Quantum Feild Theory Ryder
Chapter 2 Single-particle relativistic wave equations
Relativistic notation
Correspondense of $SU(2)$ and $SO(3)$ group
$$U=\mathrm{e}^{\mathrm{i} \boldsymbol{\sigma} \cdot \boldsymbol{\theta} / 2}=\cos \theta / 2+\mathrm{i} \boldsymbol{\sigma} \cdot \mathbf{n} \sin \theta / 2 \leftrightarrow R=\mathrm{e}^{\mathrm{i} \cdot \boldsymbol{J} \boldsymbol{\theta}}$$Generator of Lorentz Group
$$ x^{0 \prime}=\gamma\left(x^{0}+\beta x^{1}\right), \quad x^{1 \prime}=\gamma\left(\beta x^{0}+x^{1}\right), \quad x^{2 \prime}=x^{2}, \quad x^{3 \prime}=x^{3} $$Observing that $\gamma^{2}-\beta^{2} \gamma^{2}=1$, we may put
$$ \gamma=\cosh \phi, \quad \gamma \beta=\sinh \phi, $$thus parametrising the transformation in terms of the variable $\phi$, with $\tanh \phi=v / c$, and we have
$$ \left(\begin{array}{l} x^{0 \prime} \\ x^{1 \prime} \\ x^{2 \prime} \\ x^{3 \prime} \end{array}\right)=\left(\begin{array}{cccc} \cosh \phi & \sinh \phi & 0 & 0 \\ \sinh \phi & \cosh \phi & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)\left(\begin{array}{l} x^{0} \\ x^{1} \\ x^{2} \\ x^{3} \end{array}\right) $$$$ K_{x}=\left.\frac{1 \partial B}{\mathrm{i} \partial \phi}\right|_{\phi=0}=-\mathrm{i}\left(\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) $$Similarly, the other boost generators are
$$ K_{y}=-\mathrm{i}\left(\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right), \quad K_{z}=-\mathrm{i}\left(\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right) $$Rotation Matrix in SO(3)
$$ R_{z}(\theta)=\left(\begin{array}{ccc} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array}\right) $$$$ \begin{aligned} &R_{x}(\phi)=\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \phi & \sin \phi \\ 0 & -\sin \phi & \cos \phi \end{array}\right) \\ &R_{y}(\psi)=\left(\begin{array}{ccc} \cos \psi & 0 & -\sin \psi \\ 0 & 1 & 0 \\ \sin \psi & 0 & \cos \psi \end{array}\right) \end{aligned} $$$$ \begin{aligned} &J_{z}=\left.\frac{1}{\mathrm{i}} \frac{\mathrm{d} R_{z}(\theta)}{\mathrm{d} \theta}\right|_{\theta=0}=\left(\begin{array}{rrr} 0 & -\mathrm{i} & 0 \\ \mathrm{i} & 0 & 0 \\ 0 & 0 & 0 \end{array}\right) \\ &J_{x}=\left.\frac{1}{\mathrm{i}} \frac{\mathrm{d} R_{x}(\phi)}{\mathrm{d} \phi}\right|_{\phi=0}=\left(\begin{array}{rrr} 0 & 0 & 0 \\ 0 & 0 & -\mathrm{i} \\ 0 & \mathrm{i} & 0 \end{array}\right) \\ &J_{y}=\left.\frac{1}{\mathrm{i}} \frac{\mathrm{d} R_{y}(\psi)}{\mathrm{d} \psi}\right|_{\psi=0}=\left(\begin{array}{rrr} 0 & 0 & \mathrm{i} \\ 0 & 0 & 0 \\ -\mathrm{i} & 0 & 0 \end{array}\right) \end{aligned} $$in $4\times 4$ representation
$$ \begin{gathered} J_{x}=-\mathrm{i}\left(\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{array}\right), \\ J_{y}=-\mathrm{i}\left(\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right), \\ J_{z}=-\mathrm{i}\left(\begin{array}{rrrr} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) . \end{gathered} $$commutation relations
$\left[K_{x}, K_{y}\right]=-\mathrm{i} J_{z}$ and cyclic perms $\left[J_{x}, K_{x}\right]=0$ etc., $\left[J_{x}, K_{y}\right]=\mathrm{i} K_{z}$ and cyclic perms, $]$
$$ \left.\begin{array}{l} \mathbf{A}=\frac{1}{2}(\mathbf{J}+\mathrm{iK}) \\ \mathbf{B}=\frac{1}{2}(\mathbf{J}-\mathrm{iK}) . \end{array}\right\} $$$\left[A_{x}, A_{y}\right]=\mathrm{i} A_{z}$ and cyclic perms, $\left[B_{x}, B_{y}\right]=\mathrm{i} B_{z}$ and cyclic perms, $\left.\left[A_{i}, B_{j}\right]=0(i, j=x, y, z) .\right]$
