Spinor and Vector Calculations

For spin-1/2 field, we have the representation of left-handed Weyl spinor $\psi_L = (\psi_1, \psi_2)^T$, and the Lorentz transformation $x\rightarrow \Lambda x ,~ \psi_L(x) \rightarrow S_L(\Lambda)\psi_L(x)$, where $S(\Lambda) = \exp[ -i\boldsymbol\theta\cdot\mathbf J -i\boldsymbol\xi\cdot\mathbf K ] = \exp( -i\boldsymbol\theta\cdot\frac{\boldsymbol\sigma}{2} -\boldsymbol\xi\cdot\frac{\boldsymbol\sigma}{2})$. Together with the right-handed Weyl spinor, we have the 4-dim Dirac spinor $\Psi = (\psi_L, \psi_R)^T$.

As the derivative of the spinor fields is so complicated, their calculus also follows specific rules. We will clarify $\varepsilon_{ij}$, $\sigma^{\mu}_{a\dot a}$, and $\gamma^{\mu}$, and derive the Dirac equation in the end.

Towards Dirac Equation

For a certain Weyl spinor, we have $\psi_a \to A^{b}_a \psi_b$, then we can try to construct Lorentz-invariant term like $\varphi\varphi,~V^\mu W_\mu=g_{\mu\nu}V^\mu W^\nu$. However, we could soon find that $\psi^{\dagger} \chi$ is not Lorentz-invariant, we need to define $\psi^T M\chi$ which satisfies $(A\psi)^T M(A\chi)=\psi^TM\chi$. This gives $\varepsilon_{ab}$ and $\psi_\alpha =\varepsilon_{\alpha\beta}\psi^\beta,~ \psi^T\varepsilon\chi = \varepsilon_{\alpha\beta}\psi^\alpha\chi^\beta =\psi_1\chi_2-\psi_2\chi_1$, which is also anti-symmetric. In this way, we can write the mass scalar term $\mathcal L_M =-\frac12 m (\varepsilon_{\alpha\beta}\psi^\alpha\psi^\beta)$. $\varepsilon_{\alpha\beta} =\begin{pmatrix}0&1\\-1&0\end{pmatrix}, \qquad \varepsilon^{\alpha\beta} =\begin{pmatrix}0&-1\\1&0\end{pmatrix}$

Then we consider the kinetic term $\partial_\mu\psi_a$, we need new matrix to connect these index $\mu,~ a$ to construct Lorentz-invariant scalar. As $(\frac12,0)\otimes (0,\frac12) = (\frac12,\frac12)$, we have $V_{a\dot a} \leftrightarrow V_{\mu}$ and link their indices with $V_{a \dot a} = V_{\mu}\sigma^\mu_{a\dot a}$, while the $2\times2$ space can be represented in a 4-dim basis $\sigma^\mu_{a\dot a} = (I,\sigma^1,\sigma^2,\sigma^3)$. We want to notice that for left-handed Weyl spinor, the Lorentz boost on z-axis is $\psi_L\rightarrow A_L\psi_L,~ A_L=e^{-\frac{\varphi}{2}\sigma^3}$, which yields $j_L^0=\psi_L^\dagger\psi_L \to j_L^3=-\psi_L^\dagger\sigma^3\psi_L$, while the right-handed Weyl spinor gives $j_R^0=\psi_R^\dagger\psi_R \to j_R^3=\psi_R^\dagger\sigma^3\psi_R$. To unify the boost transformation, we define $\bar \sigma^\mu_{a\dot a} = (I,-\sigma^1,-\sigma^2,-\sigma^3)$ for left-handed, and use $X(x)=x_\mu\bar\sigma^\mu$ for later derivation. Under Lorentz transformation$X'=AXA^\dagger,~ x^\mu=\Lambda^\mu{}_\nu(A)x^\nu$, we have $A\bar\sigma_\nu A^\dagger = \Lambda^\mu{}_\nu\bar\sigma_\mu$. Then we can construct a Lorentz vector $j_L^\mu=\psi_L^\dagger\bar\sigma^\mu\psi_L$, and combine it with $\partial_{\mu}$ to write the Lorentz-invariant kinetic term $\mathcal L_L =i\psi_L^\dagger\bar\sigma^\mu\partial_\mu\psi_L$. The $i$ is to keep $S=\int d^4x\,\mathcal L_L$ real with just a total derivative in the conjugate $\mathcal{L}^{\dagger}$.

