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I understand why the cobalt-60 decay experiment shows that parity is violated in the weak interaction.

However, my lecture notes also say that 'Note that the outcomes of this experiment can also be interpreted as a violation of charge conjugation invariance and that the combination of parity and charge conjugation is still obeyed in this experiment'.

How can one say charge conjugation is violated too from this experiment, if there were no anti-cobalt-60 nuclei or positron tests being done?

My only thoughts are that if you postulate that CP is always conserved, then C must always be violated whenever P is, so the fact that you have measured P violated here means you expect C to be violated here too (eventhough you can't see that from the experiment directly). I know that in reality only CPT is conserved, but maybe this is all they mean?

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    $\begingroup$ I think you're last paragraph has it. Without assuming CP is conserved (at least approximately), this experiment in and of itself doesn't rule out that C is a valid symmetry but CP isn't. $\endgroup$ – jwimberley Jan 3 at 16:40
  • $\begingroup$ This might be useful, even if it isn't a dirct answer. Time reversal and CPT symmetry (III) $\endgroup$ – mmesser314 Jan 3 at 17:16
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Looking at:

$$ \vec n \rightarrow p + e^- + \bar{\nu}_R $$

it violates parity because the electron (odd) momentum depends on the neutron's (even) spin. (You can sub a cobalt nucleus if you want).

Under charge conjugation:

$$ \bar n \rightarrow \bar p + e^+ + \nu_R $$

The only reason this would violate charge conjugation is because $\nu_R$ doesn't exists, or, because the the neutrino does have a small mass, the rate is suppressed, and hence different.

Of course, applying $\hat P$ to that:

$$ \bar n \rightarrow \bar p + e^+ + \nu_L $$

is perfectly fine, and probably is the main decay mode of anti-neutrons.

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  • $\begingroup$ I don't understand how this answer addresses the core question. Yes, that is presumably how the decay of anti-neutrons in anti-cobalt-60 would work, but the experiment does not directly test any of that. $\endgroup$ – Emilio Pisanty Jan 3 at 17:24
  • $\begingroup$ @EmilioPisanty The core question is, and I quote: "but maybe this is all they mean?", so yeah ^^^that^^^ is all they mean. $\endgroup$ – JEB Jan 3 at 17:37

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