I meant, how can you use synchrotron radiation to polarize electrons? Would any polarized light source work (e.g. a laser)? Or is there something special about the synchrotron radiation?
Hello! how does one produced big ensembles of polarized charged particles (electrons, protons, muons etc.) for certain experiments? In the case of neutral particles (for example the nucleus in an atom) this could be done using a magnetic field, but I guess this won't work that straightforward...
I am not sure what you mean. In the spectra, you would get 3 peaks, a main one and 2 secondary (with amplitudes half of the main one), which are stationary (in terms of position and amplitude). This is assuming the laser is on resonance (this is a Mollow triplet). So I am not sure I understand...
Hello! Can someone explain to me in an intuitive way (or a nice mathematical demonstration) or point me towards some accessible papers about the AC Stark effect (Autler-Townes effect)? I am mainly confused by how can one start from a 2 level system (consider a 2 level system for simplicity) add...
@vanhees71 I have one more questions about this topic. I see that in all the papers I read (see e.g. the Appendix of this) they start with the normal Hamiltonian that one would write down assuming a nuclear EDM, which is:
$$H = \sum_i{(K_i+V_i+e\phi(r_i)-eE_0r_i)}-dE_0$$ where the sum is over...
Thank you again for the reply. I know that in reality the nucleus is not a point like object, but isn't this precisely the assumption we are making in this derivation, that IT IS a point object? Isn't this the core reason as to why we are getting the delta function, because we are assuming that...
I am really sorry for being so dumb, but I am still a bit confused. In this formula: $$\rho_{\ell,m}=\int_V \mathrm{d}^3 r' r^{\prime \ell} \text{Y}_{lm}^*(\vartheta',\varphi') \rho(\vec{r}')$$
what is ##r'## referring to and what is the integration volume? I would normally assume that the...
Thank you for this (and sorry for the late reply)! So in this derivation if we take the limit ##r_{<} \to 0## we would get the results we got originally? However, I am still confused by this statement: "It's just that you approximate the charge distribution entirely as a point-like object"...
I came across the Schiff theorem and I am a bit confused about cases in which it doesn't apply. The statement for the theorem I found in all the papers/slides I read is of the form: "For a non relativistic system made up of point, charged particles which interact electrostatically with each...
Ohhh I see! Thank you so much! So this has nothing to do with QM, atomic physics, EDM etc. That Schiff term is supposed to be there (and it arises naturally) in classical EM, too. The reason why it is ignored and I didn't notice it, is that it is zero anywhere except at the origin but in...
Right! To be honest I didn't do the math of it yet (trusted them with the separation of terms). Could it be that in classical EM books they just, somehow, keep the reducible tensor, without doing this separation (I should also go back in more details in my EM derivation, too). Please let me know...
Exactly! This is my confusion. So did Vanadium 50 answer help you understand? Could you explain it to me in more details? I am not sure I see how the fact that we have an atom matter (assuming d=0)?
I am not sure why the fact that we have atoms or not matters. As I said my confusions starts from ##\rho(r)##. I could have just posted this question (even in the classical mechanics forum) from the beginning by writing down equations 1-5 in the original paper (with d=0), without mentioning...
Thank you for this! So I read a bit more about it and I actually realized that I am confused about the point 2. i.e. finite size effects. In the Schiff original paper, when he writes down the Hamiltonian of the system, equation 1, he uses ##\rho(r)## to describe the nuclear charge density (and a...
Thank you for your reply. I am a bit confused about this statement: "Only with the additional term involving a possible electric dipole moment of the nucleus you get such P- and T-odd contributions, and that's calculated in detail in Eqs. (1)-(5)." If I don't missunderstand it, they state in the...
I looked at that reference. That is the derivation as to why one can't see the EDM of a nucleus (in the case of a point like nucleus). But my questions is purely classical, and it has to do only with the derivation between equations 1 and 5. It seems like the term in equation 5 appears naturally...
A: Sorry, it seems like the PRL link is not working. I replaced it with the arXiv one.
B: It is not high energy physics. The paper I mentioned above is a nuclear physics paper, so I assumed this is the best place to post the question. I am not sure where it would fit better.
Hello! I just found out about the Schiff moment. This is the paper where I encountered it, equations 3 and 4. The paper covers other things, too, that are not related to my question.
The main question I have is that, it seems like the derivation from equation 1 to 4 is purely classical (one can...
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...
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...
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)?
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...
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...
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...
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...
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...
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...
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)
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...