Ze Chen
 Index Index

## Quantum Field Theory (I)

Mathematical Prerequisites

#### Grassman Number

• Properties:
• Anticommutative: $\theta\eta = -\eta\theta.$
• Nilpotent: $\theta^2 = 0.$
• Integration: $\int \dd{\theta} f(\theta) = \int \dd{\theta} \qty(A+B\theta) = B.$
• Rule of proximity: $\int \dd{\theta} \int \dd{\eta} \eta\theta = 1.$
• Rule of proximity for derivatives: $\dv{}{\eta} \theta\eta = -\dv{}{\eta} \eta\theta = -\theta.$
• Complex conjugation: $(\theta\eta)^* = \eta^* \theta^* = -\theta^* \eta^*.$
• Integration of complex Grassman numbers: $\theta = \frac{\theta_1 + i\theta_2}{\sqrt{2}},\quad \theta^* = \frac{\theta_1 - i\theta_2}{\sqrt{2}}$ treated as independent.

Gaussian integral $\qty(\prod_i \int \dd{\theta^*_i} \dd{\theta}_i) e^{-\theta^*_i B_{ij} \theta_j} = \qty(\prod_i \int \dd{\theta^*_i} \dd{\theta_i}) e^{-\sum_i \theta^*_i b_i \theta_i} = \det B.$ $\qty(\prod_i \int \dd{\theta^*_i} \dd{\theta_i}) \theta_k \theta^*_l e^{-\theta^*_i B_{ij} \theta_j} = (\det B) (B^{-1})_{kl}.$

#### Gamma Matrices

##### Properties
• $$\gamma_a$$ satisfies the commutation relations $\color{orange}\qty{\gamma_a,\gamma_b} = 2\delta_{ab} \mathbb{1}.$
• A series of product of Gamma matrices may be simplified using the anti-commutation relation. These products form the group $$\Gamma_N$$.
• $$\gamma_a$$'s are Hermitian and Unitary.
• $\gamma^{(N)}_\chi = \gamma_1 \gamma_2 \cdots \gamma_N,\quad \qty(\gamma^{(N)}_\chi)^2 = (-1)^{N(N-1)/2}\mathbb{1}.$
• For odd $$N$$, the representation may be given by a simple modification of the case of even $$N$$.
• For $$N=2l$$, the order $$\abs{\Gamma_N} = 2^{2l+1}$$.
• The dimension is given by $$d^{(2l)} = 2^l$$.
• We may define $\gamma^{2l}_f = (-1)^l \gamma_1 \cdots \gamma_{2l}$ such that $\qty(\gamma^{(2l)}_f)^2 = \mathbb{1}.$
• Let $$\Gamma'_{2l} = \Gamma_{2l}/(x\sim -x)$$. Then elements of $$\Gamma'$$ are linearly independent.
• There are $$2^{2l}+1$$ conjugacy classes. $$+\mathbb{1}$$, $$-\mathbb{1}$$, as well as $$2^{2l}-1$$ other $$\qty{\pm S}$$. Therefore, there are $$2^{2l}+1$$ inequivalent irreducible representations. $$2^{2l}$$ of which are one-dimensional unfaithful representations.
• We are left with a unique faithful representation of $$\Gamma_{2l}$$, of dimension $$2^l$$.

Theorem. If $$N=2l$$, then representations of dimension $$2^l$$ are equivalent.

##### Four-Dimensional Case

We define $\color{orange}\gamma^5 = i\gamma^0 \gamma^1 \gamma^2 \gamma^3,$ which satisfies the same conditions as $$\gamma_0$$ to $$\gamma_4$$:

• $(\gamma^5)^\dagger = \gamma^5;$
• $(\gamma^5)^2 = \mathbb{1}.$
• $\qty{\gamma^5,\gamma^\mu} = 0.$
##### Weyl (Chiral) Representation, Pauli Matrices, etc.

We define

• $\sigma^\mu = (\mathbb{1},\vb*{\sigma}),$
• $\overline{\sigma}^\mu = (\mathbb{1}, -\vb*{\sigma}).$

The Weyl or Chiral representation is given by $\gamma^0 = \begin{pmatrix} 0 & \mathbb{1} \\ \mathbb{1} & 0 \end{pmatrix},\quad \gamma^i = \begin{pmatrix} 0 & \sigma^i \\ -\sigma^i & 0 \end{pmatrix}.$

Or, $\gamma^\mu = \begin{pmatrix} 0 & \sigma^\mu \\ \overline{\sigma}^\mu & 0 \end{pmatrix}.$

Under this representation, $\gamma_5 = i\gamma^0\gamma^1\gamma^2\gamma^3 = \begin{pmatrix} -\mathbb{1} & 0 \\ 0 & \mathbb{1} \end{pmatrix}.$

Useful identities of $$\sigma$$:

• $\color{darkcyan} \sigma^2 \vb*{\sigma}^* = -\vb*{\sigma}\sigma^2.$
• $(p\cdot \sigma)(p\cdot \overline{\sigma}) = p^2 = m^2.$

