# Tag Archives: interpretation of the wavefunction

# God’s Number explained

My posts on the fine-structure constant – God’s Number as it is often referred to – have always attracted a fair amount of views. I think that’s because I have always tried to clarify this or that relation by showing *how and why exactly *it pops us in this or that formula (e.g. Rydberg’s energy formula, the ratio of the various radii of an electron (Thomson, Compton and Bohr radius), the coupling constant, the anomalous magnetic moment, etcetera), as opposed to what most seem to try to do, and that is to further mystify it. You will probably not want to search through all of my writing so I will just refer you to my summary of these efforts on the viXra.org site: “*Layered Motions: the Meaning* *of the Fine-Structure Constant.*”

However, I must admit that – till now – I wasn’t quite able to answer this very simple question: what *is* that fine-structure constant? Why *exactly *does it appear as a scaling constant or a coupling constant in almost any equation you can think of but *not *in, say, Einstein’s mass-energy equivalence relation, or the *de Broglie *relations?

I finally have a final answer (pun intended) to the question, and it’s surprisingly easy: it is the radius of the naked charge in the electron expressed in terms of the natural distance unit that comes out of our realist interpretation of what an electron actually is. [For those who haven’t read me before, this realist interpretation is based on Schrödinger’s discovery of the *Zitterbewegung *of an electron.] That natural distance unit is the *Compton *radius of the electron: it is the effective radius of an electron as measured in inelastic collisions between high-energy photons and the electron. I like to think of it as a quantum of space in which interference happens but you will want to think that through for yourself.* *

The point is: that’s it. That’s all. All the other calculations follow from it. Why? It would take me a while to explain that but, if you carefully look at the logic in my classical calculations of the anomalous magnetic moment, then you should be able to understand why these calculations are somewhat more fundamental than the others and why we can, therefore, get everything else out of them. 🙂

**Post scriptum**: I quickly checked the downloads of my papers on Phil Gibbs’ site, and I am extremely surprised my very first paper (the quantum-mechanical wavefunction as a gravitational wave) of mine still gets downloads. To whomever is interested in this paper, I would say: the realist interpretation we have been pursuing – based on the Zitterbewegung model of an electron – is based on the idea of a naked charge (with zero rest mass) orbiting around some center. The energy in its motion – a perpetual current ring, really – gives the electron its (equivalent) mass. That’s just Wheeler’s idea of ‘mass without mass’. But the force is *definitely not* gravitational. It cannot be. The force has to grab onto something, and all it can grab onto here is that naked charge. The force is, therefore, electromagnetic. It must be. I now look at my very first paper as a first immature essay. It did help me to develop some basic intuitive ideas on what any realist interpretation of QM should look like, but the quantum-mechanical wavefunction has nothing to do with gravity. Quantum mechanics is electromagnetics: we just add the quantum. The idea of an elementary cycle. Gravity is dealt with by general relativity theory: energy – or its equivalent mass – bends spacetime. That’s very significant, but it doesn’t help you when analyzing the QED sector of physics. I should probably pull this paper of the site – but I won’t. Because I think it shows where I come from: very humble origins. 🙂

# Philosophy and Science: Dirac’s Principles

# Interpreting quantum mechanics

My book is moving forward. I just produced a very first promotional video. Have a look and let me know what you think of it ! 🙂

# The electron as a quantum-mechanical oscillator

# The metaphysics of physics

I realized that my last posts were just some crude and rude soundbites, so I thought it would be good to briefly summarize them into something more coherent. Please let me know what you think of it.

# The Uncertainty Principle: epistemology versus physics

Anyone who has read anything about quantum physics will know that its concepts and principles are very non-intuitive. Several interpretations have therefore emerged. The mainstream interpretation of quantum mechanics is referred to as the Copenhagen interpretation. It mainly distinguishes itself from more frivolous interpretations (such as the many-worlds and the pilot-wave interpretations) because it is… Well… Less frivolous. Unfortunately, the Copenhagen interpretation itself seems to be subject to interpretation.

One such interpretation may be referred to as radical skepticism – or radical empiricism[1]: we can only say something meaningful about Schrödinger’s cat if we open the box and observe its state. According to this rather particular viewpoint, we cannot be sure of its reality if we don’t make the observation. All we can do is describe its reality by a superposition of the two *possible *states: dead or alive. That’s Hilbert’s logic[2]: the two states (dead or alive) are mutually exclusive but we add them anyway. If a tree falls in the wood and no one hears it, then it is both standing and not standing. Richard Feynman – who may well be the most eminent representative of mainstream physics – thinks this epistemological position is nonsensical, and I fully agree with him:

“A real tree falling in a real forest makes a sound, of course, even if nobody is there. Even if no one is present to hear it, there are other traces left. The sound will shake some leaves, and if we were careful enough we might find somewhere that some thorn had rubbed against a leaf and made a tiny scratch that could not be explained unless we assumed the leaf were vibrating.” (*Feynman’s Lectures*, III-2-6)

So what is the mainstream physicist’s interpretation of the Copenhagen interpretation of quantum mechanics then? To fully answer that question, I should encourage the reader to read all of Feynman’s Lectures on quantum mechanics. But then you are reading this because you don’t want to do that, so let me quote from his introductory Lecture on the Uncertainty Principle: “Making an observation affects the phenomenon. *The point is that the effect cannot be disregarded or minimized or decreased arbitrarily by rearranging the apparatus. When we look for a certain phenomenon we cannot help but disturb it in a certain minimum way.*” (ibidem)

It has nothing to do with consciousness. Reality and consciousness are two very different things. After having concluded the tree did make a noise, even if no one was there to hear it, he wraps up the philosophical discussion as follows: “We might ask: was there a *sensation* of sound? No, sensations have to do, presumably, with consciousness. And whether ants are conscious and whether there were ants in the forest, or whether the tree was conscious, we do not know. Let us leave the problem in that form.” In short, I think we can all agree that the cat is dead *or *alive, or that the tree is standing or not standing¾regardless of the observer. It’s a binary situation. Not something in-between. The box obscures our view. That’s all. There is nothing more to it.

Of course, in quantum physics, we don’t study cats but look at the behavior of photons and electrons (we limit our analysis to quantum electrodynamics – so we won’t discuss quarks or other *sectors *of the so-called Standard Model of particle physics). The question then becomes: what can we reasonably say about the electron – or the photon – before we observe it, or before we make any measurement. Think of the Stein-Gerlach experiment, which tells us that we’ll always measure the angular momentum of an electron – along any axis we choose – as either +ħ/2 or, else, as -ħ/2. So what’s its *state *before it enters the apparatus? Do we have to assume it has some *definite* angular momentum, and that its value is as binary as the state of our cat (dead or alive, *up *or *down*)?

We should probably explain what we mean by a *definite *angular momentum. It’s a concept from classical physics, and it assumes a precise *value *(or magnitude) along some precise *direction*. We may challenge these assumptions. The direction of the angular momentum may be changing all the time, for example. If we think of the electron as a pointlike charge – whizzing around in its own space – then the concept of a precise direction of its angular momentum becomes quite fuzzy, because it changes all the time. And if its direction is fuzzy, then its value will be fuzzy as well. In classical physics, such fuzziness is not allowed, because angular momentum is conserved: it takes an outside force – or *torque *– to change it. But in quantum physics, we have the Uncertainty Principle: some energy (force over a distance, remember) can be borrowed – so to speak – as long as it’s swiftly being returned – within the quantitative limits set by the Uncertainty Principle: ΔE·Δt = ħ/2.

Mainstream physicists – including Feynman – do not try to think about this. For them, the Stern-Gerlach apparatus is just like Schrödinger’s box: it obscures the view. The cat is dead *or *alive, and each of the two states has some probability – but they must add up to one – and so they will write the *state *of the electron before it enters the apparatus as the superposition of the *up *and *down *states. I must assume you’ve seen this before:

|ψ〉 = *C*_{up}|up〉 + *C*_{down}|down〉

It’s the so-called *Dirac *or *bra-ket *notation. *C*_{up} is the amplitude for the electron spin to be equal to +ħ/2 along the chosen direction – which we refer to as the *z*-direction because we will choose our reference frame such that the *z*-axis coincides with this chosen direction – and, likewise, *C*_{up} is the amplitude for the electron spin to be equal to -ħ/2 (along the same direction, obviously). *C*_{up} and *C*_{up} will be functions, and the associated probabilities will vary sinusoidally – with a phase difference so as to make sure both add up to one.

The model is consistent, but it feels like a mathematical trick. This description of reality – if that’s what it is – does *not *feel like a model of a *real *electron. It’s like reducing the cat in our box to the mentioned fuzzy state of being alive and dead at the same time. Let’s try to come up with something more exciting. 😊

[1] Academics will immediately note that radical empiricism and radical skepticism are very different epistemological positions but we are discussing some basic principles in physics here rather than epistemological theories.

[2] The reference to Hilbert’s logic refers to Hilbert spaces: a Hilbert space is an abstract vector space. Its properties allow us to work with quantum-mechanical states, which become *state vectors*. You should not confuse them with the real or complex vectors you’re used to. The only thing state vectors have in common with real or complex vectors is that (1) we also need a *base *(aka as a *representation* in quantum mechanics) to define them and (2) that we can make linear combinations.

# The ‘flywheel’ electron model

Physicists describe the reality of electrons by a *wavefunction*. If you are reading this article, you know how a wavefunction looks like: it is a superposition of *elementary *wavefunctions. These elementary wavefunctions are written as A* _{i}*·exp(-

*i*θ

*), so they have an amplitude A*

_{i}*and an argument θ*

_{i}*= (E*

_{i}*/ħ)·t – (p*

_{i}*/ħ)·x. Let’s forget about uncertainty, so we can drop the index (*

_{i}*i*) and think of a geometric interpretation of A·exp(-

*i*θ) = A·

*e*

^{–i}

^{θ}.

Here we have a weird thing: physicists think the minus sign in the exponent (-*i*θ) should always be there: the convention is that we get the *imaginary unit *(*i*) by a 90° rotation of the real unit (1) – but the rotation is *counterclockwise *rotation. I like to think a rotation in the *clockwise *direction must also describe something real. Hence, if we are seeking a geometric interpretation, then we should explore the two mathematical possibilities: A·*e*^{–i}^{θ} and A·*e*^{+i}^{θ}. I like to think these two wavefunctions describe the same electron but with opposite spin. How should we visualize this? I like to think of A·*e*^{–i}^{θ} and A·*e*^{+i}^{θ} as two-dimensional harmonic oscillators:

A·*e*^{–i}^{θ} = cos(-θ) + *i*·sin(-θ) = cosθ – *i*·sinθ

A·*e*^{+i}^{θ} = cosθ + *i*·sinθ

So we may want to imagine our electron as a pointlike electric charge (see the green dot in the illustration below) to spin around some center in either of the two possible directions. The cosine keeps track of the oscillation in one dimension, while the sine (plus or minus) keeps track of the oscillation in a direction that is perpendicular to the first one.

**Figure 1: A pointlike charge in orbit**

So we have a weird oscillator in two dimensions here, and we may calculate the energy in this oscillation. To calculate such energy, we need a mass concept. We only have a charge here, but a (moving) charge has an *electromagnetic* mass. Now, the electromagnetic mass of the electron’s charge may or may not explain all the mass of the electron (most physicists think it doesn’t) but let’s assume it does for the sake of the model that we’re trying to build up here. The point is: the theory of electromagnetic mass gives us a very simple explanation for the concept of mass here, and so we’ll use it for the time being. So we have some mass oscillating in two directions simultaneously: we basically assume space is, somehow, elastic. We have worked out the V-2 engine *metaphor *before, so we won’t repeat ourselves here.

**Figure 2: A perpetuum mobile?**

Previously unrelated but *structurally similar* formulas may be related here:

- The energy of an oscillator: E = (1/2)·m·
*a*^{2}ω^{2} - Kinetic energy: E = (1/2)·m·
*v*^{2} - The rotational (kinetic) energy that’s stored in a flywheel: E = (1/2)·I·ω
^{2}= (1/2)·m·*r*^{2}·ω^{2} - Einstein’s energy-mass equivalence relation: E = m·
*c*^{2}

Of course, we are mixing relativistic and non-relativistic formulas here, and there’s the 1/2 factor – but these are minor issues. For example, we were talking not one but *two *oscillators, so we should add their energies: (1/2)·m·*a*^{2}·ω^{2} + (1/2)·m·*a*^{2}·ω^{2} = m·*a*^{2}·ω^{2}. Also, one can show that the classical formula for kinetic energy (i.e. E = (1/2)·m·*v*^{2}) morphs into E = m·*c*^{2} when we use the relativistically correct force equation for an oscillator. So, yes, our metaphor – or our suggested physical interpretation of the wavefunction, I should say – makes sense.

If you know something about physics, then you know the concept of the electromagnetic mass – its mathematical derivation, that is – gives us the classical electron radius, aka as the *Thomson *radius. It’s the smallest of a trio of radii that are relevant when discussing electrons: the other two radii are the Bohr radius and the Compton scattering radius respectively. The Thomson radius is used in the context of elastic scattering: the frequency of the incident particle (usually a photon), and the energy of the electron itself, do not change. In contrast, Compton scattering does change the frequency of the photon that is being scattered, and also impacts the energy of our electron. [As for the Bohr radius, you know that’s the radius of an electron orbital, roughly speaking – or the size of a hydrogen atom, I should say.]

Now, if we combine the E = m·*a*^{2}·ω^{2} and E = m·*c*^{2} equations, then *a*·ω must be equal to *c*, right? Can we show this? Maybe. It is easy to see that we get the desired equality by substituting the amplitude of the oscillation (*a*) for the Compton scattering radius *r *= ħ/(m·c), and ω (the (angular) frequency of the oscillation) by using the Planck relation (ω = E/ħ): * *

*a*·ω = [ħ/(m·*c*)]·[E/ħ] = E/(m·*c*) = m·*c*^{2}/(m·*c*) = *c*

We get a wonderfully simple geometric model of an electron here: an electric charge that spins around in a plane. Its radius is the *Compton *electron radius – which makes sense – and the radial velocity of our spinning charge is the speed of light – which may or may not make sense. Of course, we need an explanation of why this spinning charge doesn’t radiate its energy away – but then we don’t have such explanation anyway. All we can say is that the electron charge seems to be spinning in its own space – that it’s racing along a geodesic. It’s just like mass creates its own space here: according to Einstein’s general relativity theory, gravity becomes a *pseudo*-force—literally: no *real *force. How? I am not sure: the model here assumes the medium – empty space – is, somehow, perfectly elastic: the electron constantly borrows energy from one direction and then returns it to the other – so to speak. A crazy model, yes – but is there anything better? We only want to present a metaphor here: a possible *visualization *of quantum-mechanical models.

However, if this model is to represent anything real, then many more questions need to be answered. For starters, let’s think about an interpretation of the results of the Stern-Gerlach experiment.

# Precession

A spinning charge is a tiny magnet – and so it’s got a *magnetic moment*, which we need to explain the Stern-Gerlach experiment. But it doesn’t explain the *discrete* nature of the electron’s angular momentum: it’s either +ħ/2 or -ħ/2, nothing in-between, and that’s the case *along any direction *we choose. How can we explain this? Also, space is three-dimensional. Why would electrons spin in a perfect plane? The answer is: they don’t.

Indeed, the corollary of the above-mentioned binary value of the angular momentum is that the angular momentum – or the electron’s spin – is never completely along any direction. This may or may not be explained by the *precession *of a spinning charge in a field, which is illustrated below (illustration taken from *Feynman’s Lectures*, II-35-3).

**Figure 3: Precession of an electron in a magnetic field**

So we do have an oscillation in three dimensions here, really – even if our wavefunction is a two-dimensional mathematical object. Note that the measurement (or the Stein-Gerlach apparatus in this case) establishes a line of sight and, therefore, a reference frame, so ‘up’ and ‘down’, ‘left’ and ‘right’, and ‘in front’ and ‘behind’ get meaning. In other words, we establish a *real *space. The question then becomes: how and why does an electron sort of *snap into place*?

The geometry of the situation suggests the logical angle of the angular momentum vector should be 45°. Now, if the value of its *z*-component (i.e. its *projection* on the *z*-axis) is to be equal to ħ/2, then the magnitude of ** J** itself should be

*larger*. To be precise, it should be equal to ħ/√2 ≈ 0.7·ħ (just apply Pythagoras’ Theorem). Is that value compatible with our flywheel model?

Maybe. Let’s see. The *classical *formula for the magnetic moment is μ = I·A, with I the (effective) current and A the (surface) area. The notation is confusing because I is also used for the moment of inertia, or rotational mass, but… Well… Let’s do the calculation. The effective current is the electron charge (q_{e}) divided by the *period *(T) of the orbital revolution: : I = q_{e}/T. The period of the orbit is the *time *that is needed for the electron to complete one loop. That time (T) is equal to the circumference of the loop (2π·*a*) divided by the tangential velocity (*v*_{t}). Now, we suggest *v*_{t} = *r*·ω = *a*·ω = *c*, and the circumference of the loop is 2π·*a*. For *a*, we still use the Compton radius *a *= ħ/(m·*c*). Now, the formula for the area is A = π·*a*^{2}, so we get:

μ = I·A = [q_{e}/T]·π·*a*^{2} = [q_{e}·*c*/(2π·*a*)]·[π·*a*^{2}] = [(q_{e}·*c*)/2]·*a* = [(q_{e}·*c*)/2]·[ħ/(m·*c*)] = [q_{e}/(2m)]·ħ

In a classical analysis, we have the following relation between angular momentum and magnetic moment:

μ = (q_{e}/2m)·J

Hence, we find that the angular momentum J is equal to ħ, so that’s *twice *the measured value. We’ve got a problem. We would have hoped to find ħ/2 or ħ/√2. Perhaps it’s because *a* = ħ/(m·c) is the so-called *reduced* Compton scattering radius…

Well… No.

Maybe we’ll find the solution one day. I think it’s already quite nice we have a model that’s accurate up to a factor of 1/2 or 1/√2. 😊

**Post scriptum**: I’ve turned this into a small article which may or may not be more readable. You can link to it here. Comments are more than welcome.

# A Survivor’s Guide to Quantum Mechanics?

When modeling electromagnetic waves, the notion of left versus right circular polarization is quite clear and fully integrated in the mathematical treatment. In contrast, quantum math sticks to the very conventional idea that the imaginary unit (*i*) is – ** always!** – a counter-clockwise rotation by 90 degrees. We all know that –

*i*would do just as an imaginary unit as

*i*, because the

*definition*of the imaginary unit says the only requirement is that its square has to be equal to –1, and (–

*i*)

^{2}is also equal to –1.

So we actually have *two *imaginary units: *i* and –*i*. However, physicists stubbornly think there is only one direction for measuring angles, and that is counter-clockwise. *That’s a mathematical convention, Professor: it’s something in your head only. **It is not real.* Nature doesn’t care about our conventions and, therefore, I feel the spin ‘up’ versus spin ‘down’ should correspond to the two

*mathematical*possibilities: if the ‘up’ state is represented by some complex function, then the ‘down’ state should be represented by its complex conjugate.

This ‘additional’ rule wouldn’t change the basic quantum-mechanical rules – which are written in terms of *state* vectors in a Hilbert space (and, yes, a Hilbert space is an unreal as it sounds: its rules just say you should separate cats and dogs while adding them – which is very sensible advice, of course). However, they would, most probably (just my intuition – I need to prove it), avoid these crazy 720 degree symmetries which inspire the likes of Penrose to say there is no physical interpretation on the wavefunction.

Oh… As for the title of my post… I think it would be a great title for a book – because I’ll need some space to work it all out. 🙂

# Quantum math: garbage in, garbage out?

This post is basically a continuation of my previous one but – as you can see from its title – it is much more aggressive in its language, as I was inspired by a very thoughtful comment on my previous post. Another advantage is that it avoids all of the math. 🙂 It’s… Well… I admit it: it’s just a rant. 🙂 [Those who wouldn’t appreciate the casual style of what follows, can download my paper on it – but that’s much longer and also has a lot more math in it – so it’s a much harder read than this ‘rant’.]

My previous post was actually triggered by an attempt to re-read Feynman’s Lectures on Quantum Mechanics, but in reverse order this time: from the last chapter to the first. [In case you doubt, I did follow the correct logical order when working my way through them for the first time because… Well… There is no other way to get through them otherwise. 🙂 ] But then I was looking at Chapter 20. It’s a Lecture on quantum-mechanical operators – so that’s a topic which, in other textbooks, is usually tackled earlier on. When re-reading it, I realize why people quickly turn away from the topic of physics: it’s a lot of mathematical formulas which are supposed to reflect reality but, in practice, few – if any – of the mathematical concepts are actually being explained. Not in the first chapters of a textbook, not in its middle ones, and… Well… Nowhere, really. Why? Well… To be blunt: I think most physicists themselves don’t really understand what they’re talking about. In fact, as I have pointed out a couple of times already, Feynman himself admits so much:

“Atomic behavior appears peculiar and mysterious to everyone—both to the novice and to the experienced physicist. *Even the experts do not understand it the way they would like to*.”

So… Well… If you’d be in need of a rather spectacular acknowledgement of the shortcomings of physics as a science, here you have it: if you don’t understand what physicists are trying to tell you, don’t worry about it, because they don’t really understand it themselves. 🙂

Take the example of a *physical state*, which is represented by a *state vector*, which we can combine and re-combine using the properties of an abstract *Hilbert space*. Frankly, I think the word is very misleading, because it actually doesn’t describe an *actual* physical state. Why? Well… If we look at this so-called physical state from another angle, then we need to *transform *it using a complicated set of transformation matrices. You’ll say: that’s what we need to do when going from one reference frame to another in classical mechanics as well, isn’t it?

Well… No. In classical mechanics, we’ll describe the physics using geometric vectors in three dimensions and, therefore, the *base *of our reference frame doesn’t matter: because we’re using *real *vectors (such as the electric of magnetic field vectors **E** and **B**), our orientation *vis-á-vis* the object – the *line of sight*, so to speak – doesn’t matter.

In contrast, in quantum mechanics, it does: Schrödinger’s equation – and the wavefunction – has only two degrees of freedom, so to speak: its so-called real and its imaginary dimension. Worse, physicists refuse to give those two dimensions any *geometric *interpretation. Why? I don’t know. As I show in my previous posts, it would be easy enough, right? We know both dimensions must be perpendicular to each other, so we just need to decide if *both *of them are going to be perpendicular to our line of sight. That’s it. We’ve only got two possibilities here which – in my humble view – explain why the matter-wave is different from an electromagnetic wave.

I actually can’t quite believe the craziness when it comes to interpreting the wavefunction: we get everything we’d want to know about our particle through these operators (momentum, energy, position, and whatever else you’d need to know), but mainstream physicists still tell us that the wavefunction is, somehow, not representing anything real. It might be because of that weird 720° symmetry – which, as far as I am concerned, confirms that those state vectors are not the right approach: you can’t represent a complex, asymmetrical shape by a ‘flat’ mathematical object!

* Huh? *Yes. The wavefunction is a ‘flat’ concept: it has two dimensions only, unlike the

*real*vectors physicists use to describe electromagnetic waves (which we may interpret as the wavefunction of the photon). Those have three dimensions, just like the mathematical space we project on events. Because the wavefunction is flat (think of a rotating disk), we have those cumbersome transformation matrices: each time we shift position

*vis-á-vis*the object we’re looking at (

*das Ding an sich*, as Kant would call it), we need to change our description of it. And our description of it – the wavefunction – is all we have, so that’s

*our*reality. However, because that reality changes as per our line of sight, physicists keep saying the wavefunction (or

*das Ding an sich*itself) is, somehow, not real.

Frankly, I do think physicists should take a basic philosophy course: you can’t describe what goes on in three-dimensional space if you’re going to use flat (two-dimensional) concepts, because the objects we’re trying to describe (e.g. non-symmetrical electron orbitals) aren’t flat. Let me quote one of Feynman’s famous lines on philosophers: “These philosophers are always with us, struggling in the periphery to try to tell us something, but they never really understand the subtleties and depth of the problem.” (Feynman’s Lectures, Vol. I, Chapter 16)

Now, I *love *Feynman’s Lectures but… Well… I’ve gone through them a couple of times now, so I do think I have an appreciation of the subtleties and depth of the problem now. And I tend to agree with some of the smarter philosophers: if you’re going to use ‘flat’ mathematical objects to describe three- or four-dimensional reality, then such approach will only get you where we are right now, and that’s a lot of mathematical* mumbo-jumbo* for the poor uninitiated. *Consistent* mumbo-jumbo, for sure, but mumbo-jumbo nevertheless. 🙂 So, yes, I do think we need to re-invent quantum math. 🙂 The description may look more complicated, but it would make more sense.

I mean… If physicists themselves have had continued discussions on the reality of the wavefunction for almost a hundred years now (Schrödinger published his equation in 1926), then… Well… Then the physicists have a problem. Not the philosophers. 🙂 As to how that new description might look like, see my papers on viXra.org. I firmly believe it can be done. This is just a hobby of mine, but… Well… That’s where my attention will go over the coming years. 🙂 Perhaps quaternions are the answer but… Well… I don’t think so either – for reasons I’ll explain later. 🙂

**Post scriptum**: There are many nice videos on Dirac’s belt trick or, more generally, on 720° symmetries, but this links to one I particularly like. It clearly shows that the 720° symmetry requires, in effect, a special relation between the observer and the object that is being observed. It is, effectively, like there is a leather belt between them or, in this case, we have an arm between the glass and the person who is holding the glass. So it’s not like we are walking around the object (think of the glass of water) and making a full turn around it, so as to get back to where we were. No. *We are turning it around by 360°! *That’s a very different thing than just looking at it, walking around it, and then looking at it again. That explains the 720° symmetry: we need to turn it around twice to get it back to its original state. So… Well… The description is more about us and what we do with the object than about the object itself. That’s why I think the quantum-mechanical description is defective.

# Should we reinvent wavefunction math?

**Preliminary note**: This post may cause brain damage. 🙂 If you haven’t worked yourself through a good introduction to physics – including the math – you will probably not understand what this is about. So… Well… Sorry. 😦 But if you *have*… Then this should be *very* interesting. Let’s go. 🙂

If you know one or two things about quantum math – Schrödinger’s equation and all that – then you’ll agree the math is anything but straightforward. Personally, I find the most annoying thing about wavefunction math are those transformation matrices: every time we look at the same thing from a different direction, we need to transform the wavefunction using one or more rotation matrices – and that gets quite complicated !

Now, if you have read any of my posts on this or my other blog, then you know I firmly believe the wavefunction represents something *real* or… Well… Perhaps it’s just the next best thing to reality: we cannot know *das Ding an sich*, but the wavefunction gives us everything we would want to know about it (linear or angular momentum, energy, and whatever else we have an *operator* for). So what am I thinking of? Let me first quote Feynman’s summary interpretation of Schrödinger’s equation (*Lectures*, III-16-1):

“We can think of Schrödinger’s equation as describing the diffusion of the probability amplitude from one point to the next. […] But the imaginary coefficient in front of the derivative makes the behavior completely different from the ordinary diffusion such as you would have for a gas spreading out along a thin tube. Ordinary diffusion gives rise to real exponential solutions, whereas the solutions of Schrödinger’s equation are complex waves.”

