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Electrically neutral particles such as neutrinos can have nonvanishing magnetic dipole moments. Spin-1 particles, e.g., deuterium nuclei, can also have dipole moments. Googling seems to show that the Z boson has a magnetic moment.

So is there an elementary argument that explains why the photon has a zero magnetic moment? By analogy with the Bohr magneton and nuclear magneton, we might actually expect that zero mass would produce infinite dipole moment. Altschul 2007 has some discussion of empirical bounds and difficulties in creating a theory in which the moment doesn't vanish, but I'm having trouble translating anything in the paper into an elementary argument for why it's so much easier to have the moment vanish.

Altschul, "Astrophysical Bounds on the Photon Charge and Magnetic Moment." Astropart. Phys. 29 no. 4, pp. 290–298 (2008), arXiv:0711.2038.

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    $\begingroup$ arxiv.org/abs/hep-th/0609008 $\endgroup$ – John Rennie Aug 16 '13 at 18:28
  • $\begingroup$ Interesting, never thought about that. Are there any other particles with spin 1 but zero $\mu$? Also, might it have to do with the speed of the photon? Photons also have indeterminate relativistic mass. Or something like that. $\endgroup$ – Manishearth Aug 16 '13 at 19:08
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    $\begingroup$ Ben, I agree with you that it's a fundamental question. If there were a simple explanation I doubt so many experimentalists would be be trying to measure it. But the best I can come up with is that the spin of the electron and photon are fundamentally different. The electron's corresponds to magnetic dipole moment while the photon's corresponds to angular momentum. $\endgroup$ – user27777 Aug 16 '13 at 19:32
  • $\begingroup$ @JohnRennie: The Villalba-Chavez paper you linked to on researchgate is available on arxiv: arxiv.org/abs/hep-th/0609008 . They show that there is such a dipole moment in an external magnetic field, but I don't see any easy way to extract an elementary argument that it has to vanish when the external field goes to zero. $\endgroup$ – Ben Crowell Aug 19 '13 at 16:42
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    $\begingroup$ The circular-polarized photon doesn't actually spin. It's like an arrow with one set of flights behind the other. It isn't spinning like a bullet. However something else is. $\endgroup$ – John Duffield Dec 1 '16 at 22:49
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Here is a possible elementary and almost completely classical answer to my own question, but I don't know if it's right.

Hnizdo 2011 discusses the field of a dipole moving at $v\ll c$. He gives references to papers that discuss the ultrarelativistic case, but those are all paywalled. However, he points out that the electric and magnetic polarizations $(-\textbf{P},\textbf{M})$ transform in exactly the same way as the fields $(\textbf{E},\textbf{B})$. This means that in the special case of a Lorentz boost with $\textbf{v}\parallel\textbf{M}$, $\textbf{M}$ is invariant. Suppose we have a uniformly polarized body with some volume, and we do a Lorentz transformation out of the body's rest frame, parallel to the polarization. The polarization stays the same, but the volume shrinks by a factor of $\gamma$ due to Lorentz contraction. Therefore the dipole moment is reduced by a factor of $\gamma$, $\textbf{m}'=\textbf{m}/\gamma$, relative to the rest frame. I'm not completely confident of this reasoning, but it does agree with Hnizdo's low-velocity limit, which says that for motion parallel to the dipole, the moment is not affected to first order in the velocity.

Now let the dipole have mass $m$. In the dipole's rest frame, there is no preferred orientation other than the one set by the dipole moment $\textbf{m}$, and therefore it's not possible to have any constraint on the direction of $\textbf{m}$. But in the limit $m\rightarrow 0$, the dipole is required to move at the speed of light, so the component of the dipole moment $\textbf{m}'_{\parallel}$ goes to zero. This means that a massless dipole must have its dipole moment perpendicular to its direction of motion. The result is purely classical, and my argument (assuming it's right) is valid regardless of the nature of the object.

Since photons are massless, this means that if a photon did have a dipole moment, it would have to be oriented perpendicular to the photon's direction of motion. But that seems implausible for symmetry reasons: the spin is parallel to the direction of motion, so within the plane perpendicular to the motion, there is no preferred direction for the dipole moment.

Since there's essentially no quantum mechanics in this argument, I doubt that it's capable of telling us anything about the anomalous dipole moment.

[EDIT] After some discussion in comments on this question, it sounds to me like there must be a hole in this argument in the case of the magnetic dipole, although it still seems correct for an electric dipole.

