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BillKet
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Hello! I read some papers about searching for induced atomic EDM. Finding such an EDM would imply a violation of the P and T-invariance (and hence CP). The way the derivation works (very roughly) is by assuming you have a PT-odd interaction in the hamiltonian (coming from a possible nuclear EDM or electron EDM, both of with would imply T-violations), call it ##V_{PT}##. If you calculate the expectation value of this small perturbation, in the first order perturbation theory, you would get a value of zero, because in the absence of this perturbation the energy levels of an atom are parity eigenstates. So what you do is to apply an electric field and induce a stark shift, which leads to the mixing of energy level. Assuming we consider 2 levels with opposite parities, ##|+>## and ##|->##, the electric field will change the (say) positive eigenstate to $$|+'>=|+>+\frac{<-|-erE|+>}{E_+-E_-}|->$$ with ##erE## being the zeeman hamiltonian. Now if you add on top of this the ##V_{PT}##, given that the energy level in the presence of the electric field are not parity eigenstates anymore, you get a nonzero correction of the form: $$<+'|V_{PT}|+'>=2\frac{<-|-erE|+>}{E_+-E_-}<+|V_{PT}|->=E\cdot 2\frac{<-|-er|+>}{E_+-E_-}<+|V_{PT}|->$$ By definition the dipole moment is a the change in energy divided by the electric field, so in the end you get that the induced electric dipole moment is $$2\frac{<-|-er|+>}{E_+-E_-}<+|V_{PT}|->$$ This is the derivation made in basically all papers talking about EDM searches, and the steps seems to make sense. However, we know that and EDM implies PT violation. Just a P-violation term wouldn't lead to an induced EDM. However in the derivations above, I can't seem to see where the argument for a P-odd hamiltonian, ##V_P##, would fail. This would imply that P-violation, without T-violation would lead to an EDM, which is not true. Does anyone know what am I missing in the derivation above? Why do you need both P and T odd hamiltonian for the derivation to hold? Thank you!
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