Intrinsic Spin

Tuesday, March 8, 2022

Magnetic Dipole Moments

The separate components of the angular momentum vector, $\boldsymbol{\overrightarrow{L}}$ , and $l$ of an atom can be determined from the interactions between an external magnetic field and the atom's magnetic dipole moment. However, performing this experiment will results in an unexpected property of the electron, called intrinsic spin.

Orbital magnetic dipole moments

A classical magnetic dipole moment can be produced by a current loop or the orbital motion of a charged object. The magnetic dipole moment, denoted $\boldsymbol{\overrightarrow{\mu}}$ , is a vector whose magnitude is equal to the product of the circulating current, $i$ , and the area enclosed by the orbital loop. The direction of the vector is perpendicular to the orbital plane, using the right hand rule with the direction of conventional current.

Relation to the angular momentum vector, $\boldsymbol{\overrightarrow{L}}$

Since quantum mechanics forbids exact knowledge about the angular momentum vector, it also forbids exact knowledge of the magnetic dipole moment vector. Only the $z$ components of these vectors can be known exactly. Note: since the electron has negative charge, $\boldsymbol{\overrightarrow{L}}$ and $\boldsymbol{\overrightarrow{\mu}}$ point in opposite directions.

Using the Bohr circular orbit model (which happens to be consistent with the quantum mechanical reality), we have a loop of current $i=dq/dt=q/T$ , where $q$ is the charge of the particle ( $-e$ for an electron) and $T$ is the time for one orbit. Assuming the electron moves with $v=p/m$ around a loop of radius $r$ , its $T$ value will be $T=2\pi r/v=2\pi rm/p$ . This means

$\mu=iA=\frac{q}{2\pi rm/p}\pi r^2=\frac{q}{2m}rp=\frac{q}{2m}\left|\boldsymbol{\overrightarrow{L}}\right|$

since $\left|\boldsymbol{\overrightarrow{L}}\right|=rp$ . In vector notation, we get

$\boldsymbol{\overrightarrow{\mu}}_{\text{L}}=-\frac{e}{2m}\boldsymbol{\overrightarrow{L}}$

Note: the subscript on the dipole vector is a reminder that the vector comes from the orbital angular momentum $\boldsymbol{\overrightarrow{L}}$ . The $z$ component of the vector is

$\mu_{\text{L,z}}=-\frac{e}{2m}L_z=-\frac{e}{2m}m_l\hbar=-\frac{e\hbar}{2m}m_l=-m_l\mu_{\text{B}}$

where

$\mu_{\text{B}}=\frac{e\hbar}{2m}$

This quantity, $\mu_{\text{B}}$ , is called the Bohr magneton and is equal to about $9.274\times10^{-24}\text{J/T}$ .

Dipoles in External Fields

An electric dipole is two equal but opposite charges $q$ separated by a distance of $r$ . The electric dipole moment, denoted $\boldsymbol{\overrightarrow{p}}$ , has a magnitude equal to $qr$ and points from the negative charge to the positive one.

In a uniform external electric field, vertical forces $\boldsymbol{\overrightarrow{F}}_+$ and $\boldsymbol{\overrightarrow{F}}_-$ act on the charges. While the net force on the dipole is $0$ , there is a torque applied to the system to make the dipole align with the field.

If the field is not uniform, the forces will not be equal, meaning there will be a net force on the dipole in addition to the torque. Another way of describing this is that if the dipole has $p_z\gt 0$ (assuming the electric field is pointing in the $z$ direction), its force will be in the negative $z$ direction and vice versa.

The same is true for a magnetic field and a magnetic dipole moment. A nonuniform magnetic field acting on the magnetic moments gives an unbalanced force that causes a displacement. In fact, if $\mu_z$ is positive, the force on the dipole is negative, and if $\mu_z$ is negative, the force on the dipole is positive.

The Stern-Gerlach Experiment

The setup

Imagine a beam of hydrogen atoms in the $n=2$ , $l=1$ state ( $m_l$ can be $-1$ , $0$ , or $+1$ ) incident on a screen. They pass through a slit then a nonuniform external magnetic field before hitting the screen. Assuming the experiment can be done before the atoms decay to the $n=1$ state, we would expect three lines to appear on the screen.

There should be one undeflected line for the $m_l=0$ atoms since they do not have a magnetic moment $\mu$ . The atoms with $m_l=+1$ have $\mu_{\text{L,z}}=-\mu_{\text{B}}$ , so they should be deflected upwards while the atoms with $m_l=-1$ have $\mu_{\text{L,z}}=\mu_{\text{B}}$ , so they should be deflected downwards.

In short, we should always expect an odd number of lines on the screen (representing different $m_l$ values) since there are always $2l+1$ values $m_l$ can take on.

Experimental results

Actually performing the experiment with hydrogen atoms in the $l=1$ state results in six lines appearing on the screen! Perhaps more confusingly, doing the experiment with atoms in the $l=0$ state results in two lines rather than the predicted one. In the $l=0$ state, $\boldsymbol{\overrightarrow{L}}$ has length $0$ , so there should be no magnetic moment. However, this cannot be true since the atoms are deflected!

What is going on here?

Following from the Schrödinger equation, atoms would need to have $l=1/2$ in order to produce two images. This is not true from how we have defined angular momentum thus far, but we can define another contributor to resolve the issue.

The intrinsic angular momentum of an electron is the hidden factor producing this oddity! An electron therefore has two kinds of angular momentum: its orbital angular momentum, $\boldsymbol{\overrightarrow{L}}$ , and its intrinsic angular momentum, $\boldsymbol{\overrightarrow{S}}$ . This new term, $\boldsymbol{\overrightarrow{S}}$ , is typically called the spin of the electron.

Spin

To resolve the strange results of the Stern-Gerlach experiment, we assign the spin quantum number, $s$ , of an electron to $1/2$ . With this spin, we also have the angular momentum vector $\boldsymbol{\overrightarrow{S}}$ , a $z$ component, $\boldsymbol{S}_z$ , an associated magnetic moment, $\boldsymbol{\overrightarrow{\mu}}_{\boldsymbol{S}}$ , and a spin magnetic quantum number, $m_S$ .

	Orbital	Spin
Quantum number	$l=0,1,2,...$	$s=1/2$
Length of vector	$\sqrt{l\left(l+1\right)}\hbar$	$\sqrt{s\left(s+1\right)}\hbar=\sqrt{3/4}\hbar$
$z$ -component	$L_z=m_l\hbar$	$S_z=m_s\hbar$
Magnetic quantum number	$m_l=0,\pm 1, \pm 2,...,\pm l$	$m_s=\pm 1/2$
Magnetic moment	$\boldsymbol{\overrightarrow{\mu}}_{\text{L}} = -\left(e/2m\right)\boldsymbol{\overrightarrow{L}}$	$\boldsymbol{\overrightarrow{\mu}}_{\text{S}} = -\left(e/m\right)\boldsymbol{\overrightarrow{S}}$

Using spin, we have an explanation for the results of the Stern-Gerlach and similar experiments. Every fundamental particle has a characteristic intrinsic spin and corresponding spin magnetic moment. The proton and neutron also have spin $1/2$ and the photon has a spin of $1$ . Other particles such as pions (pi mesons) has spin $0$ .