Comment on “Spin correlations in elastic e$^{+}$e$^{-}$ scattering in QED”

Abstract

In the previous work, “Spin correlations in elastic $e^+ e^-$ scattering in QED” (Yongram in Eur Phys J 7:71–74, 2008), spin correlations for entangled electrons and positrons as emergent particles of electron–positron scattering (also known as Bhabha scattering) were calculated at tree level in QED. When trying to reproduce the author’s work, we have found different results. In this work, we show the calculation for fully (initial and final polarized states) and partially (just final polarized states) polarized probability amplitudes for electron–positron scattering at all energies. While Yongram claims that violation of the Clauser–Horne inequality (CHI) occurs at all energies for both mentioned cases, for fully polarized scattering we found violation of the CHI for speeds $\beta \gtrsim 0.696$, including the high energy limit, supporting the agreement between QED and foundations of quantum mechanics. However, for initially unpolarized particles we found no violation of the CHI. We argue that averaging over the initial polarizations washes out the non-local correlations needed to violate a Bell-type inequality.

Graphic abstract

Appendices

Appendix A: Spinors for section II

In the Dirac representation of the gamma matrices, the spinors for particles and antiparticles normalized as $\overline{u}{(p)}^su{(p)}^r = 2\,m \delta _{sr}$ and $\overline{v}{(p)}^sv{(p)}^r =- 2\,m \delta _{sr}$, are given by

$$\begin{aligned}{} & {} u^{s}{(p)}=\sqrt{E+m} \begin{pmatrix} \varphi ^s \\ \frac{\vec {\sigma } \cdot \vec {p}}{E+m} \varphi ^s \end{pmatrix}, \end{aligned}$$

(20)

$$\begin{aligned}{} & {} v^{s}{(p)}=-\sqrt{E+m} \begin{pmatrix} \frac{\vec {\sigma } \cdot \vec {p}}{E+m} \eta ^{s} \\ \eta ^{s} \end{pmatrix}, \end{aligned}$$

(21)

where $\varphi ^s$ and $\eta ^s$ are Weyl spinors encoding spin direction. For spin along x, y or z axis they are eigenstates of the Pauli sigma matrices $\sigma ^j$, if the spin direction is quantized along an arbitrary axis $\hat{n}$, then they are eigenstates of the linear combination $\vec {\sigma } \cdot \hat{n}$.

Working in the center of mass reference frame, lets consider an initial electron with spin up along z axis and four-momentum $p^\mu _1$ and similarly a initial positron also with spin up along z axis and four-momentum $p^\mu _2$, with their momenta given by

$$\begin{aligned} p^\mu _1 = ( E,0,\gamma m\beta ,0 )&,p^\mu _2 = ( E,0, -\gamma m\beta ,0 ) . \end{aligned}$$

(22)

For the electron, physical particle spin is the same as spinor $u{(p_1)}^s$ spin. Pointing in the $+z$ direction, the Weyl spinor is

$$\begin{aligned} \varphi ^1 = \left( \begin{array}{c} 1\\ 0\\ \end{array}\right) . \end{aligned}$$

(23)

For the positron, physical particle spin is opposite to the spinor $v{(p_2)}^r$ spin. For physical spin along the z direction, the corresponding Weyl spinor is

$$\begin{aligned} \eta ^2 = \left( \begin{array}{c} 0\\ 1\\ \end{array}\right) , \end{aligned}$$

(24)

plugging this into (20) we get the initial spinors

$$\begin{aligned} u{(p_1)} = \sqrt{E + m} \left( \begin{array}{c} 1 \\ 0 \\ 0 \\ i\rho \end{array}\right) , v{(p_2)} = - \sqrt{E + m} \left( \begin{array}{c} i\rho \\ 0 \\ 0 \\ 1 \end{array}\right) . \end{aligned}$$

(25)

Now for the final electron and positron with momenta $k^\mu _1$ and $k^\mu _2$ respectively

$$\begin{aligned} k^\mu _1 = ( E,0,0,\gamma m\beta )&,k^\mu _2 = ( E,0,0,-\gamma m\beta ) , \end{aligned}$$

