# Transforming amplitudes for spin-1/2 particles

Some say it is not possible to fully understand quantum-mechanical spin. Now, I do agree it is difficult, but I do not believe it is impossible. That’s why I wrote so many posts on it. Most of these focused on elaborating how the classical view of how a rotating charge precesses in a magnetic field might translate into the weird world of quantum mechanics. Others were more focused on the corollary of the quantization of the angular momentum, which is that, in the quantum-mechanical world, the angular momentum is never quite all in one direction only—so that explains some of the seemingly inexplicable randomness in particle behavior.

Frankly, I think those explanations help us quite a bit already but… Well… We need to go the extra mile, right? In fact, that’s drives my search for a geometric (or physical) interpretation of the wavefunction: the extra mile. 🙂

Now, in one of these many posts on spin and angular momentum, I advise my readers – you, that is – to try to work yourself through Feynman’s 6th Lecture on quantum mechanics, which is highly abstract and, therefore, usually skipped. [Feynman himself told his students to skip it, so I am sure that’s what they did.] However, if we believe the physical (or geometric) interpretation of the wavefunction that we presented in previous posts is, somehow, true, then we need to relate it to the abstract math of these so-called transformations between representations. That’s what we’re going to try to do here. It’s going to be just a start, and I will probably end up doing several posts on this but… Well… We do have to start somewhere, right? So let’s see where we get today. 🙂

The thought experiment that Feynman uses throughout his Lecture makes use of what Feynman’s refers to as modified or improved Stern-Gerlach apparatuses. They allow us to prepare a pure state or, alternatively, as Feynman puts it, to analyze a state. In theory, that is. The illustration below present a side and top view of such apparatus. We may already note that the apparatus itself—or, to be precise, our perspective of it—gives us two directions: (1) the up direction, so that’s the positive direction of the z-axis, and (2) the direction of travel of our particle, which coincides with the positive direction of the y-axis. [This is obvious and, at the same time, not so obvious, but I’ll talk about that in my next post. In this one, we basically need to work ourselves through the math, so we don’t want to think too much about philosophical stuff.] The kind of questions we want to answer in this post are variants of the following basic one: if a spin-1/2 particle (let’s think of an electron here, even if the Stern-Gerlach experiment is usually done with an atomic beam) was prepared in a given condition by one apparatus S, say the +S state, what is the probability (or the amplitude) that it will get through a second apparatus T if that was set to filter out the +T state?

The result will, of course, depend on the angles between the two apparatuses S and T, as illustrated below. [Just to respect copyright, I should explicitly note here that all illustrations are taken from the mentioned Lecture, and that the line of reasoning sticks close to Feynman’s treatment of the matter too.] We should make a few remarks here. First, this thought experiment assumes our particle doesn’t get lost. That’s obvious but… Well… If you haven’t thought about this possibility, I suspect you will at some point in time. So we do assume that, somehow, this particle makes a turn. It’s an important point because… Well… Feynman’s argument—who, remember, represents mainstream physics—somehow assumes that doesn’t really matter. It’s the same particle, right? It just took a turn, so it’s going in some other direction. That’s all, right? Hmm… That’s where I part ways with mainstream physics: the transformation matrices for the amplitudes that we’ll find here describe something real, I think. It’s not just perspective: something happened to the electron. That something does not only change the amplitudes but… Well… It describes a different electron. It describes an electron that goes in a different direction now. But… Well… As said, these are reflections I will further develop in my next post. 🙂 Let’s focus on the math here. The philosophy will follow later. 🙂 Next remark.

Second, we assume the (a) and (b) illustrations above represent the same physical reality because the relative orientation between the two apparatuses, as measured by the angle α, is the same. Now that is obvious, you’ll say, but, as Feynman notes, we can only make that assumption because experiments effectively confirm that spacetime is, effectively, isotropic. In other words, there is no aether allowing us to establish some sense of absolute direction. Directions are relativerelative to the observer, that is… But… Well… Again, in my next post, I’ll argue that it’s not because directions are relative that they are, somehow, not real. Indeed, in my humble opinion, it does matter whether an electron goes here or, alternatively, there. These two different directions are not just two different coordinate frames. But… Well… Again. The philosophy will follow later. We need to stay focused on the math here.

Third and final remark. This one is actually very tricky. In his argument, Feynman also assumes the two set-ups below are, somehow, equivalent. You’ll say: Huh? If not, say it! Huh? 🙂 Yes. Good. Huh? Feynman writes equivalentnot the same because… Well… They’re not the same, obviously:

1. In the first set-up (a), T is wide open, so the apparatus is not supposed to do anything with the beam: it just splits and re-combines it.
2. In set-up (b) the T apparatus is, quite simply, not there, so… Well… Again. Nothing is supposed to happen with our particles as they come out of S and travel to U.

The fundamental idea here is that our spin-1/2 particle (again, think of an electron here) enters apparatus U in the same state as it left apparatus S. In both set-ups, that is! Now that is a very tricky assumption, because… Well… While the net turn of our electron is the same, it is quite obvious it has to take two turns to get to U in (a), while it only takes one turn in (b). And so… Well… You can probably think of other differences too. So… Yes. And no. Same-same but different, right? 🙂

Right. That is why Feynman goes out of his way to explain the nitty-gritty behind: he actually devotes a full page in small print on this, which I’ll try to summarize in just a few paragraphs here. [And, yes, you should check my summary against Feynman’s actual writing on this.] It’s like this. While traveling through apparatus T in set-up (a), time goes by and, therefore, the amplitude would be different by some phase factor δ. [Feynman doesn’t say anything about this, but… Well… In the particle’s own frame of reference, this phase factor depend on the energy, the momentum and the time and distance traveled. Think of the argument of the elementary wavefunction here: θ = (E∙t – px)/ħ).] Now, if we believe that the amplitude is just some mathematical construct—so that’s what mainstream physicists (not me!) believe—then we could effectively say that the physics of (a) and (b) are the same, as Feynman does. In fact, let me quote him here:

“The physics of set-up (a) and (b) should be the same but the amplitudes could be different by some phase factor without changing the result of any calculation about the real world.”