two types of vector
$$ \left.\begin{array}{ll} (j, 0) \rightarrow \mathbf{J}^{(j)}=\mathrm{i} \mathbf{K}^{j} & (\mathbf{B}=0), \\ (0, j) \rightarrow \mathbf{J}^{(j)}=-\mathrm{i} \mathbf{K}^{(j)} & (\mathbf{A}=0), \end{array}\right\} $$and this in fact corresponds to the two possibilities in $(2.69)$. We may now define two types of spinor:
Type I: $\left(\frac{1}{2}, 0\right): \quad J^{(1 / 2)}=\sigma / 2, \quad \mathbf{K}^{(1 / 2)}=-\mathrm{i} \sigma / 2$. We denote the spinor $\xi$. If $(\boldsymbol{\theta}, \boldsymbol{\phi})$ are the parameters of a rotation and pure Lorentz transformation, $\xi$ transforms as
$$ \begin{aligned} \xi & \rightarrow \exp \left(\mathrm{i} \frac{\sigma}{2} \cdot \boldsymbol{\theta}+\frac{\boldsymbol{\sigma}}{2} \cdot \boldsymbol{\phi}\right) \xi \\ &=\exp \left[\mathrm{i} \frac{\boldsymbol{\sigma}}{2} \cdot(\boldsymbol{\theta}-\mathrm{i} \boldsymbol{\phi})\right] \xi \equiv M \xi \end{aligned} $$Type II: $\quad\left(0, \frac{1}{2}\right): \quad \mathbf{J}^{(1 / 2)}=\sigma / 2, \quad \mathbf{K}^{(1 / 2)}=\mathrm{i} \sigma / 2$. This spinor is denoted $\eta$ and transforms like
$$ \eta \rightarrow \exp \left[\mathrm{i} \frac{\sigma}{2} \cdot(\boldsymbol{\theta}+\mathrm{i} \phi)\right] \eta \equiv N \eta $$Relationship betwen $M$ and $N$ and $M$
$$ \begin{aligned} M=\exp \left[\mathrm{i} \frac{\boldsymbol{\sigma}}{2} \cdot(\boldsymbol{\theta}-\mathrm{i} \boldsymbol{\phi})\right]\\ N=\exp \left[\mathrm{i} \frac{\sigma}{2} \cdot(\boldsymbol{\theta}+\mathrm{i} \phi)\right] \end{aligned} $$$$ N=\zeta M^{*} \zeta^{-1} \quad \text { with } \zeta=-\mathrm{i} \sigma_{2} $$$$ \begin{gathered} \sigma_{2} \sigma^{*} \sigma_{2}=-\sigma_{2}^{2} \sigma=-\sigma_{1} \\ \zeta M^{*} \zeta^{-1}=\sigma_{2} \exp \left[-\frac{\mathrm{i}}{2} \sigma^{*} \cdot(\boldsymbol{\theta}+\mathrm{i} \boldsymbol{\phi})\right] \sigma_{2} \\ =\sigma_{2}^{2} \exp \left[\frac{\mathrm{i}}{2} \boldsymbol{\sigma} \cdot(\boldsymbol{\theta}+\mathrm{i} \boldsymbol{\varphi})\right] \\ =N . \end{gathered} $$Parity
$\mathbf{v} \rightarrow-\mathbf{v}$
$\mathbf{K} \rightarrow-\mathbf{K}$
$\mathbf{J} \rightarrow+\mathbf{J}$
$(j, 0) \leftrightarrow(0, j), \quad$ under parity
$\xi \leftrightarrow \eta .$
Irreducible representation of Lorentz Group
$$ \begin{aligned} \left(\begin{array}{l} \xi \\ \eta \end{array}\right) & \rightarrow\left(\begin{array}{cc} \mathrm{e}^{1 / 2 \sigma \cdot(\theta-\mathrm{i} \phi)} & 0 \\ 0 & \mathrm{e}^{1 / 2 \sigma \cdot(\theta+\mathrm{i} \phi)} \end{array}\right)\left(\begin{array}{l} \xi \\ \eta \end{array}\right) \\ &=\left(\begin{array}{cc} D(\Lambda) & 0 \\ 0 & \bar{D}(\Lambda) \end{array}\right)\left(\begin{array}{l} \xi \\ \eta \end{array}\right) \end{aligned} $$$$ \bar{D}(\Lambda)=\zeta D^{*}(\Lambda) \zeta^{-1} $$$\zeta = -i\sigma^2$
not Unitary, and not compact
Derive of Dirac equation
$\theta=0$ and relable
$$ \xi \rightarrow \phi_{\mathrm{R}}, \quad \eta \rightarrow \phi_{\mathrm{L}} $$$$ \begin{aligned} \phi_{\mathrm{R}} & \rightarrow \mathrm{e}^{1 / 2 \boldsymbol{\sigma} \cdot \boldsymbol{\varphi}} \phi_{\mathrm{R}} \\ &=[\cosh (\phi / 2)+\boldsymbol{\sigma} \cdot \mathbf{n} \sinh (\phi / 2)] \phi_{\mathrm{R}} \end{aligned} $$$$ \begin{aligned} \cosh(\phi/2)=[(\gamma+1)/2]^{1/2}\\ \sinh(\phi/2)=[(\gamma-1)/2]^{1/2} \end{aligned} $$$$ \phi_{\mathrm{R}}(\mathbf{p})=\left[\left(\frac{\gamma+1}{2}\right)^{1 / 2}+\sigma \cdot \hat{\mathbf{p}}\left(\frac{\gamma-1}{2}\right)^{1 / 2}\right] \phi_{\mathrm{R}}(0) $$$\gamma=E / m$
