue}\psi_1} \psi_2 {\color{blue}\overline{\psi}_3} \overline{\psi}_4) = -{\color{blue}\psi_1}{\color{blue}\overline{\psi}_3} N(\psi_2 \overline{\psi}_4) = -S_F(x_1 - x_3)N(\psi_2 \overline{\psi}_4).$

With the normal ordering we consider only either the $$a_{\vb*{p}}$$ or $$a_{\vb*{p}}^\dagger$$ part in the expansion of $$\color{blue}\phi_I$$.

• Contracting field with creation and annihilation operators:
• ${\color{blue}\phi_I}(x)\ket{\color{blue}\vb*{p}} = {\color{blue}\phi_I}(x){\color{blue}a^\dagger_{\vb*{p}}}\ket{0} = e^{-ip\cdot x}\ket{0}{ }.$
• $\bra{\color{blue}\vb*{p}}{\color{blue}\phi_I}(x) = \bra{0}{\color{blue} a_{\vb*{p}}}{\color{blue}\phi_I}(x) = \bra{0} e^{+ip\cdot x}{ }.$
• Mixing different kinds of fields: Do I contract a Klein-Gordon field with a Dirac field?
• I don't know. I am fabricating shit here. No. If the two fields decouple in the unperturbed Hamiltonian, then their creation and annihilation operators commutes, and therefore we don't give a fuck to their relative ordering. See also

Wick's theorem: \begin{align*} & \color{red} T\qty{\phi(x_1)\phi(x_2)\cdots \phi(x_m)} \\ & \color{red} = N\qty{\phi(x_1)\phi(x_2)\cdots \phi(x_m) + \text{all possible contractions}}. \end{align*}

For $$m=4$$ we have \begin{align*} & T\qty{\phi_1\phi_2\phi_3\phi_4} \\ &= N\big\{ \phi_1\phi_2\phi_3\phi_4 \\ & + {\color{blue}\phi_1}{\color{blue}\phi_2}\phi_3\phi_4 + {\color{blue}\phi_1}\phi_2{\color{blue}\phi_3}\phi_4 + {\color{blue}\phi_1}\phi_2\phi_3{\color{blue}\phi_4} \\ & + \phi_1{\color{blue}\phi_2}{\color{blue}\phi_3}\phi_4 + \phi_1{\color{blue}\phi_2}\phi_3{\color{blue}\phi_4} + \phi_1\phi_2{\color{blue}\phi_3}{\color{blue}\phi_4} \\ & + {\color{blue}\phi_1}{\color{blue}\phi_2}{\color{magenta}\phi_3}{\color{magenta}\phi_4} + {\color{blue}\phi_1}{\color{magenta}\phi_2}{\color{blue}\phi_3}{\color{magenta}\phi_4} + {\color{blue}\phi_1}{\color{magenta}\phi_2}{\color{magenta}\phi_3}{\color{blue}\phi_4}\big\}. \end{align*}

Since the normal order put annihilation on the right, we have \begin{align*} &\bra{0}T\qty{\phi_1\phi_2\phi_3\phi_4}\ket{0} \\ &= D_F(x_1 - x_2) D_F(x_3 - x_4) \\ &+ D_F(x_1 - x_3) D_F(x_2 - x_4) \\ &+ D_F(x_1 - x_4) D_F(x_2 - x_3). \end{align*}

Only fully paired contractions are present in$\color{red}\bra{0}T\qty{\cdots}\ket{0}.$

#### Feynman Diagrams: Wick Contractions / Functional Derivatives Visualized

The $$m=4$$ example may be rewritten using Feynman diagrams as

$$\bra{0}T\qty{\phi_1\phi_2\phi_3\phi_4}\ket{0} =$$ $$+$$ $$+$$ .