Hmm… It’s one of those mysterious short passages where we’d all like geniuses like Feynman (or Einstein, or whomever) to be more explicit on their world view: if the amplitudes are different, can the physics really be the same? I mean… Exactly the same? It all boils down to that unfathomable belief that, somehow, the particle is real but the wavefunction that describes it, is not. Of course, I admit that it’s true that choosing another zero point for the time variable would also change all amplitudes by a common phase factor and… Well… That’s something that I consider to be not real. But… Well… The time and distance traveled in the apparatus is the time and distance traveled in the apparatus, right?

Bon… I have to stay away from these questions as for now—we need to move on with the math here—but I will come back to it later. But… Well… Talking math, I should note a very interesting mathematical point here. We have these transformation matrices for amplitudes, right? Well… Not yet. In fact, the coefficient of these matrices are exactly what we’re going to try to derive in this post, but… Well… Let’s assume we know them already. 🙂 So we have a 2-by-2 matrix to go from S to T, from T to U, and then one to go from S to U without going through T, which we can write as RSTRTU,  and RSU respectively. Adding the subscripts for the base states in each representation, the equivalence between the (a) and (b) situations can then be captured by the following formula: So we have that phase factor here: the left- and right-hand side of this equation is, effectively, same-same but different, as they would say in Asia. 🙂 Now, Feynman develops a beautiful mathematical argument to show that the eiδ factor effectively disappears if we convert our rotation matrices to some rather special form that is defined as follows: I won’t copy his argument here, but I’d recommend you go over it because it is wonderfully easy to follow and very intriguing at the same time. [Yes. Simple things can be very intriguing.] Indeed, the calculation below shows that the determinant of these special rotation matrices will be equal to 1. So… Well… So what? You’re right. I am being sidetracked here. The point is that, if we put all of our rotation matrices in this special form, the eiδ factor vanishes and the formula above reduces to: So… Yes. End of excursion. Let us remind ourselves of what it is that we are trying to do here. As mentioned above, the kind of questions we want to answer will be variants of the following basic one: if a spin-1/2 particle was prepared in a given condition by one apparatus (S), say the +S state, what is the probability (or the amplitude) that it will get through a second apparatus (T) if that was set to filter out the +T state?

We said the result would depend on the angles between the two apparatuses S and T. I wrote: angles—plural. Why? Because a rotation will generally be described by the three so-called Euler angles:  α, β and γ. Now, it is easy to make a mistake here, because there is a sequence to these so-called elemental rotations—and right-hand rules, of course—but I will let you figure that out. 🙂

The basic idea is the following: if we can work out the transformation matrices for each of these elemental rotations, then we can combine them and find the transformation matrix for any rotation. So… Well… That fills most of Feynman’s Lecture on this, so we don’t want to copy all that. We’ll limit ourselves to the logic for a rotation about the z-axis, and then… Well… You’ll see. 🙂

So… The z-axis… We take that to be the direction along which we are measuring the angular momentum of our electron, so that’s the direction of the (magnetic) field gradient, so that’s the up-axis of the apparatus. In the illustration below, that direction points out of the page, so to speak, because it is perpendicular to the direction of the x– and the y-axis that are shown. Note that the y-axis is the initial direction of our beam. Now, because the (physical) orientation of the fields and the field gradients of S and T is the same, Feynman says that—despite the angle—the probability for a particle to be up or down with regard to and T respectively should be the same. Well… Let’s be fair. He does not only say that: experiment shows it to be true. [Again, I am tempted to interject here that it is not because the probabilities for (a) and (b) are the same, that the reality of (a) and (b) is the same, but… Well… You get me. That’s for the next post. Let’s get back to the lesson here.] The probability is, of course, the square of the absolute value of the amplitude, which we will denote as C+C, C’+, and C’ respectively. Hence, we can write the following: Now, the absolute values (or the magnitudes) are the same, but the amplitudes may differ. In fact, they must be different by some phase factor because, otherwise, we would not be able to distinguish the two situations, which are obviously different. As Feynman, finally, admits himself—jokingly or seriously: “There must be some way for a particle to know that it has turned the corner at P1.” [P1 is the midway point between and in the illustration, of course—not some probability.]

So… Well… We write:

C’+ = eiλ ·C+ and C’ = eiμ ·C

Now, Feynman notes that an equal phase change in all amplitudes has no physical consequence (think of re-defining our t0 = 0 point), so we can add some arbitrary amount to both λ and μ without changing any of the physics. So then we can choose this amount as −(λ + μ)/2. We write: Now, it shouldn’t you too long to figure out that λ’ is equal to λ’ = λ/2 + μ/2 = −μ’. So… Well… Then we can just adopt the convention that λ = −μ. So our C’+ = eiλ ·C+ and C’ = eiμ ·C equations can now be written as:

C’+ = eiλ ·C+ and C’ = eiλ·C

The absolute values are the same, but the phases are different. Right. OK. Good move. What’s next?

Well… The next assumption is that the phase shift λ is proportional to the angle (α) between the two apparatuses. Hence, λ is equal to λ = m·α, and we can re-write the above as:

C’+ = ei·C+ and C’ = ei·C

Now, this assumption may or may not seem reasonable. Feynman justifies it with a continuity argument, arguing any rotation can be built up as a sequence of infinitesimal rotations and… Well… Let’s not get into the nitty-gritty here. [If you want it, check Feynman’s Lecture itself.] Back to the main line of reasoning. So we’ll assume we can write λ as λ = m·α. The next question then is: what is the value for m? Now, we obviously do get exactly the same physics if we rotate by 360°, or 2π radians. So we might conclude that the amplitudes should be the same and, therefore, that ei = eim·2π has to be equal to one, so C’+ = C+ and C’ = C . That’s the case if m is equal to 1. But… Well… No. It’s the same thing again: the probabilities (or the magnitudes) have to be the same, but the amplitudes may be different because of some phase factor. In fact, they should be different. If m = 1/2, then we also get the same physics, even if the amplitudes are not the same. They will be each other’s opposite: Huh? Yes. Think of it. The coefficient of proportionality (m) cannot be equal to 1. If it would be equal to 1, and we’d rotate by 180° only, then we’d also get those C’+ = −C+ and C’ = −C equations, and so these coefficients would, therefore, also describe the same physical situation. Now, you will understand, intuitively, that a rotation of the apparatus by 180° will not give us the same physical situation… So… Well… In case you’d want a more formal argument proving a rotation by 180° does not give us the same physical situation, Feynman has one for you. 🙂

