Consider the massless limit of a process in which an electron-positron pair annihilates into a virtual photon - the final state doesn't matter. If the electron is massless (or if the energy is high enough), helicity and chirality become the same, and they are conserved. My problem is that I'm getting contradictory results: the math says that the amplitude is nonzero only when the electron and the positron have the same helicity, while every book on the subject (and physical common sense) claims otherwise.

The amplitude is proportional to $\bar{v}\gamma^\mu u$, where $u$ is the electron's spinor and $v$ the positron's. Let's go to the center of mass frame, and take the electron's momentum to be $p^\mu = (p, 0, 0, p)$ and the positron's to be $p'^\mu = (p, 0, 0, -p)$. Using the Dirac basis, I have the following definite helicity spinors (following the Wikipedia article on spinors):

$$u_R = \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix}\ \ \ u_L = \begin{pmatrix} 0 \\ 1 \\ 0 \\ -1 \end{pmatrix}$$

$$v_R = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 1 \end{pmatrix}\ \ \ v_L = \begin{pmatrix} -1 \\ 0 \\ 1 \\ 0 \end{pmatrix}$$

Suppose the electron has positive helicity and the positron has negative helicity; in other words, both have spin up along the z axis. Books like Thomson's Modern Particle Physics or Halzen and Martin's Quarks and Leptons say that the annihilation should take place in this case, and it makes sense: the initial state has total spin 1, just what you need to create the virtual photon.

The problem is that I can calculate $\bar{v}_L \gamma^\mu u_R$ explicitly, and I get zero. I can even show it abstractly: Defining $P_R = \frac12 (1+\gamma^5)$ and $P_L = \frac12 (1-\gamma^5)$ and noting that $u_R = P_R u_R$ and so on, it can be shown quite generally that $\bar{v} \gamma^\mu u$ vanishes unless both spinors have the same helicity.

What is going on here? My best guess is that somehow the helicity assignments for antiparticles are reversed, but I don't see how that can be: I just followed the Wikipedia article and every book I could find, not to mention that I've checked that my spinors satisfy the Dirac equation with the proper momentum, and that the spins are right and that $P_L v_R = 0$ and $P_R v_L = 0$.

  • $\begingroup$ Just checking because I'm too lazy to check the calculations myself right now: You remembered that there's a $\gamma^0$ in the definition of the conjugate $\bar v$, right? $\endgroup$ – ACuriousMind Feb 18 '16 at 0:56
  • $\begingroup$ @ACuriousMind: Yes, I did. As I said, there's a proof that doesn't rely on concrete calculations. $\endgroup$ – Javier Feb 18 '16 at 1:04

It is not true that "$\overline{v}_L\gamma^\mu u_R$ is zero". While $u_R \equiv P_R u$, if you check carefully you will find that $\overline{v}_L \equiv \overline{v}P_L$. And if you "take $P_L$ to the other side of the $\gamma^\mu$" using the usual anticommutation relations, you get $P_R$. And, of course, $P_R^2 = P_R$.

The problem may have been originally caused by forgetting that $\overline{v}_L$ has the opposite sign for $\gamma^5$ in the projection operators than $\overline{u}_L$.

  • $\begingroup$ You mean that $P_L$ and $P_R$ are reversed when acting on $v$? That wouldn't surprise me, except that I've checked with the Dirac equation to make sure that my positrons have negative z-momentum, and I've checked with the spin matrix $\mathbf{\Sigma} = \text{diag}(\mathbf{\sigma}, \mathbf{\sigma})$ to make sure $v_L$ has spin up and $v_R$ has spin down. And it fits with the fact that $P_R v_L = 0$. So how can the helicity projectors be reversed? $\endgroup$ – Javier Feb 18 '16 at 13:37
  • $\begingroup$ > You mean that $P_L$ and $P_R$ are reversed when acting on $v$? Yes, absolutely. See for instance Table 9.2 of Griffiths's "Intro to Elem Particles", 2nd edition. $\endgroup$ – TimeVariant Feb 19 '16 at 22:21
  • $\begingroup$ You're absolutely right, but how do I reconcile this with my calculation of momentum of spin? You're saying that what I have called $v_L$ in fact has right handed helicity, but I can check that $\Sigma_3 v_L = v_L$, meaning it has spin up and that $\not p' v_L = 0$, which according to my definition of $p'$ means that its z-momentum is negative. So why does it have right handed helicity? Does it have to do with the spin and/or momentum being reversed? (I get now why people abandoned the whole "antiparticles are particles going backwards in time" business) $\endgroup$ – Javier Feb 20 '16 at 16:23
  • $\begingroup$ Yes, the flipped helicity has to do with momentum being "reversed from what you expect" for antiparticles. Another way to convince yourself might be to use the charge conjugation operator on the particle spinors to obtain the antiparticle spinors, a matter that's explained in Halzen and Martin's text. $\endgroup$ – TimeVariant Feb 21 '16 at 4:52

Your problem is when you consider $$v = \begin{pmatrix} 0 \\ 1 \\ 0\\ 1 \end{pmatrix}$$ as right-handed (chirality or helicity). Actually, it is associated to a left-handed component. Consider the helicity operator: $$\hat{h} = \frac{1}{|\vec{p}|} \vec{\Sigma}.\hat{\vec{p}}$$ with $$\vec{\Sigma} = \begin{pmatrix} \vec{\sigma} & 0 \\ 0 & \vec{\sigma} \\ \end{pmatrix}$$ ($\vec{\sigma}$ being the Pauli matrices) and: $$ \hat{\vec{p}} = -i \vec{\nabla}$$ The $v$-spinors are associated to a propagation term $e^{i(Et-\vec{p}.\vec{x})}$ (the sign in the exponential is opposite to the case of $u$-spinor). Hence, applying $\hat{h}$ to the whole term $\psi = v e^{i(Et-\vec{p}.\vec{x})}$, it gives: $$ \hat{h} \psi = \frac{1}{|\vec{p}|}\vec{\Sigma} v . (-\vec{p}) e^{i(Et-\vec{p}.\vec{x})} $$ Since you chose a momentum along the z-axis: $\vec{p} = -|\vec{p}| \vec{u}_z$, we have: $$ \hat{h} \psi = \frac{1}{|\vec{p}|}\Sigma_z v |\vec{p}| e^{i(Et-\vec{p}.\vec{x})} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 &0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \\ 0\\ 1 \end{pmatrix} e^{i(Et-\vec{p}.\vec{x})} = -\psi $$ Thus $\psi$ is Left-handed! Your spinor $v$ is associated to a left-handed antiparticle.


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