I Diamagnetism vs. paramagnetism

1. Mar 22, 2016

pierce15

I am studying some solid state physics, in particular magnetism; I have a few questions about the explanations.

My understanding is that paramagnetism is primarily an effect of the electron's spin-angular momentum: an unpaired electron will have a dipole moment which preferentially aligns with an external field, so the magnetization is proportional to the applied field.

On the other hand, diamagnetism is an effect of orbital angular momentum of electrons. What I don't understand is why the above argument cannot be applied to an electron with orbital angular momentum: surely an orbiting electron creates a dipole moment which aligns preferentially with an applied field? Or is it simply that the Lorentz force is much more powerful? (But if the Lorentz force were more powerful, then unpaired electrons [which have orbital angular momentum] should also exhibit diamagnetism, in contradiction of the prior explanation).

Thanks for bearing with me.

2. Mar 22, 2016

My understanding of diamagnetism (I'm no expert here-I'm speaking as someone who still has some learning to do on the subject), is that diamagnetism is mostly from free conduction electrons and it will obey LeChatlier's principle in that the response of free electrons will respond in such a way as to reduce or screen any applied magnetic field. Meanwhile paramagetism (and ferromagnetism) are both the result of electrons bound to the atoms, whether from orbital or spin angular momentum and will have a tendency to align with the applied field. In diamagnetism, I believe the electron paths can be macroscopic in size.

3. Mar 23, 2016

DrDu

Yes, the orbital momentum can also contribute to paramagnetism, especially in gasses. However in condensed phases, the orbital angular momentum is mostly "quenched". I.e. the atomic orbitals which are occupied by unpaired electrons are not eigenstates of angular momentum but combinations of states with m and -m. The degeneracy of the orbitals is lifted by the chemical bonding environment. You may want to read the article about crystal field theory on wikipedia: https://en.wikipedia.org/wiki/Crystal_field_theory

4. Mar 23, 2016

DrDu

Not quite. A text book example for the calculation of the diamagnetic response is for the Helium atom which certainly has no free electrons.
Both dia- and paramagnetism can be due to both free and bound electrons. For free electrons, the theory of paramagnetism was worked out by Landau. Here, orbital angular momentum is quite important, but it turns out that the total effect is quite small.

5. Mar 23, 2016

Perhaps one item of interest regarding diamagnetism is the superconductor, which, if I understand it correctly, is the extreme case of diamagnetism where the magnetic field is completely screened from the interior of the superconductor by the Meissner effect. Meanwhile, at least in homogeneous solids, the magnetic field enhancement that occurs in paramagnetism and ferromagnetism (that come from the internal magnetic moments lining up with the applied magnetic field) can be attributed to (magnetic) surface currents. (see e.g. Griffiths E&M textbook. He doesn't seem to emphasize this a great deal, but it follows from his magnetic surface current equation.)

6. Mar 23, 2016

pierce15

I don't really understand this, can you explain in a little more detail? Here's what I'm asking phrased differently: electrons orbiting the nucleus have a dipole moment, why shouldn't the dipole want to align with the field?

I am also having trouble seeing how this relates to crystal field theory.

7. Mar 23, 2016

I think I can offer a clue on some of this that may be helpful. I think the paramagnetism, diamagnetism, and ferromagnetism will at times have tremendous variations in the observed magnetic fields that portions of individual atoms experience at the microscopic/atomic level even though at the macroscopic level, the magnetic field can be treated as a uniform field. To understand some of the phenomena completely, it is probably necessary to introduce the variation at a microscopic level. In those cases, instead of a moment aligning with the macroscopic field, it might align itself to a microscopic/atomic scale portion of it.

Last edited by a moderator: Mar 24, 2016
8. Mar 24, 2016

DrDu

In atoms in the gas phase, you can observe this alignment. However, in condensed matter, electrons get reflected by other atoms, so the expectation value of angular momentum vanishes. In crystal field theory you can show that e.g. none of the orbitals from the $E_g$ or $T_{2g}$ set carries angular momentum.

9. Mar 24, 2016

DrDu

Here some example:
Consider a single electron on an atom in a d-shell. Among others, it can be in $d_2$ and $d_{-2}$ orbitals which are energetically degenerate in the absence of a magnetic field and carry magnetic momenta of $m_z=\pm \hbar \mu$.
In the absence of a magnetic field, we can as well form the two real functions $d_{x^2-y^2}=d_2+d_{-2}$ and $d_{xy}=(d_2-d_{-2})/i$. But in the solid state or in a molecule where each atom is surrounded quadratically by 4 ligands with positive charges, the degeneracy will be lifted and the $d_{x^2-y^2}$ orbital will be lower than the $d_{xy}$ orbital as its lobes point in the direction of the positive charges. When these states are brought into a magnetic field, they will admix only very little of the other wavefunction and the paramagnetism will vanish. This is what is called quenching and how it is related to ligand field theory, see also this scetch here:

10. Mar 24, 2016

One feature of paramagnetism that is worth mentioning (to the OP), is that these magnetic moments will have a tendency to align in the same direction as the magnetic field, (because energy $E=-u*B$ where u is the magnetic moment with a vector dot product). This energy is quite small for the moments from single electrons (orbital and/or spin), and thermal effects(e.g. the Botzmann factor of statistical physics, $exp(-E/(kT))$, will result in only a slight difference between the populations of the "+" and "-" states (e.g. $m_s=+1/2$ and $m_s=-1/2$ ). This is typical of paramagnetism. With ferromagnetism, there is much more extensive alignment of the magnetic moments with the magnetic field. With the strength of these magnetic fields (e.g. inside permanent magnets), the Boltzmann factor would predict that ferromagnetism should only be possible at extremely low temperatures(i.e. T<10 K). Experimentally it is found that the ferromagnetic materials become paramagnetic at unexpectedly high temperatures (the Curie temperature ($T_C =1042 K$ for iron). The explanation is that the exchange effect makes for adjacent moments to couple to each other and point in the same direction. This exchange effect is so strong that thermal effects even at room temperature do very little to disrupt this exchange coupling.