# The geometry of the wavefunction, electron spin and the form factor

Pre-script (dated 26 June 2020): Our ideas have evolved into a full-blown realistic (or classical) interpretation of all things quantum-mechanical. In addition, I note the dark force has amused himself by removing some material. So no use to read this. Read my recent papers instead. ðŸ™‚

Original post:

Our previous posts showed how a simple geometric interpretation of the elementary wavefunction yielded the (Compton scattering) radius of an elementary particleâ€”for an electron, at least: for the proton, we only got the order of magnitude rightâ€”but then a proton is not an elementary particle.Â We got lots of other interesting equations as well… But… Well… When everything is said and done, it’s that equivalence between theÂ E =Â mÂ·a2Â·Ï‰2Â andÂ E =Â mÂ·c2Â relations that we… Well… We need to be moreÂ specific about it.

Indeed, I’ve been ambiguous here and thereâ€”oscillatingÂ between various interpretations, so to speak. ðŸ™‚ In my own mind, I refer to my unanswered questions, or my ambiguous answers to them, as the form factorÂ problem.Â So… Well… That explains the title of my post. But so… Well… I do want to be somewhat moreÂ conclusiveÂ in this post. So let’s go and see where we end up. ðŸ™‚

To help focus our mind, let us recall the metaphor of the V-2 perpetuum mobile, as illustrated below. With permanently closed valves, the air inside the cylinder compresses and decompresses as the pistons move up and down. It provides, therefore, a restoring force. As such, it will store potential energy, just like a spring, and the motion of the pistons will also reflect that of a mass on a spring: it is described by a sinusoidal function, with the zero point at the center of each cylinder. We can, therefore, think of the moving pistons as harmonic oscillators, just like mechanical springs. Of course, instead of two cylinders with pistons, one may also think of connecting two springs with a crankshaft, but then that’s not fancy enough for me. ðŸ™‚

At first sight, the analogy between our flywheel model of an electron and the V-twin engine seems to be complete: the 90 degree angle of ourÂ V-2 engine makes it possible to perfectly balance the pistons and we may, therefore, think of the flywheel as a (symmetric) rotating mass, whose angular momentum is given by the product of the angular frequency and the moment of inertia: L =Â Ï‰Â·I. Of course,Â the moment of inertia (aka the angular mass) will depend on theÂ formÂ (orÂ shape) of our flywheel:

1. I = mÂ·a2Â for a rotating pointÂ mass m or, what amounts to the same, for a circular hoop of mass m and radiusÂ rÂ =Â a.
2. For a rotating (uniformly solid)Â disk, we must add a 1/2 factor: IÂ =Â mÂ·a2/2.

How can we relate those formulas to the E =Â mÂ·a2Â·Ï‰2Â formula? TheÂ kinetic energy that is being stored in a flywheel is equal EkineticÂ = IÂ·Ï‰2/2, so that is only halfÂ of theÂ E =Â mÂ·a2Â·Ï‰2Â product if we substitute I forÂ I = mÂ·a2. [For a disk, we get a factor 1/4, so that’s even worse!] However, our flywheel model of an electron incorporates potential energy too. In fact, theÂ E =Â mÂ·a2Â·Ï‰2Â formula just adds the (kinetic and potential) energy of two oscillators: we do not really consider the energy in the flywheel itself because… Well… The essence of our flywheel model of an electron is not the flywheel: the flywheel justÂ transfersÂ energy from one oscillator to the other, but so… Well… We don’tÂ includeÂ it in our energy calculations. The essence of our model is thatÂ two-dimensional oscillation whichÂ drivesÂ the electron, and which is reflected in Einstein’sÂ E =Â mÂ·c2Â formula.Â That two-dimensional oscillationâ€”theÂ a2Â·Ï‰2Â = c2Â equation, reallyâ€”tells us that theÂ resonantÂ (orÂ natural) frequencyÂ of the fabric of spacetime is given by theÂ speed of lightâ€”but measured in units ofÂ a. [If you don’t quite get this, re-write theÂ a2Â·Ï‰2Â = c2Â equation asÂ Ï‰ = c/a: the radius of our electron appears as a naturalÂ distance unit here.]

Now, we were extremely happy with this interpretation not only because of the key results mentioned above, but also because it has lots of other nice consequences. Think of our probabilities as being proportional to energy densities, for exampleâ€”and all of the other stuff I describe in my published paper on this. But there is even more on the horizon: a follower of this blog (a reader with an actual PhD in physics, for a change) sent me an article analyzing elementary particles as tiny black holes because… Well… If our electron is effectively spinning around, then its tangential velocity is equal toÂ vÂ =Â aÂ·Ï‰Â =Â c. Now, recent research suggest black holes are also spinning at (nearly) the speed of light. Interesting, right? However, in order to understand what she’s trying to tell me, I’ll first need to get a better grasp of general relativity, so I can relate what I’ve been writing here and in previous posts to the Schwarzschild radiusÂ and other stuff.

Let me get back to the lesson here. In the reference frame of our particle, the wavefunction really looks like the animation below: it has two components, and the amplitude of the two-dimensional oscillation is equal to a, which we calculated asÂ aÂ =Â Ä§Â·/(mÂ·c) = 3.8616Ã—10âˆ’13Â m, so that’s the (reduced) Compton scattering radius of an electron.

In my original article on this, I used a more complicated argument involving the angular momentum formula, but I now prefer a more straightforward calculation:

cÂ = aÂ·Ï‰Â =Â aÂ·E/Ä§ =Â aÂ·mÂ·c2/Ä§Â Â â‡”Â aÂ =Â Ä§/(mÂ·c)

The question is: whatÂ is that rotating arrow? I’ve been vague and not so vague on this. The thing is: I can’tÂ proveÂ anything in this regard. But myÂ hypothesisÂ is that it is, in effect, aÂ rotatingÂ field vector, so it’s just like the electric field vector of a (circularly polarized) electromagnetic wave (illustrated below).

There are a number of crucial differences though:

1. The (physical) dimension of the field vector of the matter-wave is different: I associate the real and imaginary component of the wavefunction with a force per unit massÂ (as opposed to the force per unit charge dimension of the electric field vector). Of course, the newton/kg dimension reduces to the dimension of acceleration (m/s2), so that’s the dimension of a gravitational field.
2. I do believe this gravitational disturbance, so to speak, does cause an electron to move about some center, and I believe it does so at the speed of light. In contrast, electromagnetic waves doÂ notÂ involve any mass: they’re just an oscillating field. Nothing more. Nothing less. In contrast, as Feynman puts it: “When you do find the electron some place, the entire charge is there.” (Feynman’s Lectures, III-21-4)
3. The third difference is one that I thought of only recently: theÂ planeÂ of the oscillation cannotÂ be perpendicular to the direction of motion of our electron, because then we can’t explain the direction of its magnetic moment, which is either up or down when traveling through a Stern-Gerlach apparatus.

