Helicities in electron-positron annihilation

Question

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 $P_R v_L = 0$.

Answer 1

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 $\overline{u}_L$.

Answer 2

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.