$$ \phi_{\mathrm{R}}(\mathbf{p})=\frac{E+m+\sigma \cdot \mathbf{p}}{[2 m(E+m)]^{1 / 2}} \phi_{\mathrm{R}}(0) $$$$ \phi_{\mathrm{L}}(\mathbf{p})=\frac{E+m-\varphi \cdot \mathbf{p}}{[2 m(E+m)]^{1 / 2}} \phi_{\mathrm{L}}(0) $$$$ \phi_{\mathrm{R}}(\mathbf{p})=\frac{E+\sigma \cdot \mathbf{p}}{m} \phi_{\mathrm{L}}(\mathbf{p}) $$and hence
$$ \phi_{\mathrm{L}}(\mathbf{p})=\frac{E-\sigma \cdot \mathbf{p}}{m} \phi_{\mathrm{R}}(\mathbf{p}) $$We may rewrite these equations as
$$ \left.\begin{array}{r} -m \phi_{\mathrm{R}}(\mathbf{p})+\left(p_{0}+\boldsymbol{\sigma} \cdot \mathbf{p}\right) \phi_{\mathrm{L}}(\mathbf{p})=0 \\ \left(p_{0}-\boldsymbol{\sigma} \cdot \mathbf{p}\right) \phi_{\mathrm{R}}(\mathbf{p})-m \phi_{\mathrm{L}}(\mathbf{p})=0 \end{array}\right\} $$or, in matrix form,
$$ \left(\begin{array}{cc} -m & p_{0}+\boldsymbol{\sigma} \cdot \mathbf{p} \\ p_{0}-\boldsymbol{\sigma} \cdot \mathbf{p} & -m \end{array}\right)\left(\begin{array}{l} \phi_{\mathrm{R}}(\mathbf{p}) \\ \phi_{\mathrm{L}}(\mathbf{p}) \end{array}\right)=0 $$Defining the 4 -spinor
$$ \psi(p)=\left(\begin{array}{l} \phi_{\mathrm{R}}(\mathbf{p}) \\ \phi_{\mathrm{L}}(\mathbf{p}) \end{array}\right) $$and the $4 \times 4$ matrices
$$ \color{red}\gamma^{0}=\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right), \quad \gamma^{i}=\left(\begin{array}{cc} 0 & -\sigma^{i} \\ \sigma^{i} & 0 \end{array}\right) $$equation $(2.90)$ becomes
$$ \left(\gamma^{0} p_{0}+\gamma^{i} p_{i}-m\right) \psi(p)=0 $$(note that $p_{\mu}=(E,-\mathbf{p})($ see $(2.11))$, so $\left.\gamma^{0} p_{0}+\gamma^{i} p_{i}=\gamma^{0} p_{0}-\gamma \cdot \mathbf{p}\right)$, or
$$ \left(\gamma^{\mu} p_{\mu}-m\right) \psi(p)=0 $$helicity
$m=0$
$$ \begin{aligned} &\left(p_{0}+\boldsymbol{\sigma} \cdot \mathbf{p}\right) \phi_{\mathrm{L}}(\mathbf{p})=0 \\ &\left(p_{0}-\boldsymbol{\sigma} \cdot \mathbf{p}\right) \phi_{\mathrm{R}}(\mathbf{p})=0 \end{aligned} $$$$ \boldsymbol{\sigma} \cdot \hat{\mathbf{p}} \phi_{\mathrm{L}}=-\phi_{\mathrm{L}}, \quad \boldsymbol{\sigma} \cdot \hat{\mathbf{p}} \phi_{\mathrm{R}}=\phi_{\mathrm{R}} $$2.4 Prediction of antiparticles
Algebra of $\gamma$ matrix
$$ \color{red}\left(\mathrm{i} \gamma^{\mu} \partial_{\mu}-m\right) \psi=0 $$$$ \begin{array}{r} {\left[-\left(\gamma^{\mu} \partial_{\mu}\right)\left(\gamma^{v} \partial_{v}\right)-\mathrm{i}\left(\gamma^{\mu} \partial_{\mu}\right) m\right] \psi=0} \\ \left(\gamma^{\mu} \gamma^{\prime} \partial_{\mu} \partial_{v}+m^{2}\right) \psi=0 \end{array} $$Now $\partial_{\mu} \partial_{v}=\partial_{v} \partial_{\mu}$, so $\gamma^{\mu} \gamma^{v}$ may be replaced by the symmetric combination
$$ \frac{1}{2}\left(\gamma^{\mu} \gamma^{v}+\gamma^{v} \gamma^{\mu}\right) \equiv \frac{1}{2}\left\{\gamma^{\mu}, \gamma^{v}\right\} $$to give
$$ \frac{1}{2}\left\{\gamma^{\mu}, \gamma^{v}\right\} \partial_{\mu} \partial_{v} \psi+m^{2} \psi=0 $$$$ \left(\square+m^{2}\right) \psi(x)=0 $$It therefore follows that the coefficient of $\partial_{\mu} \partial_{v}$ is $g^{\mu v}$, so
$$ \color{red}\left\{\gamma^{\mu}, \gamma^{v}\right\}=2 g^{\mu v} . $$This is the general relation which the coefficients $\gamma^{\mu}$ must satisfy. Taking in turn $\mu=v=0, \mu=v=i$ and $\mu \neq v$, we have
$$ \left(\gamma^{0}\right)^{2}=1, \quad\left(\gamma^{i}\right)^{2}=-1, \quad \gamma^{\mu} \gamma^{v}=-\gamma^{v} \gamma^{\mu} \quad(v \neq \mu) . $$probability current $j^\mu$