I mentioned that in my previous post but, for your convenience, I’ll repeat what I wrote there.Â The basic idea here is illustrated below (credit for this illustration goes toÂ another blogger on physics). As for the Stern-Gerlach experiment itself, let me refer you to aÂ YouTube videoÂ from theÂ Quantum Made SimpleÂ site.

The point is: the direction of the angular momentum (and the magnetic moment) of an electronâ€”or, to be precise, its component as measured in the direction of the (inhomogeneous) magnetic field through which our electron isÂ travelingâ€”cannotÂ be parallel to the direction of motion. On the contrary, it isÂ perpendicularÂ to the direction of motion. In other words, if we imagine our electron as spinning around some center, then the disk it circumscribes will compriseÂ the direction of motion.

However, we need to add an interesting detail here. As you know, we don’t really have a precise direction of angular momentum in quantum physics. [If you don’t know this… Well… Just look at one of my many posts on spin and angular momentum in quantum physics.] Now, we’ve explored a number of hypotheses but, when everything is said and done, a rather classical explanation turns out to be the best: an object with an angular momentum JÂ and a magnetic momentÂ Î¼Â (I used bold-face because these areÂ vector quantities) that is parallel to some magnetic field B, will notÂ line up, as you’d expect a tiny magnet to do in a magnetic fieldâ€”or not completely, at least: it willÂ precess. I explained that in another post on quantum-mechanical spin, which I advise you to re-read if you want to appreciate the point that I am trying to make here. That post integrates some interesting formulas, and so one of the things on my ‘to do’ list is to prove that these formulas are, effectively, compatible with the electron model we’ve presented in this and previous posts.

Indeed, when one advances a hypothesis like this, it’s not enough to just sort ofÂ showÂ that the general geometry of the situation makes sense: we also need to show the numbers come out alright. So… Well… Whatever weÂ thinkÂ our electronâ€”or its wavefunctionâ€”might be, it needs to be compatible with stuff like the observedÂ precession frequencyÂ of an electron in a magnetic field.

Our model also needs to be compatible with the transformation formulas for amplitudes. I’ve been talking about this for quite a while now, and so it’s about time I get going on that.

Last but not least, those articles that relate matter-particles to (quantum) gravityâ€”such as the one I mentioned aboveâ€”are intriguing too and, hence, whatever hypotheses I advance here, I’d better check them against those more advanced theories too, right? ðŸ™‚ Unfortunately, that’s going to take me a few more years of studying… But… Well… I still have many years aheadâ€”I hope. ðŸ™‚

Post scriptum: It’s funny how one’s brain keeps working when sleeping. When I woke up this morning, I thought: “But itÂ isÂ that flywheel that matters, right? That’s the energy storage mechanism and also explains how photons possibly interact with electrons. The oscillatorsÂ driveÂ the flywheel but, without the flywheel, nothing is happening. It is really theÂ transferÂ of energyâ€”through the flywheelâ€”which explains why our flywheel goes round and round.”

It may or may not be useful to remind ourselves of the math in this regard.Â The motionÂ ofÂ our first oscillator is given by the cos(Ï‰Â·t) = cosÎ¸ function (Î¸ = Ï‰Â·t), and its kinetic energy will be equal toÂ sin2Î¸. Hence, the (instantaneous)Â changeÂ in kinetic energy at any point in time (as a function of the angle Î¸) isÂ equal to:Â d(sin2Î¸)/dÎ¸ = 2âˆ™sinÎ¸âˆ™d(sinÎ¸)/dÎ¸ = 2âˆ™sinÎ¸âˆ™cosÎ¸. Now, the motion of theÂ second oscillator (just look at that second piston going up and down in the V-2 engine) is given by theÂ sinÎ¸ function, which is equal to cos(Î¸ âˆ’ Ï€ /2). Hence, its kinetic energy is equal toÂ sin2(Î¸ âˆ’ Ï€ /2), and how itÂ changesÂ (as a function of Î¸ again) is equal toÂ 2âˆ™sin(Î¸ âˆ’ Ï€ /2)âˆ™cos(Î¸ âˆ’ Ï€ /2) =Â = âˆ’2âˆ™cosÎ¸âˆ™sinÎ¸ = âˆ’2âˆ™sinÎ¸âˆ™cosÎ¸. So here we have our energy transfer: the flywheel organizes the borrowing and returning of energy, so to speak. That’s the crux of the matter.

So… Well… WhatÂ if the relevant energy formula isÂ E =Â mÂ·a2Â·Ï‰2/2 instead ofÂ E =Â mÂ·a2Â·Ï‰2? What are the implications? Well… We get aÂ âˆš2 factor in our formula for the radiusÂ a, as shown below.

Now that isÂ notÂ so nice. For the tangential velocity, we getÂ vÂ =Â aÂ·Ï‰ =Â âˆš2Â·c. This is alsoÂ notÂ so nice. How can we save our model? I am not sure, but here I am thinking of the mentioned precessionâ€”theÂ wobbling of our flywheel in a magnetic field. Remember we may think of Jzâ€”the angular momentum or, to be precise, its component in theÂ z-direction (the direction in which weÂ measureÂ itâ€”as the projection of theÂ realÂ angular momentumÂ J. Let me insert Feynman’s illustration here again (Feynman’s Lectures, II-34-3), so you get what I am talking about.

Now, all depends on the angle (Î¸) betweenÂ JzÂ andÂ J, of course. We did a rather obscure post on these angles, but the formulas there come in handy now. Just click the link and review it if and when you’d want to understand the following formulas for theÂ magnitudeÂ of theÂ presumedÂ actualÂ momentum:In this particular case (spin-1/2 particles),Â j is equal to 1/2 (in units ofÂ Ä§, of course). Hence,Â JÂ is equal toÂ âˆš0.75Â â‰ˆ 0.866. Elementary geometry then tells us cos(Î¸) =Â (1/2)/âˆš(3/4) =Â  = 1/âˆš3. Hence,Â Î¸Â â‰ˆ 54.73561Â°. That’s a big angleâ€”larger than the 45Â° angle we had secretly expected because… Well… The 45Â° angle has thatÂ âˆš2 factor in it:Â cos(45Â°) =Â sin(45Â°) = 1/âˆš2.

