Technical question about AMO experiment in a paper

In summary, the conversation discusses the possibility of detecting an electric dipole moment (eEDM) in YbAg molecules, as well as the challenges in differentiating between eEDM and parity-violating effects involving the nucleus. The use of YbAg is favored due to its larger electronegativity and sensitivity to eEDM compared to Yb-alkali molecules. However, Yb-174, which has no nuclear spin, may have a smaller interaction between the electron and the Ag nucleus. This requires second order perturbation theory to determine the effective hamiltonian, which may not be the same as the one for eEDM. Further research and experimentation is needed to accurately distinguish between eEDM and parity-viol
  • #1
BillKet
312
29
Hello! This is quite technical, but any advice would be greatly appreciated (@Twigg ?). It is about this paper: https://arxiv.org/pdf/1909.02650.pdf. In principle, beside the EDM, we also have spin dependent parity violating (time-reversal conserving) effects. This is always true, as we need a nuclear spin for this method to work. The formula for these effects is usually ##H_P = W_P \ n\cdot (I\times S)##. Following the same notation as in the paper the effective hamiltonian becomes: ##H_P = W_P \xi\cdot (I\times S)_z##, assuming again that the electric field is in the z direction. By doing the math, and working in the 2 level system spanned by ##|g>## and ##|e>##, this is equivalent to ##H_P = i W_P \xi S_z##. So it is basically the same as the PT-violating effect, except for the complex factor i. Now equation (3) will have in addition: ##\frac{\Omega_P}{2}(\sin\beta\sigma_x+\cos\beta\sigma_y)##, and in (4) the term inside the sine squared becomes: ##\Omega_B+\Omega_{PT}\cos\beta+\Omega_P\sin\beta##, where ##\Omega_P \equiv \frac{1}{2}W_P\xi##. But for most (all?) systems, ##\Omega_P >>> \Omega_{PT}##, so unless ##\beta<<<1##, ##\Omega_P\sin\beta## will dominate and any EDM signal will be hidden by the P-odd, T-even signal. This seems like a non-reducible background. Am I doing something wrong in my calculations? Thank you!
 
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  • #2
You might be right. As far as nuclear P-violating effects, you probably know more than I do. What I can say is that there is a common understanding in eEDM experiments that you can't distinguish between an eEDM signal and a parity-violating interaction that involves the nucleus. If one experiment reports a non-zero eEDM, then they'll wait for someone else to measure the eEDM at the same statistical level with a different molecule. If they measure the same number on different molecules (and thus different nuclei), then it's an eEDM. If not, then it is this parity-violating effect (which I don't understand the physics of, to be honest).

Also, I haven't been involved in eEDM measurement for years, so my information may be almost certainly is out of date. Take everything I say on this topic with a big grain of salt.
 
  • #3
Twigg said:
You might be right. As far as nuclear P-violating effects, you probably know more than I do. What I can say is that there is a common understanding in eEDM experiments that you can't distinguish between an eEDM signal and a parity-violating interaction that involves the nucleus. If one experiment reports a non-zero eEDM, then they'll wait for someone else to measure the eEDM at the same statistical level with a different molecule. If they measure the same number on different molecules (and thus different nuclei), then it's an eEDM. If not, then it is this parity-violating effect (which I don't understand the physics of, to be honest).

Also, I haven't been involved in eEDM measurement for years, so my information may be almost certainly is out of date. Take everything I say on this topic with a big grain of salt.
Thank you for your reply! I might miss understand what you are saying. I agree that in one single measurement one can't distinguish between eEDM and nuclear EDM. However in normal EDM experiments (such as ACME), they are not sensitive to P-odd, T-even effects. Basically the polarizing field ensures that you need a T-odd effect to get a signal, so the T-even effects are not a background. However here I am talking about P-odd, T-even background for a P-odd, T-odd measurement.

For example for eEDM the operator is ##n\cdot S##, while for nuclear EDM it is ##n \cdot I##. However for a P-odd, T-even (e.g. anapole moment), the operator is ##n\cdot (S\times I)##. This latter one would not be a background for ACME (if they had nuclear spin).
 