• Line $$=$$ propagator $$=$$ contraction $$=$$ whatever you multiply.
• Vertex:
• In position space: vertex $$=$$ integration, $$\displaystyle \int \dd{^4 x}$$.
• In momentum space: vertex $$=$$ intergration $$\displaystyle \int \dd{^4 p}$$ and $$\delta(\sum p)$$.
##### Correlation Amplitude to Diagram (Canonical Quantization)

How do we convert the expression for $$\bra{\Omega}T\qty{\phi(x)\phi(y)}\ket{\Omega}{}$$ into diagram?

• Converting the numerator to diagram:
• The numerator is given by $\bra{0}T\qty{\phi_I(x)\phi_I(y)\exp\qty[-i\int_{-T}^{T}\dd{t}H_I(t)]}\ket{0}.$
• Expand the $$\exp$$ via Taylor series. Note that $$H_I$$ consists of fields ($$\varphi, \psi$$, etc.): $\bra{0} T\qty{ \phi(x)\phi(y) + \phi(x)\phi(y)\qty[-i\int \dd{t} H_I(t)] + \cdots } \ket{0}.$
• Apply Wick's theorem to each item after doing the Taylor expansion.
• Retain only full contractions.
• Convert contractions into Feynman propagators.
• Conclusion: $\color{red} \bra{0}T\qty[\phi(x_1)\phi(x_2)\cdots \phi(x_n)]\ket{0} = \qty(\sum \text{ all possible diagrams with } n \text{ external points}).$
• Taking the denominator into account:
• It can be shown that the disconnected diagrams (disconnected from the two external points) contribute a factor that cancels the denominator in the equation for $$\bra{\Omega}T\qty{\phi(x)\phi(y)}\ket{\Omega}{}$$.

Conclusion. \begin{align*} &\phantom{{}={}} \color{red}\bra{\Omega}T\qty[\phi(x_1)\phi(x_2)\cdots \phi(x_n)]\ket{\Omega} \\ &\color{red}= \qty(\sum \text{ connected diagrams with } n \text{ external points}). \end{align*}

For the $$\phi^4$$ perturbation to the first order we find \begin{align*} & \bra{0}T\qty{\phi(x)\phi(y)\qty(\frac{-i\lambda}{4!})\int \dd{^4 z}\phi(z)\phi(z)\phi(z)\phi(z)}\ket{0} \\ & = 3\cdot \qty(\frac{-i\lambda}{4!})D_F(x-y) \int \dd{^4 z} D_F(z-z)D_F(z-z) \\ & + 12 \cdot \qty(\frac{-i\lambda}{4!}) \int \dd{^4 z} D_F(x-z)D_F(y-z)D_F(z-z). \end{align*} This may be rewritten using Feynman diagrams as $$+$$ .

One of the contractions in the $$\lambda^3$$ term is given by \begin{align*} & \bra{0}{\color{blue}\phi}(x){\color{magenta}\phi}(y) \frac{1}{3!} \qty(\frac{-i\lambda}{4!})^3 \int \dd{^4 z} {\color{blue}\phi}{\color{cyan}\phi}{\color{cyan}\phi}{\color{green}\phi} \int \dd{^4 w} {\color{green}\phi}{\color{orange}\phi}{\color{red}\phi}{\color{magenta}\phi} \int \dd{^4 u} {\color{red}\phi}{\color{orange}\phi}{\color{purple}\phi}{\color{purple}\phi} \ket{0} \\ &= \frac{1}{3!}\qty(\frac{-i\lambda}{4!})^3 \int \dd{^4 z} \dd{^4 w} \dd{^4 u} D_F(x-z) D_F(z-z) D_F (z-w) D_F(w-y) D_F^2(w-u) D_F(u-u), \end{align*} which accounts for $10368 = 3! \times 4\cdot 3 \times 4\cdot 3\cdot 2 \times 4\cdot 3 \times 1/2$ contractions. The Feynman diagram is .