Hmm… As you can see, there is no easy fix here. Those damn 1/2 factors! They pop up everywhere, don’t they? ðŸ™‚ We’ll solve the puzzle. One day… But not today, I am afraid. I’ll call it the form factor problem… Because… Well… It sounds better than the 1/2 orÂ âˆš2 problem, right? ðŸ™‚

Note: If you’re into quantum math, you’ll noteÂ aÂ =Â Ä§/(mÂ·c) is theÂ reducedÂ Compton scattering radius. The standard Compton scattering radius is equal toÂ aÂ·2Ï€Â = (2Ï€Â·Ä§)/(mÂ·c) =Â  h/(mÂ·c) = h/(mÂ·c). It doesn’t solve theÂ âˆš2 problem. Sorry. The form factor problem. ðŸ™‚

To be honest, I finished my published paper on all of this with a suggestion that, perhaps, we should think of twoÂ circularÂ oscillations, as opposed to linear ones. Think of a tiny ball, whose center of mass stays where it is, as depicted below. Any rotation â€“ around any axis â€“ will be some combination of a rotation around the two other axes. Hence, we may want to think of our two-dimensionalÂ oscillation as an oscillation of a polar and azimuthal angle. It’s just a thought but… Well… I am sure it’s going to keep me busy for a while. ðŸ™‚They are oscillations, still, so I am not thinking ofÂ twoÂ flywheels that keep going around in the same direction. No. More like a wobbling object on a spring. Something like the movement of a bobblehead on a spring perhaps. ðŸ™‚

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Some content on this page was disabled on June 17, 2020 as a result of a DMCA takedown notice from Michael A. Gottlieb, Rudolf Pfeiffer, and The California Institute of Technology. You can learn more about the DMCA here:

# Spin

Pre-script (dated 26 June 2020): This post has become less relevant (even irrelevant, perhaps) because my views on all things quantum-mechanical have evolved significantly as a result of my progression towards a more complete realist (classical) interpretation of quantum physics. In addition, some of the material was removed by a dark force (that also created problems with the layout, I see now). In any case, we recommend you read our recent papers. I keep blog posts like these mainly because I want to keep track of where I came from. I might review them one day, but I currently don’t have the time or energy for it. ðŸ™‚

Original post:

In the previous posts, I showed how the ‘real-world’ properties of photons and electrons emerge out of very simple mathematical notions and shapes. The basic notions are time and space. The shape is the wavefunction.

Let’s recall the story once again.Â Space is an infinite number of three-dimensional points (x, y, z), and time is a stopwatch hand going round and roundâ€”a cyclicalÂ thing. All points in space are connected by an infinite number of pathsÂ â€“Â straight or crooked, whateverÂ Â â€“Â of which we measure the length.Â And then we have ‘photons’ that move from A to B, but so we don’t know whatÂ is actually moving in space here. We just associate each and every possible path (in spacetime) between A and B with an amplitude: an ‘arrow‘ whose length and direction depends on (1) the length of the path lÂ (i.e. the ‘distance’ in space measured along the path, be it straight or crooked), and (2) the difference in time between the departure (at point A) and the arrival (at point B) of our photon (i.e. the ‘distance in time’ as measured by that stopwatch hand).

Now, in quantum theory, anything is possible and, hence, not only do we allow for crooked paths, but we also allow for the difference in time to differ from l/c. Hence, our photon may actually travel slower or faster than the speed of light c! There is one lucky break, however, that makes all come out alright: the arrows associated withÂ the odd paths and strange timings cancel each other out. Hence, what remains, are the nearby paths in spacetime onlyâ€”the ‘light-like’ intervals only: a small core of space which our photon effectively uses as it travels through empty space.Â And when it encounters an obstacle, like a sheet of glass, it may or may not interact with the other elementary particleâ€“the electron. And then we multiply and add the arrows â€“ or amplitudes as we call them â€“ to arrive at a final arrow, whose square is what physicists want to find, i.e. the likelihood of the event that we are analyzing (such a photon going from point A to B, in empty space, through two slits, or through as sheet of glass, for example) effectively happening.

The combining of arrows leads to diffraction, refraction orÂ â€“ to use the more general description of what’s going onÂ â€“Â interferenceÂ patterns:

1. Adding two identical arrows that are ‘lined up’ yields a final arrow with twice the length of either arrow alone and, hence, a square (i.e. a probability) that is four times as large. This is referred to as ‘positive’ or ‘constructive’ interference.
2. Two arrows of the same length but with opposite direction cancel each other out and, hence, yield zero: that’s ‘negative’ or ‘destructive’ interference.

Both photons and electrons are represented byÂ wavefunctions, whose argument is the position in space (x, y, z) and time (t), and whose value is an amplitudeÂ orÂ ‘arrow’ indeed, with a specific direction and length. But here we get a bifurcation. When photons interact with other, their wavefunctions interact just like amplitudes: we simply add them.Â However, when electrons interact with each other, we have to apply a different rule: we’ll take a difference. Indeed, anything is possible in quantum mechanics and so we combine arrows (or amplitudes, or wavefunctions) in two different ways: we can either add them or, as shown below, subtract one from the other.

There are actuallyÂ four distinct logical possibilities,Â because we may also change the order of A and B in the operation, but when calculating probabilities, all we need is the square of the final arrow, so we’re interested in its final length only, not in its direction (unless we want to use that arrow in yet another calculation). And so… Well… The fundamental dualityÂ in Nature between light and matter is based on this dichotomyÂ only: identical (elementary) particles behave in one of two ways: their wavefunctions interfere eitherÂ constructively or destructively, and that’s what distinguishes bosons (i.e. force-carrying particles, such as photons) from fermions (i.e. matter-particles, such as electrons).Â The mathematical description is complete and respects Occam’s Razor. There is no redundancy. One cannot further simplify: every logical possibility in the mathematical description reflects a physical possibility in the real world.

Having said that, there is more to an electron than just Fermi-Dirac statistics, of course. What about its charge, and this weird number, its spin?,

Well… That’s what’s this post is about. As Feynman puts it: “So far we have been considering only spin-zero electrons and photons, fakeÂ electrons and fakeÂ photons.”

I wouldn’t call them ‘fake’, because they do behave like real photons and electrons already but… Yes. We can make them more ‘real’ by including charge andÂ spin in the discussion.Â Let’s go for it.

Charge and spin

From what I wrote above, it’s clear that the dichotomy between bosons and fermions (i.e. between ‘matter-particles’ and ‘force-carriers’ or, to put it simply, between light and matter) is not based on the (electric) charge. It’s true we cannot pile atoms or molecules on top of each other because of the repulsive forces between the electron cloudsâ€”but it’s not impossible, asÂ nuclear fusion proves: nuclear fusionÂ is possible because the electrostatic repulsive force can be overcome, and then theÂ nuclear force is much stronger (and, remember, no quarks are being destroyed or created: all nuclearÂ energy that’s being released or used is nuclear binding energy).

It’s also true that the force-carriers we know best, notably photons and gluons, doÂ notÂ carry any (electric) charge, as shown in the table below. So that’s another reason why we might, mistakenly, think that charge somehow defines matter-particles. However, we can see that matter-particles, first carry very different charges (positive or negative, and with very different values: 1/3, 2/3 or 1), and even be neutral, like the neutrinos. So, if there’s a relation, it’s very complex. In addition, one of the two force-carrier for the weak force, the W boson, can have positive or negative charge too, so that doesn’t make sense, does it? [I admit the weak force is a bit of a ‘special’ case, and so I should leave it out of the analysis.] The point is: the electric charge is what it is, but it’s not what defines matter. It’s justÂ one of the possible charges that matter-particles can carry. [The other charge, as you know, is the color charge but, to confuse the picture once again, that’s a charge that can also be carried by gluons, i.e. the carriers of the strong force.]

