Neutron electric dipole moment and $T$ symmetry violation Our textbook (and other sources I have found) says that non-zero electric dipole moment of neutron would violate $T$ symmetry. They prove this statement by first assuming $\boldsymbol{D}=\beta\boldsymbol{J}$, where $\boldsymbol{D}$ is the dipole moment, $\boldsymbol{J}$ is the angular momentum, and $\beta$ is a constant.
But why? Why is $\boldsymbol{D}$ proportional to $\boldsymbol{J}$? Why is $\boldsymbol{D}$ related to $\boldsymbol{J}$ at all? And how can't this argument be applied to other composite particles such as atoms and molecules, thereby breaking T symmetry for most of the world?
 A: As the neutron is not point-like, consider it has a continuous distribution of charge $\rho(\mathbf{r})$ confined in a volume $\Omega$. The dipole electric moment is then given by
$\mathbf{D}(\mathbf{r})=\int_\Omega \rho(\mathbf{r}')\delta(\mathbf{r}-\mathbf{r}')d^3r'$
where the coordinates are measured from the centre of mass of the distribution.
For a charged particle, this definition implies that for $\mathbf{D} \neq\mathbf{0}$ the "centre of charge" is displaced from the centre of mass of the distribution. For a distribution which has no net charge, that is
$Q=\int_\Omega \rho(\mathbf{r}) d^3r=0$
this definition implies that a there is a greater positive charge side of your distribution and a correspondingly greater negative charge in the other side.
Consider now that your particle has angular momentum $\mathbf{J}$ and that its orientation is given by $m$ (the eigenvalue of the $\hat{J}_z$ operator) relative to the $\hat{\mathbf{z}}$ axis. Notice that the only way to know the orientation of your charge distribution ("particle") is by the orientation of the angular momentum.
As a consequence, both $\mathbf{J}$ and $\mathbf{D}$ must transform equally under parity $P$ and time reversal $T$ if $\mathbf{D} \neq \mathbf{0}$ and if there is $P$ and $T$ symmetries. But $\mathbf{D}$ changes its sign under $P$ whereas $\mathbf{J}$ does not so $\mathbf{D}$ must vanish if there is $P$ symmetry. In a similar way, $\mathbf{D}$ does not change sign under $T$ but $\mathbf{J}$ does, so $\mathbf{D}$ has to vanish if there is $T$  symmetry. Hence if the neutron electric dipole is not zero we will have a violation of $PT$ symmetry.
Remark: This argument only applies to particles with non-zero dipole moment.
Experimental searches of the neutron electric dipole moment can be found:

*

*Smith et al. Phys. Rev. 108, 120 (1957) [link to paper].


*Baker et al. Phys. Rev. Lett. 97, 131801 (2006) [link to paper].
The upper bound in the last one for $|\mathbf{D}|$ is $2.9 \cdot 10^{-26}$ e cm.
D.
EDIT: As David said below, there is not $CPT$ violation in the, hypothetical case, of having $PT$ violation [=existence of non zero electric dipole moment].
