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?

  • $\begingroup$ 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.. $\endgroup$ – Manishearth Mar 8 '12 at 16:19
  • $\begingroup$ @Manishearth: I'm talking about electric dipole moment, where as $\mu$ is magnetic dipole moment. $\endgroup$ – Siyuan Ren Mar 9 '12 at 4:26
  • $\begingroup$ 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) $\endgroup$ – 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.


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].

  • $\begingroup$ 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!) $\endgroup$ – David Z Mar 8 '12 at 19:52
  • $\begingroup$ Well, thank you! I tried to do my best, even though I am not an expert in nuclear/high energy physics. $\endgroup$ – DaniH Mar 8 '12 at 21:49
  • 3
    $\begingroup$ "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? $\endgroup$ – Siyuan Ren Mar 9 '12 at 4:27
  • $\begingroup$ 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, $\mathbf{J}=\mathbf{L}+\mathbf{S}$. $\endgroup$ – DaniH Mar 9 '12 at 9:05
  • $\begingroup$ 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 $\mathbf{D}$. $\endgroup$ – DaniH Mar 9 '12 at 9:16

protected by ACuriousMind Feb 29 '16 at 19:28

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