So what is it, then? Well… From the table above, you can see that the property of ‘spin’ (i.e. the third number in the top left-hand corner) matches the above-mentioned dichotomy in behavior, i.e. the two different types of interference (bosons versus fermions or, to use a heavier term, Bose-Einstein statistics versus Fermi-Dirac statistics): all matter-particles are so-called spin-1/2 particles, while all force-carriers (gauge bosons) all have spin one. [Never mind the Higgs particle: that’s ‘just’ a mechanism to give (most) elementary particles some mass.]

So why is that? Why are matter-particles spin-1/2 particles and force-carries spin-1 particles?Â To answer that question, we need to answer the question: what’s thisÂ spin number? And to answer that question, we first need to answer the question: what’s spin?

Spin in the classical world

In the classical world, it’s, quite simply, the momentumÂ associated with a spinningÂ or rotating object, which is referred to as the angular momentum. We’ve analyzed the math involved in another post, and so I won’t dwell on that here, but you should note that, in classical mechanics, we distinguish two types of angular momentum:

1. Orbital angular momentum: that’s the angular momentum an object gets from circling in an orbit, like the Earth around the Sun.
2. Spin angular momentum: that’s the angular momentum an object gets from spinning around its own axis., just like the Earth, in addition to rotating around the Sun, is rotating around its own axis (which is what causes day and night, as you know).

The math involved in both is pretty similar, but it’s still useful to distinguish the two, if only because we’ll distinguish them in quantum mechanics too! Indeed, when I analyzed the math in the above-mentioned post, I showed how we represent angular momentum by a vector that’s perpendicular to the direction of rotation, with its direction given by the ubiquitous right-hand ruleâ€”as in the illustration below, which shows both the angular momentum (L) as well as the torque (Ï„) that’s produced by a rotating mass. The formulas are given too: the angular momentum L is the vector cross product of the position vectorÂ r and the linear momentumÂ p, while the magnitude of the torqueÂ Ï„ is given by the product of the length of the lever arm and the applied force. An alternative approach is to define theÂ angular velocityÂ Ï‰ and the moment of inertiaÂ I, and we get the same result: L = IÏ‰.Â

Of course, the illustration above shows orbital angular momentum only and, as you know, we no longer have a ‘planetary model’ (aka the Rutherford model) of an atom.Â So should we be looking at spin angular momentum only?

Well… Yes and no. More yes than no, actually. But it’s ambiguous. In addition, the analogy between the concept of spin in quantum mechanics, and the concept of spin in classical mechanics, is somewhat less than straightforward. Well… It’s not straightforward at all actually. But let’s get on with it and use more precise language. Let’s first explore it for light, not because it’s easier (it isn’t) but… Well… Just because. ðŸ™‚

The spin of a photon

I talked about the polarization of light in previous posts (see, for example, my post on vector analysis): when we analyze light as a traveling electromagnetic wave (so we’re still in the classical analysis here, not talking about photons as ‘light particles’), we know that the electric field vector oscillates up and down and is, in fact, likely to rotate in the xy-plane (with z being the direction of propagation). The illustration below shows the idealized (akaÂ limiting)Â case of perfectly circular polarization: if there’s polarization, it is more likely to be elliptical. The other limiting case is plane polarization: in that case, the electric field vector just goes up and down in one direction only. [In case you wonder whether ‘real’ light is polarized, it often is: there’s an easy article on that on the Physics Classroom site.]

The illustration above uses Dirac’s bra-ket notation |LâŒª and |RâŒª to distinguish the two possible ‘states’, which are left- or right-handed polarization respectively. In case you forgot about bra-ket notations, let me quickly remind you: an amplitude is usually denoted by âŒ©x|sâŒª, in whichÂ âŒ©x| is the so-called ‘bra’, i.e. the final condition, and |sâŒª is the so-called ‘ket’, i.e. the starting condition, soÂ âŒ©x|sâŒª could mean: a photon leaves at sÂ (from source) and arrives at x. It doesn’t matter much here. We could have used any notation, as we’re just describing some state, which is either |LâŒª (left-handed polarization) or |RâŒª (right-handed polarization).Â The more intriguing extras in the illustration above, besides the formulas, are the values:Â Â±Â Ä§ = Â±h/2Ï€. So that’s plus or minus the (reduced) Planck constant which, as you know, is a very tiny constant. I’ll come back to that. SoÂ whatÂ exactly is being represented here?

At first, you’ll agree it looks very much like the momentumÂ of light (p) which, in a previous post, we calculated from the (average) energy (E) as p = E/c. Now, we know that E is related to the (angular) frequency of the light through the Planck-Einstein relation E = hÎ½ = Ä§Ï‰. Now, Ï‰ is the speed of light (c) times the wave number (k), so we can write: p = Ä§Ï‰ = Ä§ck/c = Ä§k. The wave number is theÂ ‘spatial frequency’, expressed either in cycles per unit distance (1/Î») or, more usually, in radians per unit distance (k = 2Ï€/Î»), so we can also write p = Ä§k = h/Î». Whatever way we write it, we find that this momentum (p) depends on the energy and/or, what amounts to saying the same, the frequency and/or the wavelength of the light.

So… Well… The momentum of light isÂ notÂ justÂ hÂ or Ä§, i.e. what’s written in that illustration above. So it must be something different. In addition, I should remind you this momentum was calculated from the magnetic field vector, as shown below (for more details, see my post on vector calculus once again), so it had nothing to do with polarization really.

Finally, last but not least, the dimensions of Ä§ and p = h/Î»Â are also different (when one is confused, it’s always good to do a dimensional analysis in physics):

1. The dimension of Planck’s constant (bothÂ h as well as Ä§ = h/2Ï€) is energy multiplied by time (JÂ·s or eVÂ·s) or, equivalently, momentum multiplied by distance. It’s referred to as the dimension ofÂ actionÂ in physics, and h is effectively, the so-called quantum of action.
2. The dimension of (linear) momentum is… Well… Let me think… Mass times velocity (mv)… But what’s the mass in this case? Light doesn’t have any mass. However, we can use the mass-energy equivalence: 1 eVÂ = 1.7826Ã—10âˆ’36 kg. [10âˆ’36? Well… Yes. An electronvolt is aÂ veryÂ tiny measure of energy.] So we can express p in eVÂ·m/s units.

Hmm… We can check: momentum times distance gives us the dimension of Planck’s constant again â€“Â (eVÂ·m/s)Â·m = eVÂ·s. OK. That’s good… […] But… Well… All of this nonsense doesn’t make us much smarter, does it? ðŸ™‚ Well… It may or may not be more useful to note that the dimension of action is, effectively, the same as the dimension of angular momentum.Â Huh?Â Why? Well… From our classicalÂ LÂ = rÃ—pÂ formula, we find L should be expressed in mÂ·(eVÂ·m/s) = eVÂ·m2/sÂ  units, so that’s… What?Â Well… HereÂ we need to use a little trick and re-express energy in mass units. We can then write L in kgÂ·m2/s units and, because 1 Newton (N) is 1 kgâ‹…m/s2, the kgÂ·m2/s unit is equivalent to the NÂ·mÂ·s = JÂ·s unit. Done!

