3 Total annihilation cross-sections

Integrating the differential rate ${\rm d}W$ over the azimuth angle, $\phi$, and dividing the resulting rate by the current density ( $=\vert p_z/E - {p_z}^\prime/E^\prime\vert / V$) we obtain the one-photon annihilation cross-section (for any p_z):

$${\sigma _{r,r^\prime}}^{N,N^\prime} = \frac {\alpha {\rm\lambda _c}^2 B_{\rm cr}} {4 B} \frac {1} {2 {p_z}^\prime {E_{\rm o}}^\prime E E_{\rm o}}$$

$$[{\varepsilon^{(\lambda)}}_z {{\varepsilon^{(\lambda)}}_z}^* (E^\prime E + {p_z}^\prime p_z - r r^\prime {E_{\rm o}}^\prime E_{\rm o})$$

$$[(E_{\rm o} + r m)({E_{\rm o}}^\prime - r^\prime m) {I_{N-1,N^\prime-1}}^2 ({\omega _\perp}^2 / 2eB)$$

$$+ (E_{\rm o} - r m)({E_{\rm o}}^\prime + r^\prime m) {I_{N,N^\prime}}^2 ({\omega _\perp}^2 / 2eB)]$$

$$+ r r^\prime \sqrt{2eBN} (p_z {E_{\rm o}}^\prime + r r^\prime {p_z}^\prime E_{\rm o})$$

$$[({E_{\rm o}}^\prime - r^\prime m) I_{N-1,N^\prime-1} ({\omega _\perp}^2 / 2eB) I_{N,N^\prime-1} ({\omega _\perp}^2 / 2eB)$$

$$({\varepsilon^{(\lambda)}}_z {{\varepsilon^{(\lambda)}}_-}^* {\rm... ... + {\varepsilon^{(\lambda)}}_- {{\varepsilon^{(\lambda)}}_z}^* {\rm e}^{i\phi})$$

$$+({E_{\rm o}}^\prime + r^\prime m) I_{N,N^\prime} ({\omega _\perp}^2 / 2eB) I_{N-1,N^\prime} ({\omega _\perp}^2 / 2eB)$$

$$({\varepsilon^{(\lambda)}}_+ {{\varepsilon^{(\lambda)}}_z}^* {\rm... ... {\varepsilon^{(\lambda)}}_z {{\varepsilon^{(\lambda)}}_+}^* {\rm e}^{i\phi}) ]$$

$$- \sqrt{2eBN^\prime} (- p_z {E_{\rm o}}^\prime - r r^\prime {p_z}^\prime E_{\rm o})$$

$$[(E_{\rm o} + r m) I_{N-1,N^\prime-1} ({\omega _\perp}^2 / 2eB) I_{N-1,N^\prime} ({\omega _\perp}^2 / 2eB)$$

$$({\varepsilon^{(\lambda)}}_+ {{\varepsilon^{(\lambda)}}_z}^* {\rm... ...} +{\varepsilon^{(\lambda)}}_z {{\varepsilon^{(\lambda)}}_+}^* {\rm e}^{i\phi})$$

$$+(E_{\rm o} - r m) I_{N,N^\prime} ({\omega _\perp}^2 / 2eB) I_{N,N^\prime-1} ({\omega _\perp}^2 / 2eB)$$

$$({\varepsilon^{(\lambda)}}_z {{\varepsilon^{(\lambda)}}_-}^* {\rm... ... +{\varepsilon^{(\lambda)}}_- {{\varepsilon^{(\lambda)}}_z}^* {\rm e}^{i\phi})]$$

$$- 4eB\sqrt{NN^\prime} {\varepsilon^{(\lambda)}}_z {{\varepsilon^{... ...\prime E E^\prime - r r^\prime {p_z}^\prime p_z + {E_{\rm o}}^\prime E_{\rm o})$$

$$I_{N,N^\prime} ({\omega _\perp}^2 / 2eB) I_{N-1,N^\prime-1} ({\omega _\perp}^2 / 2eB)$$

$$+ (E E^\prime - {p_z}^\prime p_z + r r^\prime {E_{\rm o}}^\prime E_{\rm o})$$

$$[ {\varepsilon^{(\lambda)}}_- {{\varepsilon^{(\lambda)}}_-}^* ({E... ... - r^\prime m) (E_{\rm o} - r m) {I_{N,N^\prime-1}}^2 ({\omega _\perp}^2 / 2eB)$$

$$+ {\varepsilon^{(\lambda)}}_+ {{\varepsilon^{(\lambda)}}_+}^* ({E... ...+ r^\prime m) (E_{\rm o} + r m) {I_{N-1,N^\prime}}^2 ({\omega _\perp}^2 / 2eB)]$$

$$+ 2eB\sqrt{N N^\prime} (r r^\prime E E^\prime - r r^\prime {p_z}^\prime p_z + {E_{\rm o}}^\prime E_{\rm o})$$

$$I_{N-1,N^\prime} ({\omega _\perp}^2 / 2eB) I_{N,N^\prime-1} ({\omega _\perp}^2 / 2eB)$$

$$({\varepsilon^{(\lambda)}}_- {{\varepsilon^{(\lambda)}}_+}^* {\rm... ...\varepsilon^{(\lambda)}}_+ {{\varepsilon^{(\lambda)}}_-}^* {\rm e}^{-i2\phi})].$$ (20)

In the above $\lambda _{\rm c}= h/mc = 2.4263\ 10^{-10}$ cm is the Compton wavelength of the electron. Summing the above expression over the polarization of the photon yields the same result as given by Wunner et al. (1986).