Symmetry Factor

To obtain the overall constant of a diagram in the $$\phi^4$$ theory, we apply the following recipe:

• We drop the factor $$\displaystyle \frac{1}{k!}\qty(\frac{1}{4!})^k$$ in front of each item, i.e. multiply by $$k! (4!)^k$$, which
• copies this diagram for $$k!$$ times: taking into accound the exchange of vertices;
• and then copies for $$(4!)^k$$ times: on each vertex we arbitrarily permute the four incoming lines.
• Then we divide the item by the symmetry factor:
• The permutation above may not result in a new diagram, and therefore we should divide by a symmtry factor to prevent these diagrams being counted multiple times.
• In the above example, the following permutations does not yield a new diagram, which contribute a symmetry factor $$S = 2\cdot 2\cdot 2 = 8$$.
• exchanging the two ends of the loop on $$z$$,
• exchanging the two ends of the loop on $$u$$,
• exchanging the two propagators connecting $$w$$ and $$u$$.
• Lines in the diagrams are refered to as propagators.
• Internal points ($$z$$, $$u$$, $$w$$, etc.) are called vertices.
##### Correlation Amplitude to Diagram (Path Integral)
• We apply the Taylor expansion to \begin{align*} Z[J_A, J_B,\cdots] &= Z_0 \cdot \exp\qty[\int \dd{^4 x} ig\qty(-i\frac{\delta}{\delta J_A(x)})\qty(-i\frac{\delta}{\delta J_B(x)})]Z_0[J_A,J_B,\cdots] \\ &= Z_0 \cdot \sum_{V=0}^\infty \qty(\int \dd{^4 x} ig\qty(-i\frac{\delta}{\delta J_A(x)})\qty(-i\frac{\delta}{\delta J_B(x)}))^V \sum_{P=0}^\infty \qty( {-\frac{1}{2}\iint \dd{^4 x} \dd{^4 x'} J_A(x) D^A_F(x-x') J_A(x') -\frac{1}{2}\iint \dd{^4 x \dd{^4 x'}} J_B(x) D^B_F(x-x') J_B(x')} )^P. \end{align*}
• For each integral in $$\sum_P$$, we associate with it a propagator (line).
• For each integral in $$\sum_V$$, we associate with it an internal vertex.
• For each source not killed, we assocaite with it an external vertex.
• We are not interested in the full form of $$Z$$. With $$Z=e^{iW}$$ we identify $$iW$$ with the contribution from all connected diagrams.
• We count only connected diagrams.
• To obtain the correlation amplitude, we apply $\qty(-i \frac{\delta}{\delta J})$ on $$iW$$. Every instance of such functional derivative kills a source, giving rise to an external vertex.
##### T-Matrix to Diagram

In the equation for the $$T$$-matrix \begin{align*} &\bra{\vb*{p}_1 \cdots \vb*{p}_n} iT \ket{\vb*{p}_A \vb*{p}_B} \\ &= \qty({_0\bra{\vb*{p}_1 \cdots \vb*{p}_1}T\qty(\exp\qty[-i\int_{-T}^T \dd{t} H_I(t)])\ket{\vb*{p}_A \vb*{p}_B}_0})_{\substack{\text{connected,}\\\text{amputated}}}, \end{align*}{ } the terminologies are explained below.

• Connected: only the fully connected (i.e. all vertices are connected to each other) diagrams are counted.
• Amputated: performed an operation that starts from the tip of each external leg and cut at the last point at which removing a single propagator would separates the leg from the rest of the diagram.

Conclusion. \begin{align*} &\phantom{{}={}}\color{red} i\mathcal{M} \cdot (2\pi)^4 \delta^{(4)}\qty(p_A + p_B - \sum p_f) \\ &\color{red}= \begin{pmatrix} \text{sum of all connected, amputated Feynman} \\ \text{diagrams with } p_A, p_B \text{ incoming, } p_f \text{ outgoing} \end{pmatrix}. \end{align*}

Conclusion. $\color{red} i\mathcal{M} = \begin{pmatrix}\text{sum of all connected, amputated diagrams}\\ \text{in the momentum space}\end{pmatrix}.$

The two rules above should be revised for loop diagrams.