Together with right-handed Weyl field, we have $\psi_L'(x')=A\psi_L(x),~ \psi_R'(x')=(A^\dagger)^{-1}\psi_R(x)$ due to the opposite direction in Lorentz boost. Thus the corresponding Lorentz scalar is $\mathcal L_m =-m(\psi_L^\dagger\psi_R+\psi_R^\dagger\psi_L)$, where left-handed and right-handed Weyl spinor couple with each other. $\mathcal L_D= i\psi_L^\dagger\bar\sigma^\mu\partial_\mu\psi_L +i\psi_R^\dagger\sigma^\mu\partial_\mu\psi_R -m(\psi_L^\dagger\psi_R+\psi_R^\dagger\psi_L)$

Doing variation on both conjugate fields, we get $i\bar\sigma^\mu\partial_\mu\psi_L=m\psi_R,~ i\sigma^\mu\partial_\mu\psi_R=m\psi_L$. For express convenience, we combine them using the gamma matrix $\gamma^{\mu}$ and the Dirac field $\Psi = (\psi_L, \psi_R)^T$. In this way, the motion equation of the Weyl field can be write as $(i\gamma^\mu\partial_\mu-m)\Psi=0$, which is also called the Dirac equation. Using $\gamma^{\mu}$, we can define the Dirac adjoint $\bar\Psi\equiv\Psi^\dagger\gamma^0 = (\psi^{\dagger}_{R},\psi^{\dagger}_{L})$, and rewrite the Lagrangian as $\mathcal L_D=\bar\Psi(i\gamma^\mu\partial_\mu-m)\Psi$. $\gamma^\mu= \begin{pmatrix} 0&\sigma^\mu\\ \bar\sigma^\mu&0 \end{pmatrix}$

• We skip the quantization for now.
• Some useful calculation: define $a\!\!\!/ = a_{\mu} \gamma^{\mu}$, $a\!\!\!/ b\!\!\!/ = -ab$, $\{\gamma^\mu,\gamma^\nu\}= -2\eta^{\mu\nu}$, $S_z=\frac{i}{4}[\gamma^1, \gamma^2] = \text{diag}(\sigma^3,\sigma^3)$ (this is given by $J_{i} = \frac{1}{2}\varepsilon_{ijk}M^{jk}$ in Section 1).
• trace calculation: $Tr(\text{odd }\gamma^\mu)=0,~ Tr (\gamma^\mu\gamma^\nu)=-4g^{\mu\nu},~ Tr(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)=4\left( g^{\mu\nu}g^{\rho\sigma} -g^{\mu\rho}g^{\nu\sigma} +g^{\mu\sigma}g^{\nu\rho} \right)$; $\gamma_\mu\gamma^\mu=-4I_4,~ \gamma_\mu\gamma^\nu\gamma^\mu=2\gamma^\nu$, and some others.

Spinor Solution

Start from the Dirac equation, we note that if we multiply $(i\gamma^\nu\partial_\nu+m)$ on the left of the Dirac equation, we will get $(\Box+m^2)\Psi=0$ in the form of Klein-Gordon equation. In this way, we can consider the plane-wave solution $\Psi (x) = u(\mathbf{p}) e^{ipx} + v(\mathbf{p}) e^{-ipx}$, which then gives $(p\!\!\!/ +m)u(p)=0,~ (p\!\!\!/ -m)v(p)=0$. Then we can write the Dirac spinor as: $\Psi(x) = \sum_s\int\widetilde{dp} [ b_s(\mathbf p)u_s(p)e^{ipx} + d_s^\dagger(\mathbf p)v_s(p)e^{-ipx}]$

