Feynman starts his Volume of *Lectures *on quantum mechanics (so that’s Volume III of the whole series) with the rules we already know, so that’s the ‘special’ math involving probability *amplitudes*, rather than probabilities. However, the particles in the introductory chapters are all so-called *zero-spin *particles, which means they are not supposed to have any *angular *momentum.

That’s OK, because it really makes it *much easier *to understand the basics of quantum math. However, *real-life *elementary particles do have angular momentum, and so we have to come to terms with it. Therefore, Feynman makes it *very *clear he expects all prospective readers of his third volume to first work their way through chapter 34 and 35 of the second volume, which discusses the angular momentum of elementary particles from both a classical as well as a quantum-mechanical perspective, and so **that’s what we will do here**.

Now, while the mentioned chapters are more generous with text than other textbooks on quantum mechanics I’ve looked at, the matter remains obscure. By way of introduction, Feynman writes the following:

“The behavior of matter on a small scale—as we have remarked many times—is different from anything that you are used to and is very strange indeed. Understanding of these matters comes very slowly, if at all. One never gets a comfortable feeling that these quantum-mechanical rules are ‘natural’. Of course they *are*, but they are not natural to our own experience at an ordinary level. The attitude that we are going to take with regard to this rule about angular momentum is quite different from many of the other things we have talked about. We are not going to try to ‘explain’ it but *tell* you what happens.”

So… Well… Let’s go for it. :-) When discussing electromagnetic radiation, we introduced the concept of atomic oscillators. It was a very useful model to help us understand. Now we’re going to introduce *atomic magnets*. It is based on the classical idea of an electron orbiting around a proton. Of course, we *know *this classical idea is *wrong*: we don’t have nice circular electron orbitals, and our discussion on the radius of an the electron in our previous post makes it clear that the idea of the electron itself is rather fuzzy. Nevertheless, the classical concepts used to analyze *rotation* are also used, ** mutatis mutandis** – i.e. with necessary alterations –

*in quantum mechanics. In this post, we want to focus on these alterations, or*

*modifications*of the classical theory. In line with Feynman’s introductory remarks, we’ll focus on the

*how*mainly—i.e.

*not*on the

*why*. So… Well… Let’s go for it. :-)

The basic idea is the following: an electron in a circular orbit is a circular current and, hence, it causes a magnetic field, i.e. a magnetic flux through the area of the loop—as illustrated below.

As such, we’ll have a magnetic (dipole) moment, and you may want to review my post(s) on that topic so as to ensure you understand what follows. The magnetic moment (μ) is the product of the current (I) and the area of the loop (i.e. π·r^{2}), and its conventional direction is given by the **μ** *vector* in the illustration below, which also shows the other relevant scalar and/or vector quantities, such as the velocity **v** and the *orbital* angular momentum **J**. The orbital angular momentum is to be distinguished from the *spin *angular momentum, which results from the spin *around its own axis*, but the spin angular momentum – which is often referred to as the spin* tout court* – is *not *depicted below, and will only be discussed in a few minutes. Hence, the focus is first on the orbital angular momentum **J** and the related magnetic moment **μ**.

The magnetic moment is the current times the area of the loop. As the velocity is constant, the current is just the electron charge q times the frequency of rotation. The frequency of rotation is, of course, the velocity (i.e. the distance traveled per second) divided by the circumference of the orbit (i.e. 2π·r). Hence, we write: I = (q·v)/(2π·r) and, therefore: μ = (q/·v)·π·r^{2})/(2π·r) = q·v·r/2.

The orbital angular momentum is the angular momentum we discussed in our post on gyroscopes. We denoted the angular momentum as **L**, and noted **L** could be calculated as the vector cross product of the position vector **r** and the momentum vector **p**, as shown in the animation below, which also shows the *torque *vector **τ**.

Unlike in the animation above, the angular momentum of the electron in circular orbit will remain constant, and its magnitude is equal to |**J**| = J = |**r**×**p**| = |**r**|·|**p**|·sinθ = r·p = r·m·v. One should note this is a non-relativistic formula, but as the relativity velocity of an electron v/*c* = α ≈ 0.0073 (see my post on the fine-structure constant if you wonder where this formula comes from), it’s OK to *not *include the Lorentz factor in our formulas as for now.

