Undergrad Difference Between T-Odd & T-Violation?

[BillKet]

That doesn't answer my question. We know that nature is not T-symmetric so there is no way to assume it is. But my questions is simply is the electron dipole moment, defined the way it is in the papers I mentioned, T-odd or T-even (I tend to trust those papers with few hundred citations more than Wikipedia or some slides).

Sigh. Is this really so difficult. So let's go step by step. If you have only electromagnetism the world is P, T, C, CP symmstric. The only way some object can have an electric dipole moment then is that it is composed in the usual way by charge distributions. E.g., a water molecule has a large electric dipole moment. Such dipole moments are given as
$$\vec{d}=\int_{V} \mathrm{d}^3 r \vec{r} \rho(\vec{r}),$$
where $\rho$ is the charge distribution.

Now let's consider the time-reversal transformation. By definition the space-time variables transform as
$$t \rightarrow -t, \quad \vec{r} \rightarrow \vec{r}.$$
Electric charge
$$q \rightarrow q.$$
With the transformation properties of $q$ and $\vec{r}$ thus
$$\rho \rightarrow \rho.$$
With the definition of the dipole moment given above this implies
$$\vec{d} \rightarrow \vec{d}.$$
It's T-even, i.e., it doesn't change under T.

Now consider an elementary particle. This can have an electric dipole moment only if this dipole moment is $\propto \vec{s}$, where $\vec{s}$ is the spin, i.e., an angular momentum. Angular momentum is $\vec{r} \times \vec{p}$. From the $T$-transformation properties (we need in addition that by definition $m \rightarrow m$) it's clar that $\vec{s} \rightarrow -\vec{s}$, and thus there must be a T-odd contribution to the Hamiltonian $\propto \vec{E} \cdot \vec{s}$. Now again using the properties of $T$ and the definition of the electric field you get that $\vec{E} \rightarrow \vec{E}$ under $T$ and thus this contribution to the Hamiltonian is $T$ odd and thus violates $T$ symmetry. So you can have an electric dipole moment for an elementary particle only if $T$ (and thus CP) is broken.

Since $T$ and $CP$ is broken by the weak interaction, the electron has, within the standard model, a tiny electric dipole moment due to the weak interaction. This CP violation is too small to explain the matter-antimatter imbalance and that's why one tries to measure the electron's electric dipole moment. So far one is far in accuracy from the predicted dipole moment from the standard model. So the hope is that one finds nevertheless an electric dipole moment which with the given sensitivity then was some orders of magnitude larger than the standard-model value, and this would indicate new physics leading to larger CP violation which then in turn may explain the matter-antimatter imbalance. So far, however, no electric dipole moment of an electron has been found given the sensitivity of the experiments. There's only an upper bound, which is already very remarkable given the difficulty of this measurement.

If you need an RMP to be convinced, take that one:

https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.63.313

Thank you for this! I will check that paper. However I still have a question (again please check this paper), as I am still not sure how to reconcile these 2 definitions. So in equation (1), they use the same formula as you did: $$\vec{d}=\int\vec{r}\rho d^3r$$ According to your argument, this definition implies the electric dipole moment is T-even. Also in equation (1) of the same paper they say: $$\vec{d}=d\frac{<\vec{J}>}{J}$$ assuming that d and J are scalars (i.e. modulus of some vectors) and in the case of the electron $\vec{J}=\vec{s}$ and also given what you said that under T, $\vec{s} \to -\vec{s}$, this implies that under T, $\vec{d} \to -\vec{d}$, so the electron EDM is T-odd. So based on equation (1) and your derivations above, 2 different expressions for the same object (electron EDM), are one of them T-odd, one of the T-even. At this point this is just an equality between some vectors, and just applying the transformations under T that you mentioned, I get a contradiction. I am still not sure what is wrong with my logic here. Where is the flow in my argument?

 

[vanhees71]

The point is that when you have $T$ violation the electric dipole moment is neither even nor odd, because if there are "fundamental electric dipole moments" you have this $T$-odd contribution from these fundamental dipole moments but also the usual $T$-even contribution for induced dipole moments. So $T$ symmetry is explicitly broken, and that's the case in the Standard model for the weak interaction only. So $T$ symmetry is a very good approximation for everything where the weak interaction can be neglected.