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?

• In classical mechanics, we have this identity for spinning bodies of charge: $\frac{\mu}{L}=\frac{q}{2m}$, $\mu$ is dipole moment, $L$ is angular momentum. I dunno how this translates to particle physics, but it may help.. Commented Mar 8, 2012 at 16:19
• @Manishearth: I'm talking about electric dipole moment, where as $\mu$ is magnetic dipole moment. Commented Mar 9, 2012 at 4:26
• Aah, my bad. Didn't see the 'electric' in the question and i'm not familiar with your usage of symbols (I use p for electric dipole) Commented Mar 9, 2012 at 4:46

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].

• The neutron is composed of $\mathrm{udd}$ valence quarks so charge conjugation would switch it to $\mathrm{\bar u\bar d\bar d}$ - it's not the identity operation. So a neutron electric dipole moment wouldn't automatically violate $CPT$ symmetry. (But other than that, good answer!) Commented Mar 8, 2012 at 19:52
• Well, thank you! I tried to do my best, even though I am not an expert in nuclear/high energy physics.
– Dani
Commented Mar 8, 2012 at 21:49
• "Notice that the only way to know the orientation of your charge distribution ("particle") is by the orientation of the angular momentum." This is exactly the part I don't understand. Electric dipole moment requires only uneven charge distribution, but it does not require those charges to move. Also why are non-zero EDM of atoms not violation of T symmetry? Commented Mar 9, 2012 at 4:27
• Concerning the orientation of the charge distribution, perhaps it is more precise to say "orientation of the electric dipole moment". The electric dipole moment is a vector and to apply, for instance, a parity transformation you need to place it in a reference frame. The $\hat{\mathbf{z}}$ direction in this reference frame is given by the z- component of the angular momentum, $\mathbf{J}=\mathbf{L}+\mathbf{S}$.
– Dani
Commented Mar 9, 2012 at 9:05
• About non-zero EDM of a molecule: it is indeed interesting to think why this not violates $T$ symmetry. This is because the molecule/atoms have non-zero ground states that are invariant under parity so that $T$ needs not to be broken to give non-zero $\mathbf{D}$.
– Dani
Commented Mar 9, 2012 at 9:16