Now, since **μ** and **J** are in the same direction in this case (both are perpendicular to the plane of the orbit), we can combine the J = r·m·v and μ = q·v·r/2 formulas to write:

**μ** = (q_{e}/2m)·**J **or **μ/****J **= (q_{e}/2m) (electron *orbit*)

In other words, the *ratio *of the magnetic moment and the angular moment depends on (1) the charge (which we’ll denote by q_{e }as we’re talking an electron here, so we can reserve the q symbol to cover the general case) and (2) the mass of the electron *only**—not *on the velocity v nor on the radius r. It can be noted that the q/2m factor is often referred to as the gyromagnetic factor (not to be confused with the g-factor, which we’ll introduce shortly). It’s good to do a quick dimensional check of this relation: the magnetic moment is expressed in *ampère *per *second* times the loop *area*, so that’s (C/s)·m^{2}. On the right-hand side, we have the dimension of the gyromagnetic factor, which is C/kg, times the dimension of the angular momentum, which is m·kg·m/s, so we have the same units on both sides: C·m^{2}/s, which is often written as *joule *per *tesla* (J/T): the *joule *is the energy unit (1 J = 1 N·m), and the *tesla* measures the strength of the magnetic field (1 T = 1 (N·s)/(C·m). OK. So that works out.

So far, so good. The story is a little bit different for the *spin *angular momentum and the *spin *magnetic moment. The formula is the following:

**μ** = (q_{e}/m)·**J** (electron *spin*)

This formula says that the **μ**/**J** ratio is *twice *what it is for the orbital motion of the electron. Why is that? Frankly, I don’t know. I’ll try to find out. There must be some similar derivation as the one we did for the *orbital* angular moment but I haven’t looked for it. Feynman avoids the question altogether in his *Lectures *so I must assume the derivation is *not *so easy. Let’s just go with it as for now. * *

Now, the *total *magnetic moment and angular momentum is obviously the sum of both, and we will also want to replace the proton by a full-blown nucleus, so we can calculate the magnetic moment and the angular momentum for a whole atomic *system *involving more than just one electron. In short, we’ll want to have a more general formula relating **μ **and **J**. The more general formula is written as:

**μ** = –g·(q_{e}/2m)·**J**

Why the minus sign? Well… We need *some *convention, because we’ll have positive and negative charges and all that and… Well… It’s just convention. And why q_{e}/2m instead of q_{e}/m in the middle? Well… If we’d take q_{e}/m, then g would be −1/2 for the *orbital *angular momentum, and the initial idea with g was that it would be some integer (we’ll quickly see that’s an* idea* only). So… Well… It’s just one more convention. For our example involving the *spin *angular momentum and the *spin *magnetic moment, g will obviously have to be –2 so as to yield that **μ** = (q_{e}/m)·**J** formula.

OK. That’s clear enough. For electrons, the g-factor is referred to as **the Landé g-factor**. There is a similar g-factor for protons, which is referred to as the ** nuclear g-factor**. In fact, there is a g-factor for neutrons too, despite the fact that they do not carry a

*net*charge: the explanation for it must have something to do with the quarks that make up the neutron but that’s a complicated matter which we will

*not*get into here. Of course, there is

**a g-factor**for a whole atom or an atomic system, and it’s one of the numbers that

**is**

**characteristic of the state of the atom**.

Of course, we’re talking quantum mechanics and, therefore, J can only take on a *finite* number of values. That should not surprise us at all, because our discussion on the fine-structure constant (α) made it clear that the various radii of the electron, its velocity and its mass and/or energy are all related one to another and, hence, they can only take on certain values. Indeed, of all the relations we discussed, there’s two you should *always *remember. The first relationship is the U_{ }= (*e*^{2}/r) = α/r. So that links the energy (which we can express in equivalent mass units), the electron charge and its radius. The second thing you should remember is that the Bohr radius and the classical electron radius are also related through α: α r_{e}/r = α^{2}. These relationships suggest that the different values for J are associated with different orbitals, and they are: remember the fine-structure constant first popped up in Arnold Sommerfeld’s 1916 explanation of the atomic spectral lines!

So it all makes sense. In fact, as you’ll see in a moment, the whole thing is not unlike the quantum-mechanical explanation of the blackbody radiation problem, which it solves by assuming that the permitted energy levels (or *states*) are equally spaced and h·*f* apart, with *f *the frequency of the light that’s being absorbed and/or emitted. So the atom takes up energies only h·*f *at a time. Here we’ve got something similar. If we have an object with a given total angular momentum J in *classical mechanics*, then any of its components x, y or z, could take on any value from +J to −J. That’s not the case here. The rule is that the ‘system’ – the atom, the nucleus, or anything really – will have a characteristic number, which is referred to as the ‘spin’ of the system. It’s denoted by j, and any component of **J** (think of the z-direction, for example) can then take on only the following values:

