Skip to the Main Content

Note:These pages make extensive use of the latest XHTML and CSS Standards. They ought to look great in any standards-compliant modern browser. Unfortunately, they will probably look horrible in older browsers, like Netscape 4.x and IE 4.x. Moreover, many posts use MathML, which is, currently only supported in Mozilla. My best suggestion (and you will thank me when surfing an ever-increasing number of sites on the web which have been crafted to use the new standards) is to upgrade to the latest version of your browser. If that's not possible, consider moving to the Standards-compliant and open-source Mozilla browser.

January 8, 2022

Spinor Helicity Variables in QED

I’m teaching Quantum Field Theory this year. One of the things I’ve been trying to emphasize is the usefulness of spinor-helicity variables in dealing with massless particles. This is well-known to the “Amplitudes” crowd, but hasn’t really trickled down to the textbooks yet. Mark Srednicki’s book comes close, but doesn’t (IMHO) quite do a satisfactory job of it.

Herewith are some notes.

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=0k^2=0, we can define helicity spinors

(1)(kσ) αβ˙=λ αλ β˙ ,(kσ¯) α˙β=λ α˙λ β(k\cdot\sigma)_{\alpha\dot\beta}= -\lambda_\alpha\lambda^\dagger_{\dot\beta},\qquad (k\cdot\overline{\sigma})^{\dot\alpha\beta} = -\lambda^{\dagger\dot\alpha}\lambda^\beta

which allow us to straightforwardly canonically-quantize.


For a Weyl fermion, =iψ σ¯ψ \mathcal{L}= i\psi^\dagger\overline{\sigma}\cdot\partial \psi the general solution to the equations of motion is ψ α(x) =d 3k(2π) 32|k|λ α(ξ k e ikx+η ke ikx) ψ α˙ (x) =d 3k(2π) 32|k|λ α˙ (η k e ikx+ξ ke ikx) \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 {ψ α(x,0),ψ β˙ (x,0)}=σ αβ˙ 0δ (3)(xx) \{\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 {ξ k,ξ k } =(2π) 32|k|δ (3)(kk) {η k,η k } =(2π) 32|k|δ (3)(kk) \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 (λ i\lambda_i or λ i \lambda^\dagger_i) depending on the helicity of the i thi^{\text{th}} incoming/outgoing particle. When we take the absolute square of the amplitude, we use (1) to rewrite λ iλ i =k iσ\lambda_i\lambda^\dagger_i=-k_i\cdot\sigma, etc.


There’s a certain amount of hand-wringing associated to quantizing the free Maxwell Lagrangian, =14F μνF μν \mathcal{L} = -\tfrac{1}{4}F_{\mu\nu}F^{\mu\nu} If we take the canonical variables to be A μA^\mu and π μ=δδ 0A μ\pi_\mu =\frac{\delta \mathcal{L}}{\delta\partial_0 A^\mu}, then the gauge-invariance entails that the symplectic structure is degenerate (π 0\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 2nd 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

(2)F μν(x)=12d 3k(2π) 32|k| (λσ μνλε (k)+λ σ¯ μνλ ε + (k))e ikx+ +(λσ μνλε +(k)+λ σ¯ μνλ ε (k))e ikx\begin{aligned} F^{\mu\nu}(x)= \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}+\\ &+\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} \end{aligned}

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

(3)[F 0i(x,0),F jk(x,0)]=i(δ ikx jδ ijx k)δ (3)(xx)[F_{0i}(\vec{x},0),F_{j k}(\vec{x}',0)]=i\left(\delta_{i k}\frac{\partial}{\partial x^j}-\delta_{i j}\frac{\partial}{\partial x^k}\right)\delta^{(3)}(\vec{x}-\vec{x}')

which, in turn, give the canonical commutation relations [ε +(k),ε + (k)] =(2π) 32|k|δ (3)(kk) [ε (k),ε (k)] =(2π) 32|k|δ (3)(kk) \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 μA^\mu, not just F μνF^{\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 kˇ 2=0\check{k}^2=0 and the corresponding helicity spinors (kˇσ) αβ˙=μ αμ β˙ ,(kˇσ¯) α˙β=μ α˙μ β (\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