Feynman further formalizes this in his *Lecture on Superconductivity *(Feynman, III-21-2), in which he refers to Schrödinger’s equation as the “equation for continuity of probabilities”. His analysis there is centered on the *local *conservation of energy, which makes *me* think Schrödinger’s equation might be an energy diffusion equation. I’ve written about this *ad nauseam *in the past, and so I’ll just refer you to one of my papers here for the details, and limit this post to the basics, which are as follows.

The wave equation (so that’s Schrödinger’s equation in its non-relativistic form, which is an approximation that is good enough) is written as:The resemblance with the standard diffusion equation (shown below) is, effectively, very obvious:As Feynman notes, it’s just that imaginary coefficient that makes the behavior quite different. *How *exactly? Well… You know we get all of those complicated electron orbitals (i.e. the various wave *functions *that satisfy the equation) out of Schrödinger’s differential equation. We can think of these solutions as (complex) *standing waves*. They basically represent some *equilibrium *situation, and the main characteristic of each is their *energy level*. I won’t dwell on this because – as mentioned above – I assume you master the math. Now, you know that – if we would want to interpret these wavefunctions as something real (which is surely what *I *want to do!) – the real and imaginary component of a wavefunction will be perpendicular to each other. Let me copy the animation for the *elementary *wavefunction ψ(θ) = *a·e*^{−i∙θ} = *a·e*^{−i∙(E/ħ)·t} *= a*·cos[(E/ħ)∙t] *−** i*·a·sin[(E/ħ)∙t] once more:

So… Well… That 90° angle makes me think of the similarity with the mathematical description of an electromagnetic wave. Let me quickly show you why. For a particle moving in free space – with no external force fields acting on it – there is no potential (U = 0) and, therefore, the Vψ term – which is just the equivalent of the the *sink *or *source *term S in the diffusion equation – disappears. Therefore, Schrödinger’s equation reduces to:

∂ψ(**x**, t)/∂t = *i*·(1/2)·(ħ/m_{eff})·∇^{2}ψ(**x**, t)

Now, the key difference with the diffusion equation – let me write it for you once again: ∂φ(**x**, t)/∂t = D·∇^{2}φ(**x**, t) – is that Schrödinger’s equation gives us *two *equations for the price of one. Indeed, because ψ is a complex-valued function, with a *real *and an *imaginary *part, we get the following equations:

*Re*(∂ψ/∂t) = −(1/2)·(ħ/m_{eff})·*Im*(∇^{2}ψ)*Im*(∂ψ/∂t) = (1/2)·(ħ/m_{eff})·*Re*(∇^{2}ψ)

** Huh? **Yes. These equations are easily derived from noting that two complex numbers a +

*i*∙b and c +

*i*∙d are equal if, and

*only*if, their real and imaginary parts are the same. Now, the ∂ψ/∂t =

*i*∙(ħ/m

_{eff})∙∇

^{2}ψ equation amounts to writing something like this: a +

*i*∙b =

*i*∙(c +

*i*∙d). Now, remembering that

*i*

^{2}= −1, you can easily figure out that

*i*∙(c +

*i*∙d) =

*i*∙c +

*i*

^{2}∙d = − d +

*i*∙c. [Now that we’re getting a bit technical, let me note that the m

_{eff}is the

*effective*mass of the particle, which depends on the medium. For example, an electron traveling in a solid (a transistor, for example) will have a different effective mass than in an atom. In free space, we can drop the subscript and just write m

_{eff}= m.] 🙂 OK.

*Onwards !*🙂

The equations above make me think of the equations for an electromagnetic wave in free space (no stationary charges or currents):

- ∂
**B**/∂t = –∇×**E** - ∂
**E**/∂t =*c*^{2}∇×**B**

Now, these equations – and, I must therefore assume, the other equations above as well – effectively describe a *propagation *mechanism in spacetime, as illustrated below:

You know how it works for the electromagnetic field: it’s the interplay between circulation and flux. Indeed, circulation around some axis of rotation creates a flux in a direction perpendicular to it, and that flux causes this, and then that, and it all goes round and round and round. 🙂 Something like that. 🙂 I will let you look up how it goes, *exactly*. The principle is clear enough. Somehow, in this beautiful interplay between linear and circular motion, energy is borrowed from one place and then returns to the other, cycle after cycle.

Now, we know the wavefunction consist of a sine and a cosine: the cosine is the real component, and the sine is the imaginary component. Could they be equally real? Could each represent *half *of the total energy of our particle? I firmly believe they do. The obvious question then is the following: why wouldn’t we represent them as *vectors*, just like **E** and **B**? I mean… Representing them as vectors (I mean *real *vectors here – something with a magnitude and a direction in a *real *space – as opposed to *state *vectors from the Hilbert space) would *show *they are real, and there would be no need for cumbersome transformations when going from one representational *base *to another. In fact, that’s why vector notation was invented (sort of): we don’t need to worry about the coordinate frame. It’s much easier to write physical laws in vector notation because… Well… They’re the *real *thing, aren’t they? 🙂

What about dimensions? Well… I am not sure. However, because we are – arguably – talking about some pointlike charge moving around in those oscillating fields, I would suspect the dimension of the real and imaginary component of the wavefunction will be the same as that of the electric and magnetic field vectors **E** and **B**. We may want to recall these:

**E**is measured in*newton per coulomb*(N/C).**B**is measured in newton per coulomb divided by m/s, so that’s (N/C)/(m/s).

The weird dimension of **B** is because of the weird force law for the magnetic force. It involves a vector cross product, as shown by Lorentz’ formula:

**F** = qE + q(** v**×

**B**)

Of course, it is only *one *force (one and the same physical reality), as evidenced by the fact that we can write **B** as the following vector cross-product: **B** = (1/*c*)∙**e****_{x}**×

**E**, with

**e****the unit vector pointing in the**

_{x}*x*-direction (i.e. the direction of propagation of the wave). [Check it, because you may not have seen this expression before. Just take a piece of paper and think about the geometry of the situation.] Hence, we may associate the (1/

*c*)∙

**e****×**

_{x}*operator*, which amounts to a rotation by 90 degrees, with the s/m dimension. Now, multiplication by

*i*also amounts to a rotation by 90° degrees. Hence, if we can agree on a suitable convention for the

*direction*of rotation here, we may boldly write:

**B** = (1/*c*)∙**e****_{x}**×

**E**= (1/

*c*)∙

*i*∙

**E**

This is, in fact, what triggered my geometric interpretation of Schrödinger’s equation about a year ago now. I have had little time to work on it, but think I am on the right track. Of course, you should note that, for an electromagnetic wave, the magnitudes of **E** and **B** reach their maximum, minimum and zero point *simultaneously* (as shown below). So their *phase *is the same.

In contrast, the phase of the real and imaginary component of the wavefunction is not the same, as shown below.

In fact, because of the Stern-Gerlach experiment, I am actually more thinking of a motion like this:

But that shouldn’t distract you. 🙂 The question here is the following: could we possibly think of a new formulation of Schrödinger’s equation – using *vectors *(again, *real *vectors – not these weird *state *vectors) rather than complex algebra?

I think we can, but then I wonder why the *inventors *of the wavefunction – Heisenberg, Born, Dirac, and Schrödinger himself, of course – never thought of that. 🙂

Hmm… I need to do some research here. 🙂

**Post scriptum**: You will, of course, wonder how and why the matter-wave would be different from the electromagnetic wave if my suggestion that the dimension of the wavefunction component is the same is correct. The answer is: the difference lies in the phase difference and then, most probably, the different orientation of the angular momentum. Do we have any other possibilities? 🙂

P.S. 2: I also published this post on my new blog: https://readingeinstein.blog/. However, I thought the followers of this blog should get it first. 🙂

# Wavefunctions, perspectives, reference frames, representations and symmetries

Ouff ! This title is quite a mouthful, isn’t it? 🙂 So… What’s the topic of the day? Well… In our previous posts, we developed a few key ideas in regard to a possible physical interpretation of the (elementary) wavefunction. It’s been an interesting excursion, and I summarized it in another pre-publication paper on the open arXiv.org site.

In my humble view, one of the toughest issues to deal with when thinking about geometric (or *physical*) interpretations of the wavefunction is the fact that a wavefunction does not seem to obey the classical 360° symmetry in space. In this post, I want to muse a bit about this and show that… Well… It does and it doesn’t. It’s got to do with what happens when you change from one representational base (or representation, *tout court*) to another which is… Well… Like changing the reference frame but, at the same time, it is also *more* than just a change of the reference frame—and so that explains the weird stuff (like that 720° symmetry of the amplitudes for spin-1/2 particles, for example).

I should warn you before you start reading: I’ll basically just pick up some statements from my paper (and previous posts) and develop some more thoughts on them. As a result, this post may not be very well structured. Hence, you may want to read the mentioned paper first.

### The reality of directions

*Huh? *The *reality *of directions? Yes. I warned you. This post may cause brain damage. 🙂 The whole argument revolves around a *thought *experiment—but one whose results have been verified in zillions of experiments in university student labs so… Well… We do *not *doubt the results and, therefore, we do not doubt the basic mathematical results: we just want to try to *understand *them better.

So what is the set-up? Well… In the illustration below (Feynman, III, 6-3), Feynman compares the physics of two situations involving rather special beam splitters. Feynman calls them modified or ‘improved’ Stern-Gerlach apparatuses. The apparatus basically splits and then re-combines the two new beams along the *z*-axis. It is also possible to block one of the beams, so we filter out only particles with their spin *up* or, alternatively, with their spin *down*. Spin (or angular momentum or the magnetic moment) as measured along the *z*-axis, of course—I should immediately add: we’re talking **the z-axis of the apparatus** here.

The two situations involve a different *relative *orientation of the apparatuses: in (a), the angle is 0**°**, while in (b) we have a (right-handed) rotation of 90° about the *z*-axis. He then proves—using geometry and logic only—that the probabilities and, therefore, **the magnitudes of the amplitudes** (denoted by

*C*

_{+}and

*C*

_{−}and

*C’*

_{+}and

*C’*

_{−}in the

*S*and

*T*representation respectively)

**must be the same, but the amplitudes**, noting—in his typical style, mixing academic and colloquial language—that “there must be some way for a particle to tell that it has turned a corner in (b).”

*must*have different phasesThe various interpretations of what actually *happens* here may shed some light on the heated discussions on the *reality *of the wavefunction—and of quantum states. In fact, I should note that Feynman’s argument revolves around quantum states. To be precise, the analysis is focused on two-state systems only, and the wavefunction—which captures a continuum of possible states, so to speak—is introduced only later. However, we may look at the amplitude for a particle to be in the *up*– or *down*-state as a wavefunction and, therefore (but do note that’s my humble opinion once more), the analysis is actually *not *all that different.

We *know*, from theory *and *experiment, that the amplitudes *are *different. For example, for the given difference in the *relative *orientation of the two apparatuses (90°), we *know* that the amplitudes are given by *C’*_{+} = *e ^{i}*

^{∙φ/2}∙

*C*

_{+}=

*e*

^{ i}^{∙π/4}∙

*C*

_{+}and

*C’*

_{−}=

*e*

^{−i∙φ/2}∙

*C*

_{+}=

*e*

^{− i∙π/4}∙

*C*

_{−}respectively (the amplitude to go from the down to the up state, or vice versa, is zero). Hence, yes,

**—**

*we**not*the particle, Mr. Feynman!—

*that, in (b), the electron has, effectively, turned a corner.*

**know**The more subtle question here is the following: is the *reality* of the particle in the two setups the same? Feynman, of course, stays away from such philosophical question. He just notes that, while “(a) and (b) are different”, “the probabilities are the same”. He refrains from making any statement on the particle itself: is or is it *not *the same? The common sense answer is obvious: of course, it is! The particle is the same, right? In (b), it just took a turn—so it is just going in some other direction. That’s all.

However, common sense is seldom a good guide when thinking about quantum-mechanical realities. Also, from a more philosophical point of view, one may argue that the reality of the particle is *not *the same: something might—or *must*—have *happened* to the electron because, when everything is said and done, the particle *did* take a turn in (b). It did *not *in (a). [Note that the difference between ‘might’ and ‘must’ in the previous phrase may well sum up the difference between a deterministic and a non-deterministic world view but… Well… This discussion is going to be way too philosophical already, so let’s refrain from inserting new language here.]

Let us think this through. The (a) and (b) set-up are, *obviously*, different but… *Wait a minute…* Nothing is obvious in quantum mechanics, right? How can we *experimentally confirm *that they are different?

* Huh? *I must be joking, right? You can

*see*they are different, right? No. I am not joking. In physics, two things are different if we get different

*measurement*results. [That’s a bit of a simplified view of the ontological point of view of mainstream physicists, but you will have to admit I am not far off.] So… Well… We can’t see those amplitudes and so… Well… If we

*measure*the same thing—same

*probabilities*, remember?—why are they different? Think of this: if we look at the two beam splitters as one single tube (an

*ST*tube, we might say), then all we did in (b) was bend the tube. Pursuing the logic that says our particle is still the same

*even when it takes a turn*, we could say the tube is still the same, despite us having wrenched it over a 90° corner.

Now, I am sure you think I’ve just gone nuts, but just try* *to stick with me a little bit longer. Feynman actually acknowledges the same: we need to *experimentally **prove *(a) and (b) are different. He does so by getting **a third apparatus **in

**(**, as shown below,

*U*)**whose**, so there is no difference there.

*relative*orientation to*T*is the same in both (a) and (b)Now, the axis of *U *is not the *z*-axis: it is the *x*-axis in (a), and the *y*-axis in (b). So what? Well… I will quote Feynman here—not (only) because his words are more important than mine but also because every word matters here:

“The two apparatuses in (a) and (b) are, in fact, different, as we can see in the following way. Suppose that we put an apparatus in front of *S *which produces a pure +*x* state. Such particles would be split into +*z* and −*z* into beams in *S*, but the two beams would be recombined to give a +*x* state again at P_{1}—the exit of *S*. The same thing happens again in *T*. If we follow *T *by a third apparatus *U*, whose axis is in the +*x* direction and, as shown in (a), all the particles would go into the + beam of *U*. Now imagine what happens if *T *and *U *are swung around *together* by 90° to the positions shown in (b). Again, the *T *apparatus puts out just what it takes in, so the particles that enter *U *are in a +*x *state ** with respect to S**, which is different. By symmetry, we would now expect only one-half of the particles to get through.”

I should note that (b) shows the *U *apparatus wide open so… Well… I must assume that’s a mistake (and should alert the current editors of the *Lectures *to it): Feynman’s narrative tells us we should also imagine it with the *minus *channel shut. In *that *case, it should, effectively, filter approximately half of the particles out, while they all get through in (a). So that’s a *measurement *result which shows the direction, as we *see *it, makes a difference.

Now, Feynman would be very angry with me—because, as mentioned, he hates philosophers—but I’d say: this experiment proves that a direction is something real. Of course, the next philosophical question then is: what *is *a direction? I could answer this by pointing to the experiment above: a direction is something that alters the probabilities between the *S**T**U* tube as set up in (a) versus the *S**T**U* tube in (b). In fact—but, I admit, that would be pretty ridiculous—we could use the varying probabilities as we wrench this tube over varying angles to *define* an angle! But… Well… While that’s a perfectly logical argument, I agree it doesn’t sound very sensical.

OK. Next step. What follows may cause brain damage. 🙂 Please abandon all pre-conceived notions and definitions for a while and think through the following logic.

You know this stuff is about transformations of amplitudes (or wavefunctions), right? [And you also want to hear about those special 720° symmetry, right? No worries. We’ll get there.] So the questions all revolve around this: what happens to amplitudes (or the wavefunction) when we go from one reference frame—or *representation*, as it’s referred to in quantum mechanics—to another?

Well… I should immediately correct myself here: a reference frame and a representation are two different things. They are *related *but… Well… Different… *Quite* different. Not same-same but different. 🙂 I’ll explain why later. Let’s go for it.

Before talking representations, let us first think about what we really *mean* by changing the *reference frame*. To change it, we first need to answer the question: what *is *our reference frame? It is a mathematical notion, of course, but then it is also more than that: it is *our *reference frame. We use it to make measurements. That’s obvious, you’ll say, but let me make a more formal statement here:

**The reference frame is given by (1) the geometry **(or the *shape*, if that sounds easier to you)** of the measurement apparatus** (so that’s the experimental set-up) here) and** (2) our perspective of it.**

If we would want to sound academic, we might refer to Kant and other philosophers here, who told us—230 years ago—that the mathematical idea of a three-dimensional reference frame is grounded in our intuitive notions of up and down, and left and right. [If you doubt this, think about the necessity of the various right-hand rules and conventions that we cannot do without in math, and in physics.] But so we do not want to sound academic. Let us be practical. Just think about the following. The apparatus gives us two *directions*:

(1) The *up *direction, which *we associate* with the positive direction of the *z*-axis, and

(2) the direction of travel of our particle, which *we associate* with the positive direction of the *y*-axis.

Now, if we have two axes, then the third axis (the *x*-axis) will be given by the right-hand rule, right? So we may say the apparatus gives us the reference frame. Full stop. So… Well… Everything is relative? Is this reference frame relative? Are directions relative? That’s what you’ve been told, but think about this: relative *to what?* Here is where the object meets the subject. What’s relative? What’s absolute? Frankly, I’ve started to think that, in this particular situation, we should, perhaps, not use these two terms. I am *not *saying that our *observation* of what *physically* happens here gives these two directions any *absolute *character but… Well… You will have to admit they are more than just some mathematical construct: when everything is said and done, we will have to admit that these two directions are *real*. because… Well… They’re part of the *reality *that we are observing, right? And the third one… Well… That’s given by our perspective—by our right-hand rule, which is… Well… *Our *right-hand rule.

Of course, now you’ll say: if you think that ‘relative’ and ‘absolute’ are ambiguous terms and that we, therefore, may want to avoid them a bit more, then ‘real’ and its opposite (unreal?) are ambiguous terms too, right? Well… Maybe. What language would *you *suggest? 🙂 Just stick to the story for a while. I am not done yet. So… Yes… What *is *their *reality*? Let’s think about that in the next section.

### Perspectives, reference frames and symmetries

You’ve done some mental exercises already as you’ve been working your way through the previous section, but you’ll need to do plenty more. In fact, they may become physical exercise too: when I first thought about these things (symmetries and, more importantly, *a*symmetries in space), I found myself walking around the table with some asymmetrical everyday objects and papers with arrows and clocks and other stuff on it—effectively analyzing what right-hand screw, thumb or grip rules actually *mean*. 🙂

So… Well… **I want you to distinguish—just for a while—between the notion of a reference frame (think of the x–y–z reference frame that comes with the apparatus) and your perspective on it.** What’s our perspective on it? Well… You may be looking from the top, or from the side and, if from the side, from the left-hand side or the right-hand side—which, if you think about it, you can only

*define*in terms of the various positive and negative directions of the various axes. 🙂 If you think this is getting ridiculous… Well… Don’t. Feynman himself doesn’t think this is ridiculous, because he starts his own “long and abstract side tour” on transformations with a very simple explanation of how the top and side

*view*of the apparatus are related to the

*axes*(i.e. the reference frame) that comes with it. You don’t believe me? This is the

*very*first illustration of his

*Lecture*on this:

He uses it to explain the apparatus (which we don’t do here because you’re supposed to already know how these (modified or improved) Stern-Gerlach apparatuses work). So let’s continue this story. Suppose that we are looking in the *positive* *y*-direction—so that’s the direction in which our particle is moving—then we might imagine how it would look like when *we *would make a 180° turn and look at the situation from the other side, so to speak. We do not change the reference frame (i.e. the *orientation*) of the apparatus here: we just change our *perspective *on it. Instead of seeing particles going *away from us*, into the apparatus, we now see particles coming *towards *us, out of the apparatus.

What happens—but that’s not scientific language, of course—is that left becomes right, and right becomes left. Top still is top, and bottom is bottom. We are looking now in the *negative y*-direction, and the positive direction of the *x*-axis—which pointed right when we were looking in the positive *y*-direction—now points left. I see you nodding your head now—because you’ve heard about parity inversions, mirror symmetries and what have you—and I hear you say: “That’s the mirror world, right?”

No. It is not. I wrote about this in another post: the world in the mirror is the world in the mirror. We don’t get a mirror image of an object by going around it and looking at its back side. I can’t dwell too much on this (just check that post, and another one who talks about the same), but so don’t try to connect it to the discussions on symmetry-breaking and what have you. Just stick to *this *story, which is about transformations of amplitudes (or wavefunctions). [If you really want to know—but I know this sounds counterintuitive—the mirror world doesn’t really switch left for right. Your reflection doesn’t do a 180 degree turn: it is just reversed front to back, with no rotation at all. It’s only your brain which *mentally* adds (or subtracts) the 180 degree turn that you assume must have happened from the observed front to back reversal. So the left to right reversal is only *apparent*. It’s a common misconception, and… Well… I’ll let you figure this out yourself. I need to move on.] Just note the following:

- The
*xyz*reference frame remains a valid right-handed reference frame. Of course it does: it comes with our beam splitter, and we can’t change its*reality*, right? We’re just looking at it from another angle. Our*perspective*on it has changed. - However, if we think of the real and imaginary part of the wavefunction describing the electrons that are going through our apparatus as perpendicular oscillations (as shown below)—a cosine and sine function respectively—then our change in perspective
*might*, effectively, mess up our convention for measuring angles.

I am not saying it *does*. Not now, at least. I am just saying it *might*. It depends on the plane of the oscillation, as I’ll explain in a few moments. Think of this: we measure angles *counter*clockwise, right? As shown below… But… Well… If the thing below would be some funny clock going backwards—you’ve surely seen them in a bar or so, right?—then… Well… If they’d be transparent, and you’d go around them, you’d see them as going… Yes… Clockwise. 🙂 [This should remind you of a discussion on real versus pseudo-vectors, or polar versus axial vectors, but… Well… We don’t want to complicate the story here.]

Now, *if *we would assume this clock represents something real—and, of course, **I am thinking of the elementary wavefunction e^{i}^{θ} = cosθ + i·sinθ now**—then… Well… Then it will look different when we go around it. When going around our backwards clock above and looking at it from… Well… The back, we’d describe it, naively, as… Well…

*Think! What’s your answer? Give me the formula!*🙂

[…]

We’d see it as *e*^{−i}^{θ} = *cos*(−θ) + *i*·*sin*(−θ) = *cos*θ − *i*·*sin*θ, right? The hand of our clock now goes clockwise, so that’s the *opposite *direction of our convention for measuring angles. Hence, instead of *e*^{i}^{θ}, we write *e*^{−i}^{θ}, right? So that’s the complex conjugate. So we’ve got a different *image *of the same thing here. *Not* good. *Not good at all.*

You’ll say: *so what? *We can fix this thing easily, right? You don’t need the convention for measuring angles or for the imaginary unit (*i*) here. This particle is moving, right? So if you’d want to look at the elementary wavefunction as some sort of circularly polarized beam (which, I admit, is very much what I would like to do, but its polarization is rather particular as I’ll explain in a minute), then you just need to define *left- and right-handed angles* as per the standard right-hand screw rule (illustrated below). *To hell with the counterclockwise convention for measuring angles!*

You are right. We *could *use the right-hand rule more consistently. We could, in fact, use it as an *alternative *convention for measuring angles: we could, effectively, measure them clockwise *or* counterclockwise depending on the direction of our particle. But… Well… The fact is: *we don’t*. We do *not* use that alternative convention when we talk about the wavefunction. Physicists do use the *counterclockwise* convention ** all of the time** and just jot down these complex exponential functions and don’t realize that,

*if they are to represent something real*, our

*perspective*on the reference frame matters. To put it differently, the

*direction*in which we are looking at things matters! Hence, the direction is

*not…*Well… I am tempted to say…

*Not*relative at all but then… Well… We wanted to avoid that term, right? 🙂

[…]

I guess that, by now, your brain may suffered from various short-circuits. If not, stick with me a while longer. Let us analyze how our wavefunction model might be impacted by this symmetry—or *a*symmetry, I should say.

### The flywheel model of an electron

In our previous posts, we offered a model that interprets the real and the imaginary part of the wavefunction as oscillations which each carry half of the total energy of the particle. These oscillations are perpendicular to each other, and the interplay between both is how energy propagates through spacetime. Let us recap the fundamental premises:

- The dimension of the matter-wave field vector is force per unit
*mass*(N/kg), as opposed to the force per unit*charge*(N/C) dimension of the electric field vector. This dimension is an acceleration (m/s^{2}), which is the dimension of the gravitational field. - We assume this gravitational disturbance causes our electron (or a charged
*mass*in general) to move about some center, combining linear and circular motion. This interpretation reconciles the wave-particle duality: fields interfere but if, at the same time, they do drive a pointlike particle, then we understand why, as Feynman puts it, “when you do find the electron some place, the entire charge is there.” Of course, we cannot prove anything here, but our elegant yet simple derivation of the Compton radius of an electron is… Well… Just nice. 🙂 - Finally, and most importantly
*in the context of this discussion*, we noted that, in light of the direction of the magnetic moment of an electron in an inhomogeneous magnetic field,**the plane which circumscribes the circulatory motion of the electron should also**Hence, unlike an electromagnetic wave, the*comprise*the direction of its linear motion.*plane*of the two-dimensional oscillation (so that’s the polarization plane, really) can*not*be perpendicular to the direction of motion of our electron.

Let’s say some more about the latter point here. The illustrations below (one from Feynman, and the other is just open-source) show what we’re thinking of. The direction of the angular momentum (and the magnetic moment) of an electron—or, to be precise, its component as measured in the direction of the (inhomogeneous) magnetic field through which our electron is traveling—can*not* be parallel to the direction of motion. On the contrary, it must be *perpendicular* to the direction of motion. In other words, if we imagine our electron as spinning around some center (see the illustration on the left-hand side), then the disk it circumscribes (i.e. the *plane *of the polarization) has to *comprise *the direction of motion.

Of course, we need to add another detail here. As my readers will know, we do not really have a precise direction of angular momentum in quantum physics. While there is no fully satisfactory explanation of this, the classical explanation—combined with the quantization hypothesis—goes a long way in explaining this: an object with an angular momentum ** J** and a magnetic moment

**that is**

*μ**not exactly*parallel to some magnetic field

**B**, will

*not*line up: it will

*precess*—and, as mentioned, the quantization of angular momentum may well explain the rest. [Well… Maybe… We have detailed our attempts in this regard in various posts on this (just search for

*spin*or

*angular momentum*on this blog, and you’ll get a dozen posts or so), but these attempts are, admittedly, not

*fully satisfactory*. Having said that, they do go a long way in relating angles to spin numbers.]