I know that, by now, you’re totally tired and bored, and so you only want the grand conclusion at this point. Well… All of what I wrote above should, hopefully, help you to understand that conclusion, which – I quote Feynman here – is the following:

If we know the amplitudes C+ and C of spin one-half particles with respect to a reference frame S, and we then use new base states, defined with respect to a reference frame T which is obtained from S by a rotation α around the z-axis, the new amplitudes are given in terms of the old by the following formulas: [Feynman denotes our angle α by phi (φ) because… He uses the Euler angles a bit differently. But don’t worry: it’s the same angle.]

What about the amplitude to go from the C to the C’+ state, and from the C+ to the C’ state? Well… That amplitude is zero. So the transformation matrix is this one: Let’s take a moment and think about this. Feynman notes the following, among other things: “It is very curious to say that if you turn the apparatus 360° you get new amplitudes. [They aren’t really new, though, because the common change of sign doesn’t give any different physics.] But if something has been rotated by a sequence of small rotations whose net result is to return it to the original orientation, then it is possible to define the idea that it has been rotated 360°—as distinct from zero net rotation—if you have kept track of the whole history.”

This is very deep. It connects space and time into one single geometric space, so to speak. But… Well… I’ll try to explain this rather sweeping statement later. Feynman also notes that a net rotation of 720° does give us the same amplitudes and, therefore, cannot be distinguished from the original orientation. Feynman finds that intriguing but… Well… I am not sure if it’s very significant. I do note some symmetries in quantum physics involve 720° rotations but… Well… I’ll let you think about this. 🙂

Note that the determinant of our matrix is equal to a·b·ceiφ/2·eiφ/2 = 1. So… Well… Our rotation matrix is, effectively, in that special form! How comes? Well… When equating λ = −μ, we are effectively putting the transformation into that special form.  Let us also, just for fun, quickly check the normalization condition. It requires that the probabilities, in any given representation, add to up to one. So… Well… Do they? When they come out of S, our electrons are equally likely to be in the up or down state. So the amplitudes are 1/√2. [To be precise, they are ±1/√2 but… Well… It’s the phase factor story once again.] That’s normalized: |1/√2|2 + |1/√2|2 = 1. The amplitudes to come out of the apparatus in the up or down state are eiφ/2/√2 and eiφ/2/√2 respectively, so the probabilities add up to |eiφ/2/√2|2 + |eiφ/2/√2|2 = … Well… It’s 1. Check it. 🙂

Let me add an extra remark here. The normalization condition will result in matrices whose determinant will be equal to some pure imaginary exponential, like eiα. So is that what we have here? Yes. We can re-write 1 as 1 = ei·0 = e0, so α = 0. 🙂 Capito? Probably not, but… Well… Don’t worry about it. Just think about the grand results. As Feynman puts it, this Lecture is really “a sort of cultural excursion.” 🙂

Let’s do a practical calculation here. Let’s suppose the angle is, effectively, 180°. So the eiφ/2 and eiφ/2/√2 factors are equal to eiπ/2 = +i and eiπ/2 = −i, so… Well… What does that mean—in terms of the geometry of the wavefunction? Hmm… We need to do some more thinking about the implications of all this transformation business for our geometric interpretation of he wavefunction, but so we’ll do that in our next post. Let us first work our way out of this rather hellish transformation logic. 🙂 [See? I do admit it is all quite difficult and abstruse, but… Well… We can do this, right?]

So what’s next? Well… Feynman develops a similar argument (I should say same-same but different once more) to derive the coefficients for a rotation of ±90° around the y-axis. Why 90° only? Well… Let me quote Feynman here, as I can’t sum it up more succinctly than he does: “With just two transformations—90° about the y-axis, and an arbitrary angle about the z-axis [which we described above]—we can generate any rotation at all.”

So how does that work? Check the illustration below. In Feynman’s words again: “Suppose that we want the angle α around x. We know how to deal with the angle α α around z, but now we want it around x. How do we get it? First, we turn the axis down onto x—which is a rotation of +90°. Then we turn through the angle α around z’. Then we rotate 90° about y”. The net result of the three rotations is the same as turning around x by the angle α. It is a property of space.” Besides helping us greatly to derive the transformation matrix for any rotation, the mentioned property of space is rather mysterious and deep. It sort of reduces the degrees of freedom, so to speak. Feynman writes the following about this:

“These facts of the combinations of rotations, and what they produce, are hard to grasp intuitively. It is rather strange, because we live in three dimensions, but it is hard for us to appreciate what happens if we turn this way and then that way. Perhaps, if we were fish or birds and had a real appreciation of what happens when we turn somersaults in space, we could more easily appreciate such things.”

In any case, I should limit the number of philosophical interjections. If you go through the motions, then you’ll find the following elemental rotation matrices: What about the determinants of the Rx(φ) and Ry(φ) matrices? They’re also equal to one, so… Yes. A pure imaginary exponential, right? 1 = ei·0 = e0. 🙂

What’s next? Well… We’re done. We can now combine the elemental transformations above in a more general format, using the standardized Euler angles. Again, just go through the motions. The Grand Result is: Does it give us normalized amplitudes? It should, but it looks like our determinant is going to be a much more complicated complex exponential. 🙂 Hmm… Let’s take some time to mull over this. As promised, I’ll be back with more reflections in my next post.

# Comments on the MIT’s Stern-Gerlach lab experiment

In my previous post, I noted that I’d go through the MIT’s documentation on the Stern-Gerlach experiment that their undergrad students have to do, because we should now – after 175 posts on quantum physics 🙂 – be ready to fully understand what is said in there. So this post is just going to be a list of comments. I’ll organize it section by section.