(4)A μ(x) =12d 3k(2π) 32|k|(μ σ¯ μλμ λ ε (k)+μσ μλ μλε + (k))e ikx+ +(λ σ¯ μμλμε (k)+λσ μμ λ μ ε +(k))e ikx =12d 3k(2π) 32|k|(μ σ¯ μλμ λ ε (k)+μσ μλ μλε + (k))e ikx (μσ μλ μλε (k)+μ σ¯ μλμ λ ε +(k))e ikx \begin{aligned} A^\mu(x)&= \frac{1}{\sqrt{2}}\int\frac{d^3\vec{k}}{{(2\pi)}^3 2|\vec{k}|}\left(\frac{\mu^\dagger\overline{\sigma}^\mu\lambda}{\mu^\dagger\lambda^\dagger}\varepsilon^\dagger_{-}(\vec{k})+ \frac{\mu\sigma^\mu\lambda^\dagger}{\mu\lambda} \varepsilon^\dagger_{+}(\vec{k})\right)e^{-ik\cdot x}+\\ &\qquad+\left(\frac{\lambda^\dagger\overline{\sigma}^\mu\mu}{\lambda\mu}\varepsilon_{-}(\vec{k})+ \frac{\lambda\sigma^\mu\mu^\dagger}{\lambda^\dagger\mu^\dagger} \varepsilon_{+}(\vec{k})\right)e^{ik\cdot x}\\ &=\frac{1}{\sqrt{2}}\int\frac{d^3\vec{k}}{{(2\pi)}^3 2|\vec{k}|}\left(\frac{\mu^\dagger\overline{\sigma}^\mu\lambda}{\mu^\dagger\lambda^\dagger}\varepsilon^\dagger_{-}(\vec{k})+ \frac{\mu\sigma^\mu\lambda^\dagger}{\mu\lambda} \varepsilon^\dagger_{+}(\vec{k})\right)e^{-ik\cdot x}\\ &\qquad-\left(\frac{\mu\sigma^\mu\lambda^\dagger}{\mu\lambda}\varepsilon_{-}(\vec{k})+ \frac{\mu^\dagger\overline{\sigma}^\mu\lambda}{\mu^\dagger\lambda^\dagger} \varepsilon_{+}(\vec{k})\right)e^{ik\cdot x}\\ \end{aligned}

which satisfies A=0\partial\cdot A=0 and (exercise for the reader) μA ν νA μ =12d 3k(2π) 32|k|(λσ μνλε (k)+λ σ¯ μνλ ε + (k))e ikx+ +(λσ μνλε +(k)+λ σ¯ μνλ ε (k))e ikx =F μν(x) \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 kˇ\check{k} changes A μ(x)A^\mu(x) by a harmonic gauge transformation.

To completely justify (4), we choose R-ξ\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 μσ μλ μλ\frac{\mu\sigma^\mu\lambda^\dagger}{\mu\lambda} or a μ σ¯ μλμ λ \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 kˇ\check{k} that we want, but verifying that the final answer is independent of kˇ\check{k} is a nice check on our calculations.

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

Posted by distler at January 8, 2022 11:47 AM

TrackBack URL for this Entry:

4 Comments & 0 Trackbacks

Re: Spinor Helicity Variables in QED

How come Peskin and Schroeder is no longer being used?

Posted by: David on January 8, 2022 10:28 PM | Permalink | Reply to this

Peskin and Schroeder

Just because I’m not using it doesn’t mean it’s “not being used”. I’m sure it is still, by far, the best-selling QFT textbook.

Posted by: Jacques Distler on January 8, 2022 11:17 PM | Permalink | PGP Sig | Reply to this

Re: Peskin and Schroeder

Yeah, that is what I meant. It seems like P and S is the most popular choice, so I was just wondering why you decided to buck the tradition.

Posted by: David on January 8, 2022 11:46 PM | Permalink | Reply to this

Re: Peskin and Schroeder

The first choice (which immediately eliminates half of the possible textbooks) is the choice of signature conventions.

I learned QFT in the “mostly-minus” signature. But, when I teach String Theory, or when I teach GR, “mostly-plus” is the default signature convention, and I wouldn’t consider any other choice. So I was slightly (but only slightly) leaning towards “mostly-plus” for this QFT course.

Of the textbooks I looked at, Srednicki’s came closest to what I wanted to do in the course. And he uses “mostly-plus”, so that basically clinched things for me.

As it turns out, I’m not actually following Srednicki all that closely, either, but by then my choice was made …

Posted by: Jacques Distler on January 9, 2022 12:52 AM | Permalink | PGP Sig | Reply to this

Post a New Comment