Having said that, all of this still doesn’t answer the question: are the linear momentum of light, i.e. our p, and those two angular momentum ‘states’,Â |LâŒª and |RâŒª, related? Can we relate |LâŒª and |RâŒª to thatÂ LÂ = rÃ—pÂ formula?

The answer is simple: no. The |LâŒª andÂ |RâŒª states represent spinÂ angular momentum indeed, while the angular momentum we would derive from the linearÂ momentum of light using thatÂ LÂ = rÃ—pÂ isÂ orbitalÂ angular momentum. Let’s introduce the proper symbols: orbital angular momentum is denoted by L, while spin angular momentum is denoted by S. And then the total angular momentum is, quite simply, J = L + S.

L and S can both be calculated using either a vector cross product r Ã— pÂ (but using different values for r and p, of course) or, alternatively, using the moment of inertia tensor I and the angular velocity Ï‰. TheÂ illustrations below (which I took from Wikipedia) show how, and also shows how L and S are added to yieldÂ J = L + S.

So what? Well… Nothing much. The illustration above show that the analysis â€“ which is entirely classical, so far â€“Â is pretty complicated. [You should note, for example, that in the S = IÏ‰Â and L =Â IÏ‰Â formulas, we don’t use the simple (scalar) moment of inertia but the moment of inertia tensorÂ (so that’s a matrix denoted by I, instead of the scalar I), because SÂ (or L) and Ï‰ are not necessarily pointing in the same direction.

By now, you’re probably very confused and wondering what’s wiggling really. The answer for the orbital angular momentum is: it’s the linear momentum vector p. Now…

Hey! Stop! Why would that vector wiggle?

You’re right. Perhaps it doesn’t. The linear momentum p is supposed to be directed in the direction of travel of the wave, isn’t it? It is. In vector notation, we haveÂ p = Ä§k, and that k vector (i.e. the wavevector)Â points in the direction of travel of the wave indeed and so…Â Well… No. It’s not that simple. The wave vector is perpendicular to the surfaces of constant phase, i.e. the so-called wave fronts, as show in the illustration below (see the direction of ek, which is a unit vector in the direction of k).

So, yes, if we’re analyzing light moving in a straight one-dimensional line only, or we’re talking a plane wave, as illustrated below, then the orbital angular momentum vanishes.

But the orbital angular momentumÂ L does not vanish when we’re looking at a real light beam, like the ones below. Real waves? Well… OK… The ones below are idealized wave shapes as well, but let’s say they are somewhatÂ moreÂ real than a plane wave. ðŸ™‚

So what do we have here? We have wavefronts that are shaped as helices, except for the one in the middle (marked by m = 0) which is, once again, an example of plane waveâ€”so for that one (m = 0), we have zero orbital angular momentum indeed. But look,Â veryÂ carefully, at the m =Â Â± 1 and m =Â Â± 2 situations. ForÂ m =Â Â± 1, we have one helical surface withÂ a step length equal to the wavelength Î». ForÂ m =Â Â± 2, we have two intertwined helical surfacesÂ with the step length of each helix surface equal to 2Î». [Don’t worry too much about the second and third column: they show a beam cross-section (so that’s notÂ a wave front but a so-calledÂ phaseÂ front)Â and the (averaged) light intensity,Â again of a beam cross-section.]Â Now, we can further generalize and analyze waves composed of m helices with the step length of each helix surface equal to |m|Î». The Wikipedia article on OAM (orbital angular momentum of light), from which I got this illustration, gives the following formula to calculate the OAM:

The same article also notes that the quantum-mechanical equivalent of this formula, i.e. the orbital angular momentum of theÂ photons one would associate with the not-cylindrically-symmetric waves above (i.e. all those for which m â‰ Â 0), is equal to:

Lz = mÄ§

So what? Well… I guess we should just accept that as a very interesting result. For example, I dulyÂ note that LzÂ is along the direction of propagation of the wave (as indicated by the z subscript), and I also note the veryÂ interesting fact that, apparently, Lz Â can be either positive or negative. Now, I am not quite sure how such result is consistent with the idea of radiation pressure, but I am sure there must be some logical explanation to that. The other point you should note is that, once again, any reference to the energy (or to the frequency or wavelength) of our photon has disappeared. Hmm… I’ll come back to this, as I promised above already.

The thing is that this rather long digression on orbital angular momentum doesn’t help us much in trying to understand what thatÂ spinÂ angular momentum (SAM) is all about. So, let me just copy the final conclusion of the Wikipedia article on the orbitalÂ angular momentum of light: the OAM is the component of angular momentum of light thatÂ is dependent on the field spatial distribution, not on the polarization of light.

So, again, what’s the spin angular momentum?Â Well… The only guidance we have is that same little drawing again and, perhaps, another illustration that’s supposed to compare SAM with OAM (underneath).

Now, the Wikipedia article on SAMÂ (spin angular momentum), from which I took the illustrations above, gives a similar-looking formula for it:

When I say ‘similar-looking’, I don’t mean it’s the same. [Of course not! Spin and orbital angular momentum are two different things!]. So what’s different in the two formulas? Well… We don’t have any del operator (âˆ‡) in the SAM formula, and we also don’t have any position vector (r) in the integral kernel (or integrand, if you prefer that term). However, we do find both the electric field vector (E) as well as the (magnetic) vector potential (A) in the equation again. Hence, the SAM (also) takes both theÂ electric as well as the magnetic field into account, just like the OAM. [According to the author of the article, theÂ expression also shows that the SAM is nonzero when the light polarization is elliptical or circular, and that it vanishes if the light polarization is linear, but I think that’s muchÂ more obvious from the illustration than from the formula… However,Â I realize I really need to move on here, because this post is, once again, becoming way too long. So…]

OK. What’s the equivalent of that formula in quantum mechanics?

Well… In quantum mechanics, the SAM becomes a ‘quantum observable’, described by a corresponding operator which has only two eigenvalues:

Sz = Â±Â Ä§

So that corresponds to the two possible values for Jz, as mentioned in the illustration, and we can understand, intuitively, that these two values correspond to two ‘idealized’ photons which describe a left- and right-handed circularly polarized wave respectively.

So… Well… There we are. That’s basically all there is to say about it. So… OK. So far, so good.

But… Yes? Why do we call a photon a spin-one particle?

