# 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?

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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.. –  Manishearth Mar 8 '12 at 16:19
@Manishearth: I'm talking about electric dipole moment, where as $\mu$ is magnetic dipole moment. –  Siyuan Ren Mar 9 '12 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) –  Manishearth Mar 9 '12 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].

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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!) –  David Z Mar 8 '12 at 19:52
Well, thank you! I tried to do my best, even though I am not an expert in nuclear/high energy physics. –  DaniH Mar 8 '12 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? –  Siyuan Ren Mar 9 '12 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, $\mathbf{J}=\mathbf{L}+\mathbf{S}$. –  DaniH Mar 9 '12 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 $\mathbf{D}$. –  DaniH Mar 9 '12 at 9:16

is almost sure that the pt symmetry be broken to the speed of light appear as constant to all inertial frames in relative motions.then left-right handed rotational invariance is not conserved,locally to smooth topologcal 4-dimensional manifolds.then the antiparticles are deformations of the spacetime curvatures that do appear symmetrics the dirac-einstein relativistics equations.therefore antimater doesn't exist in the nature neither early universe and the antiparticles are bundleled locally energy or "holes generated by the asymmetry of space and time conjugated by spacetime continuos to compensate the differences of speed of light in the topological4-dimensional manifolds that contain exotics structures coupled with the constancy of the speed of light so as the torsion fields with asymmetries to left-right handed frames generated in the topological 4-dimension manifolds with metrics defined non-hermitician hamiltonian matrices with spectral operators to complex coolections that are the fundamental entities in the universe that generates the spacetime continuos.theirs has non -commutative properties and pseudo-associative as split quaternions.

the time is plitted by two opposed orientations that if conjute deforming the space as fundamental to the smooth structures of the 4-dimension manifolds.then the spacetime curvatures continuos are pseudo constant but are stable to the kahler-einstein metrics

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