G-factor of electron, proton, and neutron don't cancel out

[LostConjugate]

If you add up the g-factor for the electron, the proton and the neutron it is not exactly zero. Doesn't this calculate to a magnetic moment for every atom in the universe proportional to it's mass.

 

[Bill_K]

LostConjugate, The full explanation of this is rather lengthy and technical, but in brief:

Number one, when you couple two or more particles together, you don't just add the g's. This in fact is where the idea of the Lande g factor originated.

Number two, if an atom or molecule has a nonzero magnetic moment, there isn't any great significance attached to that.

And number three, many atoms have total angular momentum zero. For a system with J = 0, it's impossible to have a nonzero magnetic moment.

 

[LostConjugate]

Thats right. Addition of quantum angular momentum involves some heavy algebra. Thanks

 

[SpectraCat]

Thats right. Addition of quantum angular momentum involves some heavy algebra. Thanks

Not to mention that the g-factors for nuclear particles are about 2000 times smaller than the g-factors for electrons. When you combine them (as in hyperfine splitting), you need to take this difference into account as well.

 

[Dickfore]

Not to mention that the g-factors for nuclear particles are about 2000 times smaller than the g-factors for electrons.

No, they are not.

 

[SpectraCat]

And number three, many atoms have total angular momentum zero. For a system with J = 0, it's impossible to have a nonzero magnetic moment.

No, I don't think that's right. J is usually reserved for the total electronic angular momentum (i.e. J=L + S, where L is orbital angular momentum, and S is spin angular momentum). So, as long as all the electrons are paired, J will always be zero, and their electronic magnetic moment will also be zero. However, there are certainly isotopes of atoms with all their electrons paired which have non-zero nuclear magnetic moments .. He-3 is the simplest, but there are lots ... most (all?) of the rare gases have NMR-active isotopes, as do the alkali-earth elements (Mg, Ca, Sr, etc.) and the group 12 metals (Zn, Cd, Mg).

 

[SpectraCat]

No, they are not.

Well, you are right in principle, since they are just dimensionless numbers. However, in the context of this thread, we are talking about hyperfine coupling of nuclear and electronic angular momenta. So, it is really that the nuclear magneton is 2000 times smaller than the Bohr magneton for electrons that is driving the effect.

However, if you want to calculate the overall g-factor for an atom using the usual equations, it will be almost exactly equal to the electronic g-factor, because you need to scale the nuclear contribution by ~1/2000. (See below)

F=J+I

$$g_F = g_J\frac{F(F+1) - I(I+1) + J(J+1)}{2F(F+1)} + g_I\frac{F(F+1) + I(I+1) - J(J+1)}{2F(F+1)}$$

For that equation, you need to scale the g_I term by ~1/2000, so that is why it is sometimes said that g_I for the nuclear spin is 2000 times smaller than g_J.

 

[Dickfore]

F=J+I

$$g_F = g_J\frac{F(F+1) - I(I+1) + J(J+1)}{2F(F+1)} + g_I\frac{F(F+1) + I(I+1) - J(J+1)}{2F(F+1)}$$

These equations are incorrect when applied to coupling of magnetic moments of two systems that have different ratios for $q/m$, as is the case with the nucleus and the electron cloud.

Furthermore, when the total angular momentum of the electrons $J = 0$, the magnetic moment of the whole atom is solely determined by the magnetic moment of the nucleus.

 

[SpectraCat]

These equations are incorrect when applied to coupling of magnetic moments of two systems that have different ratios for $q/m$, as is the case with the nucleus and the electron cloud.

Furthermore, when the total angular momentum of the electrons $J = 0$, the magnetic moment of the whole atom is solely determined by the magnetic moment of the nucleus.

Yup .. that's what I said.

 

[Dickfore]

I don't think that equation is valid, period. Can you give a reference for it?

 

[Bill_K]

No, I don't think that's right. J is usually reserved for the total electronic angular momentum (i.e. J=L + S, where L is orbital angular momentum, and S is spin angular momentum).

Fine, so whatever you want to call the total angular momentum of the atom, that's what I meant was zero.

 

[SpectraCat]

I don't think that equation is valid, period. Can you give a reference for it?

I don't have a reference handy .. I got that from memory. I am pretty sure I can derive it .. I will do that if I have time. (For what it's worth, it also seems to be backed up by Wikipedia, http://en.wikipedia.org/wiki/Landé_g-factor but the equation there is not referenced either).

Why don't you think it is valid (with the caveat that g_I needs to be scaled by the electron/proton mass ratio)? The hyperfine interaction is an angular momentum coupling, so the algebra should be the same. In the usual usage of the Lande g-factor for spin-orbit splitting, the g-factors for the two terms aren't the same .. the g-factor for the orbital angular momentum is 1, while the g-factor for the spin angular momentum is (about 2). The Lande g-factor obtained is then appropriately scaled to calculate spin-orbit splitting of multi-electron atoms with different L and S using the Bohr magneton. The version of the Lande g-factor I provided does exactly the same thing, except takes the (small) effect of the nuclear moment into effect as well, again scaling it so that it can be used with the Bohr magneton. This would allow one to predict small differences in the Zeeman splitting of lines from the same atoms, but different isotopes. Isn't that what is observed experimentally?

However, upon further consideration, I have to say that I guess is that this version of the g-factor is not really that useful in practice, because if you have sufficient resolution to see the difference in the splittings due to this g-factor, you are probably resolving the hyperfine splitting any way, and thus will probably be using the full hyperfine Hamiltonian to analyze the spectra.