### Theory of atomic beam experiments

The theory is known – and then it isn’t, of course. The key idea is that individual atoms behave like little magnets. Why? In the simplest and most naive of models, it’s because the electrons somehow circle around the nucleus. You’ve seen the illustration below before. Note that current is, by convention, the flow of positive charge, which is, of course, opposite to the electron flow. You can check the direction by applying the right-hand rule: if you curl the fingers of your right hand in the direction of the current in the loop (so that’s opposite to v), your thumb will point in the direction of the magnetic moment (μ). So the electron orbit – in whatever way we’d want to visualize it – gives us L, which we refer to as the orbital angular momentum. We know the electron is also supposed to spin about its own axis – even if we know this planetary model of an electron isn’t quite correct. So that gives us a spin angular momentum S. In the so-called vector model of the atom, we simply add the two to get the total angular momentum J = L + S.

Of course, now you’ll say: only hydrogen has one electron, so how does it work with multiple electrons? Well… Then we have multiple orbital angular momentum li which are to be added to give a total orbital angular momentum L. Likewise, the electrons spins si can also be added to give some total spin angular momentum S. So we write:

J = L + S with L = Σi li and S = Σi si

Really? Well… If you’d google this to double-check – check the Wikipedia article on it, for example – then you’ll find this additivity property is valid only for relatively light atoms (Z ≤ 30) and only if any external magnetic field is weak enough. The way individual orbital and spin angular momenta have to be combined so as to arrive at some total L, S and J is referred to a coupling scheme: the additivity rule above is referred to as LS coupling, but one may also encounter LK coupling, or jj coupling, or other stuff. The US National Institute of Standards and Technology (NIST) has a nice article on all these models – but we need to move on here. Just note that we do assume the LS coupling scheme applies to our potassium beam – because its atomic number (Z) is 19, and the external magnetic field is assumed to be weak enough.

The vector model of the atom describes the atom using angular momentum vectors. Of course, we know that a magnetic field will cause our atomic magnet to precess – rather than line up. At this point, the classical analogy between a spinning top – or a gyroscope – and our atomic magnet becomes quite problematic. First, think about the implications for L and S when assuming,  as we usually do, that J precesses nicely about an axis that is parallel to the magnetic field – as shown in the illustration below, which I took from Feynman’s treatment of the matter If J is the sum of two other vectors L and S, then this has rather weird implications for the precession of L and S, as shown in the illustration below – which I took from the Wikipedia article on LS coupling. Think about: if L and S are independent, then the axis of precession for these two vectors should be just the same as for J, right? So their axis of precession should also be parallel to the magnetic field (B), so that’s the direction of the Jz component, which is just the z-axis of our reference frame here. More importantly, our classical model also gets into trouble when actually measuring the magnitude of Jz: repeated measurements will not yield some randomly distributed continuous variable, as one would classically expect. No. In fact, that’s what this experiment is all about: it shows that Jz will take only certain quantized values. That is what is shown in the illustration below (which once again assumes the magnetic field (B) is along the z-axis). I copied the illustration above from the HyperPhysics site, because I found it to be enlightening and intriguing at the same time. First, it also shows this rather weird implication of the vector model: if J continually changes direction because of its precession in a weak magnetic field, then L and S must, obviously, also continually change direction. However, this illustration is even more intriguing than the Wikipedia illustration because it assumes the axis of precession of L and S and L actually the same!

So what’s going on here? To better understand what’s going on, I started to read the whole HyperPhysics article on the vector model, which also includes the illustration below, with the following comments: “When orbital angular momentum L and electron spin S are combined to produce the total angular momentum of an atomic electron, the combination process can be visualized in terms of a vector model. Both the orbital and spin angular momenta are seen as precessing about the direction of the total angular momentum J. This diagram can be seen as describing a single electron, or multiple electrons for which the spin and orbital angular momenta have been combined to produce composite angular momenta S and L respectively. In so doing, one has made assumptions about the coupling of the angular momenta which are described by the LS coupling scheme which is appropriate for light atoms with relatively small external magnetic fields.” Hmm… What about those illustrations on the right-hand side – with the vector sums and those values for and mj? I guess the idea may also be illustrated by the table below: combining different values for l (±1) and (±1/2) gives four possible values, ranging from +3/2 to -1/2, for l + s. Having said that, the illustration raises a very fundamental question: the length of the sum of two vectors is definitely not the same as the sum of the length of the two vectors! So… Well… Hmm… Something doesn’t make sense here! However, I can’t dwell any longer on this. I just wanted to note you should not take all that’s published on those oft-used sites on quantum mechanics for granted. But so I need to move on. Back to the other illustration – copied once more below. We have that very special formula for the magnitude (J) of the angular momentum J:

J│= J = √(J·J) = √[j·(j+1)·ħ2] = √[j·(j+1)]·ħ

So if = 3/2, then J is equal to √[(3/2)·(3/2+1)]·ħ = √[(15/4)·ħ ≈ 1.9635·ħ, so that’s almost 2ħ. 🙂 At the same time, we know that for = 3/2, the possible values of Jz can only be +3ħ/2, +ħ/2, -ħ/2, and -3ħ/2. So that’s what’s shown in that half-circular diagram: the magnitude of J is larger than its z-component – always!

OK. Next. What’s that 3p3/2 notation? Excellent question! Don’t think this 3p denotes an electron orbital, like 1s or 3d – i.e. the orbitals we got from solving Schrödinger’s equation. No. In fact, the illustration above is somewhat misleading because the correct notation is not 3p3/2 but 3P3/2. So we have a capital P which is preceded by a superscript 3. This is the notation for the so-called term symbol for a nuclear, atomic or molecular (ground) state which – assuming our LS coupling model is valid – because we’ve got other term symbols for other coupling models – we can write, more generally, as:

2S+1LJ

The J, L and S in this thing are the following:

1. The J is the total angular momentum quantum number, so it is – the notation gets even more confusing now – the in the │J│= J = √(J·J) = √[j·(j+1)·ħ2] = √[j·(j+1)]·ħ expression. We know that number is 1/2 for electrons, but it may take on other values for nuclei, atoms or molecules. For example, it is 3/2 for nitrogen, and 2 for oxygen, for which the corresponding terms are 4S3/2 and 3P2 respectively.

