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  1. The neutron magnetic dipole moment (MDM) is the intrinsic magnetic dipole moment of the neutron.
  • Under parity P, the MDM does not change direction.
  • Under time reversal T which is equivalent to CP, the MDM does flip direction.
  1. The neutron electric dipole moment (EDM) is the distribution of positive and negative charge inside the neutron. (Is it intrinsic?)
  • Under time reversal T (or CP), the EDM does not change direction.
  • Under parity P, the EDM does flip direction.

(The P and T=CP processes on MDM and EDM are shown below.)

Questions

What are the precise arguments and logic saying that the measurement of nonzero neutron electric dipole moment (EDM) implies CP violation? (Since the EDM itself only violates P, but the EDM does not violate T = CP.)

Does this CP violation require that the neutron magnetic moment (MDM) is also nonzero? But then because the neutron MDM already breaks T = CP, then why do we need another EDM to break only P?

My guess is that because we want to have the MDM to break T = CP, while we must have the EDM to break P, so it also breaks C or what? (It does not seem to be obvious.)

Figure source Wiki page here:

enter image description here

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ann marie cœur
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1 Answers1

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What matter are not the behavior of the dipole moments $\vec{\mu}$ and $\vec{d}$ themselves, but the Hamiltonian terms $-\vec{\mu}\cdot\vec{B}$ and $-\vec{d}\cdot\vec{E}$,* showing the interactions of the moments with external fields.

Under parity, the electric field (being a polar vector) changes sign, but under time reversal, it does not. (This can be understood by visualizing inverting the locations of all charges; this reverses the vector field $\vec{E}$ defined explicitly by Coulomb's Law. However, inverting the direction of time does not reverse the explicit expression for $\vec{E}$, in which time does not appear.) In contrast, under P, the magnetic field does not reverse sign, because the explicit formula for $\vec{B}$ is the Biot-Savart Law, which contains a cross product. (Under $\vec{V}\rightarrow-\vec{V}$ and $\vec{W}\rightarrow-\vec{W}$, we hve $\vec{V}\times\vec{W}\rightarrow\vec{V}\times\vec{W}$, with the two minus signs canceling. Thus the cross product of two vectors is a pseudovector or axial vector.) However, the time reversal T inverts the directions of all currents while leaving the spatial configuration unchanged, which naturally reverses $\vec{B}$.

So the different transformation properties of $\vec{E}$ and $\vec{B}$ mean that the actual Hamiltonian/Lagrangian terms representing the dipole interactions have different discrete symmetries. For the magnetic dipole, $\vec{B}$ changes sign under exactly the same combinations of C, P, and T as the spin vector (remembering that $\vec{\mu}=\mu\vec{\sigma}$ and $\vec{d}=d\vec{\sigma}$), so the combination $-\vec{\mu}\cdot\vec{B}$ is invariant under all the discrete symmetries. However, the discrete symmetries of $\vec{\sigma}$ and $\vec{E}$ do not match up. They are both odd under C; however, $\vec{\sigma}$ is P-even and T-odd, while $\vec{E}$ is P-odd and T-even, making $-\vec{d}\cdot\vec{E}$ odd under P, T, CP, and CT.

*Actually, these are the nonrelativistic limits of the general spin interactions with $\mu\bar{\psi}\sigma^{\mu\nu}\psi F_{\mu\nu}$ and $d\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}\bar{\psi}\sigma_{\mu\nu}\psi\tilde{F}_{\rho\sigma}$. In these forms, it is relatively straightforward to see that the electric dipole moment term must be odd under P and T, because of the presence of the Levi-Civita pseudo-tensor $\epsilon^{\mu\nu\rho\sigma}$.

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