Useful identities of $$\gamma$$:

• $\color{darkcyan} \Lambda_{1/2}^{-1} \gamma^\mu \Lambda_{1/2} = \Lambda{^\mu}{_\nu} \gamma^\nu.$
##### Dicrete Transformation
• Charge conjugation $$C$$ is a unitary matrix such that $C^\dagger \gamma_a C = -\gamma_a^T.$
• We have $C^T = \lambda C,$ where $\lambda = (-1)^{l(l+1)/2}.$
• Parity $$P$$ is a unitary matrix such that $P^\dagger \gamma_a P = \gamma_a^T.$
• We have $P^T = \tau P,$ where $\tau = (-1)^{l(l+1)/2}.$
• $$C$$ and $$P$$ are related by $P = \gamma_f C.$
• For $$N=4$$ we have $C^T = -C,\quad P^T = -P.$

#### Lorentz Invariance

##### How do Tuples Transform?

Vectors are transformed in the familiar way: $\partial_\mu \phi(x) \rightarrow (\Lambda^{-1}){^\nu}{_\mu} (\partial_\nu \phi) (\Lambda^{-1} x).$

On the LHS we have quantities in the old frame while on the RHS in the new frame.

For a general $$n$$-component multiplet, the transformation law is given by an $$n$$-dimensional representation of the Lorentz group: $\color{red} \varphi_a(x) \rightarrow U^{-1}(\Lambda) \varphi_a(x) U(\Lambda) = L{_a}{^b}(\Lambda) \varphi_b(\Lambda^{-1}x).$ This is the most general transformation rule! You should fucking remember it! As well as its infinitesimal form! $\color{red} \varphi_a(x) \rightarrow (\delta{_a}{^b} + \frac{i}{2} \delta \omega_{\mu\nu} (S^{\mu\nu}){_a}{^b}) \varphi_b.$ Don't give a shit to $$S^{\mu\nu}$$ now. We don't know it until we obtain a specific representation of the Lorentz group.

In infinitesimal form,

• $\Lambda{^\mu}{_\nu} = \delta{^\mu}{_\nu} + \delta\omega{^\mu}{_\nu},$
• $U(1+\delta\omega) = \mathbb{1} + \frac{i}{2}\delta\omega_{\mu\nu} M^{\mu\nu},$
• $L{_a}{^b}(1+\delta\omega) = \delta{_a}{^b} + \frac{i}{2} \delta\omega_{\mu\nu} (S^{\mu\nu}){_a}{^b},$
• $[\varphi_a(x), M^{\mu\nu}] = \mathcal{L}^{\mu\nu}\varphi_a(x) + (S^{\mu\nu}){_a}{^b} \varphi_b(x),$ where $\color{blue} \mathcal{L}^{\mu\nu} = \frac{1}{i}(x^\mu \partial^\nu - x^\nu\partial^\mu).$
##### Representation of the Lorentz Group
• There are six generators to the Lorentz group.
• The six generators are organized into a $$4\times 4$$ anti-symmetric matrix $$M^{\mu\nu}$$.
• They satisfy the commutation relation $[M^{\mu\nu}, M^{\rho\sigma}] = i(g^{\nu\rho} M^{\mu\sigma} - (\mu\leftrightarrow \nu)) - (\rho \leftrightarrow \sigma).$
• $$M^{\mu\nu}$$, $$\mathcal{L}^{\mu\nu}$$ and $$(S^{\mu\nu}){_a}{^b}$$ should satisfy the same commutation relation.
• The six generators may be divided into
• angular momentum: $\color{blue} M^{12} = J_3,\ M^{31} = J_2,\ M^{23} = J_1,$
• boosts: $\color{blue} K_i = M^{i0}.$
• We have the commutation relations between them: \begin{align*} [J_i,J_j] &= +i\epsilon_{ijk} J_k, \\ [J_i,K_j] &= +i\epsilon_{ijk} K_k, \\ [K_i,K_j] &= -i\epsilon_{ijk} J_k. \end{align*}
• A new set of operators \begin{align*} N_i &= \frac{1}{2} (J_i - iK_i), \\ N_i^\dagger &= \frac{1}{2} (J_i + iK_i) \end{align*} yields \begin{align*} [N_i,N_j] &= i\epsilon_{ijk} N_k, \\ [N^\dagger_i, N^\dagger_j] &= i\epsilon_{ijk} N^\dagger_k, \\ [N_i, N^\dagger_j] &= 0. \end{align*}
• The representations of the Lorentz group may be specified by two half-integers.
• $$(0,0)$$: scalar or singlet representation;
• $$(1/2, 0)$$: left-handed spinor representation;
• $$(0, 1/2)$$: right-handed spinor representation;
• $$(1/2,1/2)$$: vector representation.
##### What the fuck are Spinors?