The thing is: we do assume our electron is spinning around. If we look from the *up*-direction *only*, then it will be spinning *clockwise *if its angular momentum is down (so its *magnetic moment *is *up*). Conversely, it will be spinning *counter*clockwise if its angular momentum is *up*. Let us take the *up*-state. So we have a top view of the apparatus, and we see something like this:I know you are laughing aloud now but think of your amusement as a nice reward for having stuck to the story so far. Thank you. 🙂 And, yes, do check it yourself by doing some drawings on your table or so, and then look at them from various directions as you walk around the table as—I am not ashamed to admit this—I did when thinking about this. So what do we get when we change the perspective? Let us walk around it, *counterclockwise*, let’s say, so we’re measuring our angle of rotation as some *positive *angle. Walking around it—in whatever direction, clockwise or counterclockwise—doesn’t change the counterclockwise direction of our… Well… That weird object that might—just *might—*represent an electron that has its spin up and that is traveling in the positive *y*-direction.

When we look in the direction of propagation (so that’s from left to right as you’re looking at this page), and we abstract away from its linear motion, then we could, vaguely, describe this by some wrenched *e ^{i}*

^{θ}=

*cos*θ +

*i*·

*sin*θ function, right? The

*x-*and

*y*-axes

*of the apparatus*may be used to measure the cosine and sine components respectively.

Let us keep looking from the top but walk around it, rotating ourselves over a 180° angle so we’re looking in the *negative *y-direction now. As I explained in one of those posts on symmetries, our mind will want to switch to a new reference frame: we’ll keep the *z*-axis (up is up, and down is down), but we’ll want the positive direction of the *x*-axis to… Well… Point right. And we’ll want the *y*-axis to point away, rather than towards us. In short, we have a transformation of the reference frame here: *z’* = *z*, *y’* = − *y*, and *x’* = − *x*. Mind you, this is still a regular right-handed reference frame. [That’s the difference with a *mirror *image: a *mirrored *right-hand reference frame is no longer right-handed.] So, in our new reference frame, that we choose to coincide with our *perspective*, we will now describe the same thing as some −*cos*θ − *i*·*sin*θ = −*e ^{i}*

^{θ}function. Of course, −

*cos*θ =

*cos*(θ + π) and −

*sin*θ =

*sin*(θ + π) so we can write this as:

−*cos*θ − *i*·*sin*θ = *cos*(θ + π) + *i*·*sin*θ = *e ^{i}*

^{·(}

^{θ+π)}=

*e*

^{i}^{π}·

*e*

^{i}^{θ}= −

*e*

^{i}^{θ}.

Sweet ! But… Well… First note this is *not *the complex conjugate: *e*^{−i}^{θ} = *cos*θ − *i*·*sin*θ ≠ −*cos*θ − *i*·*sin*θ = −*e ^{i}*

^{θ}. Why is that? Aren’t we looking at the same clock, but from the back? No. The plane of polarization is different. Our clock is more like those in Dali’s painting: it’s flat. 🙂 And, yes, let me lighten up the discussion with that painting here. 🙂 We need to have

*some*fun while torturing our brain, right?

So, because we assume the plane of polarization is different, we get an −*e ^{i}*

^{θ}function instead of a

*e*

^{−i}

^{θ}function.

Let us now think about the *e ^{i}*

^{·(}

^{θ+π)}function. It’s the same as −

*e*

^{i}^{θ}but… Well… We walked around the

*z*-axis taking a full 180° turn, right? So that’s π in radians. So that’s the

*phase shift*here.

*Hey!*Try the following now. Go back and walk around the apparatus once more, but let the reference frame

*rotate with us*, as shown below. So we start left and look in the direction of propagation, and then we start moving about the

*z*-axis (which points out of this page,

*toward*you, as you are looking at this), let’s say by some small angle α. So we rotate the reference frame about the

*z*-axis by α and… Well… Of course, our

*e*

^{i}^{·}

^{θ}now becomes an our

*e*

^{i}^{·(}

^{θ+α)}function, right? We’ve just derived the transformation coefficient for a rotation about the

*z*-axis, didn’t we? It’s equal to

*e*

^{i}^{·}

^{α}, right? We get the transformed wavefunction in the new reference frame by multiplying the old one by

*e*

^{i}^{·}

^{α}, right? It’s equal to

*e*

^{i}^{·}

^{α}·

*e*

^{i}^{·}

^{θ}=

*e*

^{i}^{·(}

^{θ+α)}, right?

Well…

[…]

No. The answer is: no. The transformation coefficient is not *e ^{i}*

^{·}

^{α}but

*e*

^{i}^{·}

^{α/2}. So we get an additional 1/2 factor in the

*phase shift*.

* Huh? *Yes. That’s what it is: when we change the representation, by rotating our apparatus over some angle α about the

*z*-axis, then we will, effectively, get a new wavefunction, which will differ from the old one by a phase shift that is equal to only

*half*of the rotation angle only.

** Huh? **Yes. It’s even weirder than that. For a spin

*down*electron, the transformation coefficient is

*e*

^{−i·}

^{α/2}, so we get an additional minus sign in the argument.

* Huh? *Yes.

I know you are terribly disappointed, but that’s how it is. That’s what hampers an easy geometric interpretation of the wavefunction. Paraphrasing Feynman, I’d say that, somehow, our electron not only knows whether or not it has taken a turn, but it also knows whether or not it is moving away from us or, conversely, towards us.

[…]

But… *Hey! Wait a minute! That’s it, right? *

What? Well… That’s it! The electron doesn’t know whether it’s moving away or towards us. That’s nonsense. But… Well… It’s like this:

**Our e^{i}^{·}^{α} coefficient describes a rotation of the reference frame. In contrast, the e^{i}^{·}^{α/2} and e^{−i·}^{α/2} coefficients describe what happens when we rotate the T apparatus! Now that is a very different proposition. **

Right! You got it! *Representations* and reference frames are different things. *Quite *different, I’d say: representations are *real*, reference frames aren’t—but then you don’t like philosophical language, do you? 🙂 But think of it. When we just go about the *z*-axis, a full 180°, but we don’t touch that *T*-apparatus, we don’t change *reality*. When we were looking at the electron while standing left to the apparatus, we watched the electrons going in and moving away from us, and when we go about the *z*-axis, a full 180°, looking at it from the right-hand side, we see the electrons coming out, moving towards us. But it’s still the same reality. We simply change the reference frame—from *xyz* to *x’y’z’* to be precise: we do *not *change the representation.

In contrast, **when we rotate the T apparatus over a full 180°, our electron now goes in the opposite direction. **And whether that’s away or towards us, that doesn’t matter: it was going in one direction while traveling through

*S*, and now it goes in the opposite direction—

*relative to the direction it was going in S*, that is.

So what happens, *really*, when we change the *representation*, rather than the reference frame? Well… Let’s think about that. 🙂

### Quantum-mechanical weirdness?

The transformation matrix for the amplitude of a system to be in an *up *or *down *state (and, hence, presumably, for a wavefunction) for a rotation about the *z*-axis is the following one:

Feynman derives this matrix in a rather remarkable intellectual *tour de force *in the 6th of his *Lectures on Quantum Mechanics*. So that’s pretty early on. He’s actually worried about that himself, apparently, and warns his students that “This chapter is a rather long and abstract side tour, and it does not introduce any idea which we will not also come to by a different route in later chapters. You can, therefore, skip over it, and come back later if you are interested.”

Well… That’s how *I *approached it. I skipped it, and didn’t worry about those transformations for quite a while. But… Well… You can’t avoid them. In some weird way, they are at the heart of the weirdness of quantum mechanics itself. Let us re-visit his argument. Feynman immediately gets that the whole transformation issue here is just a matter of finding an easy formula for that phase shift. Why? He doesn’t tell us. Lesser mortals like us must just assume that’s how the instinct of a genius works, right? 🙂 So… Well… Because he *knows*—from experiment—that the coefficient is *e ^{i}*

^{·}

^{α/2}instead of

*e*

^{i}^{·}

^{α}, he just says the phase shift—which he denotes by λ—must be some

*proportional*to the angle of rotation—which he denotes by φ rather than α (so as to avoid confusion with the

*Euler*angle α). So he writes:

λ = m·φ

Initially, he also tries the obvious thing: m should be one, right? So λ = φ, right? Well… No. It can’t be. Feynman shows why that can’t be the case by adding a third apparatus once again, as shown below.

Let me quote him here, as I can’t explain it any better:

“Suppose *T* is rotated by 360°; then, clearly, it is right back at zero degrees, and we should have *C’*_{+} = *C*_{+} and *C’*_{−} = *C*_{−} or, what is the same thing, *e ^{i}*

^{·m·2π}= 1. We get m = 1. [But no!]

*This argument is wrong!*To see that it is, consider that

*T*is rotated by 180°. If m were equal to 1, we would have

*C’*

_{+}=

*e*

^{i}^{·π}

*C*

_{+}= −

*C*

_{+}and

*C’*

_{−}=

*e*

^{−}

^{i}^{·π}

*C*

_{−}= −

*C*

_{−}. [Feynman works with

*states*here, instead of the wavefunction of the particle as a whole. I’ll come back to this.] However, this is just the

*original*state all over again.

**amplitudes are just multiplied by −1 which gives back the original physical system. (It is again a case of a**

*Both***phase change.) This means that if the angle between**

*common**T*and

*S*is increased to 180°, the system would be indistinguishable from the zero-degree situation, and the particles would again go through the (+) state of the

*U*apparatus. At 180°, though, the (+) state of the

*U*apparatus is the (−

*x*) state of the original

*S*apparatus. So a (+

*x*) state would become a (−

*x*) state. But we have done nothing to

*change*the original state; the answer is wrong. We cannot have m = 1. We must have the situation that a rotation by 360°, and

*no smaller angle*reproduces the same physical state. This will happen if m = 1/2.”

The result, of course, is this weird 720° symmetry. While we get the same *physics* after a 360° rotation of the *T* apparatus, we do *not *get the same amplitudes. We get the opposite (complex) number: *C’*_{+} = *e ^{i}*

^{·2π/2}

*C*

_{+}= −

*C*

_{+}and

*C’*

_{−}=

*e*

^{−}

^{i}^{·2π/2}

*C*

_{−}= −

*C*

_{−}. That’s OK, because… Well… It’s a

*common*phase shift, so it’s just like changing the origin of time. Nothing more. Nothing less. Same physics. Same

*reality.*But… Well…

*C’*

_{+}≠ −

*C*

_{+}and

*C’*

_{−}≠ −

*C*

_{−}, right? We only get our original amplitudes back if we rotate the

*T*apparatus two times, so that’s by a full 720 degrees—as opposed to the 360° we’d expect.

Now, space is isotropic, right? So this 720° business doesn’t make sense, right?

Well… It does and it doesn’t. We shouldn’t dramatize the situation. What’s the *actual* difference between a complex number and its opposite? It’s like *x* or −*x*, or *t* and −*t. *I’ve said this a couple of times already again, and I’ll keep saying it many times more: *Nature *surely can’t be bothered by how we measure stuff, right? In the positive or the negative direction—that’s just our choice, right? *Our *convention. So… Well… It’s just like that −*e ^{i}*

^{θ}function we got when looking at the

*same*experimental set-up from the other side: our

*e*

^{i}^{θ}and −

*e*

^{i}^{θ}functions did

*not*describe a different reality. We just changed our perspective. The

*reference frame*. As such, the reference frame isn’t

*real*. The experimental set-up is. And—I know I will anger mainstream physicists with this—the

*representation*is. Yes. Let me say it loud and clear here:

**A different representation describes a different reality. **

In contrast, a different perspective—or a different reference frame—does not.

### Conventions

While you might have had a lot of trouble going through all of the weird stuff above, the point is: it is *not *all that weird. We *can *understand quantum mechanics. And in a fairly intuitive way, really. It’s just that… Well… I think some of the conventions in physics hamper such understanding. Well… Let me be precise: one convention in particular, really. It’s that convention for measuring angles. Indeed, Mr. Leonhard Euler, back in the 18th century, might well be “the master of us all” (as Laplace is supposed to have said) but… Well… He couldn’t foresee how his omnipresent formula—*e*^{i}^{θ} = *cos*θ + *i*·*sin*θ—would, one day, be used to represent *something real*: an electron, or any elementary particle, really. If he *would *have known, I am sure he would have noted what I am noting here: *Nature *can’t be bothered by our conventions. Hence, if *e*^{i}^{θ} represents something real, then *e*^{−i}^{θ} must also represent something real. [Coz I admire this genius so much, I can’t resist the temptation. Here’s his portrait. He looks kinda funny here, doesn’t he? :-)]

Frankly, he would probably have understood quantum-mechanical theory as easily and instinctively as Dirac, I think, and I am pretty sure he would have *noted*—and, if he would have known about circularly polarized waves, probably *agreed* to—that *alternative *convention for measuring angles: we could, effectively, measure angles clockwise *or* counterclockwise depending on the direction of our particle—as opposed to Euler’s ‘one-size-fits-all’ counterclockwise convention. But so we did *not *adopt that alternative convention because… Well… We want to keep honoring Euler, I guess. 🙂

So… Well… If we’re going to keep honoring Euler by sticking to that ‘one-size-fits-all’ counterclockwise convention, then **I do believe that e^{i}^{θ} and e^{−i}^{θ} represent two different realities: spin up versus spin down.**

Yes. In our geometric interpretation of the wavefunction, these are, effectively, two different spin directions. And… Well… These are *real* directions: we *see *something different when they go through a Stern-Gerlach apparatus. So it’s *not* just some convention to *count *things like 0, 1, 2, etcetera versus 0, −1, −2 etcetera. It’s the same story again: different but related *mathematical *notions are (often) related to different but related *physical *possibilities. So… Well… I think that’s what we’ve got here. Think of it. Mainstream quantum math treats all wavefunctions as right-handed but… Well… A particle with *up *spin is a different particle than one with *down *spin, right? And, again, *Nature* surely can*not* be bothered about our convention of measuring phase angles clockwise or counterclockwise, right? So… Well… Kinda obvious, right? 🙂

Let me spell out my conclusions here:

**1.** The angular momentum can be positive or, alternatively, negative: *J* = +ħ/2 or −ħ/2. [Let me note that this is *not* obvious. Or less obvious than it seems, at first. In classical theory, you would expect an electron, or an atomic magnet, to line up with the field. Well… The Stern-Gerlach experiment shows they don’t: they keep their original orientation. Well… If the field is weak enough.]

**2.** Therefore, we would probably like to think that an *actual* particle—think of an electron, or whatever other particle you’d think of—comes in two *variants*: right-handed and left-handed. They will, therefore, *either* consist of (elementary) right-handed waves or, *else*, (elementary) left-handed waves. An elementary right-handed wave would be written as: ψ(θ* _{i}*)

*=*

*e*^{i}^{θi}

*= a*·(

_{i}*cos*θ

*+*

_{i}*i·sin*θ

*). In contrast, an elementary left-handed wave would be written as: ψ(θ*

_{i}*)*

_{i}*=*

*e*^{−i}^{θi}

*·(*

*=*a_{i}*cos*θ

*−*

_{i}*i·sin*θ

*). So that’s the complex conjugate.*

_{i}So… Well… Yes, I think complex conjugates are not just some *mathematical *notion: I believe they represent something real. It’s the usual thing: *Nature *has shown us that (most) mathematical possibilities correspond to *real *physical situations so… Well… Here you go. It is really just like the left- or right-handed circular polarization of an electromagnetic wave: we can have both for the matter-wave too! [As for the differences—different polarization plane and dimensions and what have you—I’ve already summed those up, so I won’t repeat myself here.] The point is: if we have two different *physical *situations, we’ll want to have two different functions to describe it. Think of it like this: why would we have *two*—yes, I admit, two *related—*amplitudes to describe the *up *or *down *state of the same system, but only one wavefunction for it? You tell me.

[…]

Authors like me are looked down upon by the so-called *professional* class of physicists. The few who bothered to react to my attempts to make sense of Einstein’s basic intuition in regard to the nature of the wavefunction all said pretty much the same thing: “Whatever your geometric (or *physical*) interpretation of the wavefunction might be, it won’t be compatible with the *isotropy *of space. You cannot *imagine *an object with a 720° symmetry. That’s *geometrically *impossible.”

Well… Almost three years ago, I wrote the following on this blog: “As strange as it sounds, a spin-1/2 particle needs *two *full rotations (2×360°=720°) until it is again in the same state. Now, in regard to that particularity, you’ll often read something like: “*There is **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 research center!]* Bollocks!* 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… Well… I am still pursuing my original dream which is… Well… Let me re-phrase what I wrote back in January 2015:

**Yes, we can actually imagine spin-1/2 particles**, and we actually do not need all that much imagination!

In fact, I am tempted to think that I’ve found a pretty good representation or… Well… A pretty good *image*, I should say, because… Well… A representation is something real, remember? 🙂

**Post scriptum** (10 December 2017): Our flywheel model of an electron makes sense, but also leaves many unanswered questions. The most obvious one question, perhaps, is: why the *up *and *down *state only?

I am not so worried about that question, even if I can’t answer it right away because… Well… Our apparatus—the way we *measure *reality—is set up to measure the angular momentum (or the *magnetic moment*, to be precise) in one direction only. If our electron is *captured* by some *harmonic *(or non-harmonic?) oscillation in multiple dimensions, then it should not be all that difficult to show its magnetic moment is going to align, somehow, in the same *or*, alternatively, the opposite direction of the magnetic field it is forced to travel through.

Of course, the analysis for the spin *up *situation (magnetic moment *down*) is quite peculiar: if our electron is a *mini*-magnet, why would it *not *line up with the magnetic field? We understand the precession of a spinning top in a gravitational field, but… *Hey**… It’s actually not that different*. Try to imagine some spinning top on the ceiling. 🙂 I am sure we can work out the math. 🙂 The electron must be some gyroscope, really: it won’t change direction. In other words, its magnetic moment won’t line up. It will precess, and it can do so in two directions, depending on its *state*. 🙂 […] At least, that’s why my instinct tells me. I admit I need to work out the math to convince you. 🙂

The second question is more important. If we just rotate the reference frame over 360°, we see the same thing: some rotating object which we, vaguely, describe by some *e*^{+i}^{·θ} function—to be precise, I should say: by some *Fourier* sum of such functions—or, if the rotation is in the other direction, by some *e*^{−i}^{·θ} function (again, you should read: a *Fourier *sum of such functions). Now, the weird thing, as I tried to explain above is the following: if we rotate the object itself, over the same 360°, we get a *different *object: our *e*^{i}^{·θ} and *e*^{−i}^{·θ} function (again: think of a *Fourier *sum, so that’s a wave *packet*, really) becomes a −*e*^{±i}^{·θ} thing. We get a *minus *sign in front of it. So what happened here? What’s the difference, *really*?

Well… I don’t know. It’s very deep. If I do nothing, and you keep watching me while turning around me, for a full 360°, then you’ll end up where you were when you started and, importantly, you’ll see the same thing. *Exactly *the same thing: if I was an *e*^{+i}^{·θ} wave packet, I am still an an *e*^{+i}^{·θ} wave packet now. Or if I was an *e*^{−i}^{·θ} wave packet, then I am still an an *e*^{−i}^{·θ} wave packet now. Easy. Logical. *Obvious*, right?

But so now we try something different: *I *turn around, over a full 360° turn, and *you *stay where you are. When I am back where I was—looking at you again, so to speak—then… Well… I am not quite the same any more. Or… Well… Perhaps I am but you *see *me differently. If I was *e*^{+i}^{·θ} wave packet, then I’ve become a −*e*^{+i}^{·θ} wave packet now. Not *hugely* different but… Well… That *minus *sign matters, right? Or If I was wave packet built up from elementary *a*·*e*^{−i}^{·θ} waves, then I’ve become a −*e*^{−i}^{·θ} wave packet now. What happened?

It makes me think of the twin paradox in special relativity. We know it’s a *paradox*—so that’s an *apparent *contradiction only: we know which twin stayed on Earth and which one traveled because of the gravitational forces on the traveling twin. The one who stays on Earth does not experience any acceleration or deceleration. Is it the same here? I mean… The one who’s turning around must experience some *force*.

Can we relate this to the twin paradox? Maybe. Note that a *minus *sign in front of the *e*^{−±i}^{·θ} functions amounts a minus sign in front of both the sine and cosine components. So… Well… The negative of a sine and cosine is the sine and cosine but with a phase shift of 180°: −*cos*θ = *cos*(θ ± π) and −*sin*θ = *sin*(θ ± π). Now, adding or subtracting a *common *phase factor to/from the argument of the wavefunction amounts to *changing *the origin of time. So… Well… I do think the twin paradox and this rather weird business of 360° and 720° symmetries are, effectively, related. 🙂

# The reality of the wavefunction

If you haven’t read any of my previous posts on the geometry of the wavefunction (this link goes to the most recent one of them), then don’t attempt to read this one. It brings too much stuff together to be comprehensible. In fact, I am not even sure if I am going to understand what I write myself. 🙂 [OK. Poor joke. Acknowledged.]

Just to recap the essentials, I part ways with mainstream physicists in regard to the *interpretation *of the wavefunction. For mainstream physicists, the wavefunction is just some mathematical construct. Nothing *real*. Of course, I acknowledge mainstream physicists have very good reasons for that, but… Well… I believe that, if there is interference, or diffraction, then *something *must be interfering, or something must be diffracting. I won’t dwell on this because… Well… I have done that too many times already. My *hypothesis *is that the wavefunction is, in effect, a *rotating **field vector*, so it’s just like the electric field vector of a (circularly polarized) electromagnetic wave (illustrated below).

Of course, it must be different, and it is. First, the (physical) dimension of the field vector of the matter-wave must be different. So what is it? Well… I am tempted to associate the real and imaginary component of the wavefunction with a force *per unit mass *(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/s^{2}), so that’s the dimension of a gravitational field.

Second, I also am tempted to think that this gravitational disturbance causes an electron (or any matter-particle) to move about some center, and I believe it does so at the speed of light. In contrast, electromagnetic waves do *not *involve any mass: they’re just an oscillating *field*. Nothing more. Nothing less. Why would I believe there must still be some pointlike particle involved? Well… As Feynman puts it: “When you do find the electron some place, the entire charge is there.” (Feynman’s *Lectures*, III-21-4) So… Well… That’s why.

The third difference is one that I thought of only recently: the *plane *of the oscillation can*not *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 am more explicit on that in the mentioned post, so you may want to check there. 🙂

I wish I mastered the software to make animations such as the one above (for which I have to credit Wikipedia), but so I don’t. You’ll just have to *imagine *it. That’s great mental exercise, so… Well… Just try it. 🙂

Let’s now think about rotating reference frames and transformations. If the *z*-direction is the direction along which we measure the angular momentum (or the magnetic moment), then the *up*-direction will be the *positive *z-direction. We’ll also assume the *y*-direction is the direction of travel of our elementary particle—and let’s just consider an electron here so we’re more real. 🙂 So we’re in the reference frame that Feynman used to derive the transformation matrices for spin-1/2 particles (or for two-state systems in general). His ‘improved’ Stern-Gerlach apparatus—which I’ll refer to as a beam splitter—illustrates this *geometry*.

So I think the magnetic moment—or the angular momentum, really—comes from an oscillatory motion in the *x*– and *y*-directions. One is the *real *component (the cosine function) and the other is the imaginary component (the sine function), as illustrated below.

So the crucial difference with the animations above (which illustrate left- and a right-handed polarization respectively) is that we, somehow, need to imagine the circular motion is *not *in the *xz*-plane, but in the *yz*-plane. Now what happens if we change the reference frame?

Well… That depends on what you mean by changing the reference frame. Suppose we’re looking in the positive *y*-direction—so that’s the direction in which our particle is moving—, then we might imagine how it would look like when *we *would make a 180° turn and look at the situation from the other side, so to speak. Now, I did a post on that earlier this year, which you may want to re-read. When we’re looking at the same thing from the other side (from the back side, so to speak), we will want to use our familiar reference frame. So we will want to keep the *z*-axis as it is (pointing upwards), and we will also want to define the *x*– and *y-*axis using the familiar right-hand rule for defining a coordinate frame. So our new *x*-axis and our new *y-*axis will the same as the old *x-* and *y-*axes but with the sign reversed. In short, we’ll have the following mini-transformation: (1) *z*‘ = *z*, (2) *x’* = −*x*, and (3) *y’* = −*y*.

So… Well… If we’re effectively looking at something *real *that was moving along the *y*-axis, then it will now still be moving along the *y’*-axis, but in the *negative *direction. Hence, our elementary wavefunction *e ^{i}*

^{θ}=

*cos*θ +

*i*·

*sin*θ will

*transform*into −

*cos*θ −

*i*·

*sin*θ = −

*cos*θ −

*i*·

*sin*θ =

*cos*θ −

*i*·

*sin*θ. It’s the same wavefunction. We just… Well… We just changed our reference frame. We didn’t change reality.

Now you’ll cry wolf, of course, because we just went through all that transformational stuff in our last post. To be specific, we presented the following transformation matrix for a rotation along the *z*-axis:

Now, if φ is equal to 180° (so that’s π in radians), then these *e*^{i}^{φ/2} and *e*^{−i}^{φ/2}/√2 factors are equal to *e*^{i}^{π/2} = *+i* and *e*^{−i}^{π/2} = −*i* respectively. Hence, our *e ^{i}*

^{θ}=

*cos*θ +

*i*·

*sin*θ becomes…

** Hey !** Wait a minute ! We’re talking about two

*very*different things here, right? The

*e*

^{i}^{θ}=

*cos*θ +

*i*·

*sin*θ is an

*elementary*wavefunction which, we presume, describes some real-life particle—we talked about an electron with its spin in the

*up*-direction—while these transformation matrices are to be applied to amplitudes describing… Well… Either an

*up*– or a

*down*-state, right?

Right. But… Well… Is it so different, really? Suppose our *e ^{i}*

^{θ}=

*cos*θ +

*i*·

*sin*θ wavefunction describes an

*up*-electron, then we still have to apply that

*e*

^{i}^{φ/2}=

*e*

^{i}^{π/2}=

*+i*factor, right? So we get a new wavefunction that will be equal to

*e*

^{i}^{φ/2}·

*e*

^{i}^{θ}=

*e*

^{i}^{π/2}·

*e*

^{i}^{θ}=

*+i*·

*e*

^{i}^{θ}=

*i*·

*cos*θ +

*i*

^{2}·

*sin*θ =

*sin*θ −

*i*·

*cos*θ, right? So how can we reconcile that with the

*cos*θ −

*i*·

*sin*θ function we thought we’d find?

We can’t. So… Well… Either *my* theory is wrong or… Well… Feynman can’t be wrong, can he? I mean… It’s not only Feynman here. We’re talking all mainstream physicists here, right?

Right. But think of it. *Our electron in that thought experiment does, effectively, make a turn of 180°, so it is going in the other direction now ! *That’s more than just… Well… Going around the apparatus and looking at stuff from the other side.