In the rest frame of single particle where $p_{\mu}=(m,0,0,0)$, for $u(\mathbf{p})$ we have $m(\gamma^0-I_4) (u_L,u_R)^T=0$, then $u_L=u_R \equiv \sqrt{m} (\xi_s, \xi_s)^T$. Consider two choices of spin, we use $\xi_\uparrow= (1,0)^T,~ \xi_\downarrow= (0,1)^T$ where z-axis is selected as the basis of the quantum states and $S_z u_{\pm} = \pm \frac12 u_{\pm}(\mathbf{0}),~ S_z v_{\pm} = \mp \frac12 v_{\pm}(\mathbf{0})$, and similarly $v_s(0)=\sqrt m (\eta_s, -\eta_s)^T$. Then with Lorentz boost generators $K_i$, we can act $D(\Lambda)=\exp(i\eta\,\hat{\mathbf p}\cdot\mathbf K)$ on them to get $u_s(p)=D(\Lambda)u_s(0),~ v_s(p)=D(\Lambda)v_s(0)$, where $\cosh \eta = \gamma,~ \sinh \eta = v$.

• some common results: $\bar u_{s'}(p)u_s(p)=2m\delta_{s's}, \bar v_{s'}(p)v_s(p)=-2m\delta_{s's}, \bar u_{s'}(p)v_s(p)=0$, where $\bar u_s(p)\equiv u_s^\dagger(p)\beta,~ \beta=\gamma^0$.
• $\bar u_{s'}(p)\gamma^\mu u_s(p)=2p^\mu\delta_{s's},~ \bar v_{s'}(p)\gamma^\mu v_s(p)=2p^\mu\delta_{s's}$, $u_s^\dagger(p)u_s(p) =v_s^\dagger(p)v_s(p)=2E_{\mathbf p}$.
• spin sums: $\sum_{s=\pm}u_s(p)\bar u_s(p)=- p\!\!\!/ + m,~ \sum_{s=\pm}v_s(p)\bar v_s(p)=- p\!\!\!/ - m$, we will use them in QED calculation.
• Gordon identity: $2m\bar u_{s'}(p')\gamma^\mu u_s(p) = \bar u_{s'}(p') \left[(p'+p)^\mu-2iS^{\mu\nu}(p'-p)_\nu\right] u_s(p)$, where $S^{\mu\nu}=\frac{i}{4}[\gamma^\mu,\gamma^\nu]$.

To further discuss fermions, we quantize the Dirac fields using the canonical anti-commutation relations. We have the canonical conjugate momentum $\Pi_\alpha(x) =\frac{\partial\mathcal L} {\partial(\partial_0\Psi_\alpha)} =i\Psi_\alpha^\dagger(x)$, and $\{\hat\Psi_\alpha(\mathbf x,t), \hat\Pi_\beta(\mathbf y,t)\} =i\delta_{\alpha\beta}\delta^3(\mathbf x-\mathbf y),~ \{\hat\Psi_\alpha,\hat\Psi_\beta\}=0$, where field and coefficients of different modes have become operators. These yield $\{\hat b_s(p),\hat b_{s'}^\dagger(p')\} =(2\pi)^3 2E_{\mathbf p} \delta_{ss'}\delta^3(\mathbf p-\mathbf p'),~ \{\hat d_s(p),\hat d_{s'}^\dagger(p')\} =(2\pi)^3 2E_{\mathbf p} \delta_{ss'}\delta^3(\mathbf p-\mathbf p')$, and in Fock space we can write $|p,s;b\rangle=\hat b_s^\dagger(p)|0\rangle,~ |p,s;d\rangle=\hat d_s^\dagger(p)|0\rangle$ with the same mass and spin. We will distinguish them with the charge, that the current is $j^\mu=\bar\Psi\gamma^\mu\Psi$ under $U(1)$ transformation $\Psi \to e^{-i\alpha}$, then the charge is $Q=\int d^3x\,j^0(x) = \sum_s\int\widetilde{dp} [b_s^\dagger(p)b_s(p) - d_s^{\dagger}(p)d_s(p)]$, which indicates that they have opposite charge and therefore called particle and anti-particle (we don’t require this conserved charge is the electromagnetic charge here).

