# T-odd vs T-violation

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## Main Question or Discussion Point

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!

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Paul Colby
Gold Member
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.

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vanhees71
Gold Member
2019 Award
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
Gold Member
page 249 in Jackson. I can't believe I read that wrong.

vanhees71
Gold Member
2019 Award
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
Paul Colby
Gold Member
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?

vanhees71
Gold Member
2019 Award
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).

• Paul Colby
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: Electric Dipole Moments of Atoms, Molecules, Nuclei and Particles (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?

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Paul Colby
Gold Member
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.

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
Gold Member
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
Gold Member
2019 Award
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
Gold Member
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.

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?

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
Gold Member
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.

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
Gold Member
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.

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
Gold Member
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.

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
Gold Member
2019 Award
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
Gold Member
2019 Award
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.

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?

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