Hmm… Interesting. Let’s think about the difference between the *sin*θ − *i*·*cos*θ and *cos*θ − *i*·*sin*θ functions. First, note that they will give us the same probabilities: the square of the absolute value of both complex numbers is the same. [It’s equal to 1 because we didn’t bother to put a coefficient in front.] Secondly, we should note that the sine and cosine functions are essentially the same. They just differ by a phase factor: *cos*θ = *sin*(θ + π/2) and −*sin*θ = *cos*(θ + π/2). Let’s see what we can do with that. We can write the following, for example:

*sin*θ − *i*·*cos*θ = −*cos*(θ + π/2) − *i*·*sin*(θ + π/2) = −[*cos*(θ + π/2) + *i*·*sin*(θ + π/2)] = −*e ^{i}*

^{·(θ + π/2)}

Well… I guess that’s something at least ! The *e ^{i}*

^{·θ}and −

*e*

^{i}^{·(θ + π/2)}functions differ by a phase shift

*and*a minus sign so… Well… That’s what it takes to reverse the direction of an electron. 🙂 Let us mull over that in the coming days. As I mentioned, these more philosophical topics are not easily exhausted. 🙂

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

Our previous posts showed how a simple geometric interpretation of the elementary wavefunction yielded the (Compton scattering) radius of an elementary particle—for an electron, at least: for the proton, we only got the order of magnitude right—but then a proton is not an elementary particle. We got lots of other interesting equations as well… But… Well… When everything is said and done, it’s that equivalence between the E = m·*a*^{2}·ω^{2} and E = m·*c*^{2} relations that we… Well… We need to be more *specific *about it.

Indeed, I’ve been ambiguous here and there—*oscillating *between various interpretations, so to speak. 🙂 In my own mind, I refer to my unanswered questions, or my ambiguous answers to them, as the *form factor *problem. So… Well… That explains the title of my post. But so… Well… I do want to be somewhat more *conclusive *in this post. So let’s go and see where we end up. 🙂

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. 🙂

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 our 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 = ω·I. Of course, the moment of inertia (aka the angular mass) will depend on the *form *(or *shape*) of our flywheel:

- I = m·
*a*^{2}for a rotating*point*mass m or, what amounts to the same, for a circular*hoop*of mass m and radius*r*=*a*. - For a rotating (uniformly solid)
*disk*, we must add a 1/2 factor: I*a*^{2}/2.

How can we relate those formulas to the E = m·*a*^{2}·ω^{2} formula? The *kinetic *energy that is being stored in a flywheel is equal E* _{kinetic}* = I·ω

*/2, so that is only*

^{2}*half*of the E = m·

*a*

^{2}·ω

^{2}product if we substitute I for I = m·

*a*

^{2}. [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, the E = m·

*a*

^{2}·ω

^{2}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 just

*transfers*energy from one oscillator to the other, but so… Well… We don’t

*include*it in our energy calculations.

**The essence of our model is that two-dimensional oscillation which**That two-dimensional oscillation—the

*drives*the electron, and which is reflected in Einstein’s E = m·*c*^{2}formula.*a*

^{2}·ω

^{2}=

*c*

^{2}equation, really—tells us that

**the**—but measured in units of

*resonant*(or*natural*)*frequency*of the fabric of spacetime is given by the speed of light*a*. [If you don’t quite get this, re-write the

*a*

^{2}·ω

^{2}=

*c*

^{2}equation as ω =

*c*/

*a*: the radius of our electron appears as a

*natural*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 example—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 to *v *= *a·*ω = *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 radius 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 as *a *= ħ·/(m·*c*) = 3.8616×10^{−13} 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:

*c* = *a*·ω = *a*·E/ħ = *a*·m·*c*^{2}/ħ ⇔ *a *= ħ/(m·*c*)

The question is: what *is *that rotating arrow? I’ve been vague and not so vague on this. The thing is: I can’t *prove *anything in this regard. But my *hypothesis *is that it is, in effect, a *rotating **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:

- 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 mass*(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/s^{2}), so that’s the dimension of a gravitational field. - 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 do
*not*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) - The third difference is one that I thought of only recently: the
*plane*of the oscillation can*not*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. The basic idea here is illustrated below (credit for this illustration goes to another blogger on physics). As for the Stern-Gerlach experiment itself, let me refer you to a YouTube video from the *Quantum Made Simple *site.

The point is: the direction of the angular momentum (and the magnetic moment) of an electron—or, to be precise, its component as measured in the direction of the (inhomogeneous) magnetic field through which our electron is *traveling*—can*not* be parallel to the direction of motion. On the contrary, it is *perpendicular* to the direction of motion. In other words, if we imagine our electron as spinning around some center, then the disk it circumscribes will *comprise *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 ** J** and a magnetic moment

**(I used bold-face because these are**

*μ**vector*quantities) that is parallel to some magnetic field

**B**, will

*not*line up, as you’d expect a tiny magnet to do in a magnetic field—or not

*completely*, at least: it will

*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 of show* *that the general geometry of the situation makes sense: we also need to show the numbers come out alright. So… Well… Whatever we *think *our electron—or its wavefunction—might be, it needs to be compatible with stuff like the *observed* precession frequency* *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) gravity—such as the one I mentioned above—are intriguing too and, hence, whatever hypotheses I advance here, I’d better check them against those more advanced theories too, right? 🙂 Unfortunately, that’s going to take me a few more years of studying… But… Well… I still have many years ahead—I hope. 🙂

**Post scriptum**: It’s funny how one’s brain keeps working when sleeping. When I woke up this morning, I thought: “But it *is *that flywheel that matters, right? That’s the energy storage mechanism and also explains how photons possibly interact with electrons. The oscillators *drive *the flywheel but, without the flywheel, nothing is happening. It is really the *transfer *of energy—through the flywheel—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. The *motion *of our first oscillator is given by the cos(ω·t) = cosθ function (θ = ω·t), and its kinetic energy will be equal to sin^{2}θ. Hence, the (instantaneous) *change *in kinetic energy at any point in time (as a function of the angle θ) is equal to: d(sin^{2}θ)/dθ = 2∙sinθ∙d(sinθ)/dθ = 2∙sinθ∙cosθ. Now, the motion of the second oscillator (just look at that second piston going up and down in the V-2 engine) is given by the sinθ function, which is equal to cos(θ − π /2). Hence, its kinetic energy is equal to sin^{2}(θ − π /2), and how it *changes *(as a function of θ again) is equal to 2∙sin(θ − π /2)∙cos(θ − π /2) = = −2∙cosθ∙sinθ = −2∙sinθ∙cosθ. 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… What *if *the relevant energy formula is E = m·*a*^{2}·ω^{2}/2 instead of E = m·*a*^{2}·ω^{2}? What are the implications? Well… We get a √2 factor in our formula for the radius *a*, as shown below.

Now that is *not *so nice. For the tangential velocity, we get *v *= *a*·ω = √2·*c*. This is also *not *so nice. How can we save our model? I am not sure, but here I am thinking of the mentioned precession—the *wobbling *of our flywheel in a magnetic field. Remember we may think of * J_{z}*—the angular momentum or, to be precise, its component in the

*z*-direction (the direction in which we

*measure*it—as the projection of the

*real*angular momentum

*. Let me insert Feynman’s illustration here again (Feynman’s*

**J***Lectures*, II-34-3), so you get what I am talking about.

Now, all depends on the angle (θ) between * J_{z}* and

**, 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 the**

*J**magnitude*of the presumed

*actual*momentum:In this particular case (spin-1/2 particles),

*j*is equal to 1/2 (in units of ħ, of course). Hence,

*J*is equal to √0.75 ≈ 0.866. Elementary geometry then tells us cos(θ) = (1/2)/√(3/4) = = 1/√3. Hence, θ ≈ 54.73561°. That’s a big angle—larger than the 45° angle we had secretly expected because… Well… The 45° angle has that √2 factor in it: cos(45°) = sin(45°) = 1/√2.

Hmm… As you can see, there is no easy fix here. Those damn 1/2 factors! They pop up everywhere, don’t they? 🙂 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 or √2 problem, right?* *🙂

**Note**: If you’re into quantum math, you’ll note *a *= *ħ*/(m·*c*) is the *reduced *Compton scattering radius. The standard Compton scattering radius is equal to *a·*2π* *= (2π·*ħ*)/(m·*c*) = *h*/(m·*c*) = *h*/(m·*c*). It doesn’t solve the √2 problem. Sorry. The form factor problem. 🙂

To be honest, I finished my published paper on all of this with a suggestion that, perhaps, we should think of two *circular *oscillations, as opposed to linear ones. Think of a tiny ball, whose center of mass stays where it is, as depicted below. Any rotation – around any axis – will be some combination of a rotation around the two other axes. Hence, we may want to think of our two-dimensional 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. 🙂They are oscillations, still, so I am not thinking of *two *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. 🙂

# The speed of light as an angular velocity

Over the weekend, I worked on a revised version of my paper on a physical interpretation of the wavefunction. However, I forgot to add the final remarks on the speed of light as an angular velocity. I know… This post is for my faithful followers only. It is dense, but let me add the missing bits here:

**Post scriptum (29 October)**: Einstein’s view on *aether* theories probably still holds true: “We may say that according to the general theory of relativity space is endowed with physical qualities; in this sense, therefore, there exists an *aether*. According to the general theory of relativity, space without *aether* is unthinkable – for in such space there not only would be no propagation of light, but also no possibility of existence for standards of space and time (measuring-rods and clocks), nor therefore any space-time intervals in the physical sense. But this *aether* may not be thought of as endowed with the quality characteristic of ponderable media, as consisting of parts which may be tracked through time. The idea of motion may not be applied to it.”

The above quote is taken from the Wikipedia article on *aether* theories. The same article also quotes Robert Laughlin, the 1998 Nobel Laureate in Physics, who said this about aether in 2005: “It is ironic that Einstein’s most creative work, the general theory of relativity, should boil down to conceptualizing space as a medium when his original premise [in special relativity] was that no such medium existed. […] The word ‘aether’ has extremely negative connotations in theoretical physics because of its past association with opposition to relativity. This is unfortunate because, stripped of these connotations, it rather nicely captures the way most physicists actually think about the vacuum. […]The modern concept of the vacuum of space, confirmed every day by experiment, is a relativistic *aether*. But we do not call it this because it is taboo.”

I really love this: a *relativistic* aether. My interpretation of the wavefunction is *very *consistent with that.

# A physical explanation for relativistic length contraction?

My last posts were all about a possible *physical *interpretation of the quantum-mechanical wavefunction. To be precise, we have been interpreting the wavefunction as a gravitational wave. In this interpretation, the real and imaginary component of the wavefunction get a *physical *dimension: force per unit mass (newton per kg). The inspiration here was the structural similarity between Coulomb’s and Newton’s force laws. They both look alike: it’s just that one gives us a force per unit *charge *(newton per *coulomb*), while the other gives us a force per unit *mass*.

So… Well… Many nice things came out of this – and I wrote about that at length – but last night I was thinking this interpretation may also offer an explanation of relativistic length contraction. Before we get there, let us re-visit our hypothesis.

**The geometry of the wavefunction**

The elementary wavefunction is written as:

ψ = *a·e*^{−i(E·t − p∙x)/ħ} = *a·cos*(**p**∙**x**/ħ – E∙t/ħ) *+** i·a·sin*(**p**∙**x**/ħ – E∙t/ħ)

Nature should not care about our conventions for measuring the phase angle clockwise or counterclockwise and, therefore, the ψ = *a·e*^{i}^{[E·t − p∙x]/ħ} function may also be permitted. We know that *cos*(θ) = *cos**(**–*θ) and *sin*θ = *–**sin**(**–*θ), so we can write: * *

ψ = *a·e*^{i}^{(E·t − p∙x)/ħ} = *a·cos*(E∙t/ħ – **p**∙**x**/ħ) *+** i·a·sin*(E∙t/ħ – **p**∙**x**/ħ)

*= **a·cos*(**p**∙**x**/ħ – E∙t/ħ) *–** i·a·sin*(**p**∙**x**/ħ – E∙t/ħ)

The vectors **p** and **x** are the the momentum and position vector respectively: **p** = (p_{x}, p_{y}, p_{z}) and **x** = (x, y, z). However, if we assume there is no uncertainty about **p** – not about the direction nor the magnitude – then we may choose an x-axis which reflects the direction of **p**. As such, **x** = (x, y, z) reduces to (x, 0, 0), and **p**∙**x**/ħ reduces to p∙x/ħ. This amounts to saying our particle is traveling along the x-axis or, if p = 0, that our particle is located somewhere on the x-axis. Hence, the analysis is one-dimensional only.

The geometry of the elementary wavefunction is illustrated below. The x-axis is the direction of propagation, and the y- and z-axes represent the real and imaginary part of the wavefunction respectively.

Note that, when applying the right-hand rule for the axes, the vertical axis is the y-axis, not the z-axis. Hence, we may associate the vertical axis with the cosine component, and the horizontal axis with the sine component. You can check this as follows: if the origin is the (x, t) = (0, 0) point, then cos(θ) = cos(0) = 1 and sin(θ) = sin(0) = 0. This is reflected in both illustrations, which show a left- and a right-handed wave respectively. We speculated this should correspond to the two possible values for the quantum-mechanical spin of the wave: +ħ/2 or −ħ/2. The cosine and sine components for the left-handed wave are shown below. Needless to say, the cosine and sine function are the same, except for a phase difference of π/2: sin(θ) = cos(θ − π/2).

As for the wave velocity, and its direction of propagation, we know that the (phase) velocity of any wave F(kx – ωt) is given by *v*_{p} = ω/k = (E/ħ)/(p/ħ) = E/p. Of course, the momentum might also be in the negative x-direction, in which case k would be equal to -p and, therefore, we would get a negative phase velocity: *v*_{p} = ω/k = *–*E/p.

**The de Broglie relations**

E/ħ = ω gives the frequency in time (expressed in radians per second), while p/ħ = k gives us the wavenumber, or the frequency in space (expressed in radians per meter). Of course, we may write: f = ω/2π and λ = 2π/k, which gives us the two de Broglie relations:

- E = ħ∙ω = h∙f
- p = ħ∙k = h/λ

The frequency in time is easy to interpret. The wavefunction of a particle with more energy, or more mass, will have a *higher density in time* than a particle with less energy.

In contrast, the second *de Broglie *relation is somewhat harder to interpret. According to the p = h/λ relation, the wavelength is *inversely *proportional to the momentum: λ = h/p. The velocity of a photon, or a (theoretical) particle with zero rest mass (m_{0} = 0), is *c* and, therefore, we find that p = m* _{v}*∙

*v*= m

*∙*

_{c}*c*= m∙

*c*(all of the energy is kinetic). Hence, we can write: p∙

*c*= m∙

*c*

^{2}= E, which we may also write as: E/p =

*c*. Hence, for a particle with zero rest mass, the wavelength can be written as:

λ = h/p = hc/E = h/mc

However, this is a limiting situation – applicable to photons only. Real-life *matter*-particles should have *some *mass[1] and, therefore, their velocity will never be *c*.[2]

Hence, if p goes to zero, then the wavelength becomes infinitely long: if p → 0 then λ* → ∞*. How should we interpret this inverse proportionality between λ and p? To answer this question, let us first see what this wavelength λ actually represents.

If we look at the ψ = *a*·cos(p∙x/ħ – E∙t/ħ) – *i*·*a*·sin(p∙x/ħ – E∙t/ħ) once more, and if we write p∙x/ħ as Δ, then we can look at p∙x/ħ as a phase factor, and so we will be interested to know for what x this phase factor Δ = p∙x/ħ will be equal to 2π. So we write:

Δ =p∙x/ħ = 2π ⇔ x = 2π∙ħ/p = h/p = λ

So now we get a meaningful interpretation for that wavelength. It is the distance between the crests (or the troughs) of the wave, so to speak, as illustrated below. Of course, this two-dimensional wave has no real crests or troughs: we measure crests and troughs against the y-axis here. Hence, our definition depend on the frame of reference.

Now we know what λ actually represents for our one-dimensional elementary wavefunction. Now, the time that is needed for one cycle is equal to T = 1/*f *= 2π·(ħ/E). Hence, we can now calculate the wave velocity:

v = λ/T = (h/p)/[2π·(ħ/E)] = E/p

Unsurprisingly, we just get the phase velocity that we had calculated already: *v *= *v*_{p} = E/p. The question remains: what if p is zero? What if we are looking at some particle at rest? It is an intriguing question: we get an infinitely long wavelength, and an infinite wave velocity.

Now, re-writing the *v *= E/p as *v *= m∙*c*^{2}/m∙*v*_{g }* *= *c*/β_{g}, in which β_{g} is the relative *classical *velocity[3] of our particle β_{g} = *v*_{g}/*c*) tells us that the *phase *velocities will effectively be superluminal (β_{g} < 1 so 1/ β_{g} > 1), but what if β_{g} approaches zero? The conclusion seems unavoidable: for a particle at rest, we only have a frequency *in time*, as the wavefunction reduces to:

ψ = a·e^{−i·E·t/ħ} = a·cos(E∙t/ħ) – i·a·sin(E∙t/ħ)

How should we interpret this?

**A physical interpretation of relativistic length contraction?**

In my previous posts, we argued that the oscillations of the wavefunction pack energy. Because the energy of our particle is finite, the wave train cannot be infinitely long. If we assume some *definite* number of oscillations, then the string of oscillations will be shorter as λ decreases. Hence, the physical interpretation of the wavefunction that is offered here may explain relativistic length contraction.

🙂

Yep. Think about it. 🙂

[1] Even neutrinos have some (rest) mass. This was first confirmed by the US-Japan Super-Kamiokande collaboration in 1998. Neutrinos oscillate between three so-called flavors: electron neutrinos, muon neutrinos and *tau *neutrinos. Recent data suggests that the *sum *of their masses is less than a millionth of the rest mass of an electron. Hence, they propagate at speeds that are very near to the speed of light.

[2] Using the Lorentz factor (γ), we can write the relativistically correct formula for the kinetic energy as KE = E − E_{0} = m_{v}*c*^{2} − m_{0}*c*^{2} = m_{0}γ*c*^{2} − m_{0}*c*^{2} = m_{0}*c*^{2}(γ − 1). As *v *approaches *c*, γ approaches infinity and, therefore, the kinetic energy would become infinite as well.

[3] Because our particle will be represented by a wave *packet*, i.e. a superimposition of elementary waves with different E and p, the classical velocity of the particle becomes the *group *velocity of the wave, which is why we denote it by *v*_{g}.

# The geometry of the wavefunction (2)

This post further builds on the rather remarkable results we got in our previous posts. Let us start with the basics once again. The elementary wavefunction is written as:

ψ = *a·e*^{−i[E·t − p∙x]/ħ} = *a·cos*(**p**∙**x**/ħ − E∙t/ħ) *+** i·a·sin*(**p**∙**x**/ħ − E∙t/ħ)

Of course, *Nature* (or God, as Einstein would put it) does not care about our conventions for measuring an angle (i.e. the phase of our wavefunction) clockwise or counterclockwise and, therefore, the ψ = *a·e*^{i}^{[E·t − p∙x]/ħ} function is also permitted. We know that *cos*(θ) = *cos*(−θ) and *sin*θ = −*sin*(*−*θ), so we can write: * *

ψ = *a·e*^{i}^{[E·t − p∙x]/ħ} = *a·cos*(E∙t/ħ − **p**∙**x**/ħ) *+** i·a·sin*(E∙t/ħ − **p**∙**x**/ħ)

*= **a·cos*(**p**∙**x**/ħ − E∙t/ħ) −* i·a·sin*(**p**∙**x**/ħ − E∙t/ħ)

The vectors **p** and **x** are the momentum and position vector respectively: **p** = (p_{x}, p_{y}, p_{z}) and **x** = (x, y, z). However, if we assume there is *no* uncertainty about **p** – not about the direction, and not about the magnitude – then the direction of **p** can be our x-axis. In this reference frame, **x** = (x, y, z) reduces to (x, 0, 0), and **p**∙**x**/ħ reduces to p∙x/ħ. This amounts to saying our particle is traveling along the x-axis or, if p = 0, that our particle is located somewhere on the x-axis. So we have an analysis in one dimension only then, which facilitates our calculations. The geometry of the wavefunction is then as illustrated below. The x-axis is the direction of propagation, and the y- and z-axes represent the real and imaginary part of the wavefunction respectively.

Note that, when applying the right-hand rule for the axes, the vertical axis is the y-axis, not the z-axis. Hence, we may associate the vertical axis with the cosine component, and the horizontal axis with the sine component. [You can check this as follows: if the origin is the (x, t) = (0, 0) point, then *cos*(θ) = *cos*(0) = 1 and *sin*(θ) = *sin*(0) = 0. This is reflected in both illustrations, which show a left- and a right-handed wave respectively.]

Now, you will remember that we speculated the two *polarizations* (left- versus right-handed) should correspond to the two possible values for the quantum-mechanical spin of the wave (+ħ/2 or −ħ/2). We will come back to this at the end of this post. Just focus on the essentials first: the cosine and sine components for the left-handed wave are shown below. Look at it carefully and try to understand. Needless to say, the cosine and sine function are the same, except for a phase difference of π/2: *sin*(θ) = *cos*(θ − π/2).

As for the wave velocity, and its direction of propagation, we know that the (phase) velocity of any waveform F(kx − ωt) is given by *v*_{p} = ω/k. In our case, we find that *v*_{p} = ω/k = (E/ħ)/(p/ħ) = E/p. Of course, the momentum might also be in the negative x-direction, in which case k would be equal to −p and, therefore, we would get a negative phase velocity: *v*_{p} = ω/k = (E/ħ)/(−p/ħ) = −E/p.

As you know, E/ħ = ω gives the *frequency in time* (expressed in radians per second), while p/ħ = k gives us the wavenumber, or the *frequency in space* (expressed in radians per meter). [If in doubt, check my post on essential wave math.] Now, you also know that *f* = ω/2π and λ = 2π/k, which gives us the two de Broglie relations:

- E = ħ∙ω = h∙
*f* - p = ħ∙k = h/λ

The frequency in time (oscillations or radians per second) is easy to interpret. A particle will always have some mass and, therefore, some energy, and it is easy to appreciate the fact that the wavefunction of a particle with more energy (or more mass) will have a *higher density in time* than a particle with less energy.

However, the second *de Broglie *relation is somewhat harder to interpret. Note that the wavelength is *inversely *proportional to the momentum: λ = h/p. Hence, if p goes to zero, then the wavelength becomes infinitely long, so we write:

If p → 0 then λ* → ∞.*

For the limit situation, a particle with zero *rest *mass (m_{0} = 0), the velocity may be *c* and, therefore, we find that p = m* _{v}*∙

*v*= m

*∙*

_{c}*c*= m∙

*c*(all of the energy is kinetic) and, therefore, p∙

*c*= m∙

*c*

^{2}= E, which we may also write as: E/p =

*c*. Hence, for a particle with zero rest mass (m

_{0}= 0), the wavelength can be written as:

λ = h/p = h*c*/E = h/m*c*

Of course, we are talking a *photon *here. We get the zero rest mass for a photon. In contrast, all *matter*-particles should have *some *mass[1] and, therefore, their velocity will *never* equal *c*.[2] The question remains: how should we interpret the inverse proportionality between *λ* and p?

Let us first see what this wavelength λ actually represents. If we look at the ψ = a·*cos*(p∙x/ħ − E∙t/ħ) − *i·a·sin*(p∙x/ħ – E∙t/ħ) once more, and if we write p∙x/ħ as Δ, then we can look at p∙x/ħ as a phase factor, and so we will be interested to know for what x this phase factor Δ = p∙x/ħ will be equal to 2π. So we write:

Δ =p∙x/ħ = 2π ⇔ x = 2π∙ħ/p = h/p = λ

So now we get a meaningful interpretation for that wavelength. It is the distance between the crests (or the troughs) of the wave, so to speak, as illustrated below. Of course, this two-dimensional wave has no real crests or troughs: they depend on your frame of reference.

So now we know what λ actually represent for our one-dimensional elementary wavefunction. Now, the time that is needed for one cycle is equal to T = 1/*f *= 2π·(ħ/E). Hence, we can now calculate the wave velocity:

*v* = λ/T = (h/p)/[2π·(ħ/E)] = E/p

Unsurprisingly, we just get the phase velocity that we had calculated already: *v *= *v*_{p} = E/p. It does not answer the question: what if p is zero? What if we are looking at some particle at rest? It is an intriguing question: we get an infinitely long wavelength, and an infinite phase velocity. Now, we know *phase *velocities can be superluminal, but they should not be infinite. So what does the mathematical inconsistency tell us? Do these infinitely long wavelengths and infinite wave velocities tell us that our particle *has to* move? Do they tell us our notion of a particle at rest is mathematically inconsistent?

Maybe. But maybe not. Perhaps the inconsistency just tells us our elementary wavefunction – or the concept of a *precise *energy, and a *precise *momentum – does not make sense. This is where the Uncertainty Principle comes in: stating that p = 0, implies zero uncertainty. Hence, the σ_{p} factor in the σ_{p}∙σ_{x} ≤ ħ/2 would be zero and, therefore, σ_{p}∙σ_{x} would be zero which, according to the Uncertainty Principle, it cannot be: it can be very small, but it cannot be zero.

It is interesting to note here that σ_{p} refers to the standard deviation from the mean, as illustrated below. Of course, the distribution may be or may not be normal – we don’t know – but a normal distribution makes intuitive sense, of course. Also, if we assume the mean is zero, then the uncertainty is basically about the *direction *in which our particle is moving, as the momentum might then be positive *or *negative.

The question of natural units may pop up. The Uncertainty Principle suggests a *numerical *value of the natural unit for momentum and distance that is equal to the *square root *of ħ/2, so that’s about 0.726×10^{−17} m for the distance unit and 0.726×10^{−17} N∙s for the momentum unit, as the product of both gives us ħ/2. To make this somewhat more real, we may note that 0.726×10^{−17} m is the attometer scale (1 am = 1×10^{−18} m), so that is very small but not unreasonably small.[3]

Hence, we need to superimpose a potentially infinite number of waves with energies and momenta centered on some mean value. It is only then that we get meaningful results. For example, the idea of a *group *velocity – which should correspond to the classical idea of the velocity of our particle – only makes sense in the context of wave *packet*. Indeed, the group velocity of a wave packet (*v*_{g}) is calculated as follows:

*v*_{g} = ∂ω* _{i}*/∂k

*= ∂(E*

_{i}*/ħ)/∂(p*

_{i}*/ħ) = ∂(E*

_{i}*)/∂(p*

_{i}*)*

_{i}This assumes the existence of a *dispersion relation* which gives us ω* _{i}* as a function of k

*– what amounts to the same – E*

_{i}*as a function of p*

_{i}*. How do we get that? Well… There are a few ways to go about it but one interesting way of doing it is to re-write Schrödinger’s equation as the following*

_{i}*pair*of equations[4]:

*Re*(∂ψ/∂t) = −[ħ/(2m_{eff})]·*Im*(∇^{2}ψ) ⇔ ω·cos(kx − ωt) = k^{2}·[ħ/(2m_{eff})]·cos(kx − ωt)*Im*(∂ψ/∂t) = [ħ/(2m_{eff})]·*Re*(∇^{2}ψ) ⇔ ω·sin(kx − ωt) = k^{2}·[ħ/(2m_{eff})]·sin(kx − ωt)

These equations imply the following dispersion relation:

ω = ħ·k^{2}/(2m)

Of course, we need to think about the subscripts now: we have ω* _{i}*, k

*, but… What about m*

_{i}_{eff}or, dropping the subscript, about m? Do we write it as m

*? If so, what is it? Well… It is the*

_{i}*equivalent*mass of E

*obviously, and so we get it from the mass-energy equivalence relation: m*

_{i}*= E*

_{i}*/*

_{i}*c*

^{2}. It is a fine point, but one most people forget about: they usually just write m. However, if there is uncertainty in the energy, then Einstein’s mass-energy relation tells us we must have some uncertainty in the (equivalent) mass too, and the two will, obviously, be related as follows: σ

_{m}= σ

_{E}/

*c*

^{2}. We are tempted to do a few substitutions here. Let’s first check what we get when doing the m

*= E*

_{i}*/*

_{i}*c*

^{2}substitution:

ω* _{i}* = ħ·k

_{i}^{2}/(2m

*) = (1/2)∙ħ·k*

_{i}

_{i}^{2}∙

*c*

^{2}/E

*= (1/2)∙ħ·k*

_{i}

_{i}^{2}∙

*c*

^{2}/(ω

*∙ħ) = (1/2)∙ħ·k*

_{i}

_{i}^{2}∙

*c*

^{2}/ω

_{i}⇔ ω_{i}^{2}/k_{i}^{2} = *c*^{2}/2 ⇔ ω* _{i}*/k

*=*

_{i}*v*

_{p}=

*c*/2 !?