• $$S$$-matrix and Feynman diagram. $\bra{\vb*{p}_1 \cdots \vb*{p}_n} S \ket{\vb*{k}_1 \vb*{k}_2} = \qty(\sqrt{Z})^{n+2} \qty[\text{Amputated}].$
##### Determination of Symmetry Factor

The symmetry factor counts the number of ways of interchanging components without changing the diagram.

The symmetry factor of is $$2\cdot 2\cdot 2 = 8$$.

##### Disconnected Diagrams
• A diagram is disconnected if it is disconnected from all external points.
• A diagram connected to some of the external points is deemed as connected.
• Disconnected diagrams are called vacuum bubbles.
• Let $$V_i$$ denote the disconnected diagrams as well as their values.
• The $$V_i$$'s are not isomorphic to each other.
• The diagrams representing the two-point correlation function yield the sum \begin{align*} &\phantom{{}={}}\sum_{\text{connected}} \sum_{\qty{n_i}} \qty(\begin{array}{c} \text{value of} \\ \text{connected piece} \end{array}) \times \qty(\prod_i \frac{1}{n_i!} {(V_i)}^{n_i}) \\ &= \qty{\sum \text{connected}} \times \sum_{\qty{n_i}} \qty(\prod_i \frac{1}{n_i!}{(V_i)^{n_i}}) \\ &= \qty{\sum \text{connected}} \times \exp\qty{\sum_i V_i}. \end{align*}
• The contribution to the denominator in the two-point correlation function is exactly $\exp\qty{\sum_i V_i}.$
• Therefore, we may take only connected diagrams in our calculation.
• Each disconnected diagram contains a factor $(2\pi)^4 \delta^{(4)}(0) = 2T\cdot V.$
##### Constructing Feynmann Rules (Position Space)
1. Write down the perturbation $$H_I$$, which may contain various kinds of fields $$\phi$$, $$\psi$$, etc.
2. Obtain the Feynman propagators: $$D_F$$, $$S_F$$, etc.
• Ideally, this is equivalent to asking (from the recipe via canonical quantization): what are $${\color{blue}\phi}(x){\color{blue}\phi}(y)$$, $${\color{blue}\psi}(x){\color{blue}\overline{\psi}}(y)$$, etc.?
• The contractions should be functions of $$(x-y)$$ only thanks to the homogeneity of spacetime.
3. Follow the procedure specified by Correlation Amplitude to Diagram to write down the diagrams.
4. For each propagator, associate with it a factor equal to the contraction it represents, $${\color{blue}\phi}(x){\color{blue}\phi}(y)$$, $${\color{blue}\psi}(x){\color{blue}\overline{\psi}}(y)$$, etc.
5. For each vertex, associate with it an integration $(-i\lambda)\int \dd{^4 z}.$
• We should have associated $-i\frac{\lambda}{4!} \int \dd{^4 z}$ for the $$\displaystyle \frac{\lambda}{4!} \phi^4$$ perturbation. However, to better take symmetries of diagrams into account, we kill the $$1/4!$$ factor to make copies of the diagram for all permutations of lines.
6. For each external point, associate a factor $$1$$.
7. Divide by the symmetry factor.
##### Constructing Feynmann Rules (Momentum Space)
1. Contruct the position space Feynman rule first.
2. Rewrite the contractions in the momentum space.
• ${\color{blue}\phi}(x){\color{blue}\phi}(y) = {\color{magenta}\int \frac{\dd{^4 p}}{(2\pi)^4}} e^{-ip\cdot({\color{darkcyan}x}-{\color{darkgreen}y})} {\color{orange}\tilde{\Delta}(p)}.$
3. The diagram is left intact. However, each propagator should be assigned a momentum (with direction). $x - y \rightarrow p.$
4. We associate each propagator with the factor $$\color{orange}\tilde{\Delta}(p)$$:
• ${\color{blue}\phi}(x){\color{blue}\phi}(y) \rightarrow {\color{orange}\tilde{\Delta}(p)}.$
5. For each vertex, kill the $$\displaystyle \int \dd{^4 z}$$ in the position space Feynmann rules: $(-i\lambda) \int \dd{^4 z} \rightarrow -i\lambda \cdots.$
6. The integral in $$z$$ should be replaced by a $$\delta$$-function representing the conservation of momentum: $(-i\lambda) \int \dd{^4 z} \rightarrow (-i\lambda) {\color{darkgreen}(2\pi)^4 \delta\qty(\sum p)}.$
7. For each external points, associate a factor $$\color{darkcyan} e^{-ip\cdot x}$$ if the momentum points towards $$x$$, and $$e^{+ip\cdot x}$$ if pointing away from $$x$$.
8. Integrate over each momentum: $\color{magenta} \int \frac{\dd{^4 p}}{(2\pi)^4}.$
9. Divide by the symmetry factor.
##### Feynman Rules for Scattering (Position Space)
1. The procedure for position space Feynman rules applies when the contraction does not involve in and out states.
2. In addition to contraction between fields, we have to obtain the factor for contraction between field and in and out states. ${\color{blue} \phi_I}(x)\ket{\color{blue} \vb*{p}} = {\color{orange} \tilde{\phi}_I(p) e^{-ip\cdot x}}\ket{0},$
• Contractions between in and out states themselves, i.e. ${\color{blue}a_{\vb*{q}}a^\dagger_{\vb*{p}}} = {\color{orange} \delta^{(4)}}(p-q),$ does not occur in (fully) connected diagrams.
• Always a line there when we are doing contraction and obtaining something as a factor!
##### Feynman Rules for Scattering (Momentum Space)
1. Construct the position space Feynman rule for scattering first.
2. The procedure for momentum space Feynman rules applies when the contraction does not involve in and out states.
3. Replace $${\color{orange} \tilde{\phi}_I(p) e^{-ip\cdot x}}$$ by its Fourier transform: ${\color{orange} \tilde{\phi}_I(p) e^{-ip\cdot x}} \rightarrow {\color{orange} \tilde{\phi}_I(p)}.$
4. External $$p$$'s should not be integrated.