If we suppose that ${\rm e}^-$ and ${\rm e}^+$ are found in the lowest Landau state ( $N=N^\prime =0$) and $p_z + {p_z}^\prime = 0$, so $\omega \cos\theta =0$, then the photon is emitted perpendicular to the magnetic field ( $\theta = \pi/2$). We allow any polarization of the photon. Then the above expression for the cross-section reduces to:

$${{\sigma _{ 1\gamma}}^{0,0}}_{r=-1,r^\prime=+1,p_z=-{p_z}^\prime}... ..._{\rm cr})} \frac {1} {2 \vert{p_z}^\prime\vert {E_{\rm o}}^\prime E E_{\rm o}}$$

$$[\varepsilon_z {\varepsilon_z}^* (E^\prime E + {p_z}^\prime p_z + {E_{\rm o}}^\prime E_{\rm o})$$

$$({E_{\rm o}}^\prime + m) (E_{\rm o} + m) {I_{0,0}}^2 (2E^2/(m^2B/B_{\rm cr}))].$$ (21)

Then using Table 2, one obtains:

$${\sigma _{1\gamma,\hat{\varepsilon}^{(2)}}}^{0,0}$$ = 0

$${\sigma _{ 1\gamma,\hat{\varepsilon}^{(\pm)}}}^{0,0} \frac {\alpha {\lambda \rm _c}^2 m^2}{2 \vert p_z\vert E (B/B_{\rm cr}) } {\rm e}^{-2E^2/(m^2B/B_{\rm cr})}$$ =

$${\sigma _{ 1\gamma,\hat{\varepsilon}^{(1)}}}^{0,0} 2 {\sigma _{ 1\gamma,\hat{\varepsilon}^{(\pm)}}}^{0,0}$$ =

$${\sigma _{ 1\gamma,\sum_{\lambda}\hat{\varepsilon}^{(\lambda)}}}^{0,0}.$$ = (22)

So the radiation obtained in 1 photon annihilation from ${\rm e}^-$ and ${\rm e}^+$ in the ground state with $p_z + {p_z}^\prime = 0$ has linear polarization. The result for summing over polarization agrees with the result of Wunner (1979).

If we take the case of an electron with zero longitudinal momentum (p_z = 0), and sum over polarization, then we find:

$${{\sigma _{ 1\gamma}}^{0,0}}_{p_{ z}=0} \frac {\alpha {\lambda _{\rm c}}^2 m}{\vert{p_{ z}}^\prime\vert (B/B_{\rm cr}) } {\rm e}^{-[1+(1+({p_{ z}}^\prime/m)^2)^{1/2}] (B_{\rm cr}/B)}$$ = (23)

which has been obtained previously by Wunner et al. (1986).

From the conservation laws: $p_z + {p_z}^\prime = \omega \cos \theta$ and $E + E^\prime = \omega$, we find the expression for the longitudinal component of the ${\rm e}^+$ momentum:

$${p_{ z}}^\prime \frac {(E \cos \theta -p_{z} ) + \sqrt{(E \cos \theta -p_{z})^2 -... ...(N^\prime- N) -2 E p_{z} \cos \theta]}} {{\rm sin}^2 \theta}\cdot \hspace*{4cm}$$ = (24)

However, if $\theta = 0$ or $180^{\rm o}$, the following applies:

$${p_{ z}}^\prime \frac { \pm 2 \gamma (N- N^\prime) \pm {p_{z}}^2 - E p_{z} } {p_{z} + E}$$ = (25)

with the upper sign for $\theta = 0$, the lower sign for $180^{\rm o}$.

If we average over the polarization of the ${\rm e}^-$ and ${\rm e}^+$ in Eq. (20) above, we obtain:

$${\sigma _{1\gamma}}^{N,N^\prime} = \frac {\alpha {\lambda _{\rm c... ...p_{z}}^\prime E} \frac{(1+\delta _{N,0})}{2} \frac{(1+\delta _{N^\prime,0})}{2}$$

$$[{\varepsilon^{(\lambda)}}_{z} {{\varepsilon^{(\lambda)}}_{z}}^* (E^\prime E + {p_{z}}^\prime p_{z} + m^2) [{I_{N-1,N^\prime-1}}^2$$

$$({\omega _\perp}^2 / 2eB)$$

$$+{I_{N,N^\prime}}^2 ({\omega _\perp}^2 / 2eB)]$$

$$+{p_{z}}^\prime \sqrt{2eBN} [ I_{N-1,N^\prime-1} ({\omega _\perp}^2 / 2eB) I_{N,N^\prime-1} ({\omega _\perp}^2 / 2eB)$$