Now we have the basic theory describing the free spin-1/2 field, we will discuss how to their propagation and interaction in QED later, especially with the photon — electromagnetic waves. Also, we will go back to charity, helicity and symmetry in later sections. For spin-1/2 field, we have the representation of left-handed Weyl spinor $\psi_L = (\psi_1, \psi_2)^T$, and the Lorentz transformation $x\rightarrow \Lambda x ,~ \psi_L(x) \rightarrow S_L(\Lambda)\psi_L(x)$, where $S(\Lambda) = \exp[ -i\boldsymbol\theta\cdot\mathbf J -i\boldsymbol\xi\cdot\mathbf K ] = \exp( -i\boldsymbol\theta\cdot\frac{\boldsymbol\sigma}{2} -\boldsymbol\xi\cdot\frac{\boldsymbol\sigma}{2})$. Together with the right-handed Weyl spinor, we have the 4-dim Dirac spinor $\Psi = (\psi_L, \psi_R)^T$.

As the derivative of the spinor fields is so complicated, their calculus also follows specific rules. We will clarify $\varepsilon_{ij}$, $\sigma^{\mu}_{a\dot a}$, and $\gamma^{\mu}$, and derive the Dirac equation in the end.

Towards Dirac Equation

For a certain Weyl spinor, we have $\psi_a \to A^{b}_a \psi_b$, then we can try to construct Lorentz-invariant term like $\varphi\varphi,~V^\mu W_\mu=g_{\mu\nu}V^\mu W^\nu$. However, we could soon find that $\psi^{\dagger} \chi$ is not Lorentz-invariant, we need to define $\psi^T M\chi$ which satisfies $(A\psi)^T M(A\chi)=\psi^TM\chi$. This gives $\varepsilon_{ab}$ and $\psi_\alpha =\varepsilon_{\alpha\beta}\psi^\beta,~ \psi^T\varepsilon\chi = \varepsilon_{\alpha\beta}\psi^\alpha\chi^\beta =\psi_1\chi_2-\psi_2\chi_1$, which is also anti-symmetric. In this way, we can write the mass scalar term $\mathcal L_M =-\frac12 m (\varepsilon_{\alpha\beta}\psi^\alpha\psi^\beta)$. $\varepsilon_{\alpha\beta} =\begin{pmatrix}0&1\\-1&0\end{pmatrix}, \qquad \varepsilon^{\alpha\beta} =\begin{pmatrix}0&-1\\1&0\end{pmatrix}$

Then we consider the kinetic term $\partial_\mu\psi_a$, we need new matrix to connect these index $\mu,~ a$ to construct Lorentz-invariant scalar. As $(\frac12,0)\otimes (0,\frac12) = (\frac12,\frac12)$, we have $V_{a\dot a} \leftrightarrow V_{\mu}$ and link their indices with $V_{a \dot a} = V_{\mu}\sigma^\mu_{a\dot a}$, while the $2\times2$ space can be represented in a 4-dim basis $\sigma^\mu_{a\dot a} = (I,\sigma^1,\sigma^2,\sigma^3)$. We want to notice that for left-handed Weyl spinor, the Lorentz boost on z-axis is $\psi_L\rightarrow A_L\psi_L,~ A_L=e^{-\frac{\varphi}{2}\sigma^3}$, which yields $j_L^0=\psi_L^\dagger\psi_L \to j_L^3=-\psi_L^\dagger\sigma^3\psi_L$, while the right-handed Weyl spinor gives $j_R^0=\psi_R^\dagger\psi_R \to j_R^3=\psi_R^\dagger\sigma^3\psi_R$. To unify the boost transformation, we define $\bar \sigma^\mu_{a\dot a} = (I,-\sigma^1,-\sigma^2,-\sigma^3)$ for left-handed, and use $X(x)=x_\mu\bar\sigma^\mu$ for later derivation. Under Lorentz transformation$X'=AXA^\dagger,~ x^\mu=\Lambda^\mu{}_\nu(A)x^\nu$, we have $A\bar\sigma_\nu A^\dagger = \Lambda^\mu{}_\nu\bar\sigma_\mu$. Then we can construct a Lorentz vector $j_L^\mu=\psi_L^\dagger\bar\sigma^\mu\psi_L$, and combine it with $\partial_{\mu}$ to write the Lorentz-invariant kinetic term $\mathcal L_L =i\psi_L^\dagger\bar\sigma^\mu\partial_\mu\psi_L$. The $i$ is to keep $S=\int d^4x\,\mathcal L_L$ real with just a total derivative in the conjugate $\mathcal{L}^{\dagger}$.

Together with right-handed Weyl field, we have $\psi_L'(x')=A\psi_L(x),~ \psi_R'(x')=(A^\dagger)^{-1}\psi_R(x)$ due to the opposite direction in Lorentz boost. Thus the corresponding Lorentz scalar is $\mathcal L_m =-m(\psi_L^\dagger\psi_R+\psi_R^\dagger\psi_L)$, where left-handed and right-handed Weyl spinor couple with each other. $\mathcal L_D= i\psi_L^\dagger\bar\sigma^\mu\partial_\mu\psi_L +i\psi_R^\dagger\sigma^\mu\partial_\mu\psi_R -m(\psi_L^\dagger\psi_R+\psi_R^\dagger\psi_L)$

Doing variation on both conjugate fields, we get $i\bar\sigma^\mu\partial_\mu\psi_L=m\psi_R,~ i\sigma^\mu\partial_\mu\psi_R=m\psi_L$. For express convenience, we combine them using the gamma matrix $\gamma^{\mu}$ and the Dirac field $\Psi = (\psi_L, \psi_R)^T$. In this way, the motion equation of the Weyl field can be write as $(i\gamma^\mu\partial_\mu-m)\Psi=0$, which is also called the Dirac equation. Using $\gamma^{\mu}$, we can define the Dirac adjoint $\bar\Psi\equiv\Psi^\dagger\gamma^0 = (\psi^{\dagger}_{R},\psi^{\dagger}_{L})$, and rewrite the Lagrangian as $\mathcal L_D=\bar\Psi(i\gamma^\mu\partial_\mu-m)\Psi$. $\gamma^\mu= \begin{pmatrix} 0&\sigma^\mu\\ \bar\sigma^\mu&0 \end{pmatrix}$

• We skip the quantization for now.
• Some useful calculation: define $a\!\!\!/ = a_{\mu} \gamma^{\mu}$, $a\!\!\!/ b\!\!\!/ = -ab$, $\{\gamma^\mu,\gamma^\nu\}= -2\eta^{\mu\nu}$, $S_z=\frac{i}{4}[\gamma^1, \gamma^2] = \text{diag}(\sigma^3,\sigma^3)$ (this is given by $J_{i} = \frac{1}{2}\varepsilon_{ijk}M^{jk}$ in Section 1).
• trace calculation: $Tr(\text{odd }\gamma^\mu)=0,~ Tr (\gamma^\mu\gamma^\nu)=-4g^{\mu\nu},~ Tr(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)=4\left( g^{\mu\nu}g^{\rho\sigma} -g^{\mu\rho}g^{\nu\sigma} +g^{\mu\sigma}g^{\nu\rho} \right)$; $\gamma_\mu\gamma^\mu=-4I_4,~ \gamma_\mu\gamma^\nu\gamma^\mu=2\gamma^\nu$, and some others.

Spinor Solution

Start from the Dirac equation, we note that if we multiply $(i\gamma^\nu\partial_\nu+m)$ on the left of the Dirac equation, we will get $(\Box+m^2)\Psi=0$ in the form of Klein-Gordon equation. In this way, we can consider the plane-wave solution $\Psi (x) = u(\mathbf{p}) e^{ipx} + v(\mathbf{p}) e^{-ipx}$, which then gives $(p\!\!\!/ +m)u(p)=0,~ (p\!\!\!/ -m)v(p)=0$. Then we can write the Dirac spinor as: $\Psi(x) = \sum_s\int\widetilde{dp} [ b_s(\mathbf p)u_s(p)e^{ipx} + d_s^\dagger(\mathbf p)v_s(p)e^{-ipx}]$

In the rest frame of single particle where $p_{\mu}=(m,0,0,0)$, for $u(\mathbf{p})$ we have $m(\gamma^0-I_4) (u_L,u_R)^T=0$, then $u_L=u_R \equiv \sqrt{m} (\xi_s, \xi_s)^T$. Consider two choices of spin, we use $\xi_\uparrow= (1,0)^T,~ \xi_\downarrow= (0,1)^T$ where z-axis is selected as the basis of the quantum states and $S_z u_{\pm} = \pm \frac12 u_{\pm}(\mathbf{0}),~ S_z v_{\pm} = \mp \frac12 v_{\pm}(\mathbf{0})$, and similarly $v_s(0)=\sqrt m (\eta_s, -\eta_s)^T$. Then with Lorentz boost generators $K_i$, we can act $D(\Lambda)=\exp(i\eta\,\hat{\mathbf p}\cdot\mathbf K)$ on them to get $u_s(p)=D(\Lambda)u_s(0),~ v_s(p)=D(\Lambda)v_s(0)$, where $\cosh \eta = \gamma,~ \sinh \eta = v$.

• some common results: $\bar u_{s'}(p)u_s(p)=2m\delta_{s's}, \bar v_{s'}(p)v_s(p)=-2m\delta_{s's}, \bar u_{s'}(p)v_s(p)=0$, where $\bar u_s(p)\equiv u_s^\dagger(p)\beta,~ \beta=\gamma^0$.
• $\bar u_{s'}(p)\gamma^\mu u_s(p)=2p^\mu\delta_{s's},~ \bar v_{s'}(p)\gamma^\mu v_s(p)=2p^\mu\delta_{s's}$, $u_s^\dagger(p)u_s(p) =v_s^\dagger(p)v_s(p)=2E_{\mathbf p}$.
• spin sums: $\sum_{s=\pm}u_s(p)\bar u_s(p)=- p\!\!\!/ + m,~ \sum_{s=\pm}v_s(p)\bar v_s(p)=- p\!\!\!/ - m$, we will use them in QED calculation.
• Gordon identity: $2m\bar u_{s'}(p')\gamma^\mu u_s(p) = \bar u_{s'}(p') \left[(p'+p)^\mu-2iS^{\mu\nu}(p'-p)_\nu\right] u_s(p)$, where $S^{\mu\nu}=\frac{i}{4}[\gamma^\mu,\gamma^\nu]$.

To further discuss fermions, we quantize the Dirac fields using the canonical anti-commutation relations. We have the canonical conjugate momentum $\Pi_\alpha(x) =\frac{\partial\mathcal L} {\partial(\partial_0\Psi_\alpha)} =i\Psi_\alpha^\dagger(x)$, and $\{\hat\Psi_\alpha(\mathbf x,t), \hat\Pi_\beta(\mathbf y,t)\} =i\delta_{\alpha\beta}\delta^3(\mathbf x-\mathbf y),~ \{\hat\Psi_\alpha,\hat\Psi_\beta\}=0$, where field and coefficients of different modes have become operators. These yield $\{\hat b_s(p),\hat b_{s'}^\dagger(p')\} =(2\pi)^3 2E_{\mathbf p} \delta_{ss'}\delta^3(\mathbf p-\mathbf p'),~ \{\hat d_s(p),\hat d_{s'}^\dagger(p')\} =(2\pi)^3 2E_{\mathbf p} \delta_{ss'}\delta^3(\mathbf p-\mathbf p')$, and in Fock space we can write $|p,s;b\rangle=\hat b_s^\dagger(p)|0\rangle,~ |p,s;d\rangle=\hat d_s^\dagger(p)|0\rangle$ with the same mass and spin. We will distinguish them with the charge, that the current is $j^\mu=\bar\Psi\gamma^\mu\Psi$ under $U(1)$ transformation $\Psi \to e^{-i\alpha}$, then the charge is $Q=\int d^3x\,j^0(x) = \sum_s\int\widetilde{dp} [b_s^\dagger(p)b_s(p) - d_s^{\dagger}(p)d_s(p)]$, which indicates that they have opposite charge and therefore called particle and anti-particle (we don’t require this conserved charge is the electromagnetic charge here).

Now we have the basic theory describing the free spin-1/2 field, we will discuss how to their propagation and interaction in QED later, especially with the photon — electromagnetic waves. Also, we will go back to charity, helicity and symmetry in later sections.

Vector Polarization

Review the expansion of vector field, and the Lagrangian $\mathcal L_{\mathrm{Proca}} =-\frac14F_{\mu\nu}F^{\mu\nu} -\frac12m^2A_\mu A^\mu$, motion equation $\partial_\mu F^{\mu\nu}-m^2A^\nu=0$, and the constrain $\partial_\mu A^\mu=0$ for $m \neq 0$, which constrains 4 components into 3 independent modes $\lambda = +1,0,-1$. In rest frame, we can write $\epsilon_+^\mu(0)=\frac1{\sqrt2}(0,1,-i,0),~ \epsilon_-^\mu(0)=\frac1{\sqrt2}(0,1,+i,0),~ \epsilon_0^\mu(0)=(0,0,0,1)$ using the eigenstates of $J_z$, which satisfy $p_\mu\epsilon^\mu(p)=0,~ \epsilon_\lambda^{\mu*}\epsilon_{\lambda'\mu} =\delta_{\lambda\lambda'}$. Then we can use Lorentz boost $\epsilon_{\lambda}^{\mu}(p) = \Lambda^{\mu}_{\nu} \epsilon_{\lambda}^{\nu}(0)$, where $\Lambda^{\mu}_{\nu}= \begin{pmatrix} E/m & \mathbf{p}^T / m\\ \mathbf{p} / m & I_3 + \mathbf{p}\mathbf{p}^T /{m(E+m)} \end{pmatrix}$. This could yield the Lorentz-invariant sum $\sum_\lambda \epsilon_\lambda^\mu(p)\epsilon_\lambda^{\nu*}(p) = g^{\mu\nu}+\frac{p^\mu p^\nu}{m^2}$.

$A^\mu(x) = \sum_{\lambda} \int\widetilde{dp}\ [ \epsilon^\mu_\lambda(p) a_\lambda(\mathbf p)e^{ipx} + \epsilon^{\mu *}_\lambda(p) a_\lambda^\dagger(\mathbf p)e^{-ipx}]$

For quantization of massive vector field, we have the canonical conjugate momentum $\Pi_0=0,~\Pi_i = \frac{\partial \mathcal{L}}{\partial \dot A_i} = \dot A_i +\partial_i A_0$, which means that $A_0$ is determined by $A_i$ and $\Pi_i$. Then we have the commutation relation $[A_i(\mathbf x,t),\Pi_j(\mathbf y,t)] =i\delta_{ij}\delta^3(\mathbf x-\mathbf y),~ [A_i,A_j]=[\Pi_i,\Pi_j]=0$, which give $[a_\lambda(p),a_{\lambda'}^\dagger(p')] =(2\pi)^3\,2E_{\mathbf p}\, \delta_{\lambda\lambda'}\delta^3(\mathbf p-\mathbf p')$. With the generators, we can define $|p,\lambda \rangle =a^{\dagger}_{\lambda}(p)|0\rangle$, and the free Hamiltonian $H=\sum_{\lambda} \int \widetilde{dp} E_{\mathbf{p}} a^{\dagger}_{\lambda}(p) a_{\lambda}(p)$. To describe vector field with charge, we will introduce the complex vector field and the corresponding anti-particle operator in later section like $\mathcal L =-\frac12W_{\mu\nu}^\dagger W^{\mu\nu} -m^2W_\mu^\dagger W^\mu$, which has similar calculation rules as the real one.

For massless particle like photon, things become more different with $\mathcal L =-\frac14F_{\mu\nu}F^{\mu\nu}$ and $\partial_{\mu} F^{\mu\nu} = 0$, which can be also derived by the Maxwell equations. We have noticed that there are only 2 modes left, the other freedom degree is transferred to gauge redundancy $A_\mu(x)\rightarrow A_\mu(x)+\partial_\mu\alpha(x)$ and constrained by the gauge choice. For the solution $A^\mu(x)=\epsilon^\mu(k)e^{ikx},~ k^2=0$, we have $k_{\mu}\epsilon^{\mu}=0$, which can be organized as $k^\mu=(\omega,0,0,\omega),~ \epsilon^0 = \epsilon^3 =0$ with Column gauge $\nabla \cdot \mathbf{A} =0$, and then we can use the circularly polarized mode $\boldsymbol\epsilon_+ =\frac1{\sqrt2}(1,-i,0),~ \boldsymbol\epsilon_- =\frac1{\sqrt2}(1,+i,0)$. Further choosing $A^0=0$, we can organize $A^i(x)= \sum_{\lambda=\pm} \int\widetilde{dk}\, \left[ \epsilon_\lambda^{i*}(\mathbf k) a_\lambda(\mathbf k)e^{ikx} + \epsilon_\lambda^i(\mathbf k) a_\lambda^\dagger(\mathbf k)e^{-ikx} \right]$ 2 with $\lambda = +,-$ 2 modes, and $\sum_{\lambda=\pm} \epsilon_\lambda^{i*}(\mathbf k) \epsilon_\lambda^j(\mathbf k) = \delta^{ij} -\frac{k^ik^j}{|\mathbf k|^2} \triangleq P_{T}^{ij}(\mathbf{k})$ (transverse projection operator $P_T^{ij}k_j=0$). Now the corresponding commutation relation can be write as $[A_i(\mathbf x,t),\Pi_j(\mathbf y,t)] =i\int\frac{d^3k}{(2\pi)^3} e^{i\mathbf k\cdot(\mathbf x-\mathbf y)} P_T^{ij}(\mathbf k)$.