We get a very interesting but nonsensical *condition* for the dispersion relation here. I wonder what mistake I made. 😦

Let us try another substitution. The group velocity is what it is, right? It is the velocity of the group, so we can write: k* _{i}* = p/ħ = m

_{i}·

*v*

*. This gives us the following result:*

_{g}ω* _{i}* = ħ·(m

*·*

_{i}*v*

_{g})

^{2}/(2m

*) = ħ·m*

_{i}*·*

_{i}*v*

_{g}

^{2}/2

It is yet another interesting *condition *for the dispersion relation. Does it make any more sense? I am not so sure. That factor 1/2 troubles us. It only makes sense when we *drop *it. Now you will object that Schrödinger’s equation gives us the electron orbitals – and many other correct descriptions of quantum-mechanical stuff – so, surely, Schrödinger’s equation cannot be wrong. You’re right. It’s just that… Well… When we are splitting in up in two equations, as we are doing here, then we are looking at *one *of the two dimensions of the oscillation only and, therefore, it’s only *half *of the mass that counts. Complicated explanation but… Well… It should make sense, because the results that come out make sense. Think of it. So we write this:

*Re*(∂ψ/∂t) = −(ħ/m_{eff})·*Im*(∇^{2}ψ) ⇔ ω·cos(kx − ωt) = k^{2}·(ħ/m_{eff})·cos(kx − ωt)*Im*(∂ψ/∂t) = (ħ/m_{eff})·*Re*(∇^{2}ψ) ⇔ ω·sin(kx − ωt) = k^{2}·(ħ/m_{eff})·sin(kx − ωt)

We then get the dispersion relation *without *that 1/2 factor:

ω* _{i}* = ħ·k

_{i}^{2}/m

_{i}The m* _{i}* = E

*/*

_{i}*c*

^{2}substitution then gives us the result we sort of expected to see:

ω* _{i}* = ħ·k

_{i}^{2}/m

*= ħ·k*

_{i}

_{i}^{2}∙

*c*

^{2}/E

*= ħ·k*

_{i}

_{i}^{2}∙

*c*

^{2}/(ω

*∙ħ) ⇔ ω*

_{i}*/k*

_{i}*=*

_{i}*v*

*=*

_{p}*c*

Likewise, the other calculation also looks more meaningful now:

ω* _{i}* = ħ·(m

*·*

_{i}*v*

_{g})

^{2}/m

*= ħ·m*

_{i}*·*

_{i}*v*

_{g}

^{2}

Sweet ! 🙂

Let us put this aside for the moment and focus on something else. If you look at the illustrations above, you see we can sort of distinguish (1) a linear velocity – the speed with which those wave crests (or troughs) move – and (2) some kind of circular or tangential velocity – the velocity along the red contour line above. We’ll need the formula for a tangential velocity: *v*_{t} = *a*∙ω.

Now, if λ is zero, then *v*_{t} = *a*∙ω = *a*∙E/ħ is just all there is. We may double-check this as follows: the distance traveled in one period will be equal to 2π*a*, and the period of the oscillation is T = 2π·(ħ/E). Therefore, *v*_{t} will, effectively, be equal to *v*_{t} = 2π*a*/(2πħ/E) = *a*∙E/ħ. However, if λ is non-zero, then the distance traveled in one period will be equal to 2π*a *+ λ. The period remains the same: T = 2π·(ħ/E). Hence, we can write:

For an electron, we did this weird calculation. We had an angular momentum formula (for an electron) which we equated with the *real-life* +ħ/2 or −ħ/2 values of its spin, and we got a *numerical *value for *a*. It was the Compton radius: the scattering radius for an electron. Let us write it out:

Using the right numbers, you’ll find the *numerical* value for *a*: 3.8616×10^{−13} m. But let us just substitute the formula itself here:

This is fascinating ! And we just calculated that *v*_{p} is equal to *c*. For the elementary wavefunction, that is. Hence, we get this amazing result:

*v*_{t} = 2*c*

This *tangential *velocity is *twice *the *linear *velocity !

Of course, the question is: what is the *physical *significance of this? I need to further look at this. Wave velocities are, essentially, *mathematical *concepts only: the wave propagates through space, but *nothing else *is really moving. However, the geometric implications are obviously quite interesting and, hence, need further exploration.

One conclusion stands out: all these results reinforce our interpretation of the speed of light as a *property *of the vacuum – or of the fabric of spacetime itself. 🙂

[1] Even neutrinos should have some (rest) mass. In fact, the mass of the known neutrino flavors was estimated to be smaller than 0.12 eV/*c*^{2}. This mass combines the three known neutrino flavors.

[2] Using the Lorentz factor (γ), we can write the relativistically correct formula for the kinetic energy as KE = E − E_{0} = m_{v}*c*^{2} − m_{0}*c*^{2} = m_{0}γ*c*^{2} − m_{0}*c*^{2} = m_{0}*c*^{2}(γ − 1). As *v *approaches *c*, γ approaches infinity and, therefore, the kinetic energy would become infinite as well.

[3] It is, of course, extremely small, but 1 am is the current sensitivity of the LIGO detector for gravitational waves. It is also thought of as the upper limit for the length of an electron, for quarks, and for fundamental strings in string theory. It is, in any case, 1,000,000,000,000,000,000 times larger than the order of magnitude of the Planck length (1.616229(38)×10^{−35} m).

[4] The m_{eff} is the *effective* mass of the particle, which depends on the medium. For example, an electron traveling in a solid (a transistor, for example) will have a different effective mass than in an atom. In free space, we can drop the subscript and just write m_{eff} = m. As for the equations, they are easily derived from noting that two complex numbers a + *i*∙b and c + *i*∙d are equal if, and only if, their real and imaginary parts are the same. Now, the ∂ψ/∂t = *i*∙(ħ/m_{eff})∙∇^{2}ψ equation amounts to writing something like this: a + *i*∙b = *i*∙(c + *i*∙d). Now, remembering that *i*^{2} = −1, you can easily figure out that *i*∙(c + *i*∙d) = *i*∙c + *i*^{2}∙d = − d + *i*∙c.

# The geometry of the wavefunction

My posts and article on the wavefunction as a gravitational wave are rather short on the exact geometry of the wavefunction, so let us explore that a bit here. By now, you know the formula for the elementary wavefunction by heart:

ψ = *a·e*^{−i[E·t − p∙x]/ħ} = *a·cos*(**p**∙**x**/ħ − E∙t/ħ) + *i·a·sin*(**p**∙**x**/ħ − E∙t/ħ)

If we assume the momentum **p** is all in the **x**-direction, then the **p** and **x** vectors will have the same direction, and **p**∙**x**/ħ reduces to p∙x/ħ. This amounts to saying our particle is traveling along the x-axis. The geometry of the wavefunction is illustrated below. The x-axis is the direction of propagation, and the y- and z-axes represent the real and imaginary part of the wavefunction respectively.

Note that, when applying the right-hand rule for the axes, the vertical axis is the y-axis, not the z-axis. Hence, we may associate the vertical axis with the cosine component, and the horizontal axis with the sine component. If the origin is the (x, t) = (0, 0) point, then cos(θ) = cos(0) = 1 and sin(θ) = sin(0) = 0. This is reflected in both illustrations, which show a left- and a right-handed wave respectively. I am convinced these correspond to the two possible values for the quantum-mechanical spin of the wave: +ħ/2 or −ħ/2. But… Well… Who am I? The *cosine *and *sine *components are shown below. Needless to say, the *cosine *and *sine *function are the same, except for a phase difference of π/2: *sin*(θ) = *cos*(θ − π/2)

Surely, Nature doesn’t care a hoot about our conventions for measuring the phase angle clockwise or counterclockwise and therefore, the ψ = *a·e*^{i}^{[E·t − p∙x]/ħ} function should, effectively, also be permitted. We know that *cos*(θ) = *cos**(**–*θ) and *sin*θ = *–**sin**(**–*θ), so we can write: * *

ψ = *a·e*^{i}^{[E·t − p∙x]/ħ} = *a·cos*(E∙t/ħ − p∙x/ħ) +* i·a·sin*(E∙t/ħ − p∙x/ħ)

= *a·cos*(p∙x/ħ − E∙t/ħ) − *i·a·sin*(p∙x/ħ − E∙t/ħ)

E/ħ = ω gives the frequency in time (expressed in radians per second), while p/ħ = k gives us the wavenumber, or the frequency in space (expressed in radians per meter). Of course, we may write: *f* = ω/2π and λ = 2π/k, which gives us the two de Broglie relations:

- E = ħ∙ω = h∙
*f* - p = ħ∙k = h/λ

The frequency in time is easy to interpret (a particle will always have some mass and, therefore, some energy), but the wavelength is *inversely *proportional to the momentum: λ = h/p. Hence, if p goes to zero, then the wavelength becomes infinitely long: if p → 0, then λ* → ∞. *For the limit situation, a particle with zero *rest *mass (m_{0} = 0), the velocity may be *c* and, therefore, we find that p = m* _{v}*∙

*v*= m∙

*c*and, therefore, p∙

*c*= m∙

*c*

^{2}= E, which we may also write as: E/p =

*c*. Hence, for a particle with zero rest mass, the wavelength can be written as:

λ = h/p = hc/E = h/mc

However, we argued that the physical dimension of the components of the wavefunction may be usefully expressed in N/kg units (force per unit *mass*), while the physical dimension of the electromagnetic wave are expressed in N/C (force per unit *charge*). This, in fact, explains the dichotomy between bosons (photons) and fermions (spin-1/2 particles). Hence, all matter-particles should have *some *mass.[1] But how we interpret the inverse proportionality between λ and p?

We should probably first ask ourselves what wavelength we are talking about. The wave only has a phase velocity here, which is equal to *v*_{p} = ω/k = (E/ħ)/(p/ħ) = E/p. Of course, we know that, *classically*, the momentum will be equal to the *group *velocity times the mass: p = m·*v*_{g}. However, when p is zero, we have a division by zero once more: if p → 0, then *v*_{p} = E/p → ∞. Infinite wavelengths and infinite phase velocities probably tell us that our particle *has to* move: our notion of a particle at rest is mathematically inconsistent. If we associate this elementary wavefunction with some particle, and if we then imagine it to move, somehow, then we get an interesting relation between the group and the phase velocity:

*v*_{p} = ω/k = E/p = E/(m·*v*_{g}) = (m·*c*^{2})/(m·*v*_{g}) = *c*^{2}/*v*_{g}

We can re-write this as *v*_{p}·*v*_{g} = *c*^{2}, which reminds us of the relationship between the electric and magnetic constant (1/ε_{0})·(1/μ_{0}) = *c*^{2}. But what is the group velocity of the *elementary *wavefunction? Is it a meaningful concept?

The phase velocity is just the ratio of ω/k. In contrast, the group velocity is the derivative of ω with respect to k. So we need to write ω as a function of k. Can we do that even if we have only one wave? We do *not *have a wave packet here, right? Just some hypothetical *building block *of a real-life wavefunction, right? Right. So we should introduce uncertainty about E and p and build up the wave packet, right? Well… Yes. But let’s wait with that, and see how far we can go in our interpretation of this *elementary *wavefunction. Let’s first get that ω = ω(k) relation. You’ll remember we can write Schrödinger’s equation – the equation that describes the *propagation *mechanism for matter-waves – as the following *pair *of equations:

*Re*(∂ψ/∂t) = −[ħ/(2m)]·*Im*(∇^{2}ψ) ⇔ ω·cos(kx − ωt) = k^{2}·[ħ/(2m)]·cos(kx − ωt)*Im*(∂ψ/∂t) = [ħ/(2m)]·*Re*(∇^{2}ψ) ⇔ ω·sin(kx − ωt) = k^{2}·[ħ/(2m)]·sin(kx − ωt)

This tells us that ω = ħ·k^{2}/(2m). Therefore, we can calculate ∂ω/∂k as:

∂ω/∂k = ħ·k/m = p/m = *v*_{g}

We learn nothing new. We are going round and round in circles here, and we always end up with a tautology: as soon as we have a non-zero momentum, we have a *mathematical* formula for the group velocity – but we don’t know what it represents – and a finite wavelength. In fact, using the p = ħ∙k = h/λ relation, we can write one as a function of the other:

λ = h/p = h/m*v*_{g} ⇔ *v*_{g} = h/mλ

What does this *mean*? It resembles the *c* = h/mλ relation we had for a particle with zero rest mass. Of course, it does: the λ = h/m*c* relation is, once again, a limit for *v*_{g} going to c. By the way, it is interesting to note that the *v*_{p}·*v*_{g} = *c*^{2} relation implies that the *phase *velocity is always superluminal. That’ easy to see when you re-write the equation in terms of *relative *velocities: (*v*_{p}/*c*)·(*v*_{g}/*c*) = β* _{phase}*·β

*= 1. Hence, if β*

_{group}*< 1, then β*

_{group}*> 1.*

_{phase}So what *is *the geometry, *really*? Let’s look at the ψ = *a·cos*(p∙x/ħ – E∙t/ħ) *–** i·a·sin*(p∙x/ħ – E∙t/ħ) formula once more. If we write p∙x/ħ as Δ, then we will be interested to know for what x this phase factor will be equal to 2π. So we write:

Δ =p∙x/ħ = 2π ⇔ x = 2π∙ħ/p = h/p = λ* *

So now we get a meaningful interpretation for that wavelength: it’s that distance between the crests of the wave, so to speak, as illustrated below.

Can we now find a *meaningful *(i.e. geometric) interpretation for the group and phase velocity? If you look at the illustration above, you see we can sort of distinguish (1) a linear velocity (the speed with which those wave crests move) and (2) some kind of circular or tangential velocity (the velocity along the red contour line above). We’ll probably need the formula for the tangential velocity: *v* = *a*∙ω. If p = 0 (so we have that weird infinitesimally long wavelength), then we have two velocities:

- The tangential velocity around the
*a·e*^{i}^{·E·t}circle, so to speak, and that will just be equal to*v*=*a*∙ω =*a*∙E/ħ. - The red contour line sort of gets stretched out, like infinitely long, and the velocity becomes… What does it do? Does it go to ∞ , or to
*c*?

Let’s think about this. For a particle at rest, we had this weird calculation. We had an angular momentum formula (for an electron) which we equated with the *real-life* +ħ/2 or −ħ/2 values of its spin. And so we got a *numerical *value for *a*. It was the Compton radius: the scattering radius for an electron. Let me copy it once again:

Just to bring this story a bit back to Earth, you should note the calculated value: *a *= 3.8616×10^{−13} m. We did then another weird calculation. We said all of the energy of the electron had to be packed in this *cylinder *that might of might not be there. The point was: the energy is finite, so that *elementary *wavefunction can*not *have an infinite length in space. Indeed, assuming that the energy was distributed uniformly, we jotted down this formula, which reflects the formula for the *volume *of a cylinder:

E = π·*a*^{2}·*l* ⇔ *l *= E/(π·*a*^{2})

Using the value we got for the Compton scattering radius (*a *= 3.8616×10^{−13} m), we got an astronomical value for *l*. Let me write it out:

*l *= (8.19×10^{−14})/(π·14.9×10^{−26}) ≈ 0.175×10^{12} m

It is, *literally*, an astronomical value: 0.175×10^{12} m is 175 *million kilo*meter, so that’s like the distance between the Sun and the Earth. We wrote, jokingly, that such space is too large to look for an electron and, hence, that we should really build a proper *packet *by making use of the Uncertainty Principle: allowing for uncertainty in the energy should, effectively, reduce the uncertainty in position.

But… Well… What if we use that value as the value for λ? We’d get that linear velocity, right? Let’s try it. The *period *is equal to T = T = 2π·(ħ/E) = h/E and λ = E/(π·*a*^{2}), so we write:We can write this as a function of m and the *c *and ħ constants only:

A weird formula but not necessarily nonsensical: we get a finite speed. Now, if the wavelength becomes somewhat less astronomical, we’ll get different values of course. I have a strange feeling that, with these formula, we should, somehow, be able to explain relativistic length contraction. But I will let you think about that as for now. Here I just wanted to *show *the geometry of the wavefunction a bit more in detail.

[1] The discussions on the mass of neutrinos are interesting in this regard. Scientists all felt the neutrino *had to *have some (rest) mass, so my instinct on this is theirs. In fact, only recently experimental confirmation came in, and the mass of the known neutrino flavors was estimated to be something like 0.12 eV/*c*^{2}. This mass combines the three known neutrino flavors. To understand this number, you should note it is the same order of magnitude of the equivalent mass of low-energy photons, like infrared or microwave radiation.

# Wavefunctions as gravitational waves

This is the paper I always wanted to write. It is there now, and I think it is good – and *that*‘s an understatement. 🙂 It is probably best to download it as a pdf-file from the viXra.org site because this was a rather fast ‘copy and paste’ job from the Word version of the paper, so there may be issues with boldface notation (vector notation), italics and, most importantly, with formulas – which I, sadly, have to ‘snip’ into this WordPress blog, as they don’t have an easy copy function for mathematical formulas.

It’s great stuff. If you have been following my blog – and many of you have – you *will* want to digest *this*. 🙂

**Abstract : **This paper explores the implications of associating the components of the wavefunction with a physical dimension: force per unit *mass* – which is, of course, the dimension of acceleration (m/s^{2}) and gravitational fields. The classical electromagnetic field equations for energy densities, the Poynting vector and spin angular momentum are then re-derived by substituting the electromagnetic N/C unit of field strength (mass per unit *charge*) by the new N/kg = m/s^{2} dimension.

The results are elegant and insightful. For example, the energy densities are proportional to the square of the absolute value of the wavefunction and, hence, to the probabilities, which establishes a *physical *normalization condition. Also, Schrödinger’s wave equation may then, effectively, be interpreted as a diffusion equation for energy, and the wavefunction itself can be interpreted as a propagating gravitational wave. Finally, as an added bonus, concepts such as the Compton scattering radius for a particle, spin angular momentum, and the boson-fermion dichotomy, can also be explained more intuitively.

While the approach offers a physical interpretation of the wavefunction, the author argues that the *core *of the Copenhagen interpretations revolves around the complementarity principle, which remains unchallenged because the interpretation of amplitude waves as traveling fields does *not *explain the particle nature of matter.

# Introduction

This is *not *another introduction to quantum mechanics. We assume the reader is already familiar with the key principles and, importantly, with the basic math. We offer an *interpretation *of wave mechanics. As such, we do *not *challenge the complementarity principle: the *physical *interpretation of the wavefunction that is offered here explains the *wave* nature of matter only. It explains diffraction and interference of amplitudes but it does *not *explain why a particle will hit the detector *not as a wave but as a particle*. Hence, the Copenhagen interpretation of the wavefunction remains relevant: we just push its boundaries.

The basic ideas in this paper stem from a simple observation: the *geometric *similarity between the quantum-mechanical wavefunctions and electromagnetic waves is remarkably similar. The components of both waves are orthogonal to the direction of propagation and to each other. Only the relative phase differs : the electric and magnetic field vectors (**E** and **B**) have the same phase. In contrast, the phase of the real and imaginary part of the (elementary) wavefunction (ψ = *a·e*^{−i∙θ} = *a*∙cosθ – *a*∙sinθ) differ by 90 degrees (π/2).[1] Pursuing the analogy, we explore the following question: if the oscillating electric and magnetic field vectors of an electromagnetic wave carry the energy that one associates with the wave, can we analyze the real and imaginary part of the wavefunction in a similar way?

We show the answer is positive and remarkably straightforward. If the physical dimension of the electromagnetic field is expressed in newton per coulomb (force per unit charge), then the physical dimension of the components of the wavefunction may be associated with force per unit mass (newton per kg).[2] Of course, force over some distance is energy. The question then becomes: what is the energy concept here? Kinetic? Potential? Both?

The similarity between the energy of a (one-dimensional) linear oscillator (E = m·*a*^{2}·ω^{2}/2) and Einstein’s relativistic energy equation E = m∙*c*^{2} inspires us to interpret the energy as a *two*-dimensional oscillation of mass. To assist the reader, we construct a two-piston engine metaphor.[3] We then adapt the formula for the electromagnetic energy density to calculate the energy densities for the wave function. The results are elegant and intuitive: the energy densities are proportional to the square of the absolute value of the wavefunction and, hence, to the probabilities. Schrödinger’s wave equation may then, effectively, be interpreted as a diffusion equation for energy itself.

As an added bonus, concepts such as the Compton scattering radius for a particle and spin angular, as well as the boson-fermion dichotomy can be explained in a fully intuitive way.[4]

Of course, such interpretation is also an interpretation of the wavefunction itself, and the immediate reaction of the reader is predictable: the electric and magnetic field vectors are, somehow, to be looked at as *real* vectors. In contrast, the real and imaginary components of the wavefunction are not. However, this objection needs to be phrased more carefully. First, it may be noted that, in a classical analysis, the magnetic force is a pseudovector itself.[5] Second, a suitable choice of coordinates may make quantum-mechanical rotation matrices irrelevant.[6]

Therefore, the author is of the opinion that this little paper may provide some fresh perspective on the question, thereby further exploring Einstein’s basic sentiment in regard to quantum mechanics, which may be summarized as follows: there must be some *physical* explanation for the calculated probabilities.[7]

We will, therefore, start with Einstein’s relativistic energy equation (E = m*c*^{2}) and wonder what it could possibly tell us.** **

# I. Energy as a two-dimensional oscillation of mass

The structural similarity between the relativistic energy formula, the formula for the *total *energy of an oscillator, and the *kinetic* energy of a moving body, is striking:

- E = m
*c*^{2} - E = mω
^{2}/2 - E = m
*v*^{2}/2

In these formulas, ω, *v *and *c *all describe some velocity.[8] Of course, there is the 1/2 factor in the E = mω^{2}/2 formula[9], but that is exactly the point we are going to explore here: can we think of an oscillation in *two *dimensions, so it stores an amount of energy that is equal to E = 2·m·ω^{2}/2 = m·ω^{2}?

That is easy enough. Think, for example, of a V-2 engine with the pistons at a 90-degree angle, as illustrated below. The 90° angle makes it possible to perfectly balance the counterweight and the pistons, thereby ensuring smooth travel at all times. With permanently closed valves, the air inside the cylinder compresses and decompresses as the pistons move up and down and 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. Hence, we can describe it 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.

**Figure 1**: Oscillations in two dimensions

If we assume there is no friction, we have a *perpetuum mobile *here. The compressed air and the rotating counterweight (which, combined with the crankshaft, acts as a flywheel[10]) store the potential energy. The moving masses of the pistons store the kinetic energy of the system.[11]

At this point, it is probably good to quickly review the relevant math. If the magnitude of the oscillation is equal to *a*, then the motion of the piston (or the mass on a spring) will be described by *x* = *a*·cos(ω·t + Δ).[12] Needless to say, Δ is just a phase factor which defines our *t* = 0 point, and ω is the *natural angular *frequency of our oscillator. Because of the 90° angle between the two cylinders, Δ would be 0 for one oscillator, and –π/2 for the other. Hence, the motion of one piston is given by *x* = *a*·cos(ω·t), while the motion of the other is given by *x* = *a*·cos(ω·t–π/2) = *a*·sin(ω·t).

The kinetic and potential energy of *one *oscillator (think of one piston or one spring only) can then be calculated as:

- K.E. = T = m·
*v*^{2}/2 = (1/2)·m·ω^{2}·*a*^{2}·sin^{2}(ω·t + Δ) - P.E. = U = k·x
^{2}/2 = (1/2)·k·*a*^{2}·cos^{2}(ω·t + Δ)

The coefficient k in the potential energy formula characterizes the restoring force: F = −k·x. From the dynamics involved, it is obvious that k must be equal to m·ω^{2}. Hence, the total energy is equal to:

E = T + U = (1/2)· m·ω^{2}·*a*^{2}·[sin^{2}(ω·t + Δ) + cos^{2}(ω·t + Δ)] = m·*a*^{2}·ω^{2}/2

To facilitate the calculations, we will briefly assume k = m·ω^{2} and *a* are equal to 1. The *motion *of our first oscillator is given by the cos(ω·t) = cosθ function (θ = ω·t), and its kinetic energy will be equal to sin^{2}θ. Hence, the (instantaneous) *change *in kinetic energy at any point in time will be equal to:

d(sin^{2}θ)/dθ = 2∙sinθ∙d(sinθ)/dθ = 2∙sinθ∙cosθ

Let us look at the second oscillator now. Just think of the second piston going up and down in the V-2 engine. Its motion is given by the sinθ function, which is equal to cos(θ−π /2). Hence, its kinetic energy is equal to sin^{2}(θ−π /2), and how it *changes *– as a function of θ – will be equal to:

2∙sin(θ−π /2)∙cos(θ−π /2) = = −2∙cosθ∙sinθ = −2∙sinθ∙cosθ

We have our *perpetuum mobile*! While transferring kinetic energy from one piston to the other, the crankshaft will rotate with a constant angular velocity: linear motion becomes circular motion, and vice versa, and the total energy that is stored in the system is T + U = m*a*^{2}ω^{2}.

We have a great *metaphor* here. Somehow, in this beautiful interplay between linear and circular motion, energy is borrowed from one place and then returns to the other, cycle after cycle. We know the wavefunction consist of a sine and a cosine: the cosine is the real component, and the sine is the imaginary component. Could they be equally real? Could each represent *half *of the total energy of our particle? Should we think of the *c *in our E = m*c*^{2} formula as an *angular *velocity?

These are sensible questions. Let us explore them.** **

# II. The wavefunction as a two-dimensional oscillation

The elementary wavefunction is written as:

ψ = *a·e*^{−i[E·t − p∙x]/ħ} = *a·e*^{−i[E·t − p∙x]/ħ} = *a·cos( p∙x/ħ *

*–*

*E∙t/ħ) + i·a·sin(*

**p**∙**x**/ħ*–*

*E∙t/ħ)*

*When *considering a particle at rest (**p** = **0**) this reduces to:

ψ = *a·e*^{−i∙E·t/ħ} = *a·cos(**–**E∙t/ħ) + i·a·sin(**–**E∙t/ħ) = a·cos(E∙t/ħ) **–** i·a·sin(E∙t/ħ) *

Let us remind ourselves of the geometry involved, which is illustrated below. Note that the argument of the wavefunction rotates *clockwise *with time, while the mathematical convention for measuring the phase angle (ϕ) is *counter*-clockwise.