OK. So far so good. We sort of ‘get’ this – I hope! – but let’s pause for a moment and analyze this—just to make sure we ‘get’ this indeed. What’s being written here? What *are *those numbers? We know ħ: it’s the Planck constant h divided by 2π, so that’s something expressed in *joule·second* *per radian*. That makes sense, because angular momentum is also the product of the *moment of inertia *(I = r^{2}·m, so that’s in m^{2}·kg = N·m·s^{2 }units) and the *angular velocity *(Ω = dθ/dt, which is expressed in *radians per second*, and related to the tangential component of the velocity by the v_{t }= r·Ω), and so we can also express the angular momentum in N·m·s = J·s. In short, we’ve got the unit of *action *once more here. Of course, ħ = h/2π is even smaller than h: it’s about 1×10^{−34} J·s ≈ 6.6×10^{−16} eV·s. In order to get an idea of the order of magnitude, you may want to compare this with the energy of a photon, which is 1.6 to 3.2 eV in the visible light spectrum, but you should note that energy does not have the time dimension, and a second is an eternity in quantum physics, so the comparison is a bit tricky. So… Well… Let’s move on. What about those coefficients? What constraints are there?

Well… The constraint is that the *difference* between +j and −j must be some integer, so *twice *j must be an integer. That implies that the spin j is always an integer or a half-integer, depending on whether j is even or odd. Let’s do a few examples:

- A
*lithium*(Li-7) nucleus has spin j = 3/2 and, therefore, the permitted values for the angular momentum around any axis (the z-axis, for example) are: 3/2, 3/2−1=1/2, 3/2−2=−1/2, and −3/2. [Note that the difference between +j and –j is 3, and each ‘step’ between those two levels is one, as we’d like it to be.] - The nucleus of the much rarer Lithium-6 isotope is one of the few stable nuclei that has spin j = 1, so the permitted values are 1, 0 and −1. [So each step is ‘one’ again, and the total difference (between +j and –j) is 2.]
- An electron is a spin-1/2 particle, and so there are only two permitted values: +1/2 and −1/2. So there is just one ‘step’ and it’s equal to the whole difference between +j and –j. In fact, this is the most common situation, because we’ll be talking elementary fermions most of the time. In short, don’t take the ‘general’ formulas too seriously: the actual situations we’ll be analyzing are often quite simple.

Indeed, we already know about the fundamental dichotomy between *fermions *and *bosons*. Fermions have half-integer spin, and all *elementary *fermions, such as protons, neutrons, electrons, neutrinos and quarks are spin-1/2 particles. [Note that a proton and a neutron are, strictly speaking, *not *elementary, as their constituent parts are quarks.] Bosons have integer spin, and the bosons we know of are actually spin-zero particles. The photon is an example, but the helium nucleus (He-4) also has zero spin, which gives rise to superfluidity when its cooled near the absolute zero point. So, in practice, we’ll be dealing almost exclusively with spin-0 and spin-1/2 particles and, occasionally, with spin-1 and spin-3/2 particles.

We know need to learn how to do a bit of math with all of this.

**The magnetic energy of atoms**

Before we start, we should, perhaps, relate the angular momentum to the magnetic moment once again. We can do that using the **μ** = (q/2m)·**J **and/or **μ** = (q/m)·**J **formula (so that’s the simple formulas for the orbital and spin angular momentum respectively) or, else, by using the more general **μ** = – g·(q/2m)·**J **formula.

Let’s use the simpler **μ** = (q_{e}/2m)·**J** formula, which is the one for the *orbital *angular momentum. What’s q_{e}/2m? It should be equal to 1.6×10^{−19} C divided by 2·9.1×10^{−31 }kg, so that’s about 0.0879×10^{12 } C/kg, or 0.0879×10^{12 }(C·m)/(N·s^{2}). Now we multiply by ħ/2 ≈ 0.527×10^{−34} J·s. We get something like 0.0463×10^{−22 }m^{2}·C/s or J/T. These numbers are ridiculously small, so they’re usually measured in terms of a so-called *natural *unit: the **Bohr magneton**, which I’ll explain in a moment but so here we’re interested in its value only, which is μ_{B} = 9.274×10^{−24 }J/T. Hence, μ/μ_{B} = 0.5 = 1/2. *What a nice number!*

Hmm… This cannot be a coincidence… […] You’re right. It isn’t. To get the full picture, we need to include the *spin *angular momentum, so we also need to see what the **μ** = (q/m)·**J** will yield. That’s easy, of course, as it’s twice the value of (q/2m)·**J**, so μ/μ_{B} = 1, and so the *total *is equal to 3/2. So the magnetic moment of an electron has the same value (when expressed in terms of the Bohr magneton) as the spin (when expressed in terms of ħ). Now *that’*s just *sweet*!