$$ \left(i \gamma^{\mu} \partial_{\mu}-m\right) \psi=0 $$$$ \psi^{\dagger}\left(-\mathrm{i} \gamma^{0} \stackrel{\leftarrow}{\partial}_{0}+\mathrm{i} \gamma^{i} \partial_{i}-m\right)=0 $$From $\gamma^i\gamma^0=-\gamma^0\gamma^i$
$$ \bar{\psi}\left(\mathrm{i} \gamma^{\mu}{\partial}_{\mu}+m\right)=0 $$$$ \color{red}\bar{\psi}=\psi^{\dagger} \gamma^{0} $$$$ \dot{j}^{\mu}=\bar{\psi} \gamma^{\mu} \psi $$$$ \begin{aligned} \partial_{\mu} j^{\mu} &=\left(\partial_{\mu} \bar{\psi}\right) \gamma^{\mu} \psi+\bar{\psi} \gamma^{\mu}\left(\partial_{\mu} \psi\right) \\ &=(i m \bar{\psi}) \psi+\bar{\psi}(-i m \psi)=0 \end{aligned} $$$$ j^{0}=\bar{\psi} \gamma^{0} \psi=\psi^{\dagger} \psi=\left|\psi_{1}\right|^{2}+\left|\psi_{2}\right|^{2}+\left|\psi_{3}\right|^{2}+\left|\psi_{4}\right|^{2} $$2.5 Construction of Dirac spinors: algebra of $\gamma$ matrix
Prove that $\bar\psi \psi$ is a scalar, $\bar\psi \gamma^5\psi$ is a pseudoscalar
$$ \psi=\left(\begin{array}{l} \phi_{\mathrm{R}} \\ \phi_{\mathrm{L}} \end{array}\right) $$$$ \phi_{\mathrm{R}} \rightarrow \exp \left[\frac{\mathrm{i}}{2} \sigma \cdot(\theta-\mathrm{i} \phi)\right] \phi_{\mathrm{R}}, \quad \phi_{\mathrm{L}} \rightarrow \exp \left[\frac{\mathrm{i}}{2} \sigma \cdot(\theta+\mathrm{i} \phi)\right] \phi_{\mathrm{L}} $$hence
$$ \phi_{\mathrm{R}}^{\dagger} \rightarrow \phi_{\mathrm{R}}^{\dagger} \exp \left[\frac{-\mathrm{i}}{2} \boldsymbol{\sigma} \cdot(\boldsymbol{\theta}+\mathrm{i} \phi)\right], \quad \phi_{\mathrm{L}}^{\dagger} \rightarrow \phi_{\mathrm{L}}^{\dagger} \exp \left[\frac{-\mathrm{i}}{2} \boldsymbol{\sigma} \cdot(\theta-\mathrm{i} \phi)\right] $$and it is immediately evident that
$$ \psi^{\dagger} \psi=\phi_{\mathrm{R}}^{\dagger} \phi_{\mathrm{R}}+\phi_{\mathrm{L}}^{\dagger} \phi_{\mathrm{L}} $$$$ \begin{aligned} &\text { is not invariant. However, the adjoint spinor } \bar{\psi} \text { defined in (2.104) has compo- } \\ &\text { nents } \\ &\qquad \bar{\psi}=\psi^{\dagger} \gamma^{0}=\left(\phi_{\mathrm{R}}^{\dagger} \phi_{\mathrm{L}}^{\dagger}\right)\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right)=\left(\phi_{\mathrm{L}}^{\dagger} \phi_{\mathrm{R}}^{\dagger}\right) \end{aligned} $$so it is easy to see that
$$ \bar{\psi} \psi=\phi_{\mathrm{L}}^{\dagger} \phi_{\mathrm{R}}+\phi_{\mathrm{R}}^{\dot{\dagger}} \phi_{\mathrm{L}} $$is invariant (i.e. a scalar) under Lorentz transformations. Moreover, under parity
$$ \phi_{\mathrm{R}} \leftrightarrow \phi_{\mathrm{L}}, $$so $\bar{\psi} \psi \rightarrow \bar{\psi} \psi$, and is a true scalar, i.e. does not change sign under space reflection.
$$ \gamma^{5}=\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right) $$$$ \gamma^{5}=\mathrm{i} \gamma^{0} \gamma^{1} \gamma^{\gamma} \gamma^{3}=\gamma_{5} . $$$$ \begin{aligned} \bar{\psi} \gamma^{5} \psi &=\left(\phi_{\mathrm{L}}^{\dagger} \phi_{\mathrm{R}}\right)\left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right)\left(\begin{array}{l} \phi_{\mathrm{R}} \\ \phi_{\mathrm{L}} \end{array}\right) \\ &=\phi_{\mathrm{L}}^{\dagger} \phi_{\mathrm{R}}-\phi_{\mathrm{R}}^{+} \phi_{\mathrm{L}} . \end{aligned} $$
$\bar{\psi} \psi$ scalar,
$\bar{\psi} \gamma_{5} \psi$ pseudoscalar,
$\bar{\psi} \gamma^{\mu} \psi$ vector,
$\bar{\psi} \gamma^{\mu} \gamma^{5} \psi$ axial vector,
$\bar{\psi}\left(\gamma^{\mu} \gamma^{v}-\gamma^{v} \gamma^{\mu}\right) \psi$ antisymmetric tensor.