**Figure 2**: Euler’s formula

If we assume the momentum **p** is all in the *x*-direction, then the **p** and **x** vectors will have the same direction, and **p***∙ x/ħ reduces to p∙x/ħ. Most illustrations – such as the one below – will either freeze x or, else, t. Alternatively, one can google web animations varying both. *The point is: we also have a two-dimensional oscillation here. These two dimensions are perpendicular to the direction of propagation of the wavefunction. For example, if the wavefunction propagates in the

*x*-direction, then the oscillations are along the

*y*– and

*z*-axis, which we may refer to as the real and imaginary axis. Note how the phase difference between the cosine and the sine – the real and imaginary part of our wavefunction – appear to give some spin to the whole. I will come back to this.

**Figure 3**: Geometric representation of the wavefunction

Hence, *if *we would say these oscillations carry half of the total energy of the particle, then we may refer to the real and imaginary energy of the particle respectively, and the interplay between the real and the imaginary *part of the wavefunction may then describe how energy propagates through space over time. *

Let us consider, once again, a particle at rest. Hence, **p** = **0** and the (elementary) wavefunction reduces to ψ = *a·e*^{−i∙E·t/ħ}. Hence, the angular velocity of both oscillations, at some point **x**, is given by ω = -E/ħ. Now, the energy of our particle includes all of the energy – kinetic, potential and rest energy – and is, therefore, equal to E = m*c*^{2}.

Can we, somehow, relate this to the m·*a*^{2}·ω^{2} energy formula for our V-2 *perpetuum mobile*? Our wavefunction has an amplitude too. Now, if the oscillations of the real and imaginary wavefunction store the energy of our particle, then their amplitude will surely matter. In fact, the energy of an oscillation is, in general, proportional to the *square *of the amplitude: E µ *a*^{2}. We may, therefore, think that the *a*^{2} factor in the E = m·*a*^{2}·ω^{2} energy will surely be relevant as well.

However, here is a complication: an *actual* particle is localized in space and can, therefore, *not *be represented by the elementary wavefunction. We must build a wave *packet* for that: a sum of wavefunctions, each with their own amplitude *a*_{k}, and their own ω* _{i}* = -E

*/ħ. Each of these wavefunctions will*

_{i}*contribute*some energy to the total energy of the wave packet. To calculate the contribution of each wave to the total, both

*a*as well as E

_{i}*will matter.*

_{i}What is E* _{i}*? E

*varies around some average E, which we can associate with some*

_{i}*average mass*m: m = E/

*c*

^{2}. The Uncertainty Principle kicks in here. The analysis becomes more complicated, but a formula such as the one below might make sense:We can re-write this as:What is the meaning of this equation? We may look at it as some sort of

*physical*normalization condition when building up the

*Fourier sum*. Of course, we should relate this to the

*mathematical*normalization condition for the wavefunction. Our intuition tells us that the probabilities must be related to the energy

*densities*, but how exactly? We will come back to this question in a moment. Let us first think some more about the enigma:

**what**

*is*mass?Before we do so, let us quickly calculate the value of *c*^{2}ħ^{2}: it is about 1´10^{–}^{51} N^{2}∙m^{4}. Let us also do a dimensional analysis: the physical dimensions of the E = m·*a*^{2}·ω^{2} equation make sense if we express m in kg, *a *in m, and ω in *rad*/s. We then get: [E] = kg∙m^{2}/s^{2} = (N∙s^{2}/m)∙m^{2}/s^{2} = N∙m = J. The dimensions of the left- and right-hand side of the physical normalization condition is N^{3}∙m^{5}.** **

# III. What is mass?

We came up, playfully, with a meaningful interpretation for energy: it is a two-dimensional oscillation of mass. But what is mass? A new *aether *theory is, of course, not an option, but then what *is* it that is oscillating? To understand the physics behind equations, it is always good to do an analysis of the physical dimensions in the equation. Let us start with Einstein’s energy equation once again. If we want to look at mass, we should re-write it as *m* = E/c^{2}:

[*m*] = [E/*c*^{2}] = J/(m/s)^{2} = N·m∙s^{2}/m^{2} = N·s^{2}/m = kg

This is not very helpful. It only reminds us of Newton’s definition of a mass: mass is that what gets accelerated by a force. At this point, we may want to think of the physical significance of the *absolute *nature of the speed of light. Einstein’s E = m*c*^{2} equation implies we can write the ratio between the energy and the mass of *any *particle is always the same, so we can write, for example:This reminds us of the ω^{2}= *C*^{–}^{1}/*L* or ω^{2} = *k*/*m* of harmonic oscillators once again.[13] The key difference is that the ω^{2}= *C*^{–}^{1}/*L* and ω^{2} = *k*/*m* formulas introduce *two *or more degrees of freedom.[14] In contrast, *c*^{2}= E/m for *any *particle, *always*. However, that is exactly the point: we can modulate the resistance, inductance and capacitance of electric circuits, and the stiffness of springs and the masses we put on them, but we live in *one *physical space only: *our *spacetime. Hence, the speed of light *c* emerges here as *the* defining property of spacetime – the resonant frequency, so to speak. We have no further degrees of freedom here.

The Planck-Einstein relation (for photons) and the *de Broglie *equation (for matter-particles) have an interesting feature: both imply that the *energy *of the oscillation is proportional to the frequency, with Planck’s constant as the constant of proportionality. Now, for *one-dimensional *oscillations – think of a guitar string, for example – we know the energy will be proportional to the *square *of the frequency. It is a remarkable observation: the two-dimensional matter-wave, or the electromagnetic wave, gives us *two* waves for the price of one, so to speak, each carrying *half* of the *total *energy of the oscillation but, as a result, we get a proportionality between E and *f* instead of between E and *f*^{2}.

However, such reflections do not answer the fundamental question we started out with: what *is *mass? At this point, it is hard to go beyond the circular definition that is implied by Einstein’s formula: energy is a two-dimensional oscillation of mass, and mass packs energy, and *c *emerges us as the property of spacetime that defines *how *exactly.

When everything is said and done, this does not go beyond stating that mass is some scalar field. Now, a scalar field is, quite simply, some real *number* that we associate with a position in spacetime. The Higgs field is a scalar field but, of course, the theory behind it goes much beyond stating that we should think of mass as some scalar field. The fundamental question is: why and how does energy, or matter, *condense *into elementary particles? That is what the Higgs *mechanism* is about but, as this paper is exploratory only, we cannot even start explaining the basics of it.

What we *can *do, however, is look at the wave *equation *again (Schrödinger’s equation), as we can now analyze it as an energy diffusion equation.** **

# IV. Schrödinger’s equation as an energy diffusion equation

The interpretation of Schrödinger’s equation as a diffusion equation is straightforward. Feynman (*Lectures*, III-16-1) briefly summarizes it as follows:

“We can think of Schrödinger’s equation as describing the diffusion of the probability amplitude from one point to the next. […] But the imaginary coefficient in front of the derivative makes the behavior completely different from the ordinary diffusion such as you would have for a gas spreading out along a thin tube. Ordinary diffusion gives rise to real exponential solutions, whereas the solutions of Schrödinger’s equation are complex waves.”[17]

Let us review the basic math. For a particle moving in free space – with no external force fields acting on it – there is no potential (U = 0) and, therefore, the Uψ term disappears. Therefore, Schrödinger’s equation reduces to:

∂ψ(**x**, t)/∂t = *i*·(1/2)·(ħ/m_{eff})·∇^{2}ψ(**x**, t)

The ubiquitous diffusion equation in physics is:

∂φ(**x**, t)/∂t = D·∇^{2}φ(**x**, t)

The *structural* similarity is obvious. The key difference between both equations is that the wave equation gives us *two *equations for the price of one. Indeed, because ψ is a complex-valued function, with a *real *and an *imaginary *part, we get the following equations[18]:

*Re*(∂ψ/∂t) = −(1/2)·(ħ/m_{eff})·*Im*(∇^{2}ψ)*Im*(∂ψ/∂t) = (1/2)·(ħ/m_{eff})·*Re*(∇^{2}ψ)

These equations make us think of the equations for an electromagnetic wave in free space (no stationary charges or currents):

- ∂
**B**/∂t = –∇×**E** - ∂
**E**/∂t =*c*^{2}∇×**B**

The above equations effectively describe a *propagation *mechanism in spacetime, as illustrated below.

**Figure 4**: Propagation mechanisms

The Laplacian operator (∇^{2}), when operating on a *scalar *quantity, gives us a flux density, i.e. something expressed per square meter (1/m^{2}). In this case, it is operating on ψ(**x**, t), so what is the dimension of our wavefunction ψ(**x**, t)? To answer that question, we should analyze the diffusion constant in Schrödinger’s equation, i.e. the (1/2)·(ħ/m_{eff}) factor:

- As a
*mathematical*constant of proportionality, it will*quantify*the relationship between both derivatives (i.e. the time derivative and the Laplacian); - As a
*physical*constant, it will ensure the*physical dimensions*on both sides of the equation are compatible.

Now, the ħ/m_{eff} factor is expressed in (N·m·s)/(N· s^{2}/m) = m^{2}/s. Hence, it does ensure the dimensions on both sides of the equation are, effectively, the same: ∂ψ/∂t is a time derivative and, therefore, its dimension is s^{–}^{1} while, as mentioned above, the dimension of ∇^{2}ψ is m^{–}^{2}. However, this does not solve our basic question: what is the dimension of the real and imaginary part of our wavefunction?

At this point, mainstream physicists will say: it does not have a physical dimension, and there is no geometric interpretation of Schrödinger’s equation. One may argue, effectively, that its argument, (**p**∙**x** – E∙t)/ħ, is just a number and, therefore, that the real and imaginary part of ψ is also just some number.

To this, we may object that ħ may be looked as a mathematical scaling constant only.** If **we do that,

**the argument of ψ will, effectively, be expressed in**

*then**action*units, i.e. in N·m·s. It then does make sense to also associate a physical dimension with the real and imaginary part of ψ. What could it be?

We may have a closer look at Maxwell’s equations for inspiration here. The electric field vector is expressed in *newton* (the unit of force) per unit of *charge* (*coulomb*). Now, there is something interesting here. The physical dimension of the magnetic field is N/C *divided* by m/s.[19] We may write **B** as the following vector cross-product: **B** = (1/*c*)∙**e****_{x}**×

**E**, with

**e****the unit vector pointing in the**

_{x}*x*-direction (i.e. the direction of propagation of the wave). Hence, we may associate the (1/

*c*)∙

**e****×**

_{x}*operator*, which amounts to a rotation by 90 degrees, with the s/m dimension. Now, multiplication by

*i*also amounts to a rotation by 90° degrees. Hence, we may boldly write:

**B**= (1/

*c*)∙

**e****×**

_{x}**E**= (1/

*c*)∙

*i*∙

**E**. This allows us to also geometrically interpret Schrödinger’s equation in the way we interpreted it above (see Figure 3).[20]

Still, we have not answered the question as to what the physical dimension of the real and imaginary part of our wavefunction should be. At this point, we may be inspired by the structural similarity between Newton’s and Coulomb’s force laws:Hence, if the electric field vector **E **is expressed in force per unit *charge *(N/C), then we may want to think of associating the real part of our wavefunction with a force per unit *mass* (N/kg). We can, of course, do a substitution here, because the mass unit (1 kg) is equivalent to 1 N·s^{2}/m. Hence, our N/kg dimension becomes:

N/kg = N/(N·s^{2}/m)= m/s^{2}

What is this: m/s^{2}? Is *that *the dimension of the *a*·*cos*θ term in the *a*·*e*^{−iθ }= *a*·*cos*θ − *i*·*a*·*sin*θ wavefunction?

My answer is: **why not?** Think of it: m/s^{2} is the physical dimension of *acceleration*: the increase or decrease in velocity (m/s) per second. It ensures the wavefunction for *any *particle – matter-particles or particles with zero rest mass (photons) – and the associated wave *equation *(which has to be the same for all, as the spacetime we live in is *one*) are mutually consistent.

In this regard, we should think of how we would model a *gravitational *wave. The physical dimension would surely be the same: force per mass unit. It all makes sense: wavefunctions may, perhaps, be interpreted as traveling distortions of spacetime, i.e. as tiny gravitational waves.

# V. Energy densities and flows

Pursuing the geometric equivalence between the equations for an electromagnetic wave and Schrödinger’s equation, we can now, perhaps, see if there is an equivalent for the energy density. For an electromagnetic wave, we know that the energy density is given by the following formula:**E** and **B** are the electric and magnetic field vector respectively. The Poynting vector will give us the directional energy flux, i.e. the energy flow per unit area per unit time. We write:Needless to say, the **∇**∙ operator is the divergence and, therefore, gives us the magnitude of a (vector) field’s *source* or *sink* at a given point. To be precise, the divergence gives us the volume density of the outward *flux *of a vector field from an infinitesimal volume around a given point. In this case, it gives us the *volume density* of the flux of ** S**.

We can analyze the dimensions of the equation for the energy density as follows:

**E**is measured in*newton per coulomb*, so [**E**∙**E**] = [E^{2}] = N^{2}/C^{2}.**B**is measured in (N/C)/(m/s), so we get [**B**∙**B**] = [B^{2}] = (N^{2}/C^{2})·(s^{2}/m^{2}). However, the dimension of our*c*^{2}factor is (m^{2}/s^{2}) and so we’re also left with N^{2}/C^{2}.- The
*ϵ*_{0}is the electric constant, aka as the vacuum permittivity. As a*physical*constant, it should ensure the dimensions on both sides of the equation work out, and they do: [ε_{0}] = C^{2}/(N·m^{2}) and, therefore, if we multiply that with N^{2}/C^{2}, we find that*u*is expressed in J/m^{3}.[21]

Replacing the *newton per coulomb* unit (N/C) by the *newton per kg* unit (N/kg) in the formulas above should give us the equivalent of the energy density for the wavefunction. We just need to substitute *ϵ*_{0} for an equivalent constant. We may to give it a try. If the energy densities can be calculated – which are also mass densities, obviously – then the probabilities should be proportional to them.

Let us first see what we get for a photon, assuming the electromagnetic wave represents its wavefunction. Substituting **B** for (1/*c*)∙*i*∙**E** or for −(1/*c*)∙*i*∙**E** gives us the following result:**Zero!?** An unexpected result! Or not? We have no stationary charges and no currents: only an electromagnetic wave in free space. Hence, the local energy conservation principle needs to be respected at all points in space and in time. The geometry makes sense of the result: for an electromagnetic wave, the magnitudes of **E** and **B** reach their maximum, minimum and zero point *simultaneously*, as shown below.[22] This is because their *phase *is the same.

**Figure 5**: Electromagnetic wave: **E** and **B**

Should we expect a similar result for the energy densities that we would associate with the real and imaginary part of the matter-wave? For the matter-wave, we have a phase difference between *a*·*cos*θ and *a*·*sin*θ, which gives a different picture of the *propagation *of the wave (see Figure 3).[23] In fact, the geometry of the suggestion suggests some inherent spin, which is interesting. I will come back to this. Let us first guess those densities. Making abstraction of any scaling constants, we may write:We get what we hoped to get: the absolute square of our amplitude is, effectively, an energy density !

|ψ|^{2 } = |*a·e*^{−i∙E·t/ħ}|^{2 }= *a*^{2 }= *u*

This is very deep. A photon has no rest mass, so it borrows and returns energy from empty space as it travels through it. In contrast, a matter-wave carries energy and, therefore, has some (*rest*) mass. It is therefore associated with an energy density, and this energy density gives us the probabilities. Of course, we need to fine-tune the analysis to account for the fact that we have a wave packet rather than a single wave, but that should be feasible.

As mentioned, the phase difference between the real and imaginary part of our wavefunction (a cosine and a sine function) appear to give some spin to our particle. We do not have this particularity for a photon. Of course, photons are bosons, i.e. spin-zero particles, while elementary matter-particles are fermions with spin-1/2. Hence, our geometric interpretation of the wavefunction suggests that, after all, there may be some more intuitive explanation of the fundamental dichotomy between bosons and fermions, which puzzled even Feynman:

“Why is it that particles with half-integral spin are Fermi particles, whereas particles with integral spin are Bose particles? We apologize for the fact that we cannot give you an elementary explanation. An explanation has been worked out by Pauli from complicated arguments of quantum field theory and relativity. He has shown that the two must necessarily go together, but we have not been able to find a way of reproducing his arguments on an elementary level. It appears to be one of the few places in physics where there is a rule which can be stated very simply, but for which no one has found a simple and easy explanation. The explanation is deep down in relativistic quantum mechanics. This probably means that we do not have a complete understanding of the fundamental principle involved.” (Feynman, *Lectures*, III-4-1)

The *physical* interpretation of the wavefunction, as presented here, may provide some better understanding of ‘the fundamental principle involved’:* the physical dimension of the oscillation is just very different*. That is all: it is force per unit charge for photons, and force per unit mass for matter-particles. We will examine the question of spin somewhat more carefully in section VII. Let us first examine the matter-wave some more.** **

# VI. Group and phase velocity of the matter-wave

The geometric representation of the matter-wave (see Figure 3) suggests a traveling wave and, yes, of course: the matter-wave effectively *travels* through space and time. But *what is traveling, exactly*? It is the pulse – or the *signal *– only: the *phase *velocity of the wave is just a mathematical concept and, even in our physical interpretation of the wavefunction, the same is true for the *group *velocity of our wave packet. The oscillation is two-dimensional, but perpendicular to the direction of travel of the wave. Hence, nothing actually moves *with *our particle.

Here, we should also reiterate that we did not answer the question as to *what *is oscillating up and down and/or sideways: we only associated a *physical *dimension with the components of the wavefunction – *newton* per *kg* (force per unit mass), to be precise. We were inspired to do so because of the physical dimension of the electric and magnetic field vectors (*newton* per *coulomb*, i.e. force per unit charge) we associate with electromagnetic waves which, for all practical purposes, we currently treat as the wavefunction for a photon. This made it possible to calculate the associated *energy densities *and a *Poynting vector *for energy dissipation. In addition, we showed that Schrödinger’s equation itself then becomes a diffusion equation for energy. However, let us now focus some more on the asymmetry which is introduced by the phase difference between the real and the imaginary part of the wavefunction. Look at the mathematical shape of the elementary wavefunction once again:

ψ = *a·e*^{−i[E·t − p∙x]/ħ} = *a·e*^{−i[E·t − p∙x]/ħ} = *a·cos*(**p**∙**x**/ħ − E∙t/ħ)* + i·a·sin*(**p**∙**x**/ħ − E∙t/ħ)

The minus sign in the argument of our sine and cosine function defines the direction of travel: an F(x−*v∙*t) wavefunction will always describe some wave that is traveling in the *positive *x-direction (with *c *the wave velocity), while an F(x+*v∙*t) wavefunction will travel in the *negative *x-direction. For a geometric interpretation of the wavefunction *in three dimensions*, we need to agree on how to define *i* or, what amounts to the same, a convention on how to define clockwise and counterclockwise directions: if we look at a clock from the back, then its hand will be moving *counter*clockwise. So we need to establish the equivalent of the right-hand rule. However, let us not worry about that now. Let us focus on the interpretation. To ease the analysis, we’ll assume we’re looking at a particle at rest. Hence, **p** = **0**, and the wavefunction reduces to:

ψ = *a·e*^{−i∙E·t/ħ} = *a·cos*(−E∙t/ħ)* + i·a·sin*(−E_{0}∙t/ħ)* = a·cos*(E_{0}∙t/ħ) −* i·a·sin*(E_{0}∙t/ħ)

E_{0} is, of course, the *rest *mass of our particle and, now that we are here, we should probably wonder *whose *time *t *we are talking about: is it *our* time, or is the proper time of our particle? Well… In this situation, we are both at rest so it does not matter: t *is*, effectively, the proper time so perhaps we should write it as t_{0}. It does not matter. You can see what we expect to see: E_{0}/ħ pops up as the *natural *frequency of our matter-particle: (E_{0}/ħ)∙t = ω∙t. Remembering the ω = 2π·*f* = 2π/T and T = 1/*f *formulas, we can associate a period and a frequency with this wave, using the ω = 2π·*f* = 2π/T. Noting that ħ = h/2π, we find the following:

T = 2π·(ħ/E_{0}) = h/E_{0} ⇔ *f *= E_{0}/h = m_{0}*c*^{2}/h

This is interesting, because we can look at the period as a *natural *unit of time for our particle. What about the wavelength? That is tricky because we need to distinguish between group and phase velocity here. The group velocity (*v*_{g}) should be zero here, because we assume our particle does not move. In contrast, the phase velocity is given by *v*_{p} = λ·*f *= (2π/k)·(ω/2π) = ω/k. In fact, we’ve got something funny here: the wavenumber k* = *p/ħ is zero, because we assume the particle is at rest, so p = 0. So we have a division by zero here, which is rather strange. What do we get assuming the particle is *not *at rest? We write:

*v*_{p} = ω/k = (E/ħ)/(p/ħ) = E/p = E/(m·*v*_{g}) = (m·*c*^{2})/(m·*v*_{g}) = *c*^{2}/*v*_{g}

This is interesting: it establishes a reciprocal relation between the phase and the group velocity, with *c *as a simple scaling constant. Indeed, the graph below shows the *shape *of the function does *not *change with the value of *c*, and we may also re-write the relation above as:

*v*_{p}/*c *= β_{p} = *c*/*v*_{p} = 1/β_{g} = 1/(*c*/*v*_{p})

**Figure 6**: Reciprocal relation between phase and group velocity

We can also write the mentioned relationship as *v*_{p}·*v*_{g} = *c*^{2}, which reminds us of the relationship between the electric and magnetic constant (1/ε_{0})·(1/μ_{0}) = *c*^{2}. This is interesting in light of the fact we can re-write this as (*c*·ε_{0})·(*c*·μ_{0}) = 1, which shows electricity and magnetism are just two sides of the same coin, so to speak.[24]

Interesting, but how do we interpret the math? What about the implications of the zero value for wavenumber k* = *p/ħ. We would probably like to think it implies the elementary wavefunction should always be associated with *some *momentum, because the concept of zero momentum clearly leads to weird math: something times *zero *cannot be equal to *c*^{2}! Such interpretation is also consistent with the Uncertainty Principle: if Δx·Δp ≥ ħ, then *neither* Δx *nor* Δp can be zero. In other words, the Uncertainty Principle tells us that the idea of a pointlike particle actually *being* at some *specific* point in time and in space does not make sense: it *has *to move. It tells us that our concept of dimensionless points in time and space are *mathematical *notions only. *Actual *particles – including photons – are always a bit spread out, so to speak, and – importantly – they *have to *move.

For a photon, this is self-evident. It has no rest mass, no rest energy, and, therefore, it is going to move at the speed of light itself. We write: p = m·*c* = m·*c*^{2}/*c *= E/*c*. Using the relationship above, we get:

*v*_{p} = ω/k = (E/ħ)/(p/ħ) = E/p = *c* ⇒ *v*_{g} = *c*^{2}/*v*_{p} = *c*^{2}/*c* = *c*

This is good: we started out with some reflections on the *matter*-wave, but here we get an interpretation of the electromagnetic wave as a wavefunction for the photon. But let us get back to our matter-wave. In regard to our interpretation of a particle *having to *move, we should remind ourselves, once again, of the fact that an *actual* particle is always localized in space and that it can, therefore, *not *be represented by the elementary wavefunction ψ = *a·e*^{−i[E·t − p∙x]/ħ} or, for a particle at rest, the ψ = *a·e*^{−i∙E·t/ħ} function. We must build a wave *packet* for that: a sum of wavefunctions, each with their own amplitude *a _{i}*, and their own ω

*= −E*

_{i}*/ħ. Indeed, in section II, we showed that each of these wavefunctions will*

_{i}*contribute*some energy to the total energy of the wave packet and that, to calculate the contribution of each wave to the total, both

*a*as well as E

_{i}*matter. This may or may not resolve the apparent paradox. Let us look at the group velocity.*

_{i}To calculate a meaningful group velocity, we must assume the *v*_{g} = ∂ω* _{i}*/∂k

*= ∂(E*

_{i}*/ħ)/∂(p*

_{i}*/ħ) = ∂(E*

_{i}*)/∂(p*

_{i}*) exists. So we must have some*

_{i}*dispersion relation*. How do we calculate it? We need to calculate ω

*as a function of k*

_{i}

_{i}*here, or E*

_{ }*as a function of p*

_{i}*. How do we do that? Well… There are a few ways to go about it but one interesting way of doing it is to re-write Schrödinger’s equation as we did, i.e. by distinguishing the real and imaginary parts of the ∂ψ/∂t =*

_{i}*i*·[ħ/(2m)]·∇

^{2}ψ wave equation and, hence, re-write it as the following

*pair*of two equations:

*Re*(∂ψ/∂t) = −[ħ/(2m_{eff})]·*Im*(∇^{2}ψ) ⇔ ω·cos(kx − ωt) = k^{2}·[ħ/(2m_{eff})]·cos(kx − ωt)*Im*(∂ψ/∂t) = [ħ/(2m_{eff})]·*Re*(∇^{2}ψ) ⇔ ω·sin(kx − ωt) = k^{2}·[ħ/(2m_{eff})]·sin(kx − ωt)

Both equations imply the following dispersion relation:

ω = ħ·k^{2}/(2m_{eff})

Of course, we need to think about the subscripts now: we have ω* _{i}*, k

*, but… What about m*

_{i}_{eff}or, dropping the subscript, m? Do we write it as m

*? If so, what is it? Well… It is the*

_{i}*equivalent*mass of E

*obviously, and so we get it from the mass-energy equivalence relation: m*

_{i}*= E*

_{i}*/*

_{i}*c*

^{2}. It is a fine point, but one most people forget about: they usually just write m. However, if there is uncertainty in the energy, then Einstein’s mass-energy relation tells us we must have some uncertainty in the (equivalent) mass too. Here, I should refer back to Section II: E

*varies around some*

_{i}*average*energy E and, therefore, the Uncertainty Principle kicks in.

# VII. Explaining spin

The elementary wavefunction *vector* – i.e. the vector sum of the real and imaginary component – rotates around the *x*-axis, which gives us the direction of propagation of the wave (see Figure 3). Its *magnitude *remains constant. In contrast, the magnitude of the electromagnetic vector – defined as the vector sum of the electric and magnetic field vectors – oscillates between zero and some maximum (see Figure 5).

We already mentioned that the *rotation *of the wavefunction vector appears to give some *spin* to the particle. Of course, a *circularly *polarized wave would also appear to have spin (think of the **E** and **B** vectors *rotating around* the direction of propagation – as opposed to oscillating up and down or sideways only). In fact, a circularly polarized light does carry angular momentum, as the *equivalent mass *of its energy may be thought of as rotating as well. But so here we are looking at a *matter*-wave.

The basic idea is the following: *if** *we look at ψ =

*a·e*

^{−i∙E·t/ħ}as some

*real*vector – as a two-dimensional oscillation of mass, to be precise –

*then*we may associate its rotation around the direction of propagation with some torque. The illustration below reminds of the math here.