Rules for scattering in the position space is for $i\mathcal{M} \cdot (2\pi)^4 \delta^{(4)}\qty(p_A + p_B - \sum p_f),$ while in the momentum space for $i\mathcal{M}.$

##### Fermion Trivia
• We denote
• scalar particles by dashed lines,
• fermions by solid lines, and
• photons by dashed lines.
• Direction of momentum:
• On internal fermion lines, the momentum must be assigned in the direction of particle-number flow.
• On external fermion lines, the momentum is assigned in the direction of particle number for fermions while oppsite for antifermions.
• Symmetry of diagrams:
• We always drop the $$\displaystyle \frac{1}{k!}$$ factor from $$\exp$$, since it cancels out the permutation of vertices.
• The diagrams of Yukawa theory has no symmetry factors, because the three fields cannot replace one for another.
• Determining the overall sign of a diagram:
• We should not allow any intruder operators between a Dirac field operator and a creation or annihilation operator.
• However, pure complex numbers ($$D_F$$, $$S_F$$, etc.) are ok.
• For a specific contraction, we permute the field operators to eliminate the intruders.
• Fully contracted operators between are ok.
• Tricks:
• $${\color{blue} \overline{\psi}\psi}$$ commutes with any operator.
• Closed loop: ${\color{darkcyan} \overline{\psi}}{\color{blue}\psi}{\color{blue} \overline{\psi}}{\color{darkcyan}\psi} = - \operatorname{tr}{[S_F S_F]}.$