Hnizdo and McDonald, "Fields and Moments of a Moving Electric Dipole," 2011, http://www.physics.princeton.edu/~mcdonald/examples/movingdipole.pdf

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  • $\begingroup$ I don't understand the symmetry argument. Why couldn't it have a moment perpendicular to motion? The emitter might not be symmetrical as well. $\endgroup$ – fffred Aug 19 '13 at 20:49
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    $\begingroup$ @fffred: It's not an ironclad symmetry argument. It would just be extremely strange if the photon were the only particle whose dipole moment was an independent degree of freedom, rather than being determined by its spin. $\endgroup$ – Ben Crowell Aug 19 '13 at 21:27
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If you want to see what is the magnetic moment of a particle, you have to couple it with an electromagnetic current. Now, suppose you have the photon wave functions $\Psi_{\mathbf{p}^{\prime}, \sigma^{\prime}}$ and $\Psi_{\mathbf{p}, \sigma}$ and you couple them to the electromagnetic current $J^{\mu}(0)$. The matrix element:

$$ (\Psi_{\mathbf{p}^{\prime}, \sigma^{\prime}}, J^{\mu}(0)\Psi_{\mathbf{p}, \sigma}) $$

is identically zero because of charge quantum number conservation. The lhs has charge quantum number -1 while the rhs has +1. So, there is no magnetic moment for the photon.

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  • $\begingroup$ What does the notation $J(0)$ mean? The lhs has charge quantum number -1 Wouldn't it be zero? $\endgroup$ – Ben Crowell Aug 19 '13 at 19:00
  • $\begingroup$ It is the electromagnetic current (see, for instance, Weinberg's QFT book, chapter 10). $\endgroup$ – Rafael Aug 19 '13 at 20:58
  • $\begingroup$ Wouldn't J be the current operator? What does the (0) mean? Please also see the second question in my first comment. $\endgroup$ – Ben Crowell Aug 19 '13 at 21:22
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    $\begingroup$ Of course in some sense it's utterly trivial that the photon doesn't have a dipole moment. For example, I can write down a vacuum solution of Maxwell's equations, and because it's a vacuum solution there's no charge anywhere, and the dipole moment is zero. The question is not so much how do we prove that its dipole moment is zero within a certain fixed theory as why it's very implausible that any theory could produce a nonzero dipole moment. $\endgroup$ – Ben Crowell Aug 19 '13 at 21:30
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Firstly I want to comment what user27777 wrote in his comment

But the best I can come up with is that the spin of the electron and photon are fundamentally different. The electron's corresponds to magnetic dipole moment while the photon's corresponds to angular momentum.

The definitions of spin for fermions and for photons are really different.

The spin of fermions is related to the magnetic dipole moment which possesses this particle. TEstablishing this relationship has to do with two phenomena:

  • Moving particles get deflected in a magnetic field (Lorentz force) and this is an interaction between the external magnetic field and the intrinsic magnetic dipole moment and the cyclic emission of photons during alignment and disalignement of the magnetic dipole moment. See How and why do accelerating charges radiate electromagnetic radiation?. At a first glance a spin has to be responsible for the deflection but at a second view it is obvious that the existence of the magnetic dipole moment and the emission of photons is enougth to describe all phenomenons of the EM induction.
  • Pauli's exclusion principle states that in an atom two fermions cannot occupy the same quantum state and the minimal difference in their states has to be their spin quantum number. To translate this in modern language it means that tow electrons in a helium atom will be flipped by their magnetic dipole moments (spin up and spin down is a little bit confusing because it means only that the two particles have opposite spins (magnetic dipole moments) to each other but this pair of spins in a amorphous material or in a gas is directed randomly in every direction in 3D. $^1)$

So the spin of a fermion is nothing else as their magnetic dipole moment.

The spin of photons has really only two values. The electric field component E and the magnetic dipole moment B of the photon both have a direction (in vacuum of course directed both perpendicular to the direction of propagation and both perpendicular on each other) and a sign. Taking a arbitrary photon and aligning our reference coordinate system so that the photons electric field component shows up there are two and only two possible directions of the magnetic dipole moment: to the left or to the right. This is what we call - I think it is confusing and not very helpful but fully established and not to undo any more - the spin of a photon.

So is there an elementary argument that explains why the photon has a zero magnetic moment?

It hasn't a zero magnetic dipole moment only from a statistical point of view over time. Having an external magnetic field oscillating with the frequency of this photon the photon will be deflected.

$^1)$ To make it more complicated I have to point out that for the Argon a distribution of his 8 electrons in one shell with 4 electrons with spin down and 4 electrons with spin up makes a lot of sense (Where do symmetries in atomic orbitals come from?).

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Here's a possible elementary argument - let me know what you think: if photons had magnetic dipole moments, then photons with parallel-aligned dipole moments would repel and photons with antiparallel-aligned moments would attract, due to their dipole-dipole interactions. This interaction between photons would mess up the usual spread-of-field-lines argument for the $1/r^2$ falloff in Coulomb's law, so Coulomb's law would need to be modified. Morever, an direct interaction between photons would lead to a violation of the superposition principle for electromagnetic fields.

Basically the same argument implies that photons must be completely uncharged, in the sense that all of their electric and magnetic multipole moments must vanish (not just the monopole moments). At the mathematical level, there are no coupling terms for the $A_\mu$ gauge field in the QED Lagrangian (beyond quadratic) to lead to direct interactions between photons.

(Of course, as with all "why" questions in physics, there's a bit of a chicken-and-egg issue here. Does Coulomb's law hold because the photon is uncharged, or is the photon uncharged because Coulomb's law holds? That's basically a philosophical question.)

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