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ClaraParticipant
I think looking at Fig.8 can help. On the absorption spectra you clearly see that the maximal absorption peaks are at around the same intensity/height. But when looking at the lower spectra of the absorption derivative, the line that crosses zero in the isotropic case is a bit longer than that for the largest line in the axial case for example. So if you read only these latter spectra, you would think that the absorption for the isotropic case is more intense, but we can actually see in the absorption spectrum that this is not the case! Thus the line that crosses zero in the absorpion derivative spectrum can give a wrong impression.
ClaraParticipantFrom reading the pdf file of the next task, I would like to add that the different principle components are determined by measuring in the corresponding principle axes of the crystal individually and changing the orientation of the magnetic field and for each magnetic field orientation you measure the current principle gfactor. Then you change the direction of the crystal to measure in the principle axis of another principle gvalue and repeat by changing the orientation of the magnetic field and measuring each time again.
ClaraParticipantg=1, I=1/2, B_0=2T
1) Without hf interaction:
E(m_J)=g_J \mu_B B_0 m_J
=> E_photon = \Delta E=g_J \mu_B B_0 \Delta m_J
Case of transition between m_J=3/2 and m_J=1/2:
E_photon,1= g_J \mu_B B_0 (1) = 0.16 meV.
Case of transition between m_J=1/2 and m_J=+1/2:
E_photon,2= g_J \mu_B B_0 (1) = 0.16 meV.ClaraParticipantReading an answer from a fellow student, I see that I indeed also forgot about the 2nd theorem.
In the case of FeII I say there are 3 ffold rotation axes which results, by this 2nd theorem, in the eletric field gradient tensor being zero.ClaraParticipantI think I have to specify that the nuclear dumbbell does not feel a different electric gradient field depending on its precise position for a specific angle theta as the distance between the nuclear dumbbell charges and the electroncharges does not change.
ClaraParticipantI realise, after reading some other answers that I read ‘neutron’ as ‘nucleus’ for some reason.
Let me try to correct then my last answer:
So we have the 2 formulae:
vec{mû}_S = (mu/S hbar) vec{\Hat{S}}
vec{mû}_S = (g e/2m) vec{\Hat{S}}For an eletron, the last formula is:
vec{mû}_S = (g u_B/ hbar) vec{\Hat{S}}
A neutron is around 1836 times heavier than an electron, and thus for the neutron:
vec{mû}_S = (g u_B/1836 hbar) vec{\Hat{S}}Equating the 2 formulae we have:
mu/S hbar =g mu_B/1836 hbar <=> mu = g mu_B S /1836
We are give that g = 3.826, and have S=1/2. So we have:
mu = 0.00104 mu_BClaraParticipantHi!
As a reply on your first question: I wonder if this has to do with the fact that the contribution to the shift from the two distributions are scalars. The size of m_1 is a scalar and the mass contribution of m_2 at the origin as well.
This is then similar to the case of the monopole moment and the monopole field both being scalar in the monopole term. So perhaps this is the standard to call the resulting energy(shift) a monopole shift?ClaraParticipantHi!
If you look at the scientific paper that is provided as optional reading material, the second page gives an overview of what is given in the different columns of the table. There it is written that the units of the electric quadrupole moment are in eb (electron barn), which would resolve your question. I guess that the units in the table takes this implicitly into account?ClaraParticipantTesting your testing of the test

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