#### Example: Quartic Interaction

$H_{\mathrm{int}} = \int \dd{^3 x} \frac{\lambda}{4!} \phi^4(\vb*{x}).$

##### Position Space Feynman Rules (Quartic Interaction)
1. For each propagator, $$=D_F(x-y)$$;
2. For each vertex, $$\displaystyle =(-i\lambda) \int \dd{^4 z}$$;
3. For each external point, $$=1$$;
4. Divide by the symmetry factor.
##### Momentum Space Feynman Rules (Quartic Interaction)
1. For each propagator, $$\displaystyle = \frac{i}{p^2 - m^2 + i\epsilon}$$;
2. For each vertex, $$\displaystyle = -i\lambda$$;
3. For each external point, $$\displaystyle = e^{-ip\cdot x}$$;
4. Impose momentum conservation at each vertex;
5. Integrate over each undetermined momentum: $$\displaystyle \int \frac{\dd{^4 p}}{(2\pi)^4}$$;
6. Divide by the symmetry factor.

In the integration on $$p$$, we should take $$p^0$$ as $$p^0\propto (1+i\epsilon)$$, i.e. using the Ferynman prescription.

##### Position Space Feynman Rules for Scattering (Quartic Interaction)
• ${\color{blue}\phi_I}(x)\ket{\color{blue}\vb*{p}} = e^{-ip\cdot x}\ket{0}{ };$ • $\bra{\color{blue}\vb*{p}}{\color{blue}\phi_I}(x) = \bra{0} e^{+ip\cdot x}{ }.$ 1. For each propagator, $$=D_F(x-y)$$;
2. For each vertex, $$\displaystyle =(-i\lambda) \int \dd{^4 z}$$;
3. For each external line, $$=e^{-ip\cdot x}$$;
4. Divide by the symmetry factor.

We evalute \begin{align*} &\bra{\vb*{p}_1 \cdots \vb*{p}_n} iT \ket{\vb*{p}_A \vb*{p}_B} \\ &= \qty({_0\bra{\vb*{p}_1 \cdots \vb*{p}_1}T\qty(\exp\qty[-i\int_{-T}^T \dd{t} H_I(t)])\ket{\vb*{p}_A \vb*{p}_B}_0})_{\substack{\text{connected,}\\\text{amputated}}}, \end{align*}{ } to the zeroth and first order below, ignoring the prescription connected and amputated first.

Zeroth order: trivial because of the conservation of momentum. \begin{align*} _0\bra{\vb*{p}_1 \vb*{p}_2}\ket{\vb*{p}_A \vb*{p}_B}_0 &= \sqrt{2E_1 2E_2 2E_A 2E_B} \bra{0} a_1 a_2 a_A^\dagger a_B^\dagger \ket{0} \\ &= 2E_A 2E_B (2\pi)^6 \qty(\delta^{(3)}\qty(\vb*{p}_A - \vb*{p}_1) \delta^{(3)}\qty(\vb*{p}_B - \vb*{p}_2) + \delta^{(3)}\qty(\vb*{p}_A - \vb*{p}_2)\delta^{(3)}\qty(\vb*{p}_B - \vb*{p}_1)). \end{align*} $$+$$ First order: using the Wick's theorem we find \begin{align*} & _0\bra{\vb*{p}_1 \vb*{p}_2} T\qty({-i\frac{\lambda}{4!}\int \dd{^4 x}\phi_I^4 (x)}) \ket{\vb*{p}_A \vb*{p}_B}_0 \\ &= {_0 \bra{\vb*{p}_1 \vb*{p}_2} N\qty({-i\frac{\lambda}{4!}\int \dd{^4 x}\phi_I^4(x) + \text{contractions}}) \ket{\vb*{p_A}\vb*{p}_B}_0}. \end{align*}

• Contraction of type $${\color{blue}\phi}{\color{blue}\phi}{\color{magenta}\phi}{\color{magenta}\phi}$$ (trivial again): $$\times \Bigg($$ $$+$$ $$\Bigg)$$

• Contraction of type $${\color{blue}\phi}{\color{blue}\phi}{\phi}{\phi}$$: the two uncontracted $$\phi$$'s should have one contracted to the bra and one to the ket. Trivial again. $$+$$ $$+$$ $$+$$ • Contraction of type $$\phi\phi\phi\phi$$, i.e. all $$\phi$$'s are contracted to the bra and ket: \begin{align*} &(4!)\cdot \qty(-i\frac{\lambda}{4!})\int \dd{^4 x} e^{-i(p_A + p_B - p_1 - p_2)\cdot x} \\ &= -i\lambda (2\pi)^4 \delta^{(4)}\qty(p_A + p_B - p_1 - p_2). \end{align*} With the definition of $$\mathcal{M}$$ we find that $$\mathcal{M} = -\lambda$$.

