Physics Bootcamp

Subsection 40.2.1 Orbital Angular Momentum and Magnetic Dipole Moment

Here, I will use fictitious current loop model of a magnetic dipole moment to show that when an electron moves in an orbit, it will develop a magnetic dipole moment due to its motion. The current loop model states that magnetic moment can be written as a product of current and an area of a fictitious closed circuit.

Suppose the circuit is formed by an electron moving uniformly with speed \(v\) in a circle of radius R. Therefore, \(I\) and \(A\) will be

This will correspond to a magnetic dipole moment

In this formula, we can replace \(vR\) by angular momentum \(L\) divided by mass of electron, \(m_e\text{.}\)

We denote this dipole moment by \(\mu_l\text{,}\) to indicate that it is from the angular momentum due to the orbital motion.

In general, for any charged particle, we expect that there will be a magnetic moment due to its orbital angular momentum \(L\text{,}\) with the proportionality depending on its charge \(q\) and mass \(m\text{.}\)

The quantity \(|q|/2m \) is called the gyromagnetic ratio of the particle, usually denoted by \(g_q\text{,}\) where the subscript is an identifier for the particle.

For an electron the value of gyromagnetic ratio is

For a proton, it is about \(1/2000\) times smaller due to the mass of proton being about \(2000\) times larger.

Since, conventional current is in the opposite direction to the motion of electron, the direction of magnetic moment vector of an electron will be in the opposite direction to its angular momentum. Thus, we will write the vector relation as

It turns out that in SI units, the value of \(\mu\) due to electron's motion in atoms are very tiny numbers. More commonly, we express it in a unit called Bohr magneton, denoted by \(\mu_B\text{.}\)

where \(\hbar\) is Planck constant \(h\) divided by \(2\pi\text{.}\)

In terms of \(\mu_B\) the magnetic dipole moment of electron that has angular momentum \(\vec L\) will be

In atoms, angular momentum \(L\) is usually a small multiple of \(\hbar\text{,}\) and hence, magnetic dipole moments are a small multiple of \(\mu_B\text{.}\)

Classically, the magnitude of angular momentum \(L\) can be any positive real number with units, of course. But, quantum mechanically, \(L\) can only have certain values given by the following formula.

That is, only \(L=0\text{,}\) \(L = \sqrt{2}\, \hbar\text{,}\) \(L = \sqrt{6}\, \hbar\text{,}\) \(L = \sqrt{12}\, \hbar\text{,}\) etc are allowed. Therefore, magnetic dipole moment will also be quantized in the same way and will have only certain values.

Furthermore, when we look at one of its Cartesian components, it can be integral values of \(\hbar\text{.}\)

This says that magnetic moment along a Cartesian axis will be integral multiples of a Bohr magneton.

Subsection 40.2.2 Spin And Magnetic Dipole Moment

Every electron has an intrinsic angular momentum called spin regardless of the state of motion of the electron. The name “spin” for this property of an electron is a misnomer and suggests that the electron is in rotational motion about an axis through its center. If that were the case, angular momentum of such a spinning motion will be zero since moment of inertia about axis through a point particle would be zero. Hence, spin of an electron is best regarded as an intrinsic property and not to be taken literally.

The spin property accounts for the fact that electrons appear to have additional angular momentum than one can account for based only on the motion of the electron.

Unlike the magnitude of the orbital angular momentum, which can take many values, depending upon integer \(l\) in Eq. (40.2.9), a measurement of the magnitude of the spin angular momentum of an electron, to be denoted by \(\vec S\text{,}\) results in only one value for the magnitude. For an electron, the magnitude \(S=\frac{\sqrt{3}}{2}\hbar\text{.}\) In general, for any particle, we can write the magnitude \(S\) as

where \(s\) is called the spin of the particle. For electron

Proton and neutrons also have \(s=\frac{1}{2}\text{.}\) As a matter of fact, \(s\) can only be positive integer multiples of \(\frac{1}{2}\text{.}\) Particles with odd multiples are called Fermions and 0 and even multiples are called Bosons.

A measurement of the spin angular momentum of an electron along any direction, say the \(z\)-axis, results in only the following two values.

similar to the way \(L_z\) works out, except now the value is \(1/2\) instead of integers.

Just as the orbital angular momentum creates a magnetic dipole moment for the electron, spin angular momentum creates additional magnetic dipole moment for the electron. However, the gyromagnetic ratio of the spin angular momentum is not the same as the gyromagnetic ratio of the orbital angular momentum. Actually, for electrons, each unit of spin angular momentum is actually about twice as effective in creating magnetic moment as orbital angular momentum.

The factor by which you will need to multiply gyromagnetic ratio of orbital angular momentum to get the g-factor of spin angular momentum is called the \(g\) factor. It is often denoted by \(g_S\text{.}\)

\(g_S\) is one of most precisely known number in physics with reported value at NIST in May/2020 to be

We will just use \(g_S=2.0\) for electron. I gave you the more precise value for you to marvel at how many decimals have actually been measured! The \(g\) factor of protons, neutrons, and other particles are different.

Therefore, magnetic moment contribution of the spin angular moment will be

with Cartesian components given by

and total spin magnetic moment to be

This much magnetic moment will be produced by orbital angular momentum between \(l=2\) and \(l=3\text{.}\)