(26)

we look for spinor spin lying on the xy plane measured relative to the x-axis for both particles. For this we take the Weyl spinor in (23) and rotate it using the spin rotation operator

$$\begin{aligned}&{e}^{-i\frac{\theta }{2} \vec {\sigma } \cdot \hat{n}} \nonumber \\&\quad = { \left( \begin{array}{cc} \cos (\theta /2)-in^{3}\sin (\theta /2) &{}-i(n^{1}-in^{2})\sin (\theta /2) \\ -i(n^{1}+in^{2})\sin (\theta /2) &{}\cos (\theta /2)+in^{3}\sin (\theta /2) \end{array}\right) } . \end{aligned}$$

(27)

Rotations are performed by an angle $\theta$ around the $\hat{n}$ axis. As illustrated in Fig. 4, looking at the electron like coming towards us, spin direction is measured clockwise, looking at the positron like coming towards us, spin direction is also measured clockwise so the Weyl spinor is the same for both spinors $u{(k_1)}$ and $v{(k_2)}$. The physical positron spin direction measured in the lab is opposite to the spinor $v{(k_2)}$ direction, so for a $\chi _2$ spinor spin direction, the lab measures a $\chi _2 + \pi$ spin direction. First rotate $\varphi ^1$ by an angle $\theta = \pi /2$ around the y-axis, then rotation again this time y an angle $\chi _1$ around the z axis. Then repeat for the positron but perform the second rotation by an angle $\chi _2$, the Weyl spinor is

$$\begin{aligned} \xi _k = \frac{1}{\sqrt{2}} \left( \begin{array}{c} {e}^{-i\chi _k /2} \\ {e}^{-i\chi _k /2} \end{array}\right) . \end{aligned}$$

(28)

Plugging this into (20) we obtain the spinor expressions for the final particles

$$\begin{aligned}&u{(k_1)} = \sqrt{\frac{E+m}{2}} \left( \begin{array}{c} {e}^{-i\chi _1 /2} \\ {e}^{i\chi _1 /2} \\ \rho {e}^{-i\chi _1 /2} \\ -\rho {e}^{i\chi _1 /2} \\ \end{array}\right) , \end{aligned}$$

(29)

$$\begin{aligned}&v{(k_2)} = - \sqrt{\frac{E+m}{2}} \left( \begin{array}{c} - \rho {e}^{-i\chi _2 /2} \\ \rho {e}^{i\chi _2 /2} \\ {e}^{-i\chi _2 /2} \\ {e}^{i\chi _2 /2} \\ \end{array}\right) . \end{aligned}$$

(30)

Appendix B: Spinors for section III

Consider the final electron and positron as having momenta $k^\mu _1 = ( E,\gamma m\beta ,0,0)$ and $k^\mu _2 = ( E,-\gamma m\beta ,0,0)$, and suppose their spins lie on the yz-plane. To describe the Weyl spinors in (20) and (21), it is enough to rotate $\varphi ^1$ from (23) around the y-axis. Denoting as $\chi _1$ and $\chi _2$ the rotation angles that describe the spin direction of the electron and positron, and using the expression (27), we obtain the Weyl spinor

$$\begin{aligned} \xi _k = \left( \begin{array}{c} \cos {(\chi _k /2)} \\ -i\sin {(\chi _k /2)} \end{array}\right) . \end{aligned}$$

(31)

Plugging this expression into (20) and (21), we obtain the final four-spinors:

$$\begin{aligned}&u{(k_1)} = \sqrt{E+m} \left( \begin{array}{c} \cos {(\chi _1 /2)} \\ -i\sin {(\chi _1 /2)} \\ -i \rho \sin {(\chi _1 /2)} \\ \rho \cos {(\chi _1 /2)} \\ \end{array}\right) , \end{aligned}$$

(32)

$$\begin{aligned}&v{(k_2)} = - \sqrt{E+m} \left( \begin{array}{c} i \rho \sin {(\chi _2 /2)} \\ -\rho \cos {(\chi _2 /2)} \\ \cos {(\chi _2 /2)} \\ -i \sin {(\chi _2 /2)} \\ \end{array}\right) . \end{aligned}$$

(33)