# Is the property of magnetism due to movement of electrons about a central nucleus or due to the spinning of electron about its own axis?

Look my teacher said that it's due to the movement about the nucleus. But there are two problems in this acc to me 1.) Motion of an electron is not AT ALL circular. 2.)When we check whether an element is diamagnetic, we write down the configuration of the element (even from the point of MOT) and see the no. of paired/unpaired electrons and spin is the main reason behind this and the revolution of electron about the nucleus has no role/effect in this process of determination. So who is the major factor in magnetizing property of a material?

• It's not really motion since, within an atom the electron is in an orbital rather than moving as such. Commented Nov 21, 2022 at 17:10
• Perhaps, your teacher said it by taking Bohr's model of the atom. However, MOT is based on a quantum mechanical model. Commented Nov 21, 2022 at 18:25
• Spinning doesn't contribute, it is introduced to compensate for lesser states available. Commented Dec 8, 2022 at 16:49

It’s both. The electron $$g$$-factor within an atom contains a term proportional to the electron spin, and also a term proportional to the electron’s orbital angular momentum quantum number. In an atom with an unpaired $$s$$-wave electron, there isn’t any orbital angular momentum, so only the spin contributes. But if a valence electron has nonzero $$L$$, the orbital state contributes to the atom’s magnetic dipole moment as well.

For example, in iron, there are four unpaired d-shell electrons which contribute to ferromagnetism. However, beware that chemists label the relevant states $$d_{z^2}$$, which has $$m=0$$, and $$d_{x^2-y^2}$$, which is a superposition of $$m=\pm2$$. The orbital magnetic moment is proportional to $$m$$, and vanishes if you use the non-complex wavefunctions in which projections like $$m=+2$$ are unavailable.

Note that ferromagnetism is a condensed-matter phenomenon, which depends on a material’s crystal structure and whose presence or absence is the subject of a phase transition. You ask in a comment whether ferromagnetism is spin or orbital in nature. I’m pretty sure the answer is “both.” The iron ground state has angular momentum $$^{2S+1}L_J = {}^5D_4$$, which corresponds to spin $$2\hbar$$, orbital $$2\hbar$$, aligned for a total of $$4\hbar$$. The spin contributes more, since the spin g-factor is twice the orbital g-factor.

However, I’m not a condensed-matter expert, and there are a couple of ways I might be wrong. It could be the iron ions which polarize, since one conduction electron is delocalized to the metal, in which case the angular momentum increases to $$^6D_{9/2}$$. But it could also be that the orbital degrees of freedom are eaten up by the interatomic binds which form the lattice, so that only the spin degrees of freedom are available to polarize. If the orbital degrees of freedom are used to define the crystal lattice, that would also explain why chemists tend to use the real-valued wavefunctions instead of the complex-valued wavefunctions with definite $$m$$.

I’m about 90% certain that, of the two magnetic materials forming the torsion pendulum in this nifty experiment, one was magnetized entirely by electron spin, and one with zero electron spin. I don’t recall whether the other material was magnetized by electron orbital momentum or (more likely) by nuclear spin, and I don’t currently have access to the paper or it’s references. However, nuclei are absolutely magnetizable. Furthermore, in nuclei, the spins and orbital angular momenta of the nucleons are not separate observables, so their magnetizability always involves both terms.

• Has anyone tried talking the chemists into using complex orbitals and dropping the polynomial labels? Commented Nov 23, 2022 at 14:14
• The complex wavefunctions have well-defined $m$, but the real wavefunctions have nice shapes in space: consider that the $p_x$ and $p_y$ wavefunctions are simple rotations of the $p_z$. The directions are nice for people who might be thinking about the shapes of molecules, instead of their sping properties.
– rob
Commented Nov 24, 2022 at 3:32
• for atoms I don’t think real orbitals have an advantage. Complex orbitals have shapes and meaningful geometry. But yeah, when you get into molecules or solids I think I agree that geometry of real orbitals might be better to focus on. Commented Nov 24, 2022 at 4:46
• Are both the factors equally contributing? Commented Nov 29, 2022 at 20:35
• Acc to me in case of paramagnetism and diamagnetism , spin is dominant .What is the case for ferromagnetism?Any idea? Commented Dec 2, 2022 at 5:29

Is the property of magnetism due to movement of electrons about a central nucleus or due to the spinning of electron about its own axis?

The simplest way to understand magnetism in the atom is to think of the subatomic particles as magnetic dipoles. The discovery of the electron as an electric elementary charge often obscures the fact that the electron is also a magnetic dipole to the same extent. See the constant of the magnetic dipole of the electron at NIST. For the magnetic moment of the neutron, by the way, this article from NIST is interesting.

Of course, you will study both spin and the distribution of charges in atomic orbitals in physics courses, but for a basic knowledge, it may not be so bad to think of subatomic particles as a unit of electric charge and magnetic dipole.

So who is the major factor in magnetizing property of a material?

The subatomic particles in the atom cause their common magnetic alignment. Unpaired electrons cause an excess magnetic moment in the atom, which may or may not lead to a common magnetic alignment of the atoms (permanent magnets). This depends on the mobility of the atoms among themselves as well as on the temperature (search for Currie temperature.

Besides the phenomenon of the permanent magnet, there is also that of magnetizability. Here, the magnetic dipoles of the atoms are simply aligned by the influence of an external magnetic field.

• And When you talk about the magnetic alingment of atoms, do you mean the alignment of spins of subatomic particles, or the alignment of magnetic dipoles which is acc to my understanding caused due to both ,(movement (though not circular) and spin)? Commented Nov 23, 2022 at 7:28
• “the electron is also a magnetic dipole to the same extent” means that it isn’t needed any explanation for the existence of the electrons magnetic field. It exist as well as exist the electrons electric field. Commented Nov 23, 2022 at 11:59

The Hamiltonian Operator has both terms:

$$H=\frac{P_i P^i}{2m} - g\vec{B}\cdot \vec{L} - \frac{g}{2}\vec{B}\cdot \vec{S}$$

The $$S$$ matrices only operate on the spin degrees of freedom, while the $$L$$ and $$P$$ matrices only operate on the position/momentum degrees of freedom of the wavefunction. Thus, the $$S$$ matrices commute with the other matrices. To figure out the time evolution of an operator, you would take its commutator with the Hamiltonian.

In the time evolution of position, i. e. $$X$$, the commutator with the spin matrices vanishes and the equation of motion looks just as if spin didn't exist.

If you follow the motion of the average value of the electron's position, it behaves as if a classical particle was experiencing force according to the formula $$\vec{F}=q\vec{v}\times \vec{B}$$. There is no contribution of the average value of spin to this force.

In the time evolution of spin, the commutator with all non-spin operators vanishes.

If you follow the time evolution of the expected value of the spin vector $$\vec{S}$$, it will behave as if there was a small current loop of the same orientation as the expected spin, with dipole momentum $$\vec{\mu}$$, experiencing a torque according to the formula $$\vec{\tau}=\vec{\mu} \times \vec{B}$$

To sum up, the coupling of spin to the magnetic field can be thought of as causing change in expected value of orientation of the electron. The coupling of orbital angular momentum to magnetic field can be thought of as causing change in expected values of position and momentum.

P. S. This answer is a special case for when the Magnetic Field is constant.