Yes, it is. All our definitions and formulas were formulated so as to make it sweet. Having said that, we do have a tiny little problem. If we use the general **μ** = −g·(q/2m)·**J **to write the result we found for the *spin *of the electron only (so we’re *not *looking at the orbital momentum here), then we’d write: **μ** = 2·(q/2m)·**J** = (q/m)·**J** and, hence, the g-factor here is −2. *Yes. We know that. You told me so already. What’s the issue?* Well… The problem is: experiments reveal the actual value of g is ** not exactly** −2: it’s −2.00231930436182(52) instead, with the last two digits (in brackets) the uncertainty in the current measurements. Just check it for yourself on the NIST website. :-) [Please do check it: it brings some

*realness*to this discussion.]

Hmm…. The accuracy of the measurement suggests we should take it seriously, even if we’re talking a difference of 0.1% only. We should. It can be explained, of course: it’s something quantum-mechanical. However, we’ll talk about this later. As for now, just try to understand the basics here. It’s complicated enough already, and so we’ll stay away from the nitty-gritty as long as we can.

Let’s now get back to the magnetic *energy *of our *atoms*. From our discussion on the *torque *on a magnetic dipole in an *external* magnetic field, we know that our magnetic atoms will have some *extra *magnetic energy when placed in an *external* field. So now we have an *external *magnetic field **B**, and we derived the formula for the energy is

U_{mag }= −μ·B·cosθ = −**μ**·**B**

I won’t explain the whole thing once again, but it might help to visualize the situation, which we do below. The loop here is not circular but square, and it’s a current-carrying wire instead of an electron in orbit, but I hope you get the point.

We need to chose some coordinate system to calculate stuff and so we’ll just choose our z-axis along the direction of the external magnetic field** B** so as to simplify those calculations. If we do that, we can just take the z-component of **μ **and then combine the interim result with our general** μ** = – g·(q/2m)·**J **formula, so we write:

U_{mag }= −μ_{z}·B = g·(q/2m)·J_{z}·B

Now, we know that the *maximum *value of J_{z} is equal to j·ħ, and so the *maximum* value of U_{mag }will be equal g(q/2m)jħB. Let’s now simplify this expression by choosing some natural unit, and that’s the unit we introduced already above: the Bohr magneton. It’s equal to (q_{e}ħ)/(2m_{e}) and its value is μ_{B} ≈ 9.274×10^{−24 }J/T. So we get the result we wanted, and that is:

Let me make a few remarks here. First on that magneton: you should note there’s also something which is known as the *nuclear *magneton which, you guessed it, is calculated using the *proton *charge and the proton mass: μ_{N} = (q_{p}ħ)/(2m_{p}) ≈ 5.05×10^{−27 }J/T. My second remark is a question: what does that formula *mean*, really? Well… Let me quote Feynman on that. The formula basically says the following:

“The energy of an atomic system is changed when it is put in a magnetic field by an amount that is proportional to the field, and proportional to J_{z}. We say that the energy of an atomic system is ‘split’ into 2*j *+ 1 ‘levels’ by a magnetic field. For instance, an atom whose energy is U_{0} *outside* a magnetic field and whose j is 3/2, will have four possible energies when placed in a field. We can show these energies by an energy-level diagram like that drawn below. Any particular atom can have only one of the four possible energies in any given field **B**. That is what quantum mechanics says about the behavior of an atomic system in a magnetic field.”

Of course, the simplest ‘atomic’ system is a single electron, which has spin 1/2 only (like most fermions really: the example in the diagram above, with spin 3/2, would be that Li-7 system or something similar). If the spin is 1/2, then there are only two energy levels, with J_{z} = ±ħ/2 and, as we mentioned already, the g-factor for an electron is −2 (again, the use of minus signs (or not) is quite confusing: I am sorry for that), and so our formula above becomes very simple:

U_{mag }= ± μ_{B}·B

The graph above becomes the graph below, and we can now speak more loosely and say that the electron either has its spin ‘up’ (so that’s along the field), or ‘down’ (so that’s opposite the field).

So… Well… That’s it for today, I think. This was (or should be) a big step forward. We’ve got all of the basics on that ‘magical’ spin number here, and so I hope it’s somewhat less ‘magical’ now. :-) Let me just copy the values of the g-factor for some elementary particles. It also shows how hard physicists have been trying to narrow down the uncertainty in the measurement. Quite impressive! The table comes from the Wikipedia article on it. I hope the explanations above will now enable you to read and understand that. :-)