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Since magnetic moment come from the circulation of charge, what is the origination of the electron's magnetic moment? Because spin of electron is not the classical spin of particle. Can we say that corresponding to every angular momentum there is a magnetic moment and since spin angular momentum is conformed from Stern-Gerlach experiment so there is a magnetic moment.

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Since magnetic moment come from the circulation of charge, what is the origination of the electron's magnetic moment?

It's not true that a magnetic moment always comes from the circulation of charge. An electron's magnetic moment cannot originate entirely from circulation of charge in some internal structure of the electron, because then its g-factor (essentially the ratio of magnetic moment to spin) would then have to equal 1, when in fact it's approximately 2. Similarly, an electron's angular momentum can't come entirely from this type of internal motion, since it would then have to be a whole-number multiple of $\hbar$.

As far as we know, the electron is a pointlike particle without any internal structure such as preons.

Can we say that corresponding to every angular momentum there is a magnetic moment

I don't think so, since photons have spin, but they have no magnetic moment. In general, QED is the theory that predicts $g$ factors, and the $g$ factors are deduced from the more fundamental principles of QED, not just postulated.

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  • $\begingroup$ Ben, how the photon does not have magnetic dipole moment? A photon has both a varying electric and a varying magnetic field component. That we can't deflect the photon with an external electric or magnetic field has to do with the frequency of this field components. But both field components exist, aren't they? $\endgroup$ Jun 26, 2016 at 13:36
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In classical mechanics, a particle with angular momentum $L$ and a charge $q$ will possess a magnetic dipole given by $$ \mu = \frac{q}{2m}L $$ Now, $L$ refers to any type of angular momentum, orbital or rotational. In quantum mechanics, its the same idea - up to an additional constant $g$. So, the intrinsic angular momentum of an electron results in an intrinsic magnetic moment.

Also, the Stern-Gerlach experiment detected the magnetic moment of the electron by observing the deflection of silver atoms in an inhomogeneous magnetic field. The concept of spin can then be introduced to explain the result.

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  • $\begingroup$ if spin of electron is not the classical analogy of spin then how we can relate the classical magnetic moment formula with electron spin magnetic moment to describe the spin angular momentum. $\endgroup$
    – Rahul
    Aug 17, 2013 at 11:19
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The magnetic moment of the electron is associated with a kind of circulation of its charge. This follows from the Dirac equation. To develop a wave-equation that is first order in derivatives, just like Maxwell's equations are for electromagnetic waves, Dirac had to introduce a four-component wave function, known as a spinor. The plane wave solution yields four independent possibilities, two of which are associated with positive energy and the other two associated with negative energy. Ignoring the negative energy solutions and using two independent solutions $\psi_1$ and $\psi_2$ it can be shown (Bjorken and Drell Relativistic Quantum Mechanics (1964) p.36) that the probability current density of the electron wave has the form $$ j^\mu \sim \frac{1}{2m} \left [ \overline \psi_2 p^\mu \psi_1 - (p^\mu \overline \psi_2 ) \psi_1 \right]- \frac{i}{2m} p_\nu (\overline \psi_2 \sigma^{\mu \nu} \psi_1 ) $$ where $\overline \psi=\psi^\dagger \gamma_0$ with $\gamma_0=\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ and $p^\mu = i \hbar \partial/\partial x_\mu$ the four-momentum operator and $\sigma_{\mu \nu}=(i/2)[\gamma_\mu,\gamma_\nu]$ the antisymmetric product of two gamma matrices.

This is all rather complicated, but the key point is that the term in square brackets $[...]$ has the same mathematical form as the probability current density in non-relativistic quantum mechanics, as associated with solutions of the Schrödinger equation. However, the second term involves rotations of the wave components and represents the spin current. It is this spin that is connected with the magnetic moment. The electric current density is just this probability current multiplied by the electron charge, which thus implies the spin gives rise to a charge circulation.

In saying the electron is a point-like particle, we really mean that in high-energy collision experiments one cannot find any internal structure. However, under usual circumstances (like low energy) the electron wave itself will always be spread out in space to some extent and its charge, likewise, will never be completely localised. Thus we can always claim there is some motion of the electron charge that gives rise to a current.

To decide whether or not a quantum field has a magnetic moment, we need to interact it with an electric or magnetic field. If we do this for the Dirac electron in the limit of low energy, the spinor wavefunction $\psi=\begin{bmatrix} \phi \\ \chi \end{bmatrix}$ can be written in terms of two-component column matrices $\phi$ and $\chi$ and the Dirac equation in an electromagnetic field with vector potential $\mathbf A$ and scalar potential $\Phi$ becomes (Bjorken and Drell Relativistic Quantum Mechanics (1964) p.12) $$ i \hbar \frac{\partial \phi}{\partial t}= \left [ \frac{(\mathbf p - (e/c) \mathbf A)^2}{2m}- \frac{e \hbar}{2 mc} \boldsymbol \sigma \cdot \mathbf B+ e\Phi \right ] \phi $$ This equation can be derived from two coupled equations associated with the Dirac equation in the limit of weak potentials and low energy, where $\chi \approx \frac{1}{2mc}\boldsymbol \sigma \cdot (\mathbf p - (e/c)\mathbf A) \phi$. The component $\chi$ depends on the vector of Pauli spin matrices $\boldsymbol\sigma$, i.e. has spin. It is this component that leads to the dependence of $\phi$ on the Pauli spin matrices that couple with the magnetic field $\mathbf B= \nabla \times \mathbf A$.

For a weak, uniform magnetic field we can write $\mathbf A = \frac{1}{2}(\mathbf B \times \mathbf r)$ which satisfies $\mathbf B = \nabla \times \mathbf A$ since $\mathbf B$ is a constant. On expanding out the wave equation with no electric potential we get $$ i \hbar \frac{\partial \phi}{\partial t}= \left [ \frac{\mathbf p^2}{2m}- \frac{e }{2 mc}\left ( \mathbf L + 2 \mathbf S \right ) \cdot \mathbf B \right ] \phi $$ where $\mathbf L=\mathbf r \times \mathbf p$ is the orbital angular momentum operator and $\mathbf S=\frac{\hbar}{2}\boldsymbol \sigma$ is the electron spin operator with eigenvalues $\pm \frac{1}{2} \hbar$. The term $ \frac{e }{2 mc} (2 \mathbf S \cdot \mathbf B) \rightarrow g \mu_B \mathbf s \cdot \mathbf B$ which expresses the interaction of a magnetic field with a magnetic moment $\mathbf m = g \mu_B \mathbf s$ with $\mu_B= e\hbar /2mc$ the Bohr magneton and the correct $g$ factor of $g=2$. This demonstrates that the combination of the electron charge $e$ and its spin $\mathbf s$ creates a magnetic moment that interacts with a magnetic field $\mathbf B$. Thus the spin combined with a charge looks just like a circulating current.

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