# Derivation of Bohr magneton; how is the angular momentum derived if not from spin?

I'm trying to understand the derivation of the Bohr magneton $m_B$. The Bohr magneton is a calculation of the magnetic moment of a charged particle with some angular momentum. Generally, magnetic moment has the following formula:

$$m_B = IA$$

where I is the current and A is the area enclosed. In the case of $m_B$, the particle is simply spinning, so the standard geometry of a circle applies. The particle has charge $q$ and rotates around the circle in $\frac{v}{2 \pi r}$, which makes the current $I=\frac{qv}{2\pi r}$. Multiplying by the area and $1=\frac{m}{m}$ we can show the following relationships:

$$m_b = \frac{qv}{2 \pi r}\left(\pi r^2\right)\left(\frac{m}{m}\right)$$ $$= \frac{qvmr}{2m}$$ $$= \frac{q}{2m}(mvr) = \frac{q}{2m}(\textrm{ angular momentum })$$

Let's apply this to the electron, which has mass $m_e$, charge $e$, and spin $\frac{1}{2}\hbar$. We can calculate the magnetic moment to be:

$$m_B = \frac{e}{2m_e}\left(\frac{1}{2}\hbar\right) = \frac{e\hbar}{4m_e}$$

Conceptually this makes sense to me. I thought this was correct, until I looked up the correct answer and found it is actually:

$$m_B = \frac{e\hbar}{2m_e}$$

along with this statement, which confused me:

Although the spin angular momentum of an electron is $\frac{1}{2}\hbar$, the intrinsic magnetic moment of the electron caused by its spin is still approximately one Bohr magneton.

Where did the factor of two go? Here I assume the spin is equivalent to the net angular momentum...

In quantum mechanics and particle physics, spin is an intrinsic form of angular momentum carried by elementary particles

Source.

...which might be where my conceptual hole lies.

Is this the case? How is the $\hbar$ angular momentum derived?

• The Bohr atom has a tense and deeply conflicted relationship with angular momentum. Its ground state has non-vanishing angular momentum, but one knows from experiment that the actual ground state angular momentum is zero. There is a good explanation in terms of phase space quantum mechanics, but this is beyond the scope of this question. – Cosmas Zachos Oct 6 at 21:35
• Check out en.wikipedia.org/wiki/… . The answer to your question basically boils down to: that's just the way electrically charged, spin-1/2 particles couple to magnetic fields. It can be derived from the Dirac equation, with additional quantum corrections to the factor of 2 calculable from loop diagrams in QED. – d_b Oct 6 at 21:40

The Bohr magneton is the magnitude of the magnetic dipole moment of an orbiting electron with an orbital angular momentum of ħ.

and

Although the spin angular momentum of an electron is 1/2 ħ, the intrinsic magnetic moment of the electron caused by its spin is still approximately one Bohr magneton. The electron spin g-factor is approximately two.

• Hi! I read that too, and I quoted it in my question, but I didn't understand what it meant. Why is the orbital angular momentum $\hbar$ if the spin is $1/2\hbar$? What is the difference between spin and orbital angular momentum? – Joseph Farah Mar 21 '18 at 20:26
• In fact, this is one of the reasons for which spin does not have any classical analogue. So you can give up trying to "explain" spin as a classical $\vec{L}$, it's a queantum angualr momentum, and behaes liek an angualr momentum, but it hasn't a classical equivalent. What's more, you can show that classical angular momenta can only be integer multiples, while momenta in general can be semi-integers. – FGSUZ Mar 21 '18 at 20:27
• So where is the $\hbar$ angular momentum derived? – Joseph Farah Mar 21 '18 at 20:28
• Also, thank you for helping me to narrow the scope of my question. I updated my question and title accordingly. – Joseph Farah Mar 21 '18 at 20:29

There were some early ideas about magnetic moments being quantized, an elementary magnetic moment. Weiss thought that he saw some evidence for it, his value was about one fifth of $$\mu_B$$.

In 1920, Pauli wrote a review article about magnetism, where he called the experimental number the Weiss magneton and the theoretical magnetic moment of the current in the Bohr atom the Bohr magneton.

It is a bit of a coincidence that Pauli's Bohr magneton is approximately equal to the value of the intrinsic spin moment of the electron; in 1920 nobody knew about this.

Ah it's correct till the angular momentum part. From Bohr's theory, angular momentum of an electron, $$L= nh/2π$$ For electron in hydrogen atom, $$n=1$$ $$L= h/2π=\hbar$$

So, magnetic moment $$M = eh/4πm$$ (where $$e$$ is charge and $$m$$ is mass of electron). And in condensed form, the Planck's constant is $$\hbar$$.

So it becomes, $$M = e\hbar/2m$$.

Hope I was able to explain.

I believe this is what you are looking for:

https://en.wikipedia.org/wiki/G-factor_(physics)

There is a difference of (approximately) two due to special relativity/quantum field theory.

The $$g$$ factor is the ratio between the naive value of the magnetic moment and the actual value. $$g \approx 2$$.

Originally people didn't know why this 2 was showing up and put it in by hand. The first person to theoretically "explain" the 2 was Dirac, with his famous Dirac equation. The Dirac equation is the correct equation wave of the electron that appropriately takes special relativity into account. Indeed, after working it out, Dirac found that his equation did predict that this factor of 2 should be present.

However, $$g$$ is not exactly 2. It is actually a little bit more than two. The reason for this comes from quantum field theory. In a fuzzy sense, the electron interacts with it's own magnetic field via the emission and absorption of virtual photons. In quantum electrodynamics, people usually approximate answers by going to higher and higher powers in $$\alpha$$, the fine structure constant, which is a dimensionless measure of the strength of the electromagnetic interaction. $$\alpha = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{\hbar c} \approx \frac{1}{137}.$$ Through a very complicated calculation in quantum field theory, Schwinger found the correction to the $$g$$ factor, to the first order in $$\alpha$$, is $$g \approx 2 + \frac{\alpha}{\pi} + \mathcal{O}(\alpha^2).$$

While Schwinger did not use Feynman diagrams, this result can be understood as coming from the following diagram: