Musings

The first step in constructing perturbation theory is to quantize the free fields. Following Weinberg and Srednicki, I’m using the “mostly-plus” signature convention (my 2-component spinor conventions are those of Dreiner et al if you define the macro \def\signofmetric{1} in the LaTeX file). For $k^2=0$, we can define helicity spinors

which allow us to straightforwardly canonically-quantize.

Spin-1/2

For a Weyl fermion, $$\mathcal{L}= i\psi^\dagger\overline{\sigma}\cdot\partial \psi$$ the general solution to the equations of motion is $$\begin{aligned} \psi_\alpha(x)&=\int\frac{d^3\vec{k}}{{(2\pi)}^3 2|\vec{k}|}\lambda_\alpha \left(\xi^\dagger_{\vec{k}}e^{-ik\cdot x}+\eta_{\vec{k}}e^{ik\cdot x}\right)\\ \psi^\dagger_{\dot\alpha}(x)&=\int\frac{d^3\vec{k}}{{(2\pi)}^3 2|\vec{k}|}\lambda^\dagger_{\dot\alpha}\left(\eta^\dagger_{\vec{k}}e^{-ik\cdot x}+\xi_{\vec{k}}e^{ik\cdot x}\right) \end{aligned}$$ The Equal-Time Anti-Commutation Relations $$\{\psi_\alpha(\vec{x},0),\psi^\dagger_{\dot\beta}(\vec{x}',0)\}=\sigma^0_{\alpha\dot\beta}\delta^{(3)}(\vec{x}-\vec{x}')$$ become the canonical anti-commutation relations $$\begin{aligned} \{\xi_{\vec{k}},\xi^\dagger_{\vec{k}'}\}&= {(2\pi)}^3 2|\vec{k}| \delta^{(3)}(\vec{k}-\vec{k}')\\ \{\eta_{\vec{k}},\eta^\dagger_{\vec{k}'}\}&= {(2\pi)}^3 2|\vec{k}| \delta^{(3)}(\vec{k}-\vec{k}')\\ \end{aligned}$$ for creation and annihilation operators for fermions of definite helicity.

The upshot, after tracking this through the LSZ reduction formula, is that external fermion lines are contracted with the corresponding helicity spinor ($\lambda_i$ or $\lambda^\dagger_i$) depending on the helicity of the $i^{\text{th}}$ incoming/outgoing particle. When we take the absolute square of the amplitude, we use (1) to rewrite $\lambda_i\lambda^\dagger_i=-k_i\cdot\sigma$, etc.

Spin-1

There’s a certain amount of hand-wringing associated to quantizing the free Maxwell Lagrangian, $$\mathcal{L} = -\tfrac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ If we take the canonical variables to be $A^\mu$ and $\pi_\mu =\frac{\delta \mathcal{L}}{\delta\partial_0 A^\mu}$, then the gauge-invariance entails that the symplectic structure is degenerate ($\pi_0$ vanishes identically). The usual approach is to fix a gauge (Weinberg and Srednicki use Coulomb gauge) and then work very hard (replacing Poisson brackets with Dirac brackets, because the constraints are 2^nd class, …).

On the other hand, if we

1. realize that the phase space is the space of classical solutions and
2. introduce spinor helicity variables, as before,

it’s easy to write down the general solution to the equations of motion

The (non-degenerate) symplectic structure on the space of classical solutions leads to the Equal-Time Commutation Relations

which, in turn, give the canonical commutation relations $$\begin{aligned} [\varepsilon_+(\vec{k}),\varepsilon^\dagger_+(\vec{k}')]&={(2\pi)}^3 2|\vec{k}| \delta^{(3)}(\vec{k}-\vec{k}')\\ [\varepsilon_-(\vec{k}),\varepsilon^\dagger_-(\vec{k}')]&={(2\pi)}^3 2|\vec{k}| \delta^{(3)}(\vec{k}-\vec{k}')\\ \end{aligned}$$ of the creation and annihilation operators for photons of definite helicity.

Unfortunately, to couple to charged matter fields, we need an expression for $A^\mu$, not just $F^{\mu\nu}$, so (2) does not quite suffice for our purposes. But, again, helicity spinors come to the rescue.

Introduce a fixed fiducial null vector $\check{k}^2=0$ and the corresponding helicity spinors $$(\check{k}\cdot\sigma)_{\alpha\dot\beta}= -\mu_\alpha\mu^\dagger_{\dot\beta},\qquad (\check{k}\cdot\overline{\sigma})^{\dot\alpha\beta} = -\mu^{\dagger\dot\alpha}\mu^\beta$$ We then can write

which satisfies $\partial\cdot A=0$ and (exercise for the reader) $$\begin{aligned} \partial^\mu A^\nu-\partial^\nu A^\mu&= \frac{1}{\sqrt{2}}\int\frac{d^3\vec{k}}{{(2\pi)}^3 2|\vec{k}|}\left(\lambda\sigma^{\mu\nu}\lambda\varepsilon^\dagger_{-}(\vec{k})+ \lambda^\dagger\overline{\sigma}^{\mu\nu}\lambda^\dagger \varepsilon^\dagger_{+}(\vec{k})\right)e^{-ik\cdot x}+\\ &\qquad+\left(\lambda\sigma^{\mu\nu}\lambda\varepsilon_{+}(\vec{k})+ \lambda^\dagger\overline{\sigma}^{\mu\nu}\lambda^\dagger \varepsilon_{-}(\vec{k})\right)e^{ik\cdot x}\\ &=F^{\mu\nu}(x) \end{aligned}$$ as before. Together, these ensure that changing the reference momentum $\check{k}$ changes $A^\mu(x)$ by a harmonic gauge transformation^†.

To completely justify (4), we choose R-$\xi$ gauge, and use BV-BRST quantization, but that’s the subject for another blog post.

Here, it suffices to say that the Feynman rules contract every external photon line with a $\frac{\mu\sigma^\mu\lambda^\dagger}{\mu\lambda}$ or a $\frac{\mu^\dagger\overline{\sigma}^\mu\lambda}{\mu^\dagger\lambda^\dagger}$, depending on the helicity of the incoming/outgoing photon. We’re free to make any choice of reference momentum $\check{k}$ that we want, but verifying that the final answer is independent of $\check{k}$ is a nice check on our calculations.

^† Notoriously, Lorentz gauge $\partial\cdot A = 0$ does not completely fix the gauge: we can still shift $A_\mu\to A_\mu+\partial_\mu f$, where $f$ is any solution to the scalar wave equation, $\square f = 0$.