Difference Between T-Odd & T-Violation?

[BillKet]

Hello! I am a bit confused by the difference between T-odd and T-violation. For example, I read that the existence of a fundamental particle EDM is a violation of time symmetry. However, placing an electric dipole in an electric field, would produce a hamiltonian (non-relativistically, which is usually the region of interest for e.g. atomic experiments): $H = -d\cdot E$, where d is the electric dipole and E is the electric field acting on the (say) electron. The dot product between d and E is odd under time reversal. But I am not sure I understand where the T-violation comes from. I thought that T-odd means just that the system changes sign under T operator, but it is still an eigenstate of it, which means that T and H commute. However, T-violation, I imagined, it means that T and H don't commute. Can someone help me clarify what odd and violation mean in this case? Thank you!

 

[Paul Colby]

Charge is even as is a separation distance between charges under time reversal. So, the dipole moment, $d$, would be even under time reversal. The energy (the hamiltonian) you wrote is clearly odd since the $E$-field is odd. Since energy must be even under time reversal, this makes the $H$ a violation since it's clearly odd. The only value that gives a consistent behavior under time reversal is $d=0$.

This is wrong, as usual.

 

[vanhees71]

Why do you think is $\vec{E}$ odd under $T$? $\vec{E}$ is a force devided by charge. The force is like $m \vec{a}$ and $m$ is even as is $\vec{a}$. So $\vec{E}$ is even. Frome $\vec{F}=q \vec{v} \times \vec{B}$ it follows that $\vec{B}$ is odd since $\vec{v}$ is odd and $\vec{F}$ is even.

 

[Paul Colby]

page 249 in Jackson. I can't believe I read that wrong.

 

[vanhees71]

If you refer to the 2nd edition, then I clearly read "even"... I'd have been very surprised had JDJ gotten that wrong ;-).

 

[Paul Colby]

Jackson gives a good discussion of P and T symmetry from the classical point of view which I "read" before replying. Perhaps there isn't a classical answer to the OP?

 

[Paul Colby]

https://en.wikipedia.org/wiki/Neutron_electric_dipole_moment

It would seem particles can't have both a magnetic and electric moments and still be time reversal symmetric since the magnetic moment flips sign under time reversal while the electric moment doesn't.

 

[vanhees71]

That's of course true, because the electric dipole moment is even and the magnetic dipole moment is odd under $T$ and thus the observable $\vec{d} \cdot \vec{m}$ flips sign under $T$, which means you can distinguish the time-reversed system from the original by this observable, and thus $T$ is broken (which is the case in the Standard Model due to the weak interaction).

 

[BillKet]

That's of course true, because the electric dipole moment is even and the magnetic dipole moment is odd under $T$ and thus the observable $\vec{d} \cdot \vec{m}$ flips sign under $T$, which means you can distinguish the time-reversed system from the original by this observable, and thus $T$ is broken (which is the case in the Standard Model due to the weak interaction).

Thank you for your reply! Related to your previous reply: I didn't say that E (the electric field) is odd under T (it is even). I said that the dot product $d\cdot E$ is odd under T. Also I don't think you are right with: "the electric dipole moment is even and the magnetic dipole moment is odd under T". They should behave the same way, as they are both proportional to the angular momentum vector of the particle (say electron). It is only when they are multiplied with the electric field and magnetic field, respectively that $\mu\cdot B$ becomes even while $d \cdot E$ becomes odd. Please see this recent review: https://journals.aps.org/rmp/pdf/10.1103/RevModPhys.91.015001?casa_token=53eUY3Z2Rh8AAAAA%3ALhYECsNOcMrBWhbL9FS9gV1oELKPf3csbqI1X4pIu37niL8EN7mOepNNaPDlevur5TcoXaX64wCdtvg(second page introduces all these things pretty well) and the references there, too. But my main questions is if T-odd and T-violation mean exactly the same thing. And if not, what is the difference?

 

[Paul Colby]

I said that the dot product d⋅E is odd under T.

