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In most introductory textbooks, the explanation of orbital magnetic moment is based on Bohr's model and electrons orbiting around the nuclues, which can be modeled as a current loop. For example, here.

But I've never seen an explanation without Bohr's model, using Schrödinger's equation. Would this be possible, or do we need some experimental hypothesis? In particular, I'd prefer avoid charge density orbitals-based explanations, if possible.

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  • $\begingroup$ It sounds a bit like a contradiction that you want to use the Schroedinger equation but without mentioning orbitals. Who is the audience that you are catering this explanation to? Atomic magnetism is an inherently quantum mechanical phenomenon. Strictly speaking, it's a relativistic phenomenon, so even the Schroedinger equation is not quite enough. $\endgroup$ – CuriousOne Oct 1 '14 at 19:04
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    $\begingroup$ Note that although the electrons do not literally orbit the nucleus, their orbital states, in general, do possess angular momentum. $\endgroup$ – ACuriousMind Oct 1 '14 at 19:05
  • $\begingroup$ @CuriousOne I guessed that line would cause problem. I mean not to consider the wavefunction as a charge density wave, the way chemists do (unless there is a rigorous association between both). $\endgroup$ – jinawee Oct 1 '14 at 19:08
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    $\begingroup$ @jinawee: I am not aware that chemists are using wavefunctions differently than physicists. It's customary to use mean field and other approximations in molecular orbital theory that remove the necessity to calculate the exact solutions to the Schroedinger problem, but that's a well motivated approach both in chemistry and physics. $\endgroup$ – CuriousOne Oct 1 '14 at 19:14
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    $\begingroup$ I am also unaware that us chemists have been using a different wavefunction :p $\endgroup$ – AngusTheMan Oct 1 '14 at 21:30
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Let's consider coupling a charged particle to a magnetic field in quantum mechanics. Assume a uniform magnetic field for simplicity. The prescription for coupling to an EM field is the substitution $\mathbf{p} \rightarrow \mathbf{p} - q\mathbf{A}$. The Hamiltonian is then \begin{equation} H = \frac{\left(\mathbf{p} - q\mathbf{A}\right)^2}{2m} + V \end{equation} Or, expanding, \begin{equation} H = \frac{1}{2m}\left[p^2 +q^2A^2 - q\left(\mathbf{p}\cdot\mathbf{A} +\mathbf{A}\cdot\mathbf{p}\right)\right] + V \end{equation} If we work in the Coulomb gauge, $\nabla\cdot\mathbf{A} =0$, then $\mathbf{p}\cdot\mathbf{A} = \mathbf{A}\cdot\mathbf{p}$ and \begin{equation} H = \frac{1}{2m}\left[p^2 +q^2A^2 - 2q\mathbf{A}\cdot\mathbf{p}\right] + V \end{equation} We still have some gauge freedom here, so let's choose explicitly $\mathbf{A} = \frac{1}{2}\mathbf{B} \times \mathbf{r}$ so that $\mathbf{A} \cdot \mathbf{p} = \frac{1}{2}\left(\mathbf{B} \times \mathbf{r}\right) \cdot \mathbf{p} = \frac{1}{2} \left(\mathbf{r} \times \mathbf{p}\right) \cdot \mathbf{B} = \frac{\mathbf{L}\cdot\mathbf{B}}{2}$ by the cyclic symmetry of the triple product.

The full Hamiltonian is \begin{equation} H = \frac{1}{2m}\left[p^2 +q^2A^2\right] - \frac{q}{2m}\mathbf{L}\cdot\mathbf{B} + V \end{equation} and by analogy to the classical energy of interaction between a magnetic dipole and magnetic field, we define $\mathbf{\mu} = \frac{q\mathbf{L}}{2m}$.

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Have you had a 3rd or 4th year QM course yet? The simple answer is that the angular momentum is in the wave-function. For spherically symmetric (3-D) potentials the solution to the Schrodinger equation is in terms of spherical harmonics (at least the radial part of the solution.) For any of the solutions you can act on it with the angular momentum operator and find the angular momentum of that state. (Since you are a physics student some of that should make sense.) (Way back when in undergrad we used McGervey's "Introduction to Modern Physics".. it's still on my shelf and he does a pretty good job of laying all that out.)

Adding some to avoid excess comments,
"What's a magnetic moment?" Well for me a magnetic moment is the magnetic analogy of a dipole moment. (dipole is two charges separated by a distance.) A magnetic moment is a loop of current with some area.

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  • $\begingroup$ I'm having a QM course, yes. My question is about orbital magnetic moment, not angular momentum. $\endgroup$ – jinawee Oct 1 '14 at 20:05
  • $\begingroup$ Oh well the electron has a charge, that and the angular momentum determine the orbital magnetic moment. (I think.. I'll get back to you) $\endgroup$ – George Herold Oct 1 '14 at 20:08
  • $\begingroup$ Oops I forgot the mass, mu_z (magnetic moment in z direction) = e*L_z/(2*m), L_z is the angular momentum in z direction. $\endgroup$ – George Herold Oct 1 '14 at 20:14
  • $\begingroup$ And the proof is? I know that formula, but the derivation I've seen is based on Bohr's model: hyperphysics.phy-astr.gsu.edu/hbase/quantum/orbmag.html. $\endgroup$ – jinawee Oct 1 '14 at 20:17
  • $\begingroup$ Wow, did you vote me down? (weird) I'm sure you can find more mathematical derivations. Angular momentum is where the orbital magnetic moment of the electron exists (at least in my view.) What's a magnetic moment? It's a charge going around in a circle (some area). (Iarea) Angular momentum is some mass going around in a circle (mv*r). If an electron is in a state with angular momentum, then there is some magnetic moment due to that motion. I don't see a reason not to see these electrons as orbiting (revolving). How else do you understand angular momentum? $\endgroup$ – George Herold Oct 1 '14 at 23:59
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At the quantum mechanical level, past the Bohr model, to every observable there corresponds an operator. It is convenient to use the semiclassical definition of the magnetic moment, and the Bohr model, but an operator format does exist.

That then makes the total magnetic dipole moment operator of a single proton equal to

magmomoper

The link goes on to explore many forms of magnetic moment at the quantized level..

This operator acting on the wave function which gives the orbitals, where the probability of finding an electron is described can be used to extract the magnetic moment of a state. Thus it is only the correspondence of "observable" to "operator" that is important. Not any hypothetical motion of an electron around the nucleus.

This is true for all observables at the quantum mechanical level.

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