2. The S in the term symbol is the total spin quantum number, and 2S+1 itself is referred to as the fine-structure multiplicity. It is not an easy concept. Just remember that the fine structure describes the splitting of the spectral lines of atoms due to electron spin. In contrast, the gross structure energy levels are those we effectively get from solving Schrödinger’s equation assuming our electrons have no spin.  We also have a hyperfine structure, which is due to the existence of a (small) nuclear magnetic moment, which we do not take into consideration here, which is why the 4S3/2 and 3P2 terms are sometimes being referred to as describing electronic ground states. In fact, the MIT lab document, which we are studying here, refers to the ground state of the potassium atoms in the beam as an electronic ground state, which is written up as 2S1/2. So is, effectively, equal to 1/2. [Are you still there? If so, just write it down: 2S+1 = 2 ⇒ = 1/2. That means the following: our potassium atom behaves like an electron: its spin is either ‘up’ or, else, it is ‘down’. There is no in-between.]

3. Finally, the in the term symbol is the total orbital angular momentum quantum number but, rather than using a number, the values of are often represented as S, P, D, F, etcetera. This number is very confusing because – as mentioned above – one would think it represents those s, p, d, f, g,… orbitals. However, that is not the case. The difference may easily be illustrated by observing that a carbon atom, for example, has six electrons, which are distributed over the 1s, 2s and 2p orbitals (one pair each). However, its ground state only gets one number: L = P. Hence, its value is 1. Of course, now you will wonder how we get that number.

Well… I wish I could give you an easy answer, but I can’t. For two electrons – think of our carbon atom once again – we can have = 0, 1 or 2, or S, P and D. They effectively correspond to different energy levels, which are related to the way these two electrons interact with each other. The phenomenon is referred to as angular momentum coupling. In fact, all of the numbers we discussed so far – J, S and L – are numbers resulting from angular momentum coupling. As Wikipedia puts it: “Angular momentum coupling refers to the construction of eigenstates of total angular momentum out of eigenstates of separate angular momentum.” [As you know, each eigenstate corresponds to an energy level, of course.]

Now that should clear some of the confusion on the 2S+1LJ notation: the capital letters J, S and L refer to some total, as opposed to the quantum numbers you are used to, i.e. n, l, m and s, i.e. the so-called principalorbitalmagnetic and spin quantum number respectively. The lowercase letters are quantum numbers that describe an electron in an atom, while those capital letters denote quantum numbers describing the atom – or a molecule – itself.

OK. Onwards. But where were we? 🙂 Oh… Yes. That J = L + S formula gives us some total electronic angular momentum, but we’ll also have some nuclear angular momentum, which our MIT paper denotes as I. Our vector model of our potassium atom allows us, once again, to simply add the two to get the total angular momentum, which is written as F = J + I = L + S + I. This, then, explains why the MIT experiment writes the magnitude of the total angular momentum as: Of course, here I don’t need to explain – or so I hope – why this quantum-mechanical formula for the calculation of the magnitude is what it is (or, equivalently, why the usual Euclidean metric – i.e. √(x2 + y2 + z2) – is not to be used here. If you do need an explanation, you’ll need to go through the basics once again.

Now, the whole point, of course, is that the z-component of F can have only the discrete values that are specified by the Fz = mf·ħ equation, with mf – i.e. the (total) magnetic quantum number – having an equally discrete value equal to mf = −f, −(f−1), …, +(f+1), f.

For the rest, I probably shouldn’t describe the experiment itself: you know it. But let me just copy the set-up below, so it’s clear what it is that we’re expecting to happen. In addition, you’ll also need the illustration below because I’ll refer to those d1 and d2 distances shown in what follows. Note the MIT documentation does spell out some additional assumptions. Most notably, it says that the potassium atoms that emerge from the oven (at a temperature of 200°) will be:

(1) almost exclusively in the ground electronic state,

(2) nearly equally distributed among the two (magnetic) sub-states characterized by f, and, finally,

(3) very nearly equally distributed among the hyperfine states, i.e. the states with the same but with different mf.

I am just noting these assumptions because it is interesting to note that – according to the man or woman who wrote this paper – we would actually have states within states here. The paper states that the hyperfine splitting of the two sub-beams we expect to come out of the magnet can only be resolved by very advanced atomic beam techniques, so… Well… That’s not the apparatus that’s being used for this experiment.

However, it’s all a bit weird, because the paper notes that the rules for combining the electronic and nuclear angular momentum – using that F = J + I = L + S + I formula – imply that our quantum number f = i ± j can be eitheror 2. These two values would be associated with the following mf and mforce values:

= 1 ⇒ Fz = mf·ħ = −ħ, 0 or +ħ (so we’d have three beams here)

= 2 ⇒ Fz = mf·ħ = −2ħ, −ħ, 0, +ħ or +2ħ (so we’d have five beams here)

Neither of the two possibilities relates to the situation at hand – which assumes two beams only. In short, I think the man or women who wrote the theoretical introduction – an assistant professor, most likely (no disrespect here: that’s how far progressed in economics – nothing more, nothing less) – might have made a mistake. Or perhaps he or she may have wanted to confuse us.

I’ll look into it over the coming days. As for now, all you need to know – please jot it down! – is that our potassium atom is fully described by 2S1/2. That shorthand notation has all the quantum number we need to know. Most importantly, it tells us is, effectively, equal to 1/2. So… Well… That 2S1/2 notation tells us our potassium atom should behave like an electron: its spin is either ‘up’ or ‘down’. No in-between. 🙂 So we should have two beams. Not three or five. No fine or hyperfine sub-structures! 🙂 In any case, the rest of the paper makes it clear the assumption is, effectively, that the angular momentum number is equal to = 1/2. So… Two beams only. 🙂

### How to calculate the expected deflection

We know that the inhomogeneous magnetic field (B), whose direction is the z-axis, will result in a force, which we have calculated a couple of times already as being equal to: In case you’d want to check this, you can check one of my posts on this. I just need to make one horrifying remark on notation here: while the same symbol is used, the force Fis, obviously, not to be confused with the z-component of the angular momentum F = J + I = L + S + I that we described above. Frankly, I hope that the MIT guys have corrected that in the meanwhile, because it’s really terribly confusing notation! In any case… Let’s move on.