**Figure 7**: Torque and angular momentum vectors

A torque on some mass about a fixed axis gives it *angular momentum*, which we can write as the vector cross-product **L** = ** r**×

**p**or, perhaps easier for our purposes here as the product of an

*angular*velocity (

**ω**) and rotational inertia (I), aka as the

*moment of inertia*or the

*angular mass*. We write:

**L** = I·**ω**

Note we can write **L** and **ω** in **boldface** here because they are (axial) vectors. If we consider their magnitudes only, we write L = I·ω (no boldface). We can now do some calculations. Let us start with the angular velocity. In our previous posts, we showed that the *period *of the matter-wave is equal to T = 2π·(ħ/E_{0}). Hence, the angular velocity must be equal to:

ω = 2π/[2π·(ħ/E_{0})] = E_{0}/ħ

We also know the distance *r*, so that is the magnitude of *r** *in the **L** = * r*×

**p**vector cross-product: it is just

*a*, so that is the

*magnitude*of ψ =

*a·e*

^{−i∙E·t/ħ}. Now, the momentum (

**p**) is the product of a

*linear*velocity (

*) – in this case, the*

**v***tangential*velocity – and some mass (m):

**p**= m·

*. If we switch to*

**v***scalar*instead of vector quantities, then the (tangential) velocity is given by

*v*=

*r*·ω. So now we only need to think about what we should use for m or, if we want to work with the

*angular*velocity (ω), the

*angular*mass (I). Here we need to make some assumption about the mass (or energy)

*distribution*. Now, it may or may not sense to assume the energy in the oscillation – and, therefore, the mass – is distributed uniformly. In that case, we may use the formula for the angular mass of a solid cylinder: I = m·

*r*

^{2}/2. If we keep the analysis non-relativistic, then m = m

_{0}. Of course, the energy-mass equivalence tells us that m

_{0}= E

_{0}/

*c*

^{2}. Hence, this is what we get:

L = I·ω = (m_{0}·*r*^{2}/2)·(E_{0}/ħ) = (1/2)·*a*^{2}·(E_{0}/*c*^{2})·(E_{0}/ħ) = *a*^{2}·E_{0}^{2}/(2·ħ·*c*^{2})

Does it make sense? Maybe. Maybe not. Let us do a dimensional analysis: that won’t check our logic, but it makes sure we made no mistakes when mapping mathematical and physical spaces. We have m^{2}·J^{2} = m^{2}·N^{2}·m^{2} in the numerator and N·m·s·m^{2}/s^{2} in the denominator. Hence, the dimensions work out: we get N·m·s as the dimension for L, which is, effectively, the physical dimension of angular momentum. It is also the *action *dimension, of course, and that cannot be a coincidence. Also note that the E = m*c*^{2} equation allows us to re-write it as:

L = *a*^{2}·E_{0}^{2}/(2·ħ·*c*^{2})

Of course, in quantum mechanics, we associate spin with the *magnetic *moment of a *charged* particle, not with its *mass *as such. Is there way to link the formula above to the one we have for the quantum-mechanical angular momentum, which is also measured in N·m·s units, and which can only take on one of two possible values: *J* = +ħ/2 and −ħ/2? It looks like a long shot, right? How do we go from (1/2)·*a*^{2}·m_{0}^{2}/ħ to ± (1/2)∙ħ? Let us do a numerical example. The energy of an electron is typically 0.510 MeV » 8.1871×10^{−14} N∙m, and *a*… What value should we take for *a*?

We have an obvious *trio* of candidates here: the Bohr radius, the classical electron radius (aka the Thompon scattering length), and the Compton scattering radius.

Let us start with the Bohr radius, so that is about 0.×10^{−10} N∙m. We get L = *a*^{2}·E_{0}^{2}/(2·ħ·*c*^{2}) = 9.9×10^{−31} N∙m∙s. Now that is about 1.88×10^{4} *times *ħ/2. That is a *huge* factor. The Bohr radius cannot be right: we are *not *looking at an electron in an orbital here. To show it does not make sense, we may want to double-check the analysis by doing the calculation in another way. We said each oscillation will always pack 6.626070040(81)×10^{−34} *joule *in energy. So our electron should pack about 1.24×10^{−20} oscillations. The angular momentum (L) we get when using the Bohr radius for *a* and the value of 6.626×10^{−34} *joule *for E_{0} and the Bohr radius is equal to 6.49×10^{−59} N∙m∙s. So that is the angular momentum per oscillation. When we multiply this with the number of oscillations (1.24×10^{−20}), we get about 8.01×10^{−51} N∙m∙s, so that is a totally different number.

The classical electron radius is about 2.818×10^{−15} m. We get an L that is equal to about 2.81×10^{−39} N∙m∙s, so now it is a tiny *fraction *of ħ/2! Hence, this leads us nowhere. Let us go for our last chance to get a meaningful result! Let us use the Compton scattering length, so that is about 2.42631×10^{−12} m.

This gives us an L of 2.08×10^{−33} N∙m∙s, which is only 20 times ħ. This is not so bad, but it is good enough? Let us calculate it the other way around: what value should we *take *for *a *so as to ensure L = *a*^{2}·E_{0}^{2}/(2·ħ·*c*^{2}) = ħ/2? Let us write it out:

In fact, this is the formula for the so-called *reduced *Compton wavelength. This is perfect. We found what we wanted to find. Substituting this value for *a *(you can calculate it: it is about 3.8616×10^{−33} m), we get what we should find:

This is a rather spectacular result, and one that would – a priori – support the interpretation of the wavefunction that is being suggested in this paper.^{ }

# VIII. The boson-fermion dichotomy

Let us do some more thinking on the boson-fermion dichotomy. Again, we should remind ourselves that an *actual* particle is localized in space and that it can, therefore, *not *be represented by the elementary wavefunction ψ = *a·e*^{−i[E·t − p∙x]/ħ} or, for a particle at rest, the ψ = *a·e*^{−i∙E·t/ħ} function. We must build a wave *packet* for that: a sum of wavefunctions, each with their own amplitude *a _{i}*, and their own ω

*= −E*

_{i}*/ħ. Each of these wavefunctions will*

_{i}*contribute*some energy to the total energy of the wave packet. Now, we can have another wild but logical theory about this.

Think of the apparent right-handedness of the elementary wavefunction: surely, *Nature* can’t be bothered about our convention of measuring phase angles clockwise or counterclockwise. Also, the angular momentum can be positive or negative: *J* = +ħ/2 or −ħ/2. Hence, we would probably like to think that an actual particle – think of an electron, or whatever other particle you’d think of – may consist of right-handed as well as left-handed elementary waves. To be precise, we may think they *either *consist of (elementary) right-handed waves or, *else*, of (elementary) left-handed waves. An elementary right-handed wave would be written as:

ψ(θ* _{i}*)

*= a*·(

_{i}*cos*θ

*+*

_{i}*i·sin*θ

*)*

_{i}In contrast, an elementary left-handed wave would be written as:

ψ(θ* _{i}*)

*= a*·(

_{i}*cos*θ

*−*

_{i}*i·sin*θ

*)*

_{i}How does that work out with the E_{0}·t argument of our wavefunction? Position is position, and direction is direction, but time? Time has only one direction, but *Nature* surely does not care how we *count *time: counting like 1, 2, 3, etcetera or like −1, −2, −3, etcetera is just the same. If we count like 1, 2, 3, etcetera, then we write our wavefunction like:

ψ = *a·cos*(E_{0}∙t/ħ)* − i·a·sin*(E_{0}∙t/ħ)

If we count time like −1, −2, −3, etcetera then we write it as:

ψ = *a·cos*(*−*E_{0}∙t/ħ)* − i·a·sin*(*−*E_{0}∙t/ħ)= *a·cos*(E_{0}∙t/ħ)* + i·a·sin*(E_{0}∙t/ħ)

Hence, it is just like the left- or right-handed circular polarization of an electromagnetic wave: we can have both for the matter-wave too! This, then, should explain why we can have *either *positive *or *negative quantum-mechanical spin (+ħ/2 or −ħ/2). It is the usual thing: we have two *mathematical *possibilities here, and so we *must *have two *physical *situations that correspond to it.

It is only natural. If we have left- and right-handed photons – or, generalizing, left- and right-handed bosons – then we should also have left- and right-handed fermions (electrons, protons, etcetera). Back to the dichotomy. The textbook analysis of the dichotomy between bosons and fermions may be epitomized by Richard Feynman’s *Lecture *on it (Feynman, III-4), which is confusing and – I would dare to say – even inconsistent: how are photons or electrons supposed to *know *that they need to interfere with a positive or a negative sign? They are not supposed to *know *anything: *knowledge *is part of our *interpretation *of whatever it is that is going on there.

Hence, it is probably best to keep it simple, and think of the dichotomy in terms of the different *physical *dimensions of the oscillation: newton per kg versus newton per coulomb. And then, of course, we should also note that matter-particles have a rest mass and, therefore, actually *carry* charge. Photons do not. But both are two-dimensional oscillations, and the point is: the so-called *vacuum *– and the *rest* *mass *of our particle (which is zero for the photon and non-zero for everything else) – give us the natural frequency for both oscillations, which is beautifully summed up in that remarkable equation for the group and phase velocity of the wavefunction, which applies to photons as well as matter-particles:

(*v _{phase}*·

*c*)·(

*v*·

_{group}*c*) = 1 ⇔

*v*·

_{p}*v*=

_{g}*c*

^{2}

The final question then is: why are photons spin-zero particles? Well… We should first remind ourselves of the fact that they do have spin when circularly polarized.[25] Here we may think of the rotation of the equivalent mass of their energy. However, if they are linearly polarized, then there is no spin. Even for circularly polarized waves, the spin angular momentum of photons is a weird concept. If photons have no (rest) mass, then they cannot carry any *charge*. They should, therefore, not have any *magnetic* moment. Indeed, what I wrote above shows an explanation of quantum-mechanical spin requires both mass *as well as *charge.[26]** **

# IX. Concluding remarks

There are, of course, other ways to look at the matter – literally. For example, we can imagine two-dimensional oscillations as *circular *rather than linear oscillations. Think of a tiny ball, whose center of mass stays where it is, as depicted below. Any rotation – around any axis – will be some combination of a rotation around the two other axes. Hence, we may want to think of a two-dimensional oscillation as an oscillation of a polar and azimuthal angle.

**Figure 8**: Two-dimensional *circular *movement

The point of this paper is not to make any definite statements. That would be foolish. Its objective is just to challenge the simplistic mainstream viewpoint on the *reality *of the wavefunction. Stating that it is a mathematical construct only without *physical significance *amounts to saying it has no meaning at all. That is, clearly, a non-sustainable proposition.

The interpretation that is offered here looks at amplitude waves as traveling fields. Their physical dimension may be expressed in force per mass unit, as opposed to electromagnetic waves, whose amplitudes are expressed in force per (electric) *charge *unit. Also, the amplitudes of matter-waves incorporate a phase factor, but this may actually explain the rather enigmatic dichotomy between fermions and bosons and is, therefore, an added bonus.

The interpretation that is offered here has some advantages over other explanations, as it explains the *how *of diffraction and interference. However, while it offers a great explanation of the wave nature of matter, it does *not *explain its particle nature: while we think of the energy as being spread out, we will still *observe *electrons and photons as pointlike particles once they hit the detector. Why is it that a detector can sort of ‘hook’ the whole blob of energy, so to speak?

The interpretation of the wavefunction that is offered here does *not *explain this. Hence, the *complementarity principle* of the Copenhagen interpretation of the wavefunction surely remains relevant.

# Appendix 1: The *de Broglie *relations and energy

The 1/2 factor in Schrödinger’s equation is related to the concept of the *effective *mass (m_{eff}). It is easy to make the wrong calculations. For example, when playing with the famous *de Broglie *relations – aka as the matter-wave equations – one may be tempted to *derive* the following energy concept:

- E = h·
*f*and p = h/λ. Therefore,*f*= E/h and λ = p/h. *v*=*f·*λ = (E/h)∙(p/h) = E/p- p = m·
*v*. Therefore, E =*v*·p = m·*v*^{2}

E = m·*v*^{2}? This *resembles *the E = m*c*^{2} equation and, therefore, one may be enthused by the discovery, especially because the m·*v*^{2} also pops up when working with the Least Action Principle in *classical *mechanics, which states that the path that is followed by a particle will minimize the following integral:Now, we can choose any reference point for the potential energy but, to reflect the energy conservation law, we can select a reference point that ensures the *sum* of the kinetic and the potential energy is zero *throughout *the time interval. If the force field is uniform, then the integrand will, effectively, be equal to KE − PE *= m·v*^{2}.[27]

However, that is *classical *mechanics and, therefore, not so relevant in the context of the *de Broglie *equations, and the apparent paradox should be solved by distinguishing between the *group *and the *phase *velocity of the matter wave.

# Appendix 2: The concept of the effective mass

The effective mass – as used in Schrödinger’s equation – is a rather enigmatic concept. To make sure we are making the right analysis here, I should start by noting you will usually see Schrödinger’s equation written as:This formulation includes a term with the potential energy (U). In free space (no potential), this term disappears, and the equation can be re-written as:

∂ψ(**x**, t)/∂t = *i*·(1/2)·(ħ/m_{eff})·∇^{2}ψ(**x**, t)

We just moved the *i*·ħ coefficient to the other side, noting that 1/*i *= –*i*. Now, in one-dimensional space, and assuming ψ is just the elementary wavefunction (so we substitute *a·e*^{−i∙[E·t − p∙x]/ħ} for ψ), this implies the following:

−*a*·*i*·(E/ħ)·*e*^{−}*i∙*^{[E·t − p∙x]/ħ} = −*i*·(ħ/2m_{eff})·*a*·(p^{2}/ħ^{2})·* e*^{−i∙[E·t − p∙x]/ħ }

⇔ E = p^{2}/(2m_{eff}) ⇔ m_{eff} = m∙(*v/c*)^{2}/2 = m∙β^{2}/2

It is an ugly formula: it *resembles *the kinetic energy formula (K.E. = m∙*v*^{2}/2) but it is, in fact, something completely different. The β^{2}/2 factor ensures the *effective *mass is always a fraction of the mass itself. To get rid of the ugly 1/2 factor, we may re-define m_{eff} as *two *times the old m_{eff} (hence, m_{eff}^{NEW} = 2∙m_{eff}^{OLD}), as a result of which the formula will look somewhat better:

m_{eff} = m∙(*v/c*)^{2} = m∙β^{2}

We know β varies between 0 and 1 and, therefore, m_{eff} will vary between 0 and m. Feynman drops the subscript, and just writes m_{eff} as m in his textbook (see Feynman, III-19). On the other hand, the electron mass as used is also the electron mass that is used to calculate the size of an atom (see Feynman, III-2-4). As such, the two mass concepts are, effectively, mutually compatible. It is confusing because the same mass is often defined as the mass of a *stationary *electron (see, for example, the article on it in the online Wikipedia encyclopedia[28]).

In the context of the derivation of the electron orbitals, we do have the potential energy term – which is the equivalent of a *source *term in a diffusion equation – and that may explain why the above-mentioned m_{eff} = m∙(*v/c*)^{2} = m∙β^{2} formula does not apply.

# References

This paper discusses general principles in physics only. Hence, references can be limited to references to physics textbooks only. For ease of reading, any reference to additional material has been limited to a more popular undergrad textbook that can be consulted online: Feynman’s Lectures on Physics (http://www.feynmanlectures.caltech.edu). References are per volume, per chapter and per section. For example, Feynman III-19-3 refers to Volume III, Chapter 19, Section 3.

# Notes

[1] Of course, an *actual* particle is localized in space and can, therefore, *not *be represented by the elementary wavefunction ψ = *a·e*^{−i∙θ} = *a·e*^{−i[E·t − p∙x]/ħ} = *a·(cosθ **–** i·a·sinθ).* We must build a wave *packet* for that: a sum of wavefunctions, each with its own amplitude *a*_{k} and its own argument θ_{k} = (E_{k}∙t – **p**_{k}∙**x**)/ħ. This is dealt with in this paper as part of the discussion on the mathematical and physical interpretation of the normalization condition.

[2] The N/kg dimension immediately, and naturally, reduces to the dimension of acceleration (m/s^{2}), thereby facilitating a direct interpretation in terms of Newton’s force law.

[3] In physics, a two-*spring *metaphor is more common. Hence, the pistons in the author’s *perpetuum mobile *may be replaced by springs.

[4] The author re-derives the equation for the Compton scattering radius in section VII of the paper.

[5] The magnetic force can be analyzed as a relativistic effect (see Feynman II-13-6). The dichotomy between the electric force as a polar vector and the magnetic force as an axial vector disappears in the relativistic four-vector representation of electromagnetism.

[6] For example, when using Schrödinger’s equation in a central field (think of the electron around a proton), the use of polar coordinates is recommended, as it ensures the symmetry of the Hamiltonian under all rotations (see Feynman III-19-3)

[7] This sentiment is usually summed up in the apocryphal quote: “God does not play dice.”The actual quote comes out of one of Einstein’s private letters to Cornelius Lanczos, another scientist who had also emigrated to the US. The full quote is as follows: “You are the only person I know who has the same attitude towards physics as I have: belief in the comprehension of reality through something basically simple and unified… It seems hard to sneak a look at God’s cards. But that He plays dice and uses ‘telepathic’ methods… is something that I cannot believe for a single moment.” (Helen Dukas and Banesh Hoffman, Albert Einstein, the Human Side: New Glimpses from His Archives, 1979)

[8] Of course, both are different velocities: ω is an *angular *velocity, while *v *is a *linear *velocity: ω is measured in *radians* per second, while *v *is measured in meter per second. However, the definition of a radian implies radians are measured in distance units. Hence, the physical dimensions are, effectively, the same. As for the formula for the total energy of an oscillator, we should actually write: E = m·*a*^{2}∙ω^{2}/2. The additional factor (*a*) is the (maximum) amplitude of the oscillator.

[9] We also have a 1/2 factor in the E = m*v*^{2}/2 formula. Two remarks may be made here. First, it may be noted this is a non-relativistic formula and, more importantly, incorporates kinetic energy only. Using the Lorentz factor (γ), we can write the relativistically correct formula for the kinetic energy as K.E. = E − E_{0} = m_{v}*c*^{2} − m_{0}*c*^{2} = m_{0}γ*c*^{2} − m_{0}*c*^{2} = m_{0}*c*^{2}(γ − 1). As for the *exclusion *of the potential energy, we may note that we may choose our reference point for the potential energy such that the kinetic and potential energy *mirror *each other. The energy concept that then emerges is the one that is used in the context of the Principle of Least Action: it equals E = m*v*^{2}. Appendix 1 provides some notes on that.

[10] Instead of two cylinders with pistons, one may also think of connecting two springs with a crankshaft.

[11] It is interesting to note that we may look at the energy in the rotating flywheel as *potential *energy because it is energy that is associated with motion, albeit *circular *motion. In physics, one may associate a rotating object with kinetic energy using the rotational equivalent of mass and linear velocity, i.e. *rotational inertia* (I) and angular velocity ω. The *kinetic *energy of a rotating object is then given by K.E. = (1/2)·I·ω^{2}.

[12] Because of the sideways motion of the connecting rods, the sinusoidal function will describe the linear motion only *approximately**,* but you can easily imagine the idealized limit situation.

[13] The ω^{2}= 1/*LC formula gives us the natural or resonant frequency for a electric circuit consisting of a resistor (R), an inductor (L), and a capacitor (C). Writing the formula as *ω^{2}= *C*^{–}^{1}/*L introduces the concept of elastance, which is the equivalent of the mechanical stiffness (k) of a spring.*

[14] The resistance in an electric circuit introduces a damping factor. When analyzing a mechanical spring, one may also want to introduce a drag coefficient. Both are usually defined as a fraction of the *inertia*, which is the mass for a spring and the inductance for an electric circuit. Hence, we would write the resistance for a spring as γ*m* and as R = γ*L* respectively.

[15] Photons are emitted by atomic oscillators: atoms going from one state (energy level) to another. Feynman (*Lectures*, I-33-3) shows us how to calculate the Q of these atomic oscillators: it is of the order of 10^{8}, which means the wave train will last about 10^{–8 }seconds (to be precise, that is the time it takes for the radiation to die out by a factor 1/*e*). For example, for sodium light, the radiation will last about 3.2×10^{–8 }seconds (this is the so-called decay time τ). Now, because the frequency of sodium light is some 500 THz (500×10^{12 }oscillations per second), this makes for some 16 million oscillations. There is an interesting paradox here: the speed of light tells us that such wave train will have a length of about 9.6 m! How is that to be reconciled with the pointlike nature of a photon? The paradox can only be explained by relativistic length contraction: in an analysis like this, one need to distinguish the reference frame of the photon – riding along the wave as it is being emitted, so to speak – and our stationary reference frame, which is that of the emitting atom.

[16] This is a general result and is reflected in the K.E. = T = (1/2)·m·ω^{2}·*a*^{2}·sin^{2}(ω·t + Δ) and the P.E. = U = k·x^{2}/2 = (1/2)· m·ω^{2}·*a*^{2}·cos^{2}(ω·t + Δ) formulas for the linear oscillator.

[17] Feynman further formalizes this in his *Lecture on Superconductivity *(Feynman, III-21-2), in which he refers to Schrödinger’s equation as the “equation for continuity of probabilities”. The analysis is centered on the *local *conservation of energy, which confirms the interpretation of Schrödinger’s equation as an energy diffusion equation.

[18] The m_{eff} is the *effective* mass of the particle, which depends on the medium. For example, an electron traveling in a solid (a transistor, for example) will have a different effective mass than in an atom. In free space, we can drop the subscript and just write m_{eff} = m. Appendix 2 provides some additional notes on the concept. As for the equations, they are easily derived from noting that two complex numbers a + *i*∙b and c + *i*∙d are equal if, and only if, their real and imaginary parts are the same. Now, the ∂ψ/∂t = *i*∙(ħ/m_{eff})∙∇^{2}ψ equation amounts to writing something like this: a + *i*∙b = *i*∙(c + *i*∙d). Now, remembering that *i*^{2} = −1, you can easily figure out that *i*∙(c + *i*∙d) = *i*∙c + *i*^{2}∙d = − d + *i*∙c.

[19] The dimension of **B** is usually written as N/(m∙A), using the SI unit for current, i.e. the *ampere *(A). However, 1 C = 1 A∙s and, hence, 1 N/(m∙A) = 1 (N/C)/(m/s). * *

[20] Of course, multiplication with* i *amounts to a *counter*clockwise rotation. Hence, multiplication by –*i* also amounts to a rotation by 90 degrees, but *clockwise*. Now, to uniquely identify the clockwise and counterclockwise directions, we need to establish the equivalent of the right-hand rule for a proper geometric interpretation of Schrödinger’s equation in three-dimensional space: if we look at a clock from the back, then its hand will be moving *counter*clockwise. When writing **B** = (1/*c*)∙*i*∙**E**, we assume we are looking in the *negative x*-direction. If we are looking in the positive *x*-direction, we should write: **B** = -(1/*c*)∙*i*∙**E**. Of course, Nature does not care about our conventions. Hence, both should give the same results in calculations. We will show in a moment they do.

[21] In fact, when multiplying C^{2}/(N·m^{2}) with N^{2}/C^{2}, we get N/m^{2}, but we can multiply this with 1 = m/m to get the desired result. It is significant that an energy density (*joule *per unit *volume*) can also be measured in *newton *(force per unit *area*.

[22] The illustration shows a linearly polarized wave, but the obtained result is general.

[23] The sine and cosine are essentially the same functions, except for the difference in the phase: sinθ = cos(θ−π /2).

[24] I must thank a physics blogger for re-writing the 1/(ε_{0}·μ_{0}) = *c*^{2} equation like this. See: http://reciprocal.systems/phpBB3/viewtopic.php?t=236 (retrieved on 29 September 2017).

[25] A circularly polarized electromagnetic wave may be analyzed as consisting of two perpendicular electromagnetic plane waves of equal amplitude and 90° difference in phase.

[26] Of course, the reader will now wonder: what about neutrons? How to explain neutron spin? Neutrons are neutral. That is correct, but neutrons are not elementary: they consist of (charged) quarks. Hence, neutron spin can (or should) be explained by the spin of the underlying quarks.

[27] We detailed the mathematical framework and detailed calculations in the following online article: https://readingfeynman.org/2017/09/15/the-principle-of-least-action-re-visited.

[28] https://en.wikipedia.org/wiki/Electron_rest_mass (retrieved on 29 September 2017).

# Playing with amplitudes

Let’s play a bit with the stuff we found in our previous post. This is going to be unconventional, or *experimental*, if you want. The idea is to give you… Well… Some ideas. So you can play yourself. 🙂 Let’s go.

Let’s first look at Feynman’s (simplified) formula for the amplitude of a photon to go from point *a* to point *b*. If we identify point *a *by the position vector ** r_{1}** and point

*b*by the position vector

**, and using Dirac’s fancy**

*r*_{2}*bra-ket*notation, then it’s written as:

So we have a vector dot product here: **p**∙*r*_{12} = |**p**|∙|*r*_{12}|· *cos*θ = p∙*r*_{12}·*cos*α. The angle here (α) is the angle between the **p **and *r*_{12} vector. All good. Well… No. We’ve got a problem. When it comes to calculating *probabilities*, the α angle doesn’t matter: |*e*^{i·θ}/*r*|^{2} = 1/*r*^{2}. Hence, for the probability, we get: P = | 〈**r**_{2}|**r**_{1}〉 |^{2} = 1/*r*_{12}^{2}. ** Always !** Now that’s strange. The θ =

**p**∙

*r*_{12}/

*ħ*argument gives us a different phase depending on the angle (α) between

**p**and

*r*_{12}. But… Well… Think of it:

*cos*α goes from 1 to 0 when α goes from 0 to ±90° and, of course, is

*negative*when

**p**and

*r*_{12}have opposite directions but… Well… According to this formula, the

*probabilities*do

*not*depend on the direction of the momentum. That’s just weird, I think. Did Feynman, in his iconic

*Lectures*, give us a meaningless formula?