##### Momentum Space Feynman Rules for Scattering (Quartic Interaction)
1. For each propagator, $$\displaystyle = \frac{i}{p^2 - m^2 + i\epsilon}$$;
2. For each vertex, $$\displaystyle = -i\lambda$$;
3. For each external point, $$\displaystyle = 1$$;
4. Impose momentum conservation at each vertex;
5. Integrate over each undetermined momentum: $$\displaystyle \int \frac{\dd{^4 p}}{(2\pi)^4}$$;
6. Divide by the symmetry factor.

#### Example: Yukawa Theory

\begin{align*} H &= H_{\text{Dirac}} + H_{\text{Klein-Gordon}} + H_{\text{int}} \\ &= H_{\text{Dirac}} + H_{\text{Klein-Gordon}} + \int \dd{^3 x} g\overline{\psi}\psi\phi. \end{align*}

##### Momentum Space Feynman Rules for Scattering (Yukawa Theory)
1. Propagators:

$${\color{blue}\phi}(x){\color{blue}\phi}(y) =$$ $$\displaystyle = \frac{i}{q^2 - m_\phi^2 + i\epsilon}{};$$

$${\color{blue}\psi}(x){\color{blue}\overline{\psi}}(y) =$$ $$\displaystyle = \frac{i(\unicode{x2215}\kern-.5em {p} + m)}{p^2 - m^2 + i\epsilon}{}.$$

1. Vertices: $$=-ig$$.
2. External leg contractions:
• $${\color{blue}\phi}\ket{\color{blue}\vb*{q}} =$$ $$=1$$;
• $$\bra{\color{blue}\vb*{q}}{\color{blue}\phi} =$$ $$=1$$;
• $${\color{blue}\psi}\underbrace{\ket{\color{blue}\vb*{p},s}}_{\text{fermion}} =$$ $$=u^s(p)$$;
• $$\underbrace{\bra{\color{blue}\vb*{p},s}}_{\text{fermion}}{\color{blue}\overline{\psi}} =$$ $$=\overline{u}^s(p)$$;
• $${\color{blue}\overline{\psi}}\underbrace{\ket{\color{blue} \vb*{k},s}}_{\text{antifermion}} =$$ $$= \overline{v}^s(k)$$;
• $$\underbrace{\bra{\color{blue}\vb*{k},s}}_{\text{antifermion}} {\color{blue}\psi} =$$ $$=v^s(k)$$.
3. Impose momentum conservation at each vertex.
4. Integrate over undetermined loop momentum.
5. Figure out the overall sign of the diagram.

We consider $\mathrm{fermion}(p) + \mathrm{fermion}(k) \longrightarrow \mathrm{fermion}(p') + \mathrm{fermion}(k').$