That has to do with convention.Â A so-called spin-zero particle has no degrees of freedom in regard to polarization. The implied ‘geometry’ is that a spin-zero particle is completely symmetric: no matter in what direction you turn it,Â it will always look the same. In short, it really behaves like a (zero-dimensional) mathematical point. As you can see from the overview of all elementary particles, it is only the Higgs boson which has spin zero. That’s why the Higgs field is referred to as a scalar field: it has no direction.Â In contrast, spin-one particles, like photons, are also ‘point particles’, but they do come with one or the other kind of polarization, as evident from all that I wrote above. To be specific, they are polarized in the xy-plane, and can have one of two directions. So, when rotating them, you need a full rotation of 360Â° if you want them to look the same again.

Now that I am here, let me exhaust the topic (to a limited extent only, of course, as I don’t want to write a book here) and mention that, in theory, we could also imagine spin-2 particles, which would look the same after half a rotation (180Â°). However, as you can see from the overview, none of the elementary particles has spin-2. A spin-2 particle could be like someÂ straight stick, as that looks the same even after it is rotated 180 degrees. I am mentioning the theoretical possibility because the graviton, if it would effectively exist, is expected to be a massless spin-2 boson. [Now why do I mention this? Not sure. I guess I am just noting this to remind you of the fact that the Higgs boson is definitely not the (theoretical) graviton, and/or that we have no quantum theory for gravity.]

Oh… That’s great, you’ll say. But what about all those spin-1/2 particles in the table? You said that all matter-particles are spin 1/2 particles, and that it’s this particular property that actually makes them matter-particles. So what’s the geometry here? What kind of ‘symmetries’ do they respect?

Well… As strange as it sounds, aÂ spin-1/2 particle needs two full rotations (2Ã—360Â°=720Â°) until it is again in the same state. Now, in regard to that particularity, you’ll often read something like: “There isÂ nothing in our macroscopic world which has a symmetry like that.” Or, worse, “Common sense tells us that something like that cannot exist, that it simply is impossible.” [I won’t quote the site from which I took this quotes, because it is, in fact, the site of a very respectable Â research center!] Bollocks!Â The Wikipedia article on spin has this wonderful animation: look at how the spirals flip between clockwise and counterclockwise orientations, and note that it’s only after spinning a full 720 degrees that this ‘point’ returns to its original configuration after spinning a full 720 degrees.

So, yes, we canÂ actually imagine spin-1/2 particles, and with not all that much imagination, I’d say. But…Â OK… This is all great fun, but we have to move on. So what’s the ‘spin’ of these spin-1/2 particles and, more in particular, what’s the concept of ‘spin’ of an electron?

The spin of an electron

When starting to read about it, I thought that the angular momentum of an electron would be easier to analyze than that of a photon. Indeed, while a photon has no mass and no electric charge, thatÂ analysis with thoseÂ E and B vectors is damn complicated, even when sticking to a strictly classical analysis. For an electron, the classical picture seems to be much more straightforwardâ€”but only at first indeed. It quickly becomes equally weird, if not more.

We can look at an electron in orbit as aÂ rotatingÂ electrically charged ‘cloud’Â indeed. Now,Â from Maxwell’s equations (or from your high school classes even), you know that aÂ rotating electric charged body creates a magnetic dipole. So an electron should behave just like a tiny bar magnet. Of course, we have to make certain assumptions about the distribution of the charge in space but, in general, we can write that the magnetic dipole moment Î¼ is equal to:

In case you want to know, in detail, where this formula comes from, let me refer you to Feynman once again, but trust me â€“ for once ðŸ™‚ â€“ it’s quite straightforward indeed: the L in this formula is the angular momentum, whichÂ may be the spin angular momentum, the orbital angular momentum, or the total angular momentum. The e and m are, of course, the charge and mass of the electron respectively.

So that’s a good and nice-looking formula, and it’s actually even correct exceptÂ for theÂ spinÂ angular momentum as measured in experiments. [You’ll wonder how we can measure orbital and spin angular momentum respectively, but I’ll talk about an 1921 experiment in a few minutes, and so that will give you some clue as to that mystery. :-)] To be precise, it turns out that one has to multiply the above formula for Î¼Â with a factor which is referred to as the g-factor. [And, no, it’s got nothing to do with the gravitational constant or… Well… Nothing. :-)] So, for theÂ spinÂ angular momentum, the formula should be:

Experimental physicists are constantly checking that value and they know measure it to be something like g = is 2.00231930419922 Â± 1.5Ã—10âˆ’12. So what’s the explanation for that g? Where does it come from? There is, in fact, a classical explanation for it, which I’ll copy hereunder (yes, from Wikipedia). This classical explanation is based on assuming that the distribution of the electric charge of the electron and its mass does not coincide:

Why do I mention this classical explanation? Well… Because, in most popular books on quantum mechanics (including Feynman’s delightful QED), you’ll read that (a) the value for g can be derived from a quantum-theoretical equation known as Dirac’s equation (or ‘Dirac theory’, as it’s referred to above) and, more importantly, that (b) physicists call the “accurate prediction of the electron g-factor” from quantum theory (i.e. ‘Dirac’s theory’ in this case) “one of the greatest triumphs” of the theory.

So what about it? Well… Whatever the merits of both explanations (classical or quantum-mechanical), they are surely not the reason why physicists abandoned the classical theory. So what was the reason then? What a stupid question!Â You know that already! The Rutherford model was, quite simply, not consistent: according to classical theory, electrons should just radiate their energy away and spiral into the nucleus.Â More in particular, there was yet another experiment that wasn’t consistent with classical theory, and it’s one that’s very relevant for the discussion at hand: it’s the so-called Stern-Gerlach experiment.

It was just as ‘revolutionary’ as the Michelson-Morley experiment (which couldn’t measure the speed of light), or the discovery of the positron in 1932. The Stern-Gerlach experiment was done in 1921, so that’s many years before quantum theory replaced classical theory and, hence, it’s not one of those experiments confirmingÂ quantum theory. No. Quite the contrary. It was, in fact, one of the experiments that triggeredÂ the so-called quantum revolution. Let me insert the experimental set-up and summarize it (below).

• The German scientists Otto Stern and Walther Gerlach produced a beam of electrically-neutral silver atoms and let it pass through a (non-uniform) magnetic field. Why silver atoms? Well… Silver atoms are easy toÂ handle (in a lab, that is) and easy to detect with a photoplate.
• These atoms came out of an oven (literally), in which the silver was being evaporated (yes, one can evaporate silver), so they had no special orientation in space and, so Stern and Gerlach thought, the magnetic momentÂ (orÂ spin) of the outer electrons in these atoms would point into all possible directions in space.
• As expected, the magnetic field did deflect the silver atoms, just like it would deflect little dipole magnets if you would shoot them through the field. However, the pattern of deflection was not the one which they expected. Instead of hitting the plate all over the place, within some contour, of course, only the contour itself was hitÂ by the atoms. There was nothing in the middle!
• And… Well… It’s a long story, but I’ll make it short. There was only one possible explanation for that behavior, and that’s that the magnetic moments â€“ and, therefore the spins â€“ had only two orientations in space, and two possible values only which â€“ Surprise, surprise!Â â€“ areÂ Â±Ä§/2 (so that’s halfÂ the value of the spin angular momentum of photons, which explains theÂ spin-1/2 terminology).

