7
$\begingroup$

Particularly in the case of Majorana neutrinos, it seems a little odd that the particle and antiparticle would have differing cross sections.

Perhaps the answer is in here, but I've missed it: http://pdg.lbl.gov/2013/reviews/rpp2013-rev-nu-cross-sections.pdf

In the caption of figure 48.1 of the PDG excerpt linked above, it says "Neutrino cross sections are typically twice as large as their corresponding antineutrino counterparts, although this difference can be larger at lower energies."

Is it common for particles to have different cross sections from their corresponding antiparticle?

Is there a reason for this difference? Can we theoretically predict the magnitude of this difference?

$\endgroup$
  • $\begingroup$ It is discussing charged current interactions, $\nu N \to \mu^- X$ and $\bar\nu N \to \mu^+ X$. $\endgroup$ – innisfree Apr 20 '14 at 15:52
9
$\begingroup$

It is "easy" to show that the cross-section of the 2 reactions: $$ (1)~~~\nu_{\mu} + d \to \mu^- + u~~~~~~(2)~~~\bar{\nu}_{\mu} + u \to \mu^+ + d$$ are different. The naive calculation gives a factor $\sigma_1/\sigma_2 = 3$ as shown below. The fact that the figure in the particle data group gives actually almost a factor 2 needs to take into account the structure function of the nucleon.

So let's check first the factor 3 and explain why: Neglecting all masses of fermions, thus assuming an energy large enough but still negligible with respect to the $W$ bosons, the amplitudes of the 2 processes are:

$$\mathcal{M}_1=\frac{G_F}{\sqrt{2}}[\bar{u}_u \gamma^{\mu}(1-\gamma^5)u_d][\bar{u}_{\mu} \gamma_{\mu} (1-\gamma^5)u_{\nu}]$$ $$\mathcal{M}_2=\frac{G_F}{\sqrt{2}}[\bar{u}_d \gamma^{\mu}(1-\gamma^5)u_u][\bar{v}_{\nu} \gamma_{\mu} (1-\gamma^5)v_{\mu}]$$ where $G_F$ is the Fermi coupling constant and $u$,$v$ the spinors for particles and anti-particle. The Cabibbo angle has been neglected ($\cos \theta_c =1)$. The differences between the 2 amplitudes is related to the presence of anti-spinors in $\mathcal{M}_2$ instead of spinors as in $\mathcal{M}_1$. First notice that for anti-spinors, the incoming anti-particle appears on the left-hand side of the $\gamma^\mu$ matrix while for spinors, the incoming particle is on the right-hand side. This will be the source of the $u$ Mandelstam variable appearing after the squaring/averaging of the amplitude $\mathcal{M}_2$ below. Second, and this is the main point, $(1-\gamma^5)$ is a chirality projector (modulo a factor 2). This is this $(1-\gamma^5)$ matrix which is responsible for the chiral nature of the weak interaction. When applied to a spinor, it selects chiral left-handed particles, while when it is applied on anti-spinor, it selects chiral right-handed anti-particles. We'll see the consequences few lines below.