Now, we assume the deflecting force is constant because of the rather particular design of the magnet pole pieces (see Appendix I of the paper). We can then use Newton’s Second Law (F = m·a) to calculate the velocity in the z-direction, which is denoted by Vz (I am not sure why a capital letter is used here, but that’s not important, of course). That velocity is assumed to go from 0 to its final value Vz while our potassium atom travels between the two magnet poles but – to be clear – at any point in time, Vz will increase linearly – not exponentially – so we can write: Vz = a·t1, with t1 the time that is needed to travel through the magnet. Now, the relevant mass is the mass of the atom, of course, which is denoted by M. Hence, it is easy to see that = Fz/M = Vz/t1. Hence, we find that V= Fz·t1/M.

Now, the vertical distance traveled (z) can be calculated by solving the usual integral: z = ∫0t1 v(t)·dt = ∫0t1 a·t·dt = a·t12/2 = (Vz/t1)·t12/2 = Vz·t1/2. Of course, once our potassium atom comes out of the magnetic field, it will continue to travel upward or downward with the same velocity Vz, which adds Vz·t2 to the total distance traveled along the z-direction. Hence, the formula for the deflection is, effectively, the one that you’ll find in the paper:

z = Vz·t1/2 + Vz·t= Vz·(t1/2 + t2)

Now, the travel times depend on the velocity of our potassium atom along the y-axis, which is approximated by equating it with │V│= V, because the y-component of the velocity is easily the largest – by far! Hence, t1 = d1/V and t2 = d2/V. Some more manipulation will then give you the expression we need, which is a formula for the deflection in terms of variables that we actually know: ### Statistical mechanics

We now need to combine this with the Maxwell-Boltzmann distribution for the velocities we gave you in our previous post: The next step is to use this formula so as to be able to calculate a distribution which would describe the intensity of the beam. Now, it’s easy to understand such intensity will be related to the flux of potassium atoms, and it’s equally easy to get that a flux is defined as the rate of flow per unit area. Hmm… So how does this get us the formula below? The tricky thing – of course – is the use of those normalized velocities because… Well… It’s easy to see that the right-hand side of the equation above – just forget about the d(V/V0 ) bit for a second, as we have it on both sides of the equation and so it cancels out anyway – is just density times velocity. We do have a product of the density of particles and the velocity with which they emerge here – albeit a normalized velocity. But then… Who cares? The normalization is just a division by V– or a multiplication by 1/V0, which is some constant. From a math point of view, it doesn’t make any difference: our variable is V/V0 instead of V. It’s just like using some other unit. No worries here – as long as you use the new variable consistently everywhere. 🙂

Alright. […] What’s next? Well… Nothing much. The only thing that we still need to explain now is that factor 2. It’s easy to see that’s just a normalization factor – just like that 4/√π factor in the first formula. So we get it from imposing the usual condition: So… What’s next… Well… We’re almost there. 🙂 As the MIT paper notes, the f(V) and I(V/V0) functions can be mapped to each other: the related transformation maps a velocity distribution to an intensity distribution – i.e. a distribution of the deflection – and vice versa.

Now, the rest of the paper is just a lot of algebraic manipulations – distinguishing the case of a quantized Fversus a continuous Fz. Here again, I must admit I am a bit shocked by the mix-up of concepts and symbols. The paper talks about a quantized deflecting force – while it’s obvious we should be talking a quantized angular momentum. The two concepts – and their units – are fundamentally different: the unit in which angular momentum is measured is the action unit: newton·meter·second (N·m·s). Force is just force: x newton.

Having said that, the mix-up does trigger an interesting philosophical question: what is quantized really? Force (expressed in N)? Energy (expressed in N·m)? Momentum (expressed in N·s)? Action (expressed in N·m·s, i.e. the unit of angular momentum)? Space? Time? Or space-time – related through the absolute speed of light (c)? Three factors (force, distance and time), six possibilities. What’s your guess?

[…]

What’s my guess? Well… The formulas tell us the only thing that’s quantized is action: Nature itself tells us we have to express it in terms of Planck units. However, because action is a product involving all of these factors, with different dimensions, the quantum-mechanical quantization of action can, obviously, express itself in various ways. 🙂

# The quantization of magnetic moments

You may not have many questions after a first read of Feynman’s Lecture on the Stern-Gerlach experiment and his more general musings on the quantization of the magnetic moment of an elementary particle. [At least I didn’t have all that many after my first reading, which I summarized in a previous post.]

However, a second, third or fourth reading should trigger some, I’d think. My key question is the following: what happens to that magnetic moment of a particle – and its spin  – as it travels through a homogeneous or inhomogeneous magnetic field? We know – or, to be precise, we assume – its spin is either “up” (Jz = +ħ/2) or “down” (Jz = −ħ/2) when it enters the Stern-Gerlach apparatus, but then – when it’s moving in the field itself – we would expect that the magnetic field would, somehow, line up the magnetic moment, right?

Feynman says that it doesn’t: from all of the schematic drawings – and the subsequent discussion of Stern-Gerlach filters – it is obvious that the magnetic field – which we denote as B, and which we assume to be inhomogeneous  – should not result in a change of the magnetic moment. Feynman states it as follows: “The magnetic field produces a torque. Such a torque you would think is trying to line up the (atomic) magnet with the field, but it only causes its precession.”

[…] OK. That’s too much information already, I guess. Let’s start with the basics. The key to a good understanding of this discussion is the force formula: We should first explain this formula before discussing the obvious question: over what time – or over what distance – should we expect this force to pull the particle up or down in the magnetic field? Indeed, if the force ends up aligning the moment, then the force will disappear!