$$({\varepsilon^{(\lambda)}}_z {{\varepsilon^{(\lambda)}}_-}^* {\rm... ... + {\varepsilon^{(\lambda)}}_- {{\varepsilon^{(\lambda)}}_z}^* {\rm e}^{i\phi})$$

$$+I_{N,N^\prime} ({\omega _\perp}^2 / 2eB) I_{N-1,N^\prime} ({\omega _\perp}^2 / 2eB)$$

$$({\varepsilon^{(\lambda)}}_+ {{\varepsilon^{(\lambda)}}_z}^* {\rm... ... {\varepsilon^{(\lambda)}}_z {{\varepsilon^{(\lambda)}}_+}^* {\rm e}^{i\phi}) ]$$

$$+ \sqrt{2eBN^\prime} p_{z} [I_{N-1,N^\prime-1} ({\omega _\perp}^2 / 2eB) I_{N-1,N^\prime} ({\omega _\perp}^2 / 2eB)$$

$$({\varepsilon^{(\lambda)}}_+ {{\varepsilon^{(\lambda)}}_z}^* {\rm... ...} +{\varepsilon^{(\lambda)}}_z {{\varepsilon^{(\lambda)}}_+}^* {\rm e}^{i\phi})$$

$$+ I_{N,N^\prime} ({\omega _\perp}^2 / 2eB) I_{N,N^\prime-1} ({\omega _\perp}^2 / 2eB)$$

$$({\varepsilon^{(\lambda)}}_z {{\varepsilon^{(\lambda)}}_-}^* {\rm... ... +{\varepsilon^{(\lambda)}}_- {{\varepsilon^{(\lambda)}}_z}^* {\rm e}^{i\phi})]$$

$$- 4eB\sqrt{NN^\prime} {\varepsilon^{(\lambda)}}_{z} {{\varepsilon^{(\lambda)}}_{z}}^*$$

$$I_{N,N^\prime} ({\omega _\perp}^2 / 2eB) I_{N-1,N^\prime-1} ({\omega _\perp}^2 / 2eB)$$

$$+ (E E^\prime - {p_{z}}^\prime p_{z} + m^2) [ {\varepsilon^{(\lam... ... {{\varepsilon^{(\lambda)}}_-}^* {I_{N,N^\prime-1}}^2 ({\omega _\perp}^2 / 2eB)$$

$$+ {\varepsilon^{(\lambda)}}_+ {{\varepsilon^{(\lambda)}}_+}^* {I_{N-1,N^\prime}}^2 ({\omega _\perp}^2 / 2eB)]$$

$$+ 2eB\sqrt{N N^\prime} I_{N-1,N^\prime} ({\omega _\perp}^2 / 2eB) I_{N,N^\prime-1} ({\omega _\perp}^2 / 2eB)$$

$$({\varepsilon^{(\lambda)}}_- {{\varepsilon^{(\lambda)}}_+}^* {\rm... ...\varepsilon^{(\lambda)}}_+ {{\varepsilon^{(\lambda)}}_-}^* {\rm e}^{-i2\phi})].$$ (26)

If we sum over polarization of the photon in the above equation and take p_z=0 (i.e. zero longitudinal momentum of the electron) we obtain:

$${\sigma _{1\gamma}}^{N,N^\prime} = \frac {\alpha {\lambda _{\rm c... ...prime E_{\rm o}} \frac{(1+\delta _{N,0})}{2} \frac{(1+\delta _{N^\prime,0})}{2}$$

$$[(E^\prime E_{\rm o} + m^2) [({I_{N-1,N^\prime-1}}^2 ({\omega _\perp}^2 / 2eB) +{I_{N,N^\prime}}^2$$

$$({\omega _\perp}^2 / 2eB)) {\rm sin}^2 \theta$$

$$+({I_{N,N^\prime-1}}^2 ({\omega _\perp}^2 / 2eB) +{I_{N-1,N^\prime}}^2 ({\omega _\perp}^2 / 2eB))$$

$$(1+{\rm cos}^2 \theta)]$$

$$- 2\sqrt{2eBN} \omega [I_{N,N^\prime} ({\omega _\perp}^2 / 2eB) I_{N-1,N^\prime} ({\omega _\perp}^2 / 2eB)$$

$$+ I_{N-1,N^\prime-1} ({\omega _\perp}^2 / 2eB) I_{N,N^\prime-1} ({\omega _\perp}^2 / 2eB)] \sin \theta {\rm cos}^2 \theta$$

$$- 4eB\sqrt{NN^\prime} [I_{N,N^\prime} ({\omega _\perp}^2 / 2eB) I_{N-1,N^\prime-1} ({\omega _\perp}^2 / 2eB)$$

$$+I_{N-1,N^\prime} ({\omega _\perp}^2 / 2eB) I_{N,N^\prime-1} ({\omega _\perp}^2 / 2eB)] {\rm sin}^2 \theta].$$ (27)

This differs by a factor of 2 in the last two terms from Eq. (7) of Wunner et al. (1986), and corrects an error in their expression.