Solutions of Dirac equation
$$ \left.\begin{array}{l} \psi(x)=u(0) \mathrm{e}^{-\mathrm{i} m t} \text { positive energy, } \\ \psi(x)=v(0) \mathrm{e}^{\mathrm{i} m t} \text { negative energy. } \end{array}\right\} $$$$ u^{(1)}(0)=\left(\begin{array}{l} 1 \\ 0 \\ 0 \\ 0 \end{array}\right), \quad u^{(2)}(0)=\left(\begin{array}{l} 0 \\ 1 \\ 0 \\ 0 \end{array}\right), \quad v^{(1)}(0)=\left(\begin{array}{l} 0 \\ 0 \\ 1 \\ 0 \end{array}\right), \quad v^{(2)}(0)=\left(\begin{array}{l} 0 \\ 0 \\ 0 \\ 1 \end{array}\right) $$$\gamma^0$ in Standard Representation
$$ \gamma^{0}=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array}\right) $$obtaioned from Chiral Representation
$$ \begin{aligned} \gamma_{\mathrm{SR}}^{0} &=S \gamma_{\mathrm{CR}}^{0} S^{-1} \\ S &=\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right) \end{aligned} $$$\psi$ in standard representation
$$ \psi=S\left(\begin{array}{l} \phi_{\mathrm{R}} \\ \phi_{\mathrm{L}} \end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{l} \phi_{\mathrm{R}}+\phi_{\mathrm{L}} \\ \phi_{\mathrm{R}}-\phi_{\mathrm{L}} \end{array}\right) $$For a Lorentz boost to a moving frame, we have, from $(2.78)$ with $\boldsymbol{\theta}=0$,
$$ \left(\begin{array}{l} \phi_{\mathrm{R}} \\ \phi_{\mathrm{L}} \end{array}\right) \rightarrow\left(\begin{array}{l} \phi_{\mathrm{R}}^{\prime} \\ \phi_{\mathrm{L}}^{\prime} \end{array}\right)=\left(\begin{array}{cc} \mathrm{e}^{1 / 2 \sigma \cdot \phi} & 0 \\ 0 & \mathrm{e}^{-1 / 2 \sigma \cdot \varphi} \end{array}\right)\left(\begin{array}{l} \phi_{\mathrm{R}} \\ \phi_{\mathrm{L}} \end{array}\right)=M\left(\begin{array}{l} \phi_{\mathrm{R}} \\ \phi_{\mathrm{L}} \end{array}\right) $$$$ M_{\mathrm{SR}}=S M_{\mathrm{CR}} S^{-1}=\left(\begin{array}{cc} \cosh (\phi / 2) & \sigma \cdot \mathrm{n} \sinh (\phi / 2) \\ \sigma \cdot \mathrm{n} \sinh (\phi / 2) & \cosh (\phi / 2) \end{array}\right) $$$$ \begin{aligned} \cosh (\phi / 2) &=\left(\frac{E+m}{2 m}\right)^{1 / 2}, \quad \sinh (\phi / 2)=\left(\frac{E-m}{2 m}\right)^{1 / 2}, \\ \tanh (\phi / 2) &=\frac{p}{E+m} \end{aligned} $$$$ M_{\mathrm{SR}}=\left(\frac{E+m}{2 m}\right)^{1 / 2}\left(\begin{array}{cccc} 1 & 0 & \frac{p_{z}}{E+m} & \frac{p_{x}-\mathrm{i} p_{y}}{E+m} \\ 0 & 1 & \frac{p_{x}+\mathrm{i} p_{y}}{E+m} & \frac{-p_{z}}{E+m} \\ \frac{p_{z}}{E+m} & \frac{p_{x}-\mathrm{i} p_{y}}{E+m} & 1 & 0 \\ \frac{p_{x}+\mathrm{i} p_{y}}{E+m} & \frac{-p_{z}}{E+m} & 0 & 1 \end{array}\right) $$$$ \begin{aligned} &u^{(1)}=\left(\frac{E+m}{2 m}\right)^{1 / 2}\left(\begin{array}{c} 1 \\ \frac{p_{z}}{E+m} \\ \frac{p_{+}}{E+m} \end{array}\right), \quad u^{(2)}=\left(\frac{E+m}{2 m}\right)^{1 / 2}\left(\begin{array}{c} 0 \\ \frac{p_{-}}{E+m} \\ \frac{-p_{z}}{E+m} \end{array}\right)\\ &v^{(1)}=\left(\frac{E+m}{2 m}\right)^{1 / 2}\left(\begin{array}{c} \frac{p_{z}}{E+m} \\ \frac{p_{+}}{E+m} \\ 1 \\ 0 \end{array}\right), \quad v^{(2)}=\left(\frac{E+m}{2 m}\right)^{1 / 2}\left(\begin{array}{c} \frac{p_{-}}{E+m} \\ \frac{-p_{z}}{E+m} \\ 0 \\ 1 \end{array}\right) \end{aligned} $$where $p_{\pm}=p_{x} \pm \mathrm{i} p_{y} .$ The normalisation of the $u$ spinors is
$$\bar{u}^{(1)} u^{(1)}= \left(\frac{E+m}{2 m}\right)\left(\begin{array}{llll}1 & 0 & \frac{p_{z}}{E+m} & \frac{p_{-}}{E+m}\end{array}\right)\left(\begin{array}{cccc}1 & & & \\ & 1 & & \\ & & -1 & \\ & & & -1\end{array}\right)\left(\begin{array}{c}1 \\ \frac{p_{z}}{E+m} \\ \frac{p_{+}}{E+m}\end{array}\right)=1$$$$ \left(\frac{E+m}{2 m}\right)\left(\begin{array}{llll} 1 & 0 & \frac{p_{z}}{E+m} & \frac{p_{-}}{E+m} \end{array}\right)\left(\begin{array}{cccc} 1 & & & \\ & 1 & & \\ & & -1 & \\ & & & -1 \end{array}\right)\left(\begin{array}{c} 1 \\ \frac{0}{E_{z}} \\ \frac{p+m}{\frac{p_{+}}{E+m}} \end{array}\right)=1 $$$$ \begin{aligned} \bar{u}^{(\alpha)}(p) u^{\left(\alpha^{\prime}\right)}(p) &=\delta_{\alpha \alpha^{\prime}} \\ \bar{v}^{(\alpha)}(p) v^{\left(\alpha^{\prime}\right)}(p) &=-\delta_{\alpha \alpha^{\prime}} \\ \bar{u}^{(\alpha)}(p) v^{\left(\alpha^{\prime}\right)}(p) &=0 \\ u^{(\alpha)+}(p) u^{\left(\alpha^{\prime}\right)}(p) &=v^{(\alpha)+}(p) v^{\left(\alpha^{\prime}\right)}(p)=\frac{E}{m} \delta_{\alpha \alpha^{\prime}} . \end{aligned} $$In addition, from (2.96) and (2.136), $u$ and $v$ satisfy and it follows that the adjoint spinors obey
$$ \begin{array}{r}\bar{u}(p)(\gamma \cdot p-m)=0 \\ \bar{v}(p)(\gamma \cdot p+m)=0\end{array} $$Projection Operator
The operator $m(\gamma)(\gamma \cdot p+m)=0 .)$
$$ P_{+}=\sum_{\alpha} u^{(\alpha)}(p) \bar{u}^{(\alpha)}(p) $$is important in many applications. It is a projection operator, since, in view of (2.139),
$$ P_{+}^{2}=\sum_{\alpha . \beta} u^{(\alpha)}(p) \bar{u}^{(\alpha)}(p) u^{(\beta)}(p) \bar{u}^{(\beta)}=\sum_{\alpha} u^{(\alpha)}(p) \bar{u}^{(\alpha)}(p)=P_{+} $$$$ (\gamma \cdot p-m) P_{+}=0 $$therefore
$$ \frac{\gamma \cdot p}{m} P_{+}=P_{+} . $$Now we assume that $P_{+}$is of the form $a+b \gamma \cdot p$. Inserting $(2.144)$ into this gives $a=m b$. Then using $P_{+}^{2}=P_{+}$gives $b=1 / 2 m$. so finally we have
$$ P_{+}=\sum_{\alpha} u^{(\alpha)}(p) \bar{u}^{(\alpha)}(p)=\frac{\gamma \cdot p+m}{2 m} . $$Similarly the projection operator for negative energy states is
$$ P_{-}=-\sum_{\alpha} v^{(\alpha)}(p) \bar{v}^{(\alpha)}(p)=\frac{-\gamma \cdot p+m}{2 m} . $$As expected, $P_{+}+P_{-}=1$.
