I Does beta decay change spin orientation?

jerry mac
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If you have a free neutron viewed from a known frame of reference which is known to be spin up and it decays, is there anything in theory or experiment which indicates whether the remaining proton will be spin up or down? Thanks Jerry
 
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no, the proton can be up or down, depending on the electron and neutrinos' momenta...
 
Thanks for the response. In popular accounts of decay, spin is not usually discussed and I am trying to understand it. I guess in order to emit a w minus particle there would be handedness change internally as the down quark changes into an up quark, but that wouldn't change the overall spin orientation of the nucleon.
 
the spin is conserved... so you don't have to think in terms of the W bosons...
initially you had +1
in the final state you must also have +1
Now the final state consists of the proton, electron and electron antineutrino
some combinations showing the up/down of proton's spin would be: +1_p + 1_e - 1_\nu or -1_p + 1_e + 1_\nu...
Now of course antineutrinos have right handed helicity, and so them having + or - spin depends on their momentum orientation...
If the antineutrino has spin -1, it should also move towards the negative z-axis, if it has +1 it will move towards the positive z-axis (so that the spin and momentum are alligned).
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
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