I have calculated the Spin Magnetic moment for all individual atoms of a 2-D material via simulation. Is the total magnetic moment of this 2-D material just the sum of all individual magnetic moments?
Spintronic
Dear Prof. Robin Singla
In addition to the right answer by Prof. Xavier Oudet , I can advice to check the monograph by Frederick Reif "Fundamentals of Statistical and Thermal Physics", McGraw Hill, 1965
First chapters are based on the analysis of a total magnetic moment M with lots of examples worked out for different cases from the perspective of Statistical Mechanics methods of distribution (binomial and gaussian).
Kind Regards.
The total magnetic moment is a combination of spin (S) and orbital magnetic moment (L) which is given by
J=L+S
If there are n number of electrons then the spin angular momentum is given by
S=s1+s2+...+sn
The orbital angular momentum is given by
L=l1+l2+...+ln
The resultant total angular momentum J is given by
J=L+S
The orbital magnetic moment is given by
µL=µBL
and the spin magnetic moment is
µS=2µBS
The resultant magnetic moment is given by
µR=µL+µS
=µB(L+2S)
=gJµB
where g is given by
g={1+J(J+1)+S(S+1)-L(L+1)}/{2J(J+1)}
When S=0
J=L and g=1 which gives the orbital contribution
When L=0
J=S and g=2 which gives the spin contribution
Hund's Rule in magnetic moment
In an atom, the set of values for S and L are governed by Hund's rule given according to the following
1) The spin of the electron follows Pauli's principle. Electrons fill vacant shells first. Each spin will contribute to 1 Bohr magneton. Therefore 5 electrons in 3d shell will give raise to 5µB
After all the electrons have occupied all the vacant shells the extra electron will pair up with the available electrons with opposite spins. The opposite spins will give raise to negative values. The resultant spin magnetic moment in such case with 6 electrons in the d shell will be 4µB since 5µB in the positive and 1µB in the negative direction will give a net value of 4µB. Completely filled d orbital will give a net moment of zero.
2) The orbital magnetic moment arises from the electrons occupying the shells according to the values given by
ml,ml-1,....mo
For instance for 5 3d electrons
mL=+2 +1 + 0 -1 -2 = 0
for 7 3d electrons
mL=+2 +1 +0 -1 -2 +2 +1 = +3
3) The total quantum number is given by
J=L+S
If the shell is more than half filled then
J=L+S
If the shell is less than half filled then
J=L - S
If the shell is half filled
J=S
If the shell is completely filled
J=0
Xavier Oudet thanks for the reply. Can you please share a link of the paper that you have mentioned?
Pedro L. Contreras E. thanks. I will definitely study it from the mentioned book.
You are most welcome, Prof. Robin Singla
For the cases of ferro & antiferro substances, a comprehensive review of how to include orbital L and spin S magnetic moments is also given in chapter VII, item & 72 ("the spin hamiltonian") of E. Lifshitz and L. Pitaevskii, Vol. IX, page 300 - 304, Pergamon Press, 1981.
Kind Regards.
If your simulation has already considered the interaction among spins, sum of all the moments is the total magnetic moment.
It depends on the type of the system that you are considering. For example, if it is a paramagnet, the magnetization( total magnetic moment per unit volume) would be zero, because of the random orientation of spins. For a ferromagnet, at absolute zero, the magnetization is the sum of all the individual magnetic moments per unit volume- this is the case of a strong ferromagnet. At finite temperatures, however, the magnetization depends on the temperature. It decreases with temperature, following a T^(3/2) law and becomes zero at T=T_c, the Curie temperature. For ferromagnets, the magnons (quanta of spin waves) are responsible for this kind of behaviour.The above explanation pertains to bulk systems. You can take these explanations as a clue to find out the situation in 2D systems. Thank you and Good Luck.
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