Trace formula
$$ \begin{aligned} \operatorname{Tr}(\gamma \cdot \alpha)(\gamma \cdot b) &=\operatorname{Tr}(\gamma \cdot b)(\gamma \cdot a) \\ &=\frac{1}{2} \operatorname{Tr} a_{\mu} b_{v}\left\{\gamma^{\mu}, \gamma^{v}\right\} \\ &=a \cdot b \operatorname{Tr} 1=4 a \cdot b \end{aligned} $$$$ \left(\gamma^{5}\right)^{2}=1, \quad\left\{\gamma^{5}, \gamma^{\mu}\right\}=0 $$$$ \operatorname{Tr} d_{1} \ldots d_{n}=0, \quad n \text { odd } $$proof by $\gamma^5$
$$ \begin{aligned} \operatorname{Tr}(\gamma \cdot a)(\gamma \cdot b)(\gamma \cdot c)(\gamma \cdot d)=&-\operatorname{Tr}(\gamma \cdot b)(\gamma \cdot a)(\gamma \cdot c)(\gamma \cdot d) \\ &+2 a \cdot b \operatorname{Tr}(\gamma \cdot c)(\gamma \cdot d) \end{aligned} $$Non-relativistic limit and the electron magnetic moment
$$ p^{\mu} \rightarrow p^{\mu}-e A^{\mu} $$or, with $p^{\mu}=(E, \mathbf{p}), A^{\mu}=(\phi, \mathbf{A})$,
$$ E \rightarrow E-e \phi, \quad \mathbf{p} \rightarrow \mathbf{p}-e \mathbf{A} . $$The Dirac equation (2.94) then becomes
$$ \gamma^{0}(E-e \phi) \psi-\gamma \cdot(\mathbf{p}-e \mathbf{A}) \psi=m \psi . $$In the standard representation of the $\gamma$ matrices (see (2.130) and (2.131))
$$ \gamma^{0}=\left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right), \quad \gamma=\left(\begin{array}{rr} 0 & \sigma \\ -\sigma & 0 \end{array}\right), \quad \psi=\left(\begin{array}{l} u \\ v \end{array}\right) $$this becomes
$$ (E-e \phi)\left(\begin{array}{r} u \\ -v \end{array}\right)-(\mathbf{p}-e \mathbf{A}) \cdot\left(\begin{array}{rr} 0 & \sigma \\ -\sigma & 0 \end{array}\right)\left(\begin{array}{l} u \\ v \end{array}\right)=m\left(\begin{array}{l} u \\ v \end{array}\right) $$which we write explicitly as two equations
$$ (E-e \phi) u-\sigma \cdot(\mathbf{p}-e \mathbf{A}) v=m u, $$$$ -(E-e \phi) v+\sigma \cdot(\mathbf{p}-e \mathbf{A}) u=m v $$The second equation gives
$$ v=(E+m-e \phi)^{-1} \sigma \cdot(\mathbf{p}-e \mathbf{A}) u . $$Note that the order of these factors is important, since $\mathbf{p}$ and $\phi(r)$ do not commute, $\mathbf{p}$ being proportional to the gradient operator. In the non-relativistic and weak field limit $E+m-e \phi \approx 2 m, p \approx m v$, so
$$ v \approx \frac{1}{2 m} \boldsymbol{\sigma} \cdot(\mathbf{p}-\mathrm{eA}) u=0\left(\frac{v}{c}\right) u $$and we see that the bottom two components of $\psi$ are much smaller than the top two. Inserting (2.160) into (2.158) gives
$$ E u=\frac{\boldsymbol{\sigma} \cdot \pi \boldsymbol{\sigma} \cdot \pi}{2 m} u+m u+e \phi u $$where $\pi=\mathbf{p}-e \mathbf{A}$, and, with $E=m+W$, we have
$$ W u=\left[\frac{1}{2 m}(\sigma \cdot \pi)(\sigma \cdot \pi)+e \phi\right] u . $$Now using $\sigma_{i} \sigma_{j}=\delta_{i j}+\mathrm{i} \varepsilon_{i j k} \sigma_{k}$ it follows that
$$ (\boldsymbol{\sigma} \cdot \mathbf{A})(\boldsymbol{\sigma} \cdot \mathbf{B})=\mathbf{A} \cdot \mathbf{B}+\mathrm{i} \boldsymbol{\sigma} \cdot(\mathbf{A} \times \mathbf{B}) $$so
$$ \begin{aligned} (\boldsymbol{\sigma} \cdot \pi)^{2} &=\pi \cdot \pi+\mathrm{i} \boldsymbol{\sigma} \cdot(\pi \times \pi) \\ &=(\mathbf{p}-e \mathbf{A})^{2}+\mathrm{i} \boldsymbol{\sigma} \cdot(\mathbf{p}-e \mathbf{A}) \times(\mathbf{p}-e \mathbf{A}) \end{aligned} $$The only non-zero part of the cross product in the last term is
$$ \mathbf{p} \times \mathbf{A}+\mathbf{A} \times \mathbf{p} . $$Using the operator equation
$$ \left[p_{i}, A_{j}\right]=-\mathrm{i} \hbar \partial_{i} A_{j} $$we have, taking the difference with the same equation with $i \leftrightarrow j$,
$$ \left(p_{i} A_{j}-p_{j} A_{i}\right)+\left(A_{i} p_{j}-A_{j} p_{i}\right)=-\mathrm{i} \hbar\left(\partial_{i} A_{j}-\partial_{j} A_{i}\right) . $$Multiplying both sides by $\varepsilon_{i j k}$ and summing over $i$ and $j$ gives the $k$ component of