I don't see this? The dipole moment, $d$, is clearly even (at least I got that part of argument correct) and E is clearly even. Nothing in the hamiltonian you gave was time reversal asymmetric. In fact, $H$ is not the issue at all. It's really the value of the observable , $d\cdot m$, where $m$ is the magnetic dipole moment which is the issue. For particles like the neutron where $m \ne 0$ is a fact, a non-zero $d$ for a neutron would then allow experiment to determine one time orientation from another. This is then a clear time symmetry violation.

 

[BillKet]

I don't see this? The dipole moment, $d$, is clearly even (at least I got that part of argument correct) and E is clearly even. Nothing in the hamiltonian you gave was time reversal asymmetric. In fact, $H$ is not the issue at all. It's really the value of the observable , $d\cdot m$, where $m$ is the magnetic dipole moment which is the issue. For particles like the neutron where $m \ne 0$ is a fact, a non-zero $d$ for a neutron would then allow experiment to determine one time orientation from another. This is then a clear time symmetry violation.

Please check the paper I put the link for. It shows why d is T-odd. Basically d is proportional to J (angular momentum), same as $\mu$, so they have the same behavior. (I could type here the derivation from there but it's easier to just check the paper)

 

[Paul Colby]

Nice review paper. I stand corrected.

Your basic question "Why is T odd not the same as T violation" seems to be more a question of observables. A quantity being T-odd isn't in general such an observable. After all velocity is T-odd as are many things. One must construct an experimental observable which is itself fixed (determined) by fundamental physics and not even under time reversal. The review paper appears to be a discussion of just those types of observables.

 

[vanhees71]

Well, Jackson also says that the ELECTRIC dipole moment is T-even. That's easy to understand, because it's given by something like $\int \mathrm{d}^3 r \rho(\vec{r}) \vec{r}$, and there's nothing $T$-odd in this expression. It's also not related to the angular momentum. That's related to the MAGNETIC dipole moment.

Could it be that you mix up the electric dipole moment (a time-even VECTOR quantity) with the Schiff moment (a time-odd pseudo scalar) $\hat{d} \propto \vec{d} \cdot \vec{J}$?

Have a look at this nice review talk:

http://www.ep.ph.bham.ac.uk/general/seminars/slides/Ben-Sauer-2018.pdf

 

[Paul Colby]

Please see this recent review: https://journals.aps.org/rmp/pdf/10.1103/RevModPhys.91.015001?casa_token=53eUY3Z2Rh8AAAAA%3ALhYECsNOcMrBWhbL9FS9gV1oELKPf3csbqI1X4pIu37niL8EN7mOepNNaPDlevur5TcoXaX64wCdtvg

Just a comment,

Equation 1 of this paper reads

$\hat{d} = \int \hat{r}\rho(r)\; d^3r = d\frac{\langle J\rangle}{J}$​

in which $J$ is the total angular momentum of the particle. This is a quantum mechanical result. Orbital angular momentum is T-odd. However, $J$ also includes spin degrees of freedom as well. In hadrons both are related to the dynamics at play. I have no reason to doubt this expression.

The classical dipole moment as defined is clearly T-even as Jackson states.

 

[BillKet]

Just a comment,

Equation 1 of this paper reads

$\hat{d} = \int \hat{r}\rho(r)\; d^3r = d\frac{\langle J\rangle}{J}$​

in which $J$ is the total angular momentum of the particle. This is a quantum mechanical result. Orbital angular momentum is T-odd. However, $J$ also includes spin degrees of freedom as well. In hadrons both are related to the dynamics at play. I have no reason to doubt this expression.

The classical dipole moment as defined is clearly T-even as Jackson states.

I am not totally sure what you mean. Of course T-violation is a quantum mechanical results. We can't have T-violation (or P, or C) at a macroscopical scale. My questions was of course QM/QFT related. But I am not sure what you mean by spin degrees of freedom. J is the angular momentum of the particle and that formula applies generally to any particle. For example, for an electron that J=S and the average of that is (in almost all experiments) $S_z$. But I am still confused. Why would the classical result be opposite to the quantum mechanical one?

 

[BillKet]

Well, Jackson also says that the ELECTRIC dipole moment is T-even. That's easy to understand, because it's given by something like $\int \mathrm{d}^3 r \rho(\vec{r}) \vec{r}$, and there's nothing $T$-odd in this expression. It's also not related to the angular momentum. That's related to the MAGNETIC dipole moment.