  • #4
"We focus on YbAg rather than a Yb-alkali molecule [29]due to the larger electronegativity of Ag compared tothe alkali atoms, which results in a more strongly polarmolecule with enhanced sensitivity to the electron EDM[24]."
The dipole moment's direction in YbLi and YbCs is probably opposite to the one in YbAg. Which one is absolutely larger I wouldn't bet based only on electronegativity.
The reference 24 is a private communication.
 
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  • #5
Yb-174 has no nuclear spin, so an electron spin, nuclear spin interaction would be restricted to the interaction of the Yb f electron and the Ag nucleus, which certainly is much smaller than the direct interaction of the spins with the external magnetic field. So to obtain the effective hamiltonian associated with this interaction, one would have to do second order perturbation theory. The resulting term in the hamiltonian will not be
##H_P = W_P \xi\cdot (I\times S)_z##
as you supposed.
 
  • #6
DrDu said:
Yb-174 has no nuclear spin, so an electron spin, nuclear spin interaction would be restricted to the interaction of the Yb f electron and the Ag nucleus, which certainly is much smaller than the direct interaction of the spins with the external magnetic field. So to obtain the effective hamiltonian associated with this interaction, one would have to do second order perturbation theory. The resulting term in the hamiltonian will not be
##H_P = W_P \xi\cdot (I\times S)_z##
as you supposed.
I am not sure I understand this. The nuclear spin-dependent parity violation Hamiltonian (operator) is ##W_P\hat{n}\cdot(I\times S)##. This has the same form as the EDM interaction, just that the vector ##S## is replaced by ##I \times S##. Basically in the most general case, the operator is of the form ##W_P\hat{n}\cdot V##, where ##V## would be an arbitrary vector (or pseudovector). In in our case ##\hat{n}## follows the electric field, so it can be replaced by the effective term, ##\xi##. So the operator in geneal would be ##W_P\xi V_z##. Why, when ##V=S## this formula applies, but when ##V = S\times I## it doesn't (I didn't make any assumptions about ##V## in this derivation so it should work for any ##V##, no?).

Also, in the standard model, the anapole moment is many (many!) orders of magnitude higher than the electron EDM effect. So I agree that the electron overlap with Ag nucleus is not large, but I still expect the anapole moment to greatly surpass the EDM effect (am I missing something?). Moreover, the electron nucleus interaction is fully included in the ##W_p## constant (which can be small) but the spin space operator doesn't care about this overlap.
 
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  • #7
My first point is that the electronic spin of Yb is due to an f-shell hole which is strongly localized on the Yb atom, while only the Ag atom has a nuclear spin. Hence the IxS term is certainly orders of magnitude smaller than the B*S and B*I splittings. The IxS term is made up from operators, which are not diagonal in the z-basis, e.g. (IxS)_z=I_xS_y-I_yS_x. If you want to construct an effective hamiltonian, this operator will contribute in second order, only. Maybe in second order it is already negligible in comparison to the EBM effect.
Anyhow, it's strength will vary as 1/B, which should make its effect distiguishable from the EBM.
 
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  • #8
DrDu said:
My first point is that the electronic spin of Yb is due to an f-shell hole which is strongly localized on the Yb atom, while only the Ag atom has a nuclear spin. Hence the IxS term is certainly orders of magnitude smaller than the B*S and B*I splittings. The IxS term is made up from operators, which are not diagonal in the z-basis, e.g. (IxS)_z=I_xS_y-I_yS_x. If you want to construct an effective hamiltonian, this operator will contribute in second order, only. Maybe in second order it is already negligible in comparison to the EBM effect.
Anyhow, it's strength will vary as 1/B, which should make its effect distiguishable from the EBM.
I am sorry I am still confused. For example in this paper: https://arxiv.org/pdf/2210.16910.pdf, they do something similar to what I am describing (but it is a different spin-spin coupling compared to my question). However they have the same operator as me, between the same states, and the final result is what I am getting. Again, the coefficient ##W_p## has different meanings in these 2 cases, but the spin space operator is exactly the same. Why would in my case the result be different, given that I also apply the operator between a singlet and triplet state formed from the 2 spins in the problem?
 

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