Go back and see the transformation rule (marked in red). Left-handed / right-handed spinors transforms according to the $$S^{\mu\nu}_{\mathrm{L}}$$ / $$S^{\mu\nu}_{\mathrm{R}}$$ specified below $\color{red} \psi_a(x) \rightarrow (\delta{_a}{^b} + \frac{i}{2} \delta \omega_{\mu\nu} (S^{\mu\nu}){_a}{^b}) \psi_b.$ where \begin{align*} \color{blue}S^{0i}_{\mathrm{L}} &\color{blue}= -\frac{i}{2}\sigma^i, \\ \color{blue}S^{0i}_{\mathrm{R}} &\color{blue}= +\frac{i}{2}\sigma^i, \\ \color{blue}S^{ij}_{\mathrm{L}/\mathrm{R}} &\color{blue}= \frac{1}{2}\epsilon^{ijk} \sigma^k.\\ \end{align*} Or, written explicitly, as \begin{align*} \color{red}\psi_{\mathrm{L}} &\color{red}\rightarrow (1 - \frac{i}{2}\vb*{\theta}\cdot \vb*{\sigma} - \frac{1}{2}\vb*{\beta}\cdot \vb*{\sigma})\psi_{\mathrm{L}}, \\ \color{red}\psi_{\mathrm{R}} &\color{red}\rightarrow (1 - \frac{i}{2}\vb*{\theta}\cdot \vb*{\sigma} + \frac{1}{2}\vb*{\beta}\cdot \vb*{\sigma})\psi_{\mathrm{R}}. \end{align*}

##### And Dirac Spinors?

Dirac spinors has four components. It transforms according to $\color{red} \psi_a(x) \rightarrow (\delta{_a}{^b} + \frac{i}{2} \delta \omega_{\mu\nu} (S^{\mu\nu}){_a}{^b}) \psi_b.$ where $\color{blue} S^{\mu\nu} = \frac{i}{4}[\gamma^\mu,\gamma^\nu] = \begin{pmatrix} S^{\mu\nu}_{\mathrm{L}} & 0 \\ 0 & S^{\mu\nu}_{\mathrm{R}} \end{pmatrix}.$

The upper two components transforms like a left-handed spinor while the lower two like a right-handed. $\color{orange} \psi = \begin{pmatrix} \psi_{\mathrm{L}} \\ \psi_{\mathrm{R}} \end{pmatrix}.$

In Peskin the exponent of the infinitesimal transform is denoted by $$\Lambda_{1/2}$$: $\delta{_a}{^b} + \frac{i}{2} \delta \omega_{\mu\nu} (S^{\mu\nu}){_a}{^b} \xrightarrow{\exp} \Lambda_{1/2} = \exp\qty(-\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu}).$

We define $\color{orange} \overline{\psi} = \psi^\dagger \gamma^0.$

The transformation laws are given by

• $\color{red} \psi(x) \rightarrow \Lambda_{1/2} \psi(\Lambda^{-1} x).$
• $\color{red} \overline{\psi}(x) \rightarrow \overline{\psi}(\Lambda^{-1}x) \Lambda^{-1}_{1/2}.$
• $$\overline{\psi}\psi$$ is a Lorentz scalar.
• $$\overline{\psi}\gamma^\mu \psi$$ is a Lorentz vector.
##### Dirac Bilinears
Definition Count
$$1$$ 1
$$\gamma^\mu$$ 4
$$\gamma^{\mu\nu} = \gamma^{[\mu}\gamma^{\nu]} = -i\sigma^{\mu\nu}{}$$ 6
$$\gamma^{\mu\nu\rho} = \gamma^{[\mu}\gamma^{\nu}\gamma^{\rho]}{}$$ 4
$$\gamma^{\mu\nu\rho\sigma} = \gamma^{[\mu}\gamma^{\nu}\gamma^{\rho}\rho^{\sigma]}{}$$ 1

$$\overline{\psi} \gamma^{\mu\nu} \psi$$ transforms like a tensor of rank 2: $\overline{\psi} \gamma^{\mu\nu} \psi \rightarrow \Lambda{^\mu}{_\alpha} \Lambda{^\nu}{_\beta} \overline{\psi} \gamma^{\alpha\beta} \psi.$

Definition Transformation Count
$$1$$ $$\overline{\psi}\psi$$: scalar 1
$$\gamma^\mu$$ $$\overline{\psi}\gamma^\mu\psi$$: vector 4
$$\displaystyle \sigma^{\mu\nu} = \frac{i}{2}[\gamma^\mu,\gamma^\nu]$$ $$\overline{\psi}\sigma^{\mu\nu}\psi$$: tensor 6
$$\gamma^\mu \gamma^5$$ $$\overline{\psi}\gamma^\mu\gamma^5\psi$$: pseudo-vector 4
$$\gamma^5$$ $$\overline{\psi}\gamma^5\psi$$: pseudo-scalar 1
 Quantum Field Theory (II) Quantization of Fields

2021/4/5 17:20:32

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