Could it be that you mix up the electric dipole moment (a time-even VECTOR quantity) with the Schiff moment (a time-odd pseudo scalar) $\hat{d} \propto \vec{d} \cdot \vec{J}$?

Have a look at this nice review talk:

http://www.ep.ph.bham.ac.uk/general/seminars/slides/Ben-Sauer-2018.pdf

Thank you for this! Did you get a chance to check the paper I suggested? They use the same formula as you for the electric dipole (eq 1) and then, in the comments after equation 3 they state that the formula is T-odd (please check the paper, it would be easier to formulate my questions once we know we are talking about the same thing). And no, Schiff moment comes later. An atom can develop an EDM through a Schiff moment, but that is way beyond my question. Based on the review paper I mentioned, the electric dipole of an electron (it's easier as there are no constituent parts, but it is the same for the other particles, with or without Schiff moments) is T-odd.

Also I am not sure how $d\cdot J$ is T-odd. Again, if you look at that paper, you will see that the magnetic moment is a constant times $J$, so $d\cdot J$ is proportional to $J \cdot J$ which is T-even.

I also looked over your slides, and on slide 11 they say that the interaction energy due to the electric dipole moment violates T, which is exactly what I said in post #9. Also, based on that expression on slide 11, $$-\eta d_e E \cdot \sigma$$ given that E is T-even (I think we agree on that), and that expression is T-odd (they clearly state that on that page) it obviously follows that the electric dipole moment $d_e \sigma$ is T-odd. I am not sure what you meant with those slides. They seem to agree with what I said (which is basically what is in that review paper, not my personal opinion obviously).

So do you agree that the electric dipole moment is T-odd? (I am definitely not an expert in the field and I really want to understand this and I am definitely open to further discussions)

 

[Paul Colby]

Why would the classical result be opposite to the quantum mechanical one?

I would guess because the classical definition of EDM makes no dynamical assumptions whereas the quantum mechanical result expressed in Equation 1 does.

 

[BillKet]

I would guess because the classical definition of EDM makes no dynamical assumptions whereas the quantum mechanical result expressed in Equation 1 does.

I am not really sure I understand. Equation 1 says, for an electron, that the EDM and the spin are parallel (or anti-parallel). There is no dynamical assumption (if by dynamical you mean actual motion of the particle). The spin of the electron is not actually the electron rotating (i.e. the spin has the same value even at 0K, where the electron doesn't move at all). The spin is a static property of a particle which implies that the EDM is too, doesn't it?

 

[Paul Colby]

There is no dynamical assumption

Then what non-dynamical assumptions went into Equation 1 we ask? Rotational symmetry - that's a dynamical assumption in that the system Lagrangian is rotationally symmetric. Parity? Likewise. All symmetry assumptions apply as constraints to the dynamics or time evolution of the system.

 

[BillKet]

Then what non-dynamical assumptions went into Equation 1 we ask? Rotational symmetry - that's a dynamical assumption in that the system Lagrangian is rotationally symmetric. Parity? Likewise. All symmetry assumptions apply as constraints to the dynamics or time evolution of the system.

What exactly do you mean by dynamics in this case? Equation 1 follows from Wigner-Eckart theorem, which basically reflects how the angular momenta can couple in a system. We didn't enforce any rotational symmetry into that equation, it follows from the fact that the electron has spin. I don't think that the fact that electron has spin has to do with rotational symmetry or parity (I am not totally sure tho; also not sure if parity counts as a dynamical symmetry).

 

[Paul Colby]

Equation 1 follows from Wigner-Eckart theorem, which basically reflects how the angular momenta can couple in a system.

As I said, rotational symmetry which the logical content of the Wigner-Eckart theorem. The charge distribution of the ground state is determined by the equations of motion for the system in question, the dynamics. For the electron these would be the relevant bits or sectors of the standard model or some such depending on the theory under test. The Dirac equation by itself doesn't yield an EDM but certainly describes electron motion. Again a dynamical assumption.

By contrast the classical expression for the EDM assumes just an arbitrary distribution of charge with no constraints on rotational symmetry or time evolution.

