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Let $\phi_a$ and $\chi_{\dot{a}}$ be two component commuting spinors, where $\chi$ is an anti-spinor.

In terms of some spinor basis, these can both be written in some arbitrary frame as $$ \phi_a(P) = \sqrt{E - |p|}\xi_a^-,~~~~~\chi_{\dot{a}}(P) = \sqrt{E+|p|}\tilde{\xi}^-_{\dot{a}} $$ Where $\xi_a^- \neq \tilde{\xi}_{\dot{a}}^-$ and $(\xi^-)^2 = (\tilde{\xi})^2 = 1$

I want to construct a Lorentz scalar from these guys, such as $$ \phi_a(P_1)\phi^a(P_2) = \sqrt{(E_1-|p_1|)(E_2 - |p_2|)},~~~~~\chi_\dot{a}(P_1)\chi^\dot{a}(P_2) = \sqrt{(E_1+|p_1|)(E_2 + |p_2|)} $$ If these are Lorentz invariant, then it should not matter what frame I choose to evaluate them in. Thus, choosing the rest frame where $P_1 = P_2 = 0$, I find that these two are equal, since now $$ \phi_1\phi_2 = \sqrt{m_1m_2} = \chi_1\chi_2 $$ However, as soon as I boost to another frame, these no longer seem to be equal, which would imply that they are not Lorentz invariants.

Any ideas?

EDIT: If the particles both have the same mass, then this can be seen to hold in the COM frame too. Imagining that both particles are travelling along the $z$ axis, then $E_1 = E_2 = E$ and $|p_1| = -|p_2| = |p|$. In that case, we have $$ \phi_1\phi_2 = \sqrt{(E-|p|)(E + |p|)} = m = \chi_1\chi_2 $$

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  • $\begingroup$ Why the downvote? $\endgroup$ – Akoben Dec 7 '17 at 11:57
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If you are trying to make an invariant entity you should use the product of the two different spinors and make use of their opposite properties to cancel the flipping due to boosts. For example the Cooper pairs do that in supraconductors. And covariant tensors couple to contravariant tensors to create an invariant. If that is not the case you need a gauge invariance.

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  • $\begingroup$ Could you elaborate a bit please? I'm not sure what you mean by 'use their opposite properties to cancel the flipping due to boosts'. $\endgroup$ – Akoben Dec 7 '17 at 13:40
  • $\begingroup$ I meant that when we boost a particle, the helicity for example, changes sign, showing it is not Lorentz invariant. But if you couple two particles with opposite helicity (the projection of their spin on the momentum axis) then this system should become invariant under Lorentz transforms and the total spin should be zero. $\endgroup$ – Gabriel Dec 7 '17 at 14:44
  • $\begingroup$ I understand this, I'm just not sure how to see that the above obeys this. Are you saying that the two spinor products should be equal provided I flip helicity? $\endgroup$ – Akoben Dec 8 '17 at 14:03
  • $\begingroup$ The mixed product should do the trick, because when one flips in some direction the other one will flip in the opposite and viceversa, keeping the over all spin zero. You do not have a mixed product there, you just have product of a spinor with itself, this is why they do their own thing and not being invariant. They must become a single entity and the only way to relate two things in this universe is to have some form of product between them, sometimes mediated by a constant showing how strong that action of one over another, really is. $\endgroup$ – Gabriel Dec 8 '17 at 14:56
  • $\begingroup$ Also, please see this interesting article about two-fermions couplings. It goes about the same way I've explained to you. arxiv.org/pdf/0801.1615.pdf $\endgroup$ – Gabriel Dec 8 '17 at 15:11

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