$$ \mathbf{p} \times \mathbf{A}+\mathbf{A} \times \mathbf{p}=-\mathrm{i} \hbar \nabla \times \mathbf{A}=-\mathrm{i} \hbar \mathbf{B}, $$so we have a value for $(2.164)$. Substituting finally in $(2.161)$ gives $W u=H u$ where
$$ H=\frac{1}{2 m}(\mathbf{p}-e \mathbf{A})^{2}+e \phi-\frac{e \hbar}{2 m} \boldsymbol{\sigma} \cdot \mathbf{B} . $$2.7 The relevance of the Poincare group: spin operators and the zero mass limit
Operator expression for angular momentum and boosts
$$ \begin{aligned} &x^{\prime}=x \cos \theta+y \sin \theta \\ &y^{\prime}=-x \sin \theta+y \cos \theta \\ &z^{\prime}=z \end{aligned} $$$$ \begin{aligned} J_{z} f(x, y, z) &=\mathrm{i} \lim _{\theta \rightarrow 0}\left[\frac{f\left(x^{\prime}, y^{\prime}, z\right)-f(x, y, z)}{\theta}\right] \\ &=\mathrm{i} \lim _{\theta \rightarrow 0}\left[\frac{f(x+y \theta, y-x \theta, z)-f(x, y, z)}{\theta}\right] \\ &=\mathrm{i}\left(y \frac{\partial f}{\partial x}-x \frac{\partial f}{\partial y}\right) \\ J_{z} &=-\mathrm{i}\left(x \frac{\partial}{\partial y}-y \frac{\partial}{\partial x}\right) . \end{aligned} $$$$ J_{x}=-\mathrm{i}\left(y \frac{\partial}{\partial z}-z \frac{\partial}{\partial y}\right), \quad J_{y}=-\mathrm{i}\left(z \frac{\partial}{\partial x}-x \frac{\partial}{\partial z}\right) $$and we can easily prove that $\left[J_{x}, J_{y}\right]=\mathrm{i} J_{z} \quad$ and cyclic perms.
Definition of generator
$$ \begin{aligned} X_{\alpha} &=\mathrm{i}\left(\left.\frac{\partial x^{\prime}}{\partial a^{\alpha}}\right|_{a=0} \frac{\partial}{\partial x}+\left.\frac{\partial y^{\prime}}{\partial a^{\alpha}}\right|_{a=0} \frac{\partial}{\partial y}+\left.\frac{\partial z^{\prime}}{\partial a^{\alpha}}\right|_{a=0} \frac{\partial}{\partial z}+\left.\frac{\partial t^{\prime}}{\partial a^{\alpha}}\right|_{a=0} \frac{\partial}{\partial t}\right) \\ &=\mathrm{i} \frac{\partial x^{\prime \mu}}{\partial a^{\alpha}} \frac{\partial}{\partial x^{\mu}}(\alpha=1, \ldots, r) \end{aligned} $$We find that the generator $K_{x}$ is given by
$$ K_{x}=\mathrm{i}\left(t \frac{\partial}{\partial x}+x \frac{\partial}{\partial t}\right) $$and similarly
$$ K_{y}=\mathrm{i}\left(t \frac{\partial}{\partial y}+y \frac{\partial}{\partial t}\right), \quad K_{z}=i\left(t \frac{\partial}{\partial z}+z \frac{\partial}{\partial t}\right) $$This gives
$$ \left.\begin{array}{rl} {\left[K_{x}, K_{y}\right]=-\mathrm{i} J_{z}} & \text { and cyclic perms, } \\ {\left[K_{x}, J_{y}\right]=\mathrm{i} K_{z}} & \text { and cyclic perms, } \\ {\left[K_{x}, J_{x}\right]=0,} & \text { etc., } \end{array}\right\} $$Chapter 3 Lagrangian formulation, symmetries and gauge fields
3.2 The real scalar field: variational principle and Noether’s theorem
Euler-Lagrange Equation
$$ \frac{\partial \mathscr{L}}{\partial \phi}-\frac{\partial}{\partial x^{\mu}}\left[\frac{\partial \mathscr{L}}{\partial\left(\partial_{\mu} \phi\right)}\right]=0 $$Lagrangian for Klein-Gordon Equation
$$ \mathscr{L}=\frac{1}{2} g^{\kappa \lambda}\left(\partial_{\kappa} \phi\right)\left(\partial_{\lambda} \phi\right)-\frac{m^{2}}{2} \phi^{2} $$This equation is written deliberately to give the Klein-Gordon Equation
Symmetry and conserved quantity(Noether’s theorem), There’s another type of conserved quantity which is topological in nature, and whose conservation has nothing to do with Nother’s theorem
Energy-Momentum tensor
3.3 Complex scalar fields and the electromagnetic field
Conservation of charge from noether’s theorem