• Lowest order: $_0\bra{\vb*{p}',\vb*{k}'} T\qty({\frac{1}{2!}(-ig) \int \dd{^4 x} \overline{\psi}_I \psi_I \phi_I (-ig) \int \dd{^4 y} \overline{\psi}_I \psi_I \phi_I}) \ket{\vb*{p},\vb*{k}}_0.$
• Implicitly we assocaite spins $$s, r, s', r'$$ to the momenta.
• We drop the factor $$1/2!$$ for any diagrams since we could interchange $$x$$ and $$y$$.
• Contractions:
• $\bra{0}{\color{blue} a_{\vb*{k}'}}{\color{darkcyan} a_{\vb*{p}'}} {\color{blue} \overline{\psi}_x} {\color{magenta} \psi_x} {\color{darkcyan} \overline{\psi}_y} {\color{orange} \psi_y} {\color{orange}a_{\vb*{p}}^\dagger} {\color{magenta} a_{\vb*{k}}^\dagger}\ket{0}.$
• Move $${\color{darkcyan} \psi_y}$$ two spaces to the left.
• Factor: $$(-1)^2 = +1$$.
• $\bra{0}{\color{darkcyan} a_{\vb*{k}'}}{\color{blue} a_{\vb*{p}'}} {\color{blue} \overline{\psi}_x} {\color{magenta} \psi_x} {\color{darkcyan} \overline{\psi}_y} {\color{orange} \psi_y} {\color{orange}a_{\vb*{p}}^\dagger} {\color{magenta} a_{\vb*{k}}^\dagger}\ket{0}.$
• Move $${\color{darkcyan} \psi_y}$$ one spaces to the left.
• Factor: $$-1$$.

Conclusion:

$$i\mathcal{M} =$$ $$+$$ \begin{align*} &= (-ig^2) \qty( \overline{u}\qty(p') u\qty(p) \frac{1}{(p'-p)^2 - m_\phi^2} \overline{u}\qty(k') u\qty(k) - \overline{u}\qty(p') u\qty(k) \frac{1}{(p'-k)^2 - m_\phi^2} \overline{u}\qty(k') u\qty(p) ). \end{align*}

Typo: wavy lines should be replaced with dashed lines here.

##### The Yukawa Potential
• For distinguishable particles, only the first diagram contributes.
• $i\mathcal{M} \approx \frac{ig^2}{\abs{\vb*{p}' - \vb*{p}}^2 + m_\phi^2} 2m\delta^{ss'} 2m\delta^{rr'}.$
• With the Born approximation we find $V(r) = -\frac{g^2}{4\pi}\frac{1}{r}e^{-m_\phi r}.$

#### Example: Quantum Electrodynamics

$H_{\text{int}} = \int \dd{^3 x} e\overline{\psi}\gamma^\mu \psi A_\mu.$

##### Momentum Space Feynman Rules for Scattering (Quantum Electrodynamics)
1. Fermion rules from the Yukawa theory apply here.
2. New vertex: $$= -ie\gamma^\mu$$.
3. Photon propagator: $$\displaystyle = \frac{-ig_{\mu\nu}}{q^2 + i\epsilon}$$.
4. External photon lines:
• $${\color{blue}A_\mu}\ket{\color{blue}\vb*{p}}=$$ $$=\epsilon_\mu(p)$$.
• $$\bra{\color{blue}\vb*{p}}{\color{blue}A_\mu}=$$ $$=\epsilon^*_\mu(p)$$.
• Initial and final state photons should be transversely polarized.
• $\epsilon^\mu = (0, \vb*{\epsilon}).$
• $\vb*{p}\cdot \vb*{\epsilon} = 0.$

We consider $\mathrm{fermion}(p) + \mathrm{fermion}(k) \longrightarrow \mathrm{fermion}(p') + \mathrm{fermion}(k').$

$$i\mathcal{M} =$$ \begin{align*} &= (-ie)^2 \overline{u}\qty(p') \gamma^\mu u\qty(p) \frac{-ig_{\mu\nu}}{\qty(p'-p)^2}\overline{u}\qty(k')\gamma^\nu u\qty(k). \end{align*}

##### The Coulomb Potential
• $i\mathcal{M} \approx \frac{-ie^2}{\abs{\vb*{p}' - \vb*{p}}^2}(2m\xi'^\dagger \xi)_p (2m\xi'^\dagger \xi)_k.$
• Comparing this to the Yukawa case we obtain a repulsive Coulomb potential $V(r) = \frac{e^2}{4\pi r} = \frac{\alpha}{r}.$
• For scattering between an electron and an anti-electron, we get an attractive potential. Quantum Field Theory (IV) Renormalization

2021/2/16 0:53:44

0%
0%

0%
0%