So let’s first explain the formula. We start by explaining the energy U. U is the potential energy of our particle, which it gets from its magnetic moment μ and its orientation in the magnetic field B. To be precise, we can write the following: Of course, μ and B are the magnitudes of μ and B respectively, and θ is the angle between μ and B: if the angle θ is zero, then Umag will be negative. Hence, the total energy of our particle (U) will actually be less than what it would be without the magnetic field: it is the energy when the magnetic moment of our particle is fully lined up with the magnetic field. When the angle is a right angle (θ = ±π/2), then the energy doesn’t change (Umag = 0). Finally, when θ is equal to π or −π, then its energy will be more than what it would be outside of the magnetic field. [Note that the angle θ effectively varies between –π and π – not between 0 and 2π!] Of course, we may already note that, in quantum mechanics, Umag will only take on a very limited set of values. To be precise, for a particle with spin number j = 1/2, the possible values of Umag will be limited to two values only. We will come back to that in a moment. First that force formula.

Energy is force over a distance. To be precise, when a particle is moved from point a to point b, then its change in energy can be written as the following line integral: Note that the minus sign is there because of the convention that we’re doing work against the force when increasing the (potential) energy of that what we’re moving. Also note that F∙ds product is a vector (dot) product: it is, obviously, equal to Ft times ds, with Ft the magnitude of the tangential component of the force. The equation above gives us that force formula: Feynman calls it the principle of virtual work, which sounds a bit mysterious – but so you get it by taking the derivative of both sides of the energy formula.

Let me now get back to the real mystery of quantum mechanics, which tells us that the magnetic moment – as measured along our z-axis – will only take one of two possible values. To be precise, we have the following formula for μz: This is a formula you just have to accept for the moment. It needs a bit of interpretation, and you need to watch out for the sign. The g-factor is the so-called Landé g-factor: it is equal to 1 for a so-called pure orbital moment, 2 for a so-called pure spin moment, and some number in-between in reality, which is always some mixture of the two: both the electron’s orbit around the nucleus as well as the electron’s rotation about its own axis contribute to the total angular momentum and, hence, to the total magnetic moment of our electron. As for the other factors, m and qe are, of course, the mass and the charge of our electron, and Jz is either +ħ/2 or −ħ/2. Hence, if we know g, we can easily calculate the two possible values for μz.

Now, that also means we could – theoretically – calculate the two possible values of that angle θ. For some reason, no handbook in physics ever does that. The reason is probably a good one: electron orbits, and the concept of spin itself, are not like the orbit and the spin of some planet in a planetary system. In fact, we know that we should not think of electrons like that at all: quantum physicists tell us we may only think of it as some kind of weird cloud around a center. That cloud has a density which is to be calculated by taking the absolute square of the quantum-mechanical amplitude of our electron.

In fact, when thinking about the two possible values for θ, we may want to remind ourselves of another peculiar consequence of the fact that the angular momentum – and, hence, the magnetic moment – is not continuous but quantized: the magnitude of the angular momentum J is not  J = √(J·J) = √J2 in quantum mechanics but J = √(J·J) = √[j·(j+1)·ħ2] = √[j·(j+1)]·ħ. For our electron, j = 1/2 and, hence, the magnitude of J is equal to J = √[(1/2)∙(3/2)]∙ ħ = √(3/4)∙ħ ≈ 0.866∙ħ. Hence, the magnitude of the angular momentum is larger than the maximum value of Jz – and not just a little bit, because the maximum value of ħ is ħ/2! That leads to that weird conclusion: in quantum mechanics, we find that the angular momentum is never completely along any one direction ! In fact, this conclusion basically undercuts the very idea of the angular momentum – and, hence, the magnetic moment – of having any precise direction at all! [This may sound spectacular, but there is actually a classical equivalent to the idea of the angular momentum having no precisely defined direction: gyroscopes may not only precess, but nutate as well. Nutation refers to a kind of wobbling around the direction of the angular momentum. For more details, see the post I wrote after my first reading of Feynman’s Lecture on the quantization of magnetic moments. :-)]

Let’s move on. So if, in quantum mechanics, we cannot associate the magnetic moment – or the angular momentum – with some specific direction, then how should we imagine it? Well… I won’t dwell on that here, but you may want to have a look at another post of mine, where I develop a metaphor for the wavefunction which may help you to sort of understand what it might be. The metaphor may help you to think of some oscillation in two directions – rather than in one only – with the two directions separated by a right angle. Hence, the whole thing obviously points in some direction but it’s not very precise. In any case, I need to move on here.

We said that the magnetic moment will take one of two values only, in any direction along which we’d want to measure it. We also said that the (maximum) value along that direction – any direction, really – will be smaller than the magnitude of the moment. [To be precise, we said that for the angular momentum, but the formulas above make it clear the conclusions also hold for the magnetic moment.] So that means that the magnetic moment is, in fact, never fully aligned with the magnetic field. Now, if it is not aligned – and, importantly, if it also does not line up – then it should precess. Now, precession is a difficult enough concept in classical mechanics, so you may think it’s going to be totally abstruse in quantum mechanics. Well… That is true – to some extent. At the same time, it is surely not unintelligible. I will not repeat Feynman’s argument here, but he uses the classical formulas once more to calculate an angular velocity and a precession frequency – although he doesn’t explain what they might actually physically represent. Let me just jot down the formula for the precession frequency: We get the same factors: g, qe and m. In addition, you should also note that the precession frequency is directly proportional  to the strength of the magnetic field, which makes sense. Now, you may wonder: what is the relevance of this? Can we actually measure any of this?

We can. In fact, you may wonder about the if I inserted above: if we can measure the Landé g-factor… Can we? We can. It’s done in a resonance experiment, which is referred to as the Rabi molecular-beam method – but then it might also be just an atomic beam, of course!

The experiment is interesting, because it shows the precession is – somehow – real. It also illustrates some other principles we have been describing above.

The set-up looks pretty complicated. We have a series of three magnets. The first magnet is just a Stern-Gerlach apparatus: a magnet with a very sharp edge on one of the pole tips so as to produce an inhomogeneous magnetic field. Indeed, a homogeneous magnetic field implies that ∂B/∂z = 0 and, hence, the force along the z-direction would be zero and our atomic magnets would not be displaced.