$$ \begin{aligned} \phi &=\left(\phi_{1}+\mathrm{i} \phi_{2}\right) / \sqrt{2} \\ \phi^{*} &=\left(\phi_{1}-\mathrm{i} \phi_{2}\right) / \sqrt{2} \end{aligned} $$$$ \mathscr{L}=\left(\partial_{\mu} \phi\right)\left(\partial^{\mu} \phi^{*}\right)-m^{2} \phi^{*} \phi $$$$ \begin{gathered} \left(\square+m^{2}\right) \phi=0 \\ \left(\square+m^{2}\right) \phi^{*}=0 \end{gathered} $$Symmetry
$$ \phi \rightarrow \mathrm{e}^{-\mathrm{i} \lambda} \phi, \quad \phi^{*} \rightarrow \mathrm{e}^{\mathrm{i} \Lambda} \phi^{*} $$where $\Lambda$ is a real constant. This is known as a gauge transformation of the first kind. Its infinitesimal form is
$$ \delta \phi=-\mathrm{i} \Lambda \phi, \quad \delta \phi^{*}=\mathrm{i} \Lambda \phi^{*} $$and so
$$ \delta\left(\partial_{\mu} \phi\right)=-\mathrm{i} \Lambda \partial_{\mu} \phi, \quad \delta\left(\partial_{\mu} \phi^{*}\right)=\mathrm{i} \Lambda \partial_{\mu} \phi^{*} $$Since the transformation (3.55) does not involve space-time (it is purely “internal”), in the notation of (3.22) we have
$$ \Phi=-\mathrm{i} \phi, \quad \Phi^{*}=\mathrm{i} \phi^{*}, \quad X=0 . $$Noether’s theorem then gives a conserved current, which is, from equation $(3.25)$
$$ J^{\mu}=\frac{\partial \mathscr{L}}{\partial\left(\partial_{\mu} \phi\right)}(-\mathrm{i} \phi)+\frac{\partial \mathscr{L}}{\partial\left(\partial_{\mu} \phi^{*}\right)}\left(\mathrm{i} \phi^{*}\right) $$(Here, the ‘internal indices’ of $\phi$ have effectively been summed over, giving separate contributions from $\phi$ and $\left.\phi^{*} .\right)$ Substituting (3.52) into (3.59) then gives
$$ J^{\mu}=\mathrm{i}\left(\phi^{*} \partial^{\mu} \phi-\phi \partial^{\mu} \phi^{*}\right) $$It follows immediately from (3.53) and (3.54) that this current has a vanishing 4-divergence, as of course it should have:
$$ \partial_{\mu} J^{\mu}=0, $$and the corresponding conserved quantity is, from (3.27),
$$ \begin{aligned} Q &=\int J^{0} \mathrm{~d} V \\ &=\mathrm{i} \int\left(\phi^{*} \frac{\partial \phi}{\partial t}-\phi \frac{\partial \phi^{*}}{\partial t}\right) \mathrm{d} V \end{aligned} $$Geometry form of guage transformation
$$ \mathscr{L}=\left(\partial_{\mu} \phi_{1}\right)\left(\partial^{\mu} \phi_{1}\right)+\left(\partial_{\mu} \phi_{2}\right)\left(\partial^{\mu} \phi_{2}\right)-m^{2}\left(\phi_{1}^{2}+\phi_{2}^{2}\right) $$$$ \phi=\mathrm{i} \phi_{1}+\mathrm{j} \phi_{2} $$and then write
$$ \phi=\mathbf{i} \phi_{1}+\mathbf{j} \phi_{2} $$as a vector in a 2-dimensional space with orthonormal basis vectors $\mathbf{i}$ and $\mathbf{j}$. The Lagrangian is now
$$ \mathscr{L}=\left(\partial_{\mu} \boldsymbol{\phi}\right) \cdot\left(\partial^{\mu} \boldsymbol{\phi}\right)-m^{2} \boldsymbol{\phi} \cdot \boldsymbol{\phi} . $$The gauge transformation (3.55) may be written
$$ \begin{aligned} &\phi_{1}^{\prime}+\mathrm{i} \phi_{2}^{\prime}=\mathrm{e}^{-\mathrm{i} \Lambda}\left(\phi_{1}+\mathrm{i} \phi_{2}\right), \\ &\phi_{1}^{\prime}-\mathrm{i} \phi_{2}^{\prime}=\mathrm{e}^{\mathrm{i} \Lambda}\left(\phi_{1}-\mathrm{i} \phi_{2}\right), \end{aligned} $$which is equivalent to
$$ \left.\begin{array}{l} \phi_{1}^{\prime}=\phi_{1} \cos \Lambda+\phi_{2} \sin \Lambda \\ \phi_{2}^{\prime}=-\phi_{1} \sin \Lambda+\phi_{2} \cos \Lambda . \end{array}\right\} $$Rotations in two dimensions form the group of $SO(2)$. On the other hand, since the transformation was equivalently represented by a unitary matrix in one dimension
$$ e^{i\Lambda}(e^{i\Lambda})^*=1 $$Then $SO(2)\approx U(1)$
Since $\Lambda$ is a constant, the gauge transformation must be the same at all points in spacetime, it is a global symmetry. We must perform the same rotations at all points at the same time, but it contradicts relativity, therefore, we adopt $\Lambda(x^\mu)$, which is called a local gauge transformation, also called a guage transformation of the second kind.