Squaring the amplitudes, averaging over the spin of the initial quark and summing over the spins of the outgoing quark and lepton yields: $$|\overline{\mathcal{M}}_1|^2 = 16 G_F^2 s^2$$ $$|\overline{\mathcal{M}}_2|^2 = 16 G_F^2 u^2$$ where $s$ and $u$ are the 2 Mandelstam variables. (Denoting by $p_1,p_2$ the 4 momenta of the 2 particles in the initial state and $p_3,p_4$ the ones of the 2 particles in the final states, we have $s=(p_1+p_2)^2$ and $u=(p_1-p_4)^2$). Using a well known formula for the differential cross-section $\frac{d\sigma}{d\Omega}=\frac{1}{64\pi^2 s} |\overline{\mathcal{M}}|^2$ (valid in our massless approximation, $\Omega$ being the solid angle in the center of mass frame) gives: $$\frac{d\sigma_1}{d\Omega}= \frac{G^2_F}{4\pi^2}s$$ $$\frac{d\sigma_2}{d\Omega}= \frac{G^2_F}{4\pi^2}\frac{u^2}{s} = \frac{G^2_F}{16\pi^2}s(1+\cos\theta)^2 $$ with $\theta$ the angle in the center of mass frame between the $\bar{\nu}_{\mu}$ and the $\mu^+$. At this stage we can better appreciate the chiral structure of the weak interaction. Indeed, weak interaction (via charged current i.e. $W$ boson exchange) involves only left-handed chirality of particles and the right-handed chirality of anti-particles. With the massless approximation, chirality is equivalent to helicity, the projection of the spin along the momentum direction. You can notice that for $\theta=\pi$, $\frac{d\sigma_2}{d\Omega}=0$. It is related to this chiral structure. Indeed the $\bar{\nu}_{\mu}$ must be right handed and the $u$ quark left handed. Thus in the center of mass frame the spin of these 2 particles are in the same direction (since they are back-to-back) giving a projection on the $z$ axis $s_z=1$ (or -1 depending on your choice). By conservation of the angular momentum, the $s_z$ of the final state must be also $s_z = 1$. But with an angle $\theta=\pi$, that means that the $\mu^+$ is going in the opposite direction of the $\bar{\nu}_{\mu}$. Since it has the same $s_z$ it must be left-handed (since the $\bar{\nu}_{\mu}$ was right-handed). But, the weak interaction only involves right-handed components of the anti-particle $\mu^+$. So necessarily this configuration cannot be possible explaining the null cross-section at this angle! We clearly see the difference between the 2 reactions at this step. We can go a bit further and integrate over the solid angle, giving: $$\sigma_1 = \frac{G^2_F}{\pi} s$$ $$\sigma_2 = \frac{G^2_F}{3\pi} s$$ We thus have the announced factor 3 for the ratio of cross-section at quark level. Moving from quark to nucleon is a bit complicated and I give here only the result assuming a target made of as many protons as neutrons: $$\sigma_{\nu N} = \frac{G^2_F}{\pi}\frac{s}{2}\int_0^1 x(q(x)+\frac{\bar{q}(x)}{3})dx$$ $$\sigma_{\bar{\nu} N} = \frac{G^2_F}{\pi}\frac{s}{2}\int_0^1 x(\bar{q}(x)+\frac{q(x)}{3})dx$$ the function $q(x)$ and $\bar{q}(x)$ being the Parton Distribution Function (PDF) of the quark and anti-quark. You do have to take into account the quarks and anti-quarks coming from the sea (quantum fluctuations) inside the nucleon. The PDF have to be measured. It is known (measured) that about half of the proton momentum is carried by quarks (the other half by gluons), meaning that $\int_0^1 x(q(x)+\bar{q}(x)) dx= 0.5$. The individual contribution of quarks and antiquarks to the proton momentum is about: $$\int_0^1 x q(x) dx = 42\%~~~~~ \int_0^1 x \bar{q}(x) dx = 9\%$$ The result of the integrals above is such that the ratio of the 2 cross-sections is pretty close to 2 (instead of 3 at the quark level).

$\endgroup$
  • $\begingroup$ Paganini: "[...] $u,v$ [denote] the spinors for particles and anti-particle. [...] Squaring the amplitudes, averaging over the spin of the initial quark and summing over the spins of the outgoing quark and lepton yields: [...]" -- It would be really terrific if you could spell out these calculations in a bit more detail, making their difference for $u$ vs. $v$ more explicit. (You should edit your post anyways wrt. spelling the surname of N. Cabibbo.) $\endgroup$ – user12262 Jun 25 '15 at 20:02
  • $\begingroup$ @user12262: thanks for the typo. I don't want to develop long calculations here for the squaring/averaging, but I've added few lines of explanations that might help. $\endgroup$ – Paganini Jun 26 '15 at 7:57
  • $\begingroup$ Paganini: "I've added few lines of explanations that might help." -- They do; thanks, +1. "thanks for the typo" -- Well, looks like I should still lay hand on it myself. Also, there's still an entire sentence left which I find difficult to grasp as it stands: "[...] So necessarily this configuration cannot be possible explaining the null cross-section at this angle!" -- Is this perhaps supposed to mean (rather, as far as I understand the argument described): "[...] So this explains that the cross-section of this process at this angle ($\theta = \pi$) is null."? $\endgroup$ – user12262 Jun 30 '15 at 21:03
  • 1
    $\begingroup$ @user12262: yes indeed, I meant "So this explains that the cross-section of this process at this angle (θ=π) is null.". My english can be sometimes approximative... $\endgroup$ – Paganini Jul 2 '15 at 16:51
3
$\begingroup$

The answer is that in general cross section of particles and antiparticles are different. For this case in specific there is one point that Dirac particle have four possible states: due two possible ranges for energy E<0, E>0, and two possible range for helicities h>0 and h<0. In the standard model of particles only the neutrino with h<0 interact and only the antineutrino with h>0. then when you compute the cross section this give different answers for neutrinos and anti-neutrinos. Answering your question if you can predict the difference between particles and antiparticles, is that you cannot predict. When you stablish your model for elementary particles choosing the group and the representation content you are going to fix the different particle and antiparticles. In our case the V-A character of standard model imposes that cros section are going to be different. But if we have a vectorial current them the cross section are equal. The difference of the cross section can be traced back to violation of parity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.