 

[BillKet]

As I said, rotational symmetry which the logical content of the Wigner-Eckart theorem. The charge distribution of the ground state is determined by the equations of motion for the system in question, the dynamics. For the electron these would be the relevant bits or sectors of the standard model or some such depending on the theory under test. The Dirac equation by itself doesn't yield an EDM but certainly describes electron motion. Again a dynamical assumption.

By contrast the classical expression for the EDM assumes just an arbitrary distribution of charge with no constraints on rotational symmetry or time evolution.

Oh, I see now what you mean by dynamics. Basically the EDM appears through some loop diagrams which in themselves reflects some dynamics of the system. I guess it makes sense, but that is really confusing, especially that Wikipedia page putting the classical image next to the quantum description. Thanks for this insight!

 

[vanhees71]

Amazing that such an obvious error could survive the review process in RMP! Eq. (1) is a contradiction in itself. The Schiff moment is the projection of the EDM to the direction of the total angular momentum but not the EDM itself. The EDM is T-even, a polar vector under parity, and the Schiff moment is T-odd and pseudoscalar under parity.

 

[vanhees71]

Thank you for this! Did you get a chance to check the paper I suggested? They use the same formula as you for the electric dipole (eq 1) and then, in the comments after equation 3 they state that the formula is T-odd (please check the paper, it would be easier to formulate my questions once we know we are talking about the same thing). And no, Schiff moment comes later. An atom can develop an EDM through a Schiff moment, but that is way beyond my question. Based on the review paper I mentioned, the electric dipole of an electron (it's easier as there are no constituent parts, but it is the same for the other particles, with or without Schiff moments) is T-odd.

Also I am not sure how $d\cdot J$ is T-odd. Again, if you look at that paper, you will see that the magnetic moment is a constant times $J$, so $d\cdot J$ is proportional to $J \cdot J$ which is T-even.

I also looked over your slides, and on slide 11 they say that the interaction energy due to the electric dipole moment violates T, which is exactly what I said in post #9. Also, based on that expression on slide 11, $$-\eta d_e E \cdot \sigma$$ given that E is T-even (I think we agree on that), and that expression is T-odd (they clearly state that on that page) it obviously follows that the electric dipole moment $d_e \sigma$ is T-odd. I am not sure what you meant with those slides. They seem to agree with what I said (which is basically what is in that review paper, not my personal opinion obviously).

So do you agree that the electric dipole moment is T-odd? (I am definitely not an expert in the field and I really want to understand this and I am definitely open to further discussions)

It's a very obvious mistake. Again: The EDM is T-even. Angular momentum is T-odd. Thus the Schiff moment is T-odd, and that's the key: You have a particle with half-integer spin (like an electron or a neutron etc.). Then its total angular momentum cannot be 0 (because it's half-integer as is the spin). Thus if the particle has an EDM the Schiff moment is non-zero and thus you have a violation of time-reversal symmetry. Due to the CPT theorem this implies that also CP-symmetry is violated, and that's very interesting with respet to the question, why there is more matter than antimatter in the universe.

 

[BillKet]

It's a very obvious mistake. Again: The EDM is T-even. Angular momentum is T-odd. Thus the Schiff moment is T-odd, and that's the key: You have a particle with half-integer spin (like an electron or a neutron etc.). Then its total angular momentum cannot be 0 (because it's half-integer as is the spin). Thus if the particle has an EDM the Schiff moment is non-zero and thus you have a violation of time-reversal symmetry. Due to the CPT theorem this implies that also CP-symmetry is violated, and that's very interesting with respet to the question, why there is more matter than antimatter in the universe.

I am really confused now. As far as I know, Schiff moment comes from a not total screening of the (possible) EDM by the electron cloud. You don't have a Schiff moment for an electron. My questions is purely about the electron EDM, so I am not sure I understand the Schiff moment argument in the case of an electron. Also as I said, based on the slides you provided, on slide 11, that equation also implies that the EDM is T-odd.