• A useful tool is the helicity $h=\frac{\mathbf{S} \cdot \mathbf{k}}{|\mathbf{k}|}$, where in $h|\mathbf k,\lambda\rangle =\lambda|\mathbf k,\lambda\rangle$ h corresponds to eigenvalues $\lambda$.

To derive the propagator for massive spin-1 particle, we can use the vacuum 2-point function $G^{\mu\nu}(x-y) = \langle0|T\{A^\mu(x)A^\nu(y)\}|0\rangle$ as well. This yields $G^{\mu\nu}(x-y) = \int\widetilde{dp}\, \left(g^{\mu\nu}+\frac{p^\mu p^\nu}{m^2}\right) e^{ip(x-y)},~ x^0>y^0$, while complex contour integral and poles further give $G_{\mu\nu}(p) = \frac{-i}{p^2+m^2-i\epsilon} \left( g_{\mu\nu}+\frac{p_\mu p_\nu}{m^2} \right)$. If we just ignore a contact term $\delta^4 (x-y)$ in calculation which has no influence in QED introduction, we can write $D_{\mu\nu}^{\mathrm{Proca}}(p) = \frac{-i}{p^2+m^2-i\epsilon} \left( g_{\mu\nu}+\frac{p_\mu p_\nu}{m^2} \right)$, and go back when necessary. Another way to derive the propagator is using the reverse of kinetic operator. As Lagrangian can be write as $\mathcal L_{\mathrm{Proca}} = \frac12 A_\mu \left[ g^{\mu\nu}(\Box-m^2) -\partial^\mu\partial^\nu \right]A_\nu$, in the momentum space where $\partial_\mu\rightarrow ip_\mu$, we can define the propagator kinetic operator $K^{\mu\nu} = -(p^2+m^2)g^{\mu\nu}+p^\mu p^\nu$ which satisfies $K^{\mu\rho}D_{\rho\nu}(p) = i\delta^\mu_{\rho}$, and then gives $D_{\mu\nu}^{\mathrm{Proca}}(p) = \frac{-i}{p^2+m^2-i\epsilon} \left( g_{\mu\nu}+\frac{p_\mu p_\nu}{m^2} \right)$.

For massless particle, there is convergence in $D_{\mu\nu}(p)$ when $m \to 0$, and the $K^{\mu\nu}(p)p_\nu=0$ can’t be reversed. To solve this, we add a gauge-fixing term $\mathcal{L}_{gf} = - \frac{1}{2\xi} (\partial_{\mu} A^{\mu})^2$ and use kinetic operator $K_\xi^{\mu\nu}(p) =-p^2g^{\mu\nu} +\left(1-\frac1\xi\right)p^\mu p^\nu$. This then gives $D_{\mu\nu}(p) = \frac{-i}{p^2-i\epsilon} \left[ g_{\mu\nu} -(1-\xi)\frac{p_\mu p_\nu}{p^2} \right]$, we can choose Feynman gauge $\xi = 1$ to write $D_{\mu\nu}^{\text{F}}(p) =\frac{-ig_{\mu\nu}}{p^2-i\epsilon}$, which is the photon propagator in later QED. Notice that this is different from the Coulomb gauge in quantization, it is the non-physical freedom corresponding to gauge redundancy arising from the equivalence of configurations in transformation $A_\mu\rightarrow A_\mu+\partial_\mu\alpha$.