The spin angular momentum of an electron is more popularly known as ‘up’ or ‘down’.

So… What about it? Well… It explains why a atomic orbital can haveÂ twoÂ electrons, rather than one only and, as such, the behavior of the electron here is the basis of the so-called periodic table, which explains all properties of the chemical elements. So… Yes. Quantum theory is relevant, I’d say. ðŸ™‚

Conclusion

This has been a terribly long post, and you may no longer remember what I promised to do. What I promised to do, is to write some more about the difference between a photon and an electron and, more in particular, I said I’d write more about theirÂ charge, and that “weird number”: their spin. I think I lived up to that promise. The summary is simple:

1. Photons have no (electric) charge, but they do have spin. Their spin is linked to their polarization in the xy-plane (if z is the direction of propagation) and, because of the strangeness of quantum mechanics (i.e. the quantizationÂ of (quantum) observables),Â the value for this spin is either +Ä§ or â€“Ä§, which explains why they are referred to as spin-one particles (because either value is one unit of the Planck constant).
2. Electrons have both electric charge as well as spin. Their spin is different and is, in fact, related to their electric charge. It can be interpreted as the magnetic dipole moment, which results from the fact we have a spinning charge here. However, again, because of the strangeness of quantum mechanics, their dipole moment is quantized and can take only one of two values: Â±Ä§/2, which is why they are referred to as spin-1/2 particles.

So now you know everything you need to know about photons and electrons, and then I mean real photons and electrons, including their properties of charge and spin. So they’re no longer ‘fake’ spin-zero photons and electrons now. Isn’t that great? You’ve just discovered the real world! ðŸ™‚

So… I am doneâ€”for the moment, that is… ðŸ™‚Â If anything, I hope this post shows that even those ‘weird’ quantum numbers are rooted in ‘physical reality’ (or in physical ‘geometry’ at least), and that quantum theory may be ‘crazy’ indeed, but that it ‘explains’ experimental results. Again, as Feynman says:

“We understand how Nature works, but not whyÂ Nature works that way. Nobody understands that. I can’t explain why Nature behave in this particular way. You may not like quantum theory and, hence, you may not accept it. But physicists have learned to realize that whether they like a theory or not isÂ notÂ the essential question. Rather, it is whether or not the theory gives predictions that agree with experiment. The theory of quantum electrodynamics describes Nature as absurd from the point of view of common sense. But it agrees fully with experiment. So I hope you can accept Nature as She isâ€”absurd.”

Frankly speaking, I am not quite prepared to accept Nature as absurd: I hope that some more familiarization with the underlying mathematical forms and shapes will make it look somewhat less absurd. More, I hope that such familiarization will, in the end, make everything look just as ‘logical’, or ‘natural’ as the two ways in which amplitudes can ‘interfere’.

Post scriptum: I said I would come back to the fact that, in the analysis of orbital and spin angular momentum of a photon (OAM and SAM), the frequency or energy variable sort of ‘disappears’. So why’s that? Let’s look at those expressions for |LâŒª and |RâŒª once again:

What’s written here really? If |LâŒª and |RâŒª are supposed to be equal toÂ either +Ä§ or â€“Ä§, then that product of ei(kzâ€“Ï‰t)Â with the 3Ã—1 matrix (which is a ‘column vector’ in this case) does not seem to make much sense, does it? Indeed, you’ll remember thatÂ ei(kzâ€“Ï‰t)Â just a regular wave function. To be precise, its phase isÂ Ï† = kzâ€“Ï‰t (with z the direction of propagation of the wave), and its real and imaginary part can be written as eiÏ†Â = cos(Ï†) + isin(Ï†) = a + bi. Multiplying it with that 3Ã—1 column vector (1, i, 0) or (1, â€“i, 0) just yields another 3Ã—1 column vector. To be specific, we get:

1. The 3Ã—1 ‘vector’ (a + bi, â€“b+ai, 0) for |LâŒª, and
2. The 3Ã—1 ‘vector’ (a + bi, bâ€“ai, 0) for |RâŒª.

So we have two new ‘vectors’ whose components are complex numbers. Furthermore, we can note that their ‘x’-component is the same, their ‘y’-component is each other’s opposite â€“b+ai = â€“(bâ€“ai), and their ‘z’-component is 0.

So… Well… In regard to their ‘y’-component, I should note that’s just the result of the multiplication with iÂ and/or â€“i: multiplying a complex number with i amounts to a 90Â° degree counterclockwise rotation, while multiplication with â€“i amounts to the same but clockwise. Hence, we must arrive at two complex numbers that are each other’s opposite. [Indeed, in complex analysis, the valueÂ â€“1 = eiÏ€Â = eâ€“iÏ€Â is a 180Â° rotation, both clockwise (Ï† = â€“Ï€) or counterclockwise (Ï† = +Ï€), of course!.]

Hmm… Still… What does it all mean really? The truth is that it takes some more advanced math to interpret the result. To be precise, pure quantum states, such |LâŒª and |RâŒª here, are represented by so-called ‘state vectors’ in a Hilbert space over complex numbers. So that’s what we’ve got here. So… Well… I can’t say much more about this right now: we’ll just need to study some more before we’ll ‘understand’ those expressions for |LâŒª and |RâŒª. So let’s not worry about it right now. We’ll get there.

Just for the record, I should note that, initially, I thought 1/âˆš2 factor in front gave some clue as to what’s going on here: 1/âˆš2 â‰ˆ 0.707 isÂ a factor that’s used to calculate the root mean square (RMS) value for a sine wave. It’s illustrated below. The RMSÂ value is a ‘special average’ one can use to calculate the energy or power (i.e. energy per time unit) of a wave. [Using the term ‘average’ is misleading, because the average of a sine wave is 1/2 over half a cycle, and 0 over a fully cycle, as you can easily see from the shape of the function. But I guess you know what I mean.]

Indeed, you’ll remember that the energy (E) of a wave is proportional to the square of its amplitude (A): EÂ âˆ¼ A2. For example, when we have a constant current I, the power P will be proportional to its square: P âˆ¼ I2. With a varying current (I) and voltage (V), the formula is more complicated but we can simply it using the rms values: PavgÂ = VRMSÂ·IRMS.

So… Looking at that formula, should we think of h and/or Ä§Â as some kind of ‘average’ energy, like the energy of a photon per cycle or per radian? That’s an interesting idea so let’s explore it. If the energy of a photon is equal to E = hÂ·Î½ = hÂ·Ï‰/2Ï€ =Â Ä§Ï‰, then we can also write:

h = E/Î½ and/or Ä§Â = E/Ï‰

So, yes:Â hÂ is the energy of a photon per cycle obviously and, because the phaseÂ covers 2Ï€ radians during each cycle, and Ä§ must be the energy of the photonÂ per radian! That’s a great result, isn’t it? It also gives a wonderfully simple interpretation to Planck’s quantum of action!