The second magnet is more complicated. Its magnetic field is uniform, so there are no vertical forces on the atoms and they go straight through. However, the magnet includes an extra set of coils that can produce an alternating horizontal field as well. I’ll come back to that in a moment. Finally, the third magnet is just like the first one, but with the field inverted. Have a look at it: It may not look very obvious but, after some thinking, you’ll agree that the atoms can only arrive at the detector if they follow the trajectories a and/or b. In fact, these trajectories are the only possible ones because of the slits S1 and S2.

Now what’s the idea of that horizontal field B’ in magnet 2? In a classical situation, we could change the angular momentum – and the magnetic moment – by applying some torque about the z-axis. The idea is shown in Figure (a) and (b) below. Figure (a) shows – or tries to show – some rotating field B’ – one that is always at right angles to both the angular momentum as well as to the (uniform) B field. That would be effective. However, Figure (b) shows another arrangement that is almost equally effective: an oscillating field that sort of pulls and pushes at some frequency ω. Classically, such fields would effectively change the angle of our gyroscope with respect to the z-axis. Is it also the case quantum-mechanically?

It turns out it sort of works the same in quantum mechanics. There is a big difference though. Classically, μz would change gradually, but in quantum mechanics it cannot: in quantum mechanics, it must jump suddenly from one value to the other, i.e. from +ħ/2 to −ħ/2, or the other way around. In other words, it must flip up or down. Now, if an atom flips, then it will, of course, no longer follow the (a) or (b) trajectories: it will follow some other path, like a’ or b’, which make it crash into the magnet. Now, it turns out that almost all atoms will flip if we get that frequency ω right. The graph below shows this ‘resonance’ phenomenon: there is a sharp drop in the ’current’ of atoms if ω is close or equal to ωp. What’s ωp? It’s that precession frequency for which we gave you that formula above. To make a long story short, from the experiment, we can calculate the Landé g-factor for that particular beam of atoms – say, silver atoms . So… Well… Now we know it all, don’t we?

Maybe. As mentioned when I started this post, when going through all of this material, I always wonder why there is no magnetization effect: why would an atom remain in the same state when it crosses a magnetic field? When it’s already aligned with the magnetic field – to the maximum extent possible, that is – then it shouldn’t flip, but what if its magnetic moment is opposite? It should lower its energy by flipping, right? And it should flip just like that. Why would it need an oscillating B’ field?

In fact, Feynman does describe how the magnetization phenomenon can be analyzed – classically and quantum-mechanically, but he does that for bulk materials: solids, or liquids, or gases – anything that involves lots of atoms that are kicked around because of the thermal motions. So that involves statistical mechanics – which I am sure you’ve skipped so far. 🙂 It is a beautiful argument – which ends with an equally beautiful formula, which tells us the magnetization (M) of a material – which is defined as the net magnetic moment per unit volume – has the same direction as the magnetic field (B) and a magnitude M that is proportional the magnitude of B: The μ in this formula is the magnitude of the magnetic moment of the individual atoms and so… Well… It’s just like the formula for the electric polarization P, which we described in some other post. In fact, the formula for P and M are same-same but different, as they would say in Thailand. 🙂 But this wonderful story doesn’t answer our question. The magnetic moment of an individual particle should not stay what it is: if it doesn’t change because of all the kicking around as a result of thermal motions, then… Well… These little atomic magnets should line up. That means atoms with their spin “up” should go into the “spin-down” state.

I don’t have an answer to my own question as for now. I suspect it’s got to do with the strength of the magnetic field: a Stern-Gerlach apparatus involves a weak magnetic field. If it’s too strong, the atomic magnets must flip. Hence, a more advanced analysis should probably include that flipping effect. When quickly googling – just now – I found an MIT lab exercise on it, which also provides a historical account of the Stern-Gerlach experiment itself. I skimmed through it – and will read all of it in the coming days – but let me just quote this from the historical background section:

“Stern predicted that the effect would be be just barely observable. They had difficulty in raising support in the midst of the post war financial turmoil in Germany. The apparatus, which required extremely precise alignment and a high vacuum, kept breaking down. Finally, after a year of struggle, they obtained an exposure of sufficient length to give promise of an observable silver deposit. At first, when they examined the glass plate they saw nothing. Then, gradually, the deposit became visible, showing a beam separation of 0.2 millimeters! Apparently, Stern could only afford cheap cigars with a high sulfur content. As he breathed on the glass plate, sulfur fumes converted the invisible silver deposit into visible black silver sufide, and the splitting of the beam was discovered.”

Isn’t this funny? And great at the same time? 🙂 But… Well… The point is: the paper for that MIT lab exercise makes me realize Feynman does cut corners when explaining stuff – and some corners are more significant than others. I note, for example, that they talk about interference peaks rather than “two distinct spots on the glass plate.” Hence, the analysis is somewhat more sophisticated than Feynman pretends it to be. So, when everything is said and done, Feynman’s Lectures may indeed be reading for undergraduate students only. Is it time to move on?

 The magnetic moment – as measured in a particular coordinate system – is equal to μ = −g·[q/(2m)]·J. The factor J in this expression is the angular momentum, and the coordinate system is chosen such that its z-axis is along the direction of the magnetic field B. The component of J along the z-axis is written as Jz. This z-component of the angular momentum is what is, rather loosely, being referred to as the spin of the particle in this context. In most other contexts, spin refers to the spin number j which appears in the formula for the value of Jz, which is Jz = j∙ħ, (j−1)∙ħ, (j−2)∙ħ,…, (−j+2)∙ħ, (−j+1), −j∙ħ. Note the separation between the possible values of Jz is equal to ħ. Hence, j itself must be an integer (e.g. 1 or 2) or a half-integer (e.g. 1/2). We usually look at electrons, whose spin number j is 1/2.

 One of the pole tips of the magnet that is used in the Stern-Gerlach experiment has a sharp edge. Therefore, the magnetic field strength varies with z. We write: ∂B/∂z ≠ 0.

 The z-direction can be any direction, really.

 The original experiment was effectively done with a beam of silver atoms. The lab exercise which MIT uses to show the effect to physics students involves potassium atoms.