Moreover, I don't understand why you say that Equation 1 in the review paper is wrong (that paper has more than 100 citations). Also in this paper, which gave the best limit on the electron EDM to data, they have the exactly same formula as in that review paper, which makes me really believe they are not wrong (see the first paragraph):$$\vec{d_e} = d_e\vec{S}/{(\hbar/2)}$$

And clearly there is no Schiff moment implication in their analysis, $d_e$ is clearly the electron EDM (they state that in the paper) and it is proportional to the spin (as in the review paper). So, are all these papers wrong, or am I miss-understanding your point?

Again: The EDM is T-even. Angular momentum is T-odd.

How can EDM be T-even, the angular momentum be T-odd, yet they are proportional to each other?

 

[nrqed]

I am really confused now. As far as I know, Schiff moment comes from a not total screening of the (possible) EDM by the electron cloud. You don't have a Schiff moment for an electron. My questions is purely about the electron EDM, so I am not sure I understand the Schiff moment argument in the case of an electron. Also as I said, based on the slides you provided, on slide 11, that equation also implies that the EDM is T-odd.

Moreover, I don't understand why you say that Equation 1 in the review paper is wrong (that paper has more than 100 citations). Also in this paper, which gave the best limit on the electron EDM to data, they have the exactly same formula as in that review paper, which makes me really believe they are not wrong (see the first paragraph):$$\vec{d_e} = d_e\vec{S}/{(\hbar/2)}$$

And clearly there is no Schiff moment implication in their analysis, $d_e$ is clearly the electron EDM (they state that in the paper) and it is proportional to the spin (as in the review paper). So, are all these papers wrong, or am I miss-understanding your point?
How can EDM be T-even, the angular momentum be T-odd, yet they are proportional to each other?

You are correct about the behaviour of the EDM Hamiltonian. It is proportional to $\vec{S} \cdot \vec{E}$. The electric field is even under time reversal whereas the spin vector is odd. So the EDM Hamiltonian is odd under time reversal.

 

[BillKet]

You are correct about the behaviour of the EDM Hamiltonian. It is proportional to $\vec{S} \cdot \vec{E}$. The electric field is even under time reversal whereas the spin vector is odd. So the EDM Hamiltonian is odd under time reversal.

Thank you for your reply. So given that the (say electron) EDM is proportional to the spin (see my previous post and the 2 papers I mentioned), it means that the EDM is also T-odd (I guess this is my only confusion right now)?

 

[vanhees71]

Again, if nature where $T$ symmetric the electric dipole moment is $T$-even. See again Jackson, Wikipedia, and the slides I quoted above.

The only intrinsic vector-like quantity of an electron is its spin (its angular momentum in its rest frame). Thus it can only have a electric dipole moment if $T$ (and thus also $CP$) is broken and then this electric dipole moment must be proportional to its spin which implies that it is $T$-odd as its spin. That's why only with $T$ (and thus $CP$) violation a non-zero EDM for the electron is possible. Indeed the Standard Model predicts an EDM due to the $T$ violation of the weak interaction, but it's too small to explain the matter-antimatter imbalance in the universe and that's why the measurements about a possible EDM of elementary particles is so interesting.

 

[BillKet]

Again, if nature where $T$ symmetric the electric dipole moment is $T$-even. See again Jackson, Wikipedia, and the slides I quoted above.

The only intrinsic vector-like quantity of an electron is its spin (its angular momentum in its rest frame). Thus it can only have a electric dipole moment if $T$ (and thus also $CP$) is broken and then this electric dipole moment must be proportional to its spin which implies that it is $T$-odd as its spin. That's why only with $T$ (and thus $CP$) violation a non-zero EDM for the electron is possible. Indeed the Standard Model predicts an EDM due to the $T$ violation of the weak interaction, but it's too small to explain the matter-antimatter imbalance in the universe and that's why the measurements about a possible EDM of elementary particles is so interesting.

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). Just to clarify what confuses me about your answer: you say that "if nature where T symmetric the electric dipole moment is T-even". But if nature was T-symmetric, we wouldn't have an electric dipole moment at all. I don't understand how we would get a T-even EDM at all, when in a T-symmetric universe the electron EDM would be zero (of course zero as a number is T-even but this is not the point)

 

[vanhees71]

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

 

[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.