Well… No. We made at least two mistakes here. The first mistake is that if we think of a photon as wave train being radiated by an atomÂ â€“ which, as we calculated in another post, lasts aboutÂ 3.2Ã—10â€“8 seconds â€“ the graph for its energy is going to resemble the graph of its amplitude, so it’s going to die out and each oscillation will carry less and less energy. Indeed, the decay time given here (Ï„ =Â 3.2Ã—10â€“8 seconds) was the time it takes for the radiation (we assumed sodium light with a wavelength of 600 nanometer) to die out by a factor 1/e. To be precise, the shape of the energy curve is E = E0eâˆ’t/Ï„, and so it’s an envelope resembling the A(t) curve below.

Indeed, remember, the energy of a wave is determined not only by its frequency (or wavelength) but also by its amplitude, and so we cannot assume the amplitude of a ‘photon wave’ is going to be the same everywhere. Just for the record: note that the energy of a wave is proportional to the frequency (doubling the frequency doubles the energy) but, when linking it to the amplitude, we should remember that the energy is proportional to theÂ squareÂ of the amplitude, so we write EÂ âˆ¼ A2.

The second mistake is that both Î½ and Ï‰ are the light frequency (expressed in cycles or radians respectively) of the light per second,Â i.e per time unit. So that’s notÂ the number of cycles or radians that we should associate with the wavetrain!Â We should use the number of cycles (or radians) packed into one photon. We can calculate that easily from the value for the decay time Ï„. Indeed, for sodium light, which which has a frequency of 500 THz (500Ã—1012Â oscillations per second) and a wavelength of 600 nm (600Ã—10â€“9Â meter), we said the radiation lasts about 3.2Ã—10â€“8Â seconds (thatâ€™s actually the time it takes for the radiation’s energy to die out by a factor 1/e, so the wavetrain will actually last (much)Â longer, but so the amplitude becomes quite small after that time), and so that makes for some 16 million oscillations, and a ‘length’ of the wavetrain of about 9.6 meter! Now, the energy of a sodium light photon is about 2eV (hÂ·Î½ â‰ˆ 4Ã—10âˆ’15Â electronvoltÂ·second times 0.5Ã—1015Â cycles/sec = 2eV) and so we could say the average energy of each of those 16 million oscillations would beÂ 2/(16Ã—106) eV = 0.125Ã—10â€“6Â eV. But, from all that I wrote above, it’s obvious that this number doesn’t mean all that much, because the wavetrain is not likely to be shaped very regularly.

So, in short, we cannot say thatÂ h is the photon energy per cycleÂ or that Ä§ isÂ the photon energy per radian!Â  That’s not only simplistic but, worse, false. Planck’s constant is what is is: a factor of proportionality for which there is no simple ‘arithmetic’ and/or ‘geometric’ explanation. It’s just there, and we’ll need to study some more math to truly understand the meaning of those two expressions forÂ |LâŒª and |RâŒª.

Having said that, and having thought about itÂ all some more, I find there’s, perhaps, a more interesting way to re-writeÂ E = hÂ·Î½:

h = E/Î½ = (Î»/c)E = TÂ·E

T? Yes. T is the period, so that’s the time needed for one oscillation: T is just the reciprocal of the frequency (T = 1/Î½ = Î»/c). It’s a very tiny number, because we divide (1) a very small number (the wavelength of light measured in meter) by (2) a very large number (the distance (in meter) traveled by light). For sodium light, T is equal to 2Ã—10â€“15Â seconds, so that’s twoÂ femtoseconds, i.e. twoÂ quadrillionths ofÂ a second.

Now, we can think of the period as a fraction of a second, and smaller fractions are, obviously, associated with higher frequencies and, what amounts to the same, shorter wavelengths (and, hence, higher energies). However, when writing T = Î»/c, we can also think of T being another kind of fraction: Î»/cÂ can also be written asÂ the ratio of the wavelength and the distance traveled by light in one second, i.e. a light-second (remember that light-seconds are measures of length, not of distance). The two fractions are the same when we express time and distance in equivalent units indeed (i.e. distance in light-second, or time in sec/cÂ units).

So that links hÂ to both time as well as distance and we may look at h as some kind of fraction or energy ‘density’ even (although the term ‘density’ in this context is not quite accurate). In the same vein, I should note that, if there’s anything that should make you think about h, is the fact that its valueÂ depends on how we measure time and distance.Â For example, if w’d measure time in other units (for example, the more ‘natural’ unit defined by the time light needs to travel one meter), then we get a different unit for h. And, of course, you also know we can relate energy to distance (1 J = 1 NÂ·m). But that’s something that’s obvious from h‘s dimension (JÂ·s), and so I shouldn’t dwell on that.

Hmm… Interesting thoughts. I think I’ll develop these things a bit further in one of my next posts. As for now, however, I’ll leave you with your own thoughts on it.

Note 1: As you’re trying to follow what I am writing above, you may have wondered whether or not the duration of the wavetrain that’s emitted by an atom is a constant, or whether or not it packs some constant number of oscillations. I’ve thought about that myself as I wrote down the following formula at some point of time:

h = (the duration of the wave)Â·(the energy of the photon)/(the number of oscillations in the wave)

As mentioned above, interpreting h as some kind of average energy per oscillation is not a great idea but, having said that, it would be a good exercise for you to try to answer that question in regard to the duration of these wavetrains, and/or the number of oscillations packed into them, yourself. There are various formulasÂ for the Q of an atomic oscillator, but the simplest one is the one expressed in terms of the so-called classical electron radius r0:

Q = 3Î»/4Ï€r0

As you can see, the Q depends onÂ Î»: higher wavelengths (so lowerÂ energy) are associated with higher Q. In fact, the relationship is directly proportional: twice the wavelength will give you twice the Q. Now, the formula for the decay timeÂ Ï„ is also dependent on the wavelength. Indeed,Â Ï„ = 2Q/Ï‰ = QÎ»/Ï€c. Combining the two formulas yields (if I am not mistaken):

Ï„ =Â 3Î»2/4Ï€2r0c.

Hence, the decay time is proportional to the square of the wavelength. Hmm… That’s an interesting result. But I really need to learn how to be a bit shorter, and so I’ll really let you think now about what all this means or could mean.

Note 2: If that 1/âˆš2 factor has nothing to do with some kind of rms calculation, where does it come from? I am not sure. It’s related to state vector math, it seems, and I haven’t started that as yet. I just copy a formula from Wikipedia here, which shows the same factor in front:

The formula above is said to represent the “superposition of joint spin states for two particles”.Â My gut instinct tells me 1/âˆš2 factor has to do with the normalization condition and/or with the fact that we have to take the (absolute) square of the (complex-valued) amplitudes to get the probability.

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