kinematics - Why the photon can't produce electron and positron in space or in vacuum?

$$\frac{hc}{\lambda} = K_e + K_p + 2m_e c^2$$ could be the energy conservation equation for a photon of wavelength $\lambda$ decaying into a electron and positron with kinetic energies $K_e$ and $K_p$ and rest mass energy $m_e c^2$.

Why does this decay not occur in space or vacuum?

Answer

You can't simultaneously conserve energy and linear momentum.

Let the photon have energy $E_{\gamma} = p_{\gamma} c$ and the electron have energy $E_{-}^{2} = p_{e}^{2}c^2 + m_{e}^{2}c^4$ and an analogous expression for the positron. Suppose the electron and positron depart from the interaction site with an angle $2\theta$ between them.

Conservation of energy.

$$p_{\gamma} c = \sqrt{(p_{e}^{2}c^2 +m_e^{2}c^4} + \sqrt{(p_{p}^{2}c^2 +m_e^{2}c^4},$$ but we know that $p_{p} = p_{e}$ from conservation of momentum perpendicular to the original photon direction. So $$p_{\gamma} = 2\sqrt{p_{e}^2 + m_e^{2}c^2}$$

Now conserving linear momentum in the original direction of the photon. $$p_{\gamma} = p_e \cos{\theta} + p_p \cos\theta = 2p_e \cos\theta$$

Equating these two expression for the photon momentum we have $$p_e \cos{\theta} = \sqrt{p_{e}^2 + m_e^{2}c^2}$$ $$\cos \theta = \sqrt{1 + m_e^{2}c^2/p_e^{2}}$$ As $\cos \theta$ cannot exceed 1 we see that this impossible.