Maybe. We may also note this function looks like the elementary wavefunction for any particle, which we wrote as:

ψ(** x**,

*t*) =

*a·e*

^{−i}^{∙θ}=

*a·e*

^{−i}^{∙}

^{(E∙t − p∙x)/ħ}=

*a*·

*e*

^{−i}

^{∙}

^{(E∙t)/ħ}·

*e*

^{i}^{∙}

^{(p∙x)/ħ}

The only difference is that the 〈**r**_{2}|**r**_{1}〉 sort of abstracts away from time, so… Well… Let’s get a feel for the quantities. Let’s think of a photon *carrying *some typical amount of energy. Hence, let’s talk visible light and, therefore, photons of a few eV only – say 5.625 eV = 5.625×1.6×10^{−19} J = 9×10^{−19} J. Hence, their momentum is equal to p = E/*c* = (9×10^{−19} N·m)/(3×10^{5} m/s) = 3×10^{−24} N·s. That’s tiny but that’s only because *newtons *and *seconds *are enormous units at the (sub-)atomic scale. As for the distance, we may want to use the thickness of a playing card as a starter, as that’s what Young used when establishing the *experimental *fact of light interfering with itself. Now, playing cards in Young’s time were obviously rougher than those today, but let’s take the smaller distance: modern cards are as thin as 0.3 mm. Still, that distance is associated with a value of θ that is equal to 13.6 *million*. Hence, the *density *of our wavefunction is enormous at this scale, and it’s a bit of a miracle that Young could see any interference at all ! As shown in the table below, we only get meaningful values (remember: θ is a phase *angle*) when we go down to the *nanometer* scale (10^{−9} m) or, even better, the *angstroms *scale ((10^{−9} m).* *

So… Well… Again: what can we do with Feynman’s formula? Perhaps he didn’t give us a *propagator* function but something that is more general (read: more *meaningful*) at our (limited) level of knowledge. As I’ve been reading Feynman for quite a while now – like three or four years 🙂 – I think… Well… Yes. That’s it. Feynman wants us to *think* about it. 🙂 *Are you joking again, Mr. Feynman? *🙂 So let’s assume the reasonable thing: let’s assume it gives us the amplitude to go from point *a *to point *b *by the position vector ** along some path r**. So, then, in line with what we wrote in our previous post, let’s say p·

*r*(momentum over a distance) is the action (

*S*) we’d associate with this particular path (

*r*) and then see where we get. So let’s write the formula like this:

ψ = *a*·*e*^{i·θ} = (1/*r*)·*e*^{i·S/ħ} = *e*^{i·p∙r/ħ}/*r*

We’ll use an index to denote the various paths: *r*_{0} is the straight-line path and *r*_{i} is any (other) path. Now, quantum mechanics tells us we should calculate this amplitude *for every possible path*. The illustration below shows the straight-line path and two nearby paths. So each of these paths is associated with some amount of action, which we measure in *Planck units*: θ = *S*/*ħ*.

The time interval is given by *t *= *t*_{0 }= *r*_{0}/*c*, *for all paths*. Why is the time interval the same *for all paths*? Because we think of a photon going from some *specific *point in space *and in time* to some other *specific *point in space *and in time*. Indeed, when everything is said and done, we do think of light as traveling from point *a *to point *b *at the speed of light (*c*). In fact, all of the weird stuff here is all about trying to explain *how *it does that. 🙂

Now, if we would think of the photon *actually *traveling along this or that path, then this implies its velocity along any of the nonlinear paths will be *larger *than *c*, which is OK. That’s just the weirdness of quantum mechanics, and you should actually *not *think of the photon actually traveling along one of these paths anyway although we’ll often put it that way. Think of something fuzzier, whatever that may be. 🙂

So the action is energy times time, or momentum times distance. Hence, the difference in action between two paths *i *and *j* is given by:

δ*S *= p·*r*_{j} − p·*r*_{i} = p·(*r*_{j} − *r*_{i}) = p·Δ*r*

I’ll explain the δ*S < *2π*ħ*/3 thing in a moment. Let’s first pause and think about the *uncertainty *and how we’re modeling it. We can effectively think of the variation in *S *as some uncertainty in the action: δ*S *= Δ*S *= p·Δ*r*. However, if *S* is also equal to energy times time (*S *= E·*t*), and we insist *t *is the same for all paths, then we must have some uncertainty in the energy, right? Hence, we can write δ*S *as Δ*S *= ΔE·*t*. But, of course, E = E = *m*·*c*^{2} = p·*c*, so we will have an uncertainty in the momentum as well. Hence, the variation in *S *should be written as:

δ*S *= Δ*S* = Δp·Δ*r*

That’s just logical thinking: if we, somehow, entertain the idea of a photon going from some *specific *point in spacetime to some other *specific *point in spacetime along various paths, then the variation, or *uncertainty*, in the action will effectively combine some uncertainty in the momentum and the distance. We can calculate Δp as ΔE/*c*, so we get the following:

δ*S *= Δ*S* = Δp·Δ*r = *ΔE·Δ*r*/*c = *ΔE·Δ*t* with Δ*t** = *Δ*r*/*c*

So we have the two expressions for the Uncertainty Principle here: Δ*S* = Δp·Δ*r* = ΔE·Δ*t*. Just be careful with the interpretation of Δ*t*: it’s just the equivalent of Δ*r*. We just express the uncertainty in distance in *seconds *using the (absolute) speed of light. We are *not *changing our spacetime interval: we’re still looking at a photon going from *a *to *b *in *t *seconds, *exactly*. Let’s now look at the δ*S < *2π*ħ*/3 thing. If we’re adding *two *amplitudes (two *arrows* or *vectors*, so to speak) and we want the magnitude of the result to be larger than the magnitude of the two contributions, then the angle between them should be smaller than 120 degrees, so that’s 2π/3 *rad*. The illustration below shows how you can figure that out geometrically.Hence, if *S*_{0} is the action for *r*_{0}, then *S*_{1} = *S*_{0} + *ħ *and *S*_{2} = *S*_{0} + 2·*ħ *are still good, but *S*_{3} = *S*_{0} + 3·*ħ* is *not *good. Why? Because the difference in the phase angles is Δθ = *S*_{1}/*ħ* − *S*_{0}/*ħ* = (*S*_{0} + *ħ*)/*ħ* − *S*_{0}/*ħ* = 1 and Δθ = *S*_{2}/*ħ* − *S*_{0}/*ħ* = (*S*_{0} + 2·*ħ*)/*ħ* − *S*_{0}/*ħ* = 2 respectively, so that’s 57.3° and 114.6° respectively and that’s, effectively, *less *than 120°. In contrast, for the next path, we find that Δθ = *S*_{3}/*ħ* − *S*_{0}/*ħ* = (*S*_{0} + 3·*ħ*)/*ħ* − *S*_{0}/*ħ* = 3, so that’s 171.9°. So that amplitude gives us a *negative *contribution.

Let’s do some calculations using a spreadsheet. To simplify things, we will assume we measure everything (time, distance, force, mass, energy, action,…) in Planck units. Hence, we can simply write: *S _{n}* =

*S*

_{0}+

*n*. Of course,

*n*= 1, 2,… etcetera, right? Well… Maybe not. We are

*measuring*action in units of

*ħ*, but do we actually think action

*comes*in units of

*ħ*? I am not sure. It would make sense, intuitively, but… Well… There’s uncertainty on the energy (E) and the momentum (

*) of our photon, right? And how accurately can we measure the distance? So there’s some randomness everywhere. 😦 So let’s leave that question open as for now.*

**p**We will also assume that the phase angle for *S*_{0} is equal to 0 (or some multiple of 2π, if you want). That’s just a matter of choosing the origin of time. This makes it really easy: Δ*S*_{n} = *S _{n}* −

*S*

_{0}=

*n*, and the associated phase angle θ

_{n}= Δθ

_{n}is the same. In short, the amplitude for each path reduces to ψ

_{n}=

*e*

^{i·n}/

*r*

_{0}. So we need to add these

*first*and

*then*calculate the magnitude, which we can then square to get a probability. Of course, there is also the issue of normalization (probabilities have to add up to one) but let’s tackle that later. For the calculations, we use Euler’s

*r*·

*e*

^{i·θ}=

*r*·(

*cos*θ +

*i*·

*sin*θ) =

*r*·

*cos*θ +

*i*·

*r*·

*sin*θ formula. Needless to say, |

*r*·

*e*

^{i·θ}|

^{2}= |

*r|*

^{2}·

*|*

*e*

^{i·θ}|

^{2}

*=*|

*r*|

^{2}·(

*cos*

^{2}θ +

*sin*

^{2}θ) =

*r*. Finally, when adding complex numbers, we add the real and imaginary parts respectively, and we’ll denote the ψ

_{0}+ ψ

_{1}+ψ

_{2}+ … sum as Ψ.

Now, we also need to see how our Δ*S* = Δp·Δ*r* works out. We may want to assume that the uncertainty in p and in *r *will both be proportional to the overall uncertainty in the action. For example, we could try writing the following: Δ*S*_{n} = Δp* _{n}*·Δ

*r*=

_{n}*n*·Δp

_{1}·Δ

*r*

_{1}. It also makes sense that you may want Δp

*and Δ*

_{n}*r*to be proportional to Δp

_{n}_{1}and Δ

*r*

_{1}respectively. Combining both, the assumption would be this:

Δp* _{n}* = √

*n*·Δp

_{1 }and Δ

*r*= √

_{n}*n*·Δ

*r*

_{1}

So now we just need to decide how we will distribute Δ*S*_{1} = *ħ *= 1 over Δp_{1} and Δ*r*_{1} respectively. For example, if we’d assume Δp_{1} = 1, then Δ*r*_{1} = *ħ*/Δp_{1} = 1/1 = 1. These are the calculations. I will let you analyze them. 🙂Well… We get a weird result. It reminds me of Feynman’s explanation of the partial reflection of light, shown below, but… Well… That doesn’t make much sense, does it?

Hmm… Maybe it does. 🙂 Look at the graph more carefully. The *peaks *sort of oscillate out so… Well… That might make sense… 🙂

Does it? Are we doing something wrong* *here? These amplitudes should reflect the ones that are reflected in those nice animations (like this one, for example, which is part of that’s part of the Wikipedia article on Feynman’s path integral formulation of quantum mechanics). So what’s wrong, if anything? Well… Our paths differ by some fixed amount of action, which doesn’t quite reflect the geometric approach that’s used in those animations. The graph below shows how the distance *r *varies as a function of *n*.

If we’d use a model in which the distance would *increase* linearly or, preferably, exponentially, then we’d get the result we want to get, right?

Well… Maybe. Let’s try it. Hmm… We need to think about the geometry here. Look at the triangle below. If *b *is the straight-line path (*r*_{0}), then *ac *could be one of the crooked paths (*r _{n}*). To simplify, we’ll assume

*isosceles*triangles, so

*a*equals

*c*and, hence,

*r*= 2·

_{n}*a*= 2·

*c*. We will also assume the successive paths are separated by the same vertical distance (

*h = h*

_{1}) right in the middle, so

*h*=

_{b}*h*

_{n}=

*n*·

*h*

_{1}. It is then easy to show the following:This gives the following graph for

*r*= 10 and

_{n}*h*

_{1 }= 0.01.

Is this the right step increase? Not sure. We can vary the values in our spreadsheet. Let’s first build it. The photon will have to travel *faster* in order to cover the extra distance in the same time, so its momentum will be higher. Let’s think about the velocity. Let’s start with the first path (*n *= 1). In order to cover the *extra *distance Δ*r*_{1}, the velocity *c*_{1} must be equal to (*r*_{0} + Δ*r*_{1})/*t *= *r*_{0}/*t *+ Δ*r*_{1}/*t =* *c *+ Δ*r*_{1}/*t *= *c*_{0}* *+ Δ*r*_{1}/*t*. We can write *c*_{1} as *c*_{1} = *c*_{0}* *+ Δ*c*_{1}, so Δ*c*_{1} = Δ*r*_{1}/*t**.* Now, the *ratio *of p_{1} and p_{0} will be equal to the *ratio *of *c*_{1} and *c*_{0} because p_{1}/p_{0 }= (m*c*_{1})/m*c*_{0}) = *c*_{1}/*c*_{0}. Hence, we have the following formula for p_{1}:

p_{1} = p_{0}·*c*_{1}/*c*_{0} = p_{0}·(*c*_{0}* *+ Δ*c*_{1})/*c*_{0} = p_{0}·[1 + Δ*r*_{1}/(*c*_{0}*·t*) = p_{0}·(1 + Δ*r*_{1}/*r*_{0})

For p* _{n}*, the logic is the same, so we write:

p* _{n}* = p

_{0}·

*c*/

_{n}*c*

_{0}= p

_{0}·(

*c*

_{0}

*+ Δ*

*c*)/

_{n}*c*

_{0}= p

_{0}·[1 + Δ

*r*/(

_{n}*c*

_{0}

*·t*) = p

_{0}·(1 + Δ

*r*/

_{n}*r*

_{0})

Let’s do the calculations, and let’s use meaningful values, so the nanometer scale and actual values for Planck’s constant and the photon momentum. The results are shown below.

Pretty interesting. In fact, this looks *really *good. The *probability *first swings around wildly, because of these zones of constructive and destructive interference, but then stabilizes. [Of course, I would need to normalize the probabilities, but you get the idea, right?] So… Well… I think we get a *very* meaningful result with this model. Sweet ! 🙂 ** I’m lovin’ it !** 🙂 And, here you go, this is (part of) the calculation table, so you can see what I am doing. 🙂

The graphs below look even better: I just changed the *h*_{1}/*r*_{0} ratio from 1/100 to 1/10. The probability stabilizes almost immediately. 🙂 So… Well… It’s not as fancy as the referenced animation, but I think the educational value of this thing here is at least as good ! 🙂

🙂 This is good stuff… 🙂

**Post scriptum **(19 September 2017): There is an obvious inconsistency in the model above, and in the calculations. We assume there is a path r_{1} = , *r*_{2}, *r*_{2},etcetera, and then we calculate the action for it, and the amplitude, and then we add the amplitude to the sum. But, surely, we should count these paths *twice*, in two-dimensional space, that is. Think of the graph: we have positive and negative interference zones that are sort of layered around the straight-line path, as shown below.

In three-dimensional space, these lines become surfaces. Hence, rather than adding *one* arrow for every δ* * having *one *contribution only, we may want to add… Well… In three-dimensional space, the formula for the surface around the straight-line path would probably look like π·*h*_{n}·*r*_{1}, right? Hmm… Interesting idea. I changed my spreadsheet to incorporate that idea, and I got the graph below. It’s a nonsensical result, because the probability does swing around, but it gradually spins out of control: it never stabilizes.That’s because we increase the *weight *of the paths that are further removed from the center. So… Well… We shouldn’t be doing that, I guess. 🙂 I’ll you look for the right formula, OK? Let me know when you found it. 🙂

# Some thoughts on the nature of reality

Some other comment on an article on my other blog, inspired me to structure some thoughts that are spread over various blog posts. What follows below, is probably the first draft of an article or a paper I plan to write. Or, who knows, I might re-write my two introductory books on quantum physics and publish a new edition soon. 🙂

## Physical dimensions and Uncertainty

The physical dimension of the quantum of action (*h *or* ħ =* *h*/2π) is force (expressed in *newton*) times distance (expressed in *meter*) times time (expressed in *seconds*): N·m·s. Now, you may think this N·m·s dimension is kinda hard to *imagine*. We can imagine its individual components, right? Force, distance and time. We know what they are. But the product of all three? What is it, *really*?

It shouldn’t be all that hard to *imagine *what it might be, right? The N·m·s unit is also the unit in which angular momentum is expressed – and you can sort of imagine what that is, right? Think of a spinning top, or a gyroscope. We may also think of the following:

- [
*h*] = N·m·s = (N·m)·s = [E]·[t] - [
*h*] = N·m·s = (N·s)·m = [p]·[x]

Hence, the physical dimension of action is that of *energy *(E) multiplied by *time* (t) or, alternatively, that of *momentum *(p) times *distance *(x). To be precise, the second dimensional equation should be written as [*h*] = [**p**]·[**x**], because both the momentum and the distance traveled will be associated with some *direction*. It’s a moot point for the discussion at the moment, though. Let’s think about the first equation first: [*h*] = [E]·[t]. What does it mean?

Energy… Hmm… In real life, we are usually not interested in the energy of a system as such, but by the energy it can *deliver*, or *absorb*, **per second**. This is referred to as the *power *of a system, and it’s expressed in J/s, or *watt*. Power is also defined as the (time) *rate *at which *work *is done. Hmm… But so here we’re *multiplying *energy and time. So what’s that? After Hiroshima and Nagasaki, we can sort of imagine the energy of an atomic bomb. We can also sort of imagine the *power *that’s being released by the Sun in light and other forms of *radiation*, which is about 385×10^{24}* joule *per *second*. But energy times time? What’s that?

I am not sure. If we think of the Sun as a huge reservoir of energy, then the physical dimension of action is just like having that reservoir of energy guaranteed for some time, *regardless of how fast or how slow we use it*. So, in short, **it’s just like the Sun – or the Earth, or the Moon, or whatever object – just being there, for some definite amount of time**. So, yes: some

*definite*amount of mass or energy (E) for some

*definite*amount of time (t).

Let’s bring the mass-energy equivalence formula in here: E = m*c*^{2}. Hence, the physical dimension of action can also be written as [*h*] = [E]·[t] = [m*c*]^{2}·[t] = (kg·m^{2}/s^{2})·s = kg·m^{2}/s. What does that say? Not all that much – for the time being, at least. We can get this [*h*] = kg·m^{2}/s through some other substitution as well. A force of one newton will give a mass of 1 kg an acceleration of 1 m/s per second. Therefore, 1 N = 1 kg·m/s^{2} and, hence, the physical dimension of *h*, or the unit of angular momentum, may also be written as 1 N·m·s = 1 (kg·m/s^{2})·m·s = 1 kg·m^{2}/s, i.e. the product of mass, velocity and distance.

Hmm… What can we do with that? Nothing much for the moment: our first reading of it is just that it reminds us of the definition of angular momentum – some *mass* with some *velocity* rotating around an axis. What about the distance? Oh… The *distance* here is just the distance from the axis, right? Right. But… Well… It’s like having some amount of linear momentum available over some distance – or in some *space*, right? That’s sufficiently significant as an interpretation for the moment, I’d think…

## Fundamental units

This makes one think about what units would be fundamental – and what units we’d consider as being derived. Formally, the *newton* is a *derived *unit in the metric system, as opposed to the units of mass, length and time (kg, m, s). Nevertheless, I personally like to think of force as being *fundamental*: a force is what causes an object to deviate from its straight trajectory in spacetime. Hence, we may want to think of the quantum of action as representing *three* fundamental physical dimensions: (1) *force*, (2) *time* and (3) distance – or *space*. We may then look at energy and (linear) momentum as physical quantities combining (1) force and distance and (2) force and time respectively.

Let me write this out:

- Force times length (think of a force that is
*acting*on some object over some distance) is energy: 1*joule*(J) = 1(N). Hence, we may think of the concept of energy as a*newton*·*meter**projection*of action in space only: we make abstraction of time. The physical dimension of the quantum of action should then be written as [*h*] = [E]·[t]. [Note the square brackets tell us we are looking at a*dimensional*equation only, so [t] is just the physical dimension of the time variable. It’s a bit confusing because I also use square brackets as parentheses.] - Conversely, the magnitude of linear momentum (p = m·
*v*) is expressed in: 1 kg·m/s = 1 (kg·m/s*newton*·*seconds*^{2})·s = 1 N·s. Hence, we may think of (linear) momentum as a projection of action in time only: we make abstraction of its spatial dimension. Think of a force that is acting on some object*during some time*. The physical dimension of the quantum of action should then be written as [*h*] = [p]·[x]

Of course, a force that is acting on some object during some time, will usually also act on the same object over some distance but… Well… Just *try*, for once, to make abstraction of one of the two dimensions here: time *or *distance.

It is a difficult thing to do because, when everything is said and done, we don’t live in space or in time alone, but in *spacetime* and, hence, such abstractions are not easy. [Of course, now you’ll say that it’s easy to think of something that moves in time only: an object that is standing still does just that – but then we know movement is relative, so there is no such thing as an object that is standing still in space *in an absolute sense*: Hence, objects never stand still in *spacetime*.] In any case, we should try such abstractions, if only because of the principle of least action is so essential and deep in physics:

- In classical physics, the path of some object in a force field will
*minimize*the total action (which is usually written as S) along that path. - In quantum mechanics, the same action integral will give us various values S – each corresponding to a particular path – and each path (and, therefore, each value of S, really) will be associated with a probability amplitude that will be proportional to some constant times
*e*^{−i·θ}=*e*^{i·(S/ħ)}. Because*ħ*is so tiny, even a small change in S will give a completely different phase angle θ. Therefore, most amplitudes will cancel each other out as we take the sum of the amplitudes over all possible paths: only the paths that*nearly*give the same phase matter. In practice, these are the paths that are associated with a variation in S of an order of magnitude that is equal to*ħ*.

The paragraph above summarizes, in essence, Feynman’s path integral formulation of quantum mechanics. We may, therefore, think of the quantum of action *expressing *itself (1) in time only, (2) in space only, or – much more likely – (3) expressing itself in both dimensions at the same time. Hence, if the quantum of action gives us the *order of magnitude *of the uncertainty – think of writing something like S ± *ħ*, we may re-write our dimensional [*ħ*] = [E]·[t] and [*ħ*] = [p]·[x] equations as the uncertainty equations:

- ΔE·Δt =
*ħ* - Δp·Δx =
*ħ*

You should note here that it is best to think of the uncertainty relations as a *pair *of equations, if only because you should also think of the concept of energy and momentum as representing different *aspects *of the same reality, as evidenced by the (relativistic) energy-momentum relation (E^{2} = p^{2}*c*^{2} – *m*_{0}^{2}*c*^{4}). Also, as illustrated below, the actual path – or, to be more precise, what we might associate with the concept of the actual path – is likely to be some mix of Δx and Δt. If Δt is very small, then Δx will be very large. In order to move over such distance, our particle will require a larger energy, so ΔE will be large. Likewise, if Δt is very large, then Δx will be very small and, therefore, ΔE will be very small. You can also reason in terms of Δx, and talk about momentum rather than energy. You will arrive at the same conclusions: the ΔE·Δt = *h *and Δp·Δx = *h *relations represent two aspects of the same reality – or, at the very least, what we might *think of *as reality.

Also think of the following: if ΔE·Δt = *h *and Δp·Δx = *h*, then ΔE·Δt = Δp·Δx and, therefore, ΔE/Δp must be equal to Δx/Δt. Hence, the *ratio *of the uncertainty about x (the distance) and the uncertainty about t (the time) equals the *ratio *of the uncertainty about E (the energy) and the uncertainty about p (the momentum).

Of course, you will note that the *actual* uncertainty relations have a factor 1/2 in them. This may be explained by thinking of both negative as well as positive variations in space and in time.

We will obviously want to do some more thinking about those physical dimensions. **The idea of a force implies the idea of some object – of some mass on which the force is acting**. Hence, let’s think about the concept of mass now. But… Well… Mass and energy are supposed to be equivalent, right? So let’s look at the concept of energy *too*.

## Action, energy and mass

What *is *energy, really? In real life, we are usually not interested in the energy of a system as such, but by the energy it can *deliver*, or *absorb*, per second. This is referred to as the *power *of a system, and it’s expressed in J/s. However, in physics, we always talk energy – not power – so… Well… What *is* the energy of a system?

According to the *de Broglie *and Einstein – and so many other eminent physicists, of course – we should not only think of the *kinetic* energy of its parts, but also of their *potential* energy, and their *rest *energy, and – for an atomic system – we may add some internal energy, which may be binding energy, or excitation energy (think of a hydrogen atom in an excited state, for example). A lot of stuff. 🙂 But, obviously, Einstein’s mass-equivalence formula comes to mind here, and summarizes it all:

E = m·*c*^{2}

The m in this formula refers to mass – not to meter, obviously. Stupid remark, of course… But… Well… What is energy, *really*? What is mass, *really*? **What’s that ***equivalence ***between mass and energy, ***really***?**

I don’t have the definite answer to that question (otherwise I’d be famous), but… Well… I do think physicists and mathematicians should invest more in exploring some basic intuitions here. As I explained in several posts, it is very tempting to think of energy as some kind of two-dimensional oscillation of mass. A force over some distance will cause a mass to accelerate. This is reflected in the dimensional analysis:

[E] = [m]·[*c*^{2}] = 1 kg·m^{2}/s^{2} = 1 kg·m/s^{2}·m = 1 N·m

The kg and m/s^{2 }factors make this abundantly clear: m/s^{2} is the physical dimension of acceleration: (the change in) velocity per time unit.

Other formulas now come to mind, such as the Planck-Einstein relation: E = h·*f* = ω·ħ. We could also write: E = h/T. Needless to say, T = 1/*f* is the *period *of the oscillation. So we could say, for example, that the energy of some particle times the period of the oscillation gives us Planck’s constant again. What does that mean? Perhaps it’s easier to think of it the other way around: E/*f* = h = 6.626070040(81)×10^{−34} J·s. Now, *f *is the number of oscillations *per second*. Let’s write it as *f *= *n*/s, so we get:

E/*f *= E/(*n*/s) = E·s/*n* = 6.626070040(81)×10^{−34} J·s ⇔ E/*n *= 6.626070040(81)×10^{−34} J

What an amazing result! Our wavicle – be it a photon or a matter-particle – will *always *pack 6.626070040(81)×10^{−34} *joule *in *one *oscillation, so that’s the *numerical *value of Planck’s constant which, of course, depends on our *fundamental *units (i.e. kg, meter, second, etcetera in the SI system).

Of course, the obvious question is: what’s *one *oscillation? If it’s a wave packet, the oscillations may not have the same amplitude, and we may also not be able to define an exact period. In fact, we should *expect* the amplitude and duration of each oscillation to be slightly different, shouldn’t we? And then…

Well… What’s an oscillation? We’re used to *counting *them: *n *oscillations per second, so that’s *per time unit*. How many do we have *in total*? We wrote about that in our posts on the shape and size of a photon. We know photons are emitted by atomic oscillators – or, to put it simply, just atoms going from one energy level to another. Feynman calculated the Q of these atomic oscillators: it’s of the order of 10^{8 }(see his *Lectures, *I-33-3: it’s a wonderfully simple exercise, and one that really shows his greatness as a physics teacher), so… Well… This wave train will last about 10^{–8 }seconds (that’s the time it takes for the radiation to die out by a factor 1/*e*). To give a somewhat more precise example, for sodium light, which has a frequency of 500 THz (500×10^{12 }oscillations per second) and a wavelength of 600 nm (600×10^{–9 }meter), the radiation will lasts about 3.2×10^{–8 }seconds. [In fact, that’s the time it takes for the radiation’s *energy* to die out *by a factor 1/e*, so(i.e. the so-called decay time τ), so the wavetrain will actually last *longer*, but so the amplitude becomes quite small after that time.] So… Well… That’s a very short time but… Still, taking into account the rather spectacular frequency (500 THz) of sodium light, that makes for some 16 million oscillations and, taking into the account the rather spectacular speed of light (3×10^{8 }m/s), that makes for a wave train with a length of, roughly, 9.6 meter. *Huh? 9.6 meter!? But a photon is supposed to be pointlike, isn’it it? It has no length, does it?*

That’s where relativity helps us out: as I wrote in one of my posts, relativistic length contraction may explain the apparent paradox. *Using the reference frame of the photon* – so if we’d be traveling at speed *c*,’ riding’ with the photon, so to say, as it’s being emitted – then we’d ‘see’ the electromagnetic transient as it’s being radiated into space.

However, while we can associate some mass *with the energy of the photon*, none of what I wrote above explains what the (rest) mass of a matter-particle could possibly be.* *There is no real answer to that, I guess. You’ll think of the Higgs field now but… Then… Well. The Higgs field is a scalar field. Very simple: some *number* that’s associated with some position in spacetime. That doesn’t explain very much, does it? 😦 When everything is said and done, the scientists who, in 2013 only, got the Nobel Price for their theory on the Higgs mechanism, simply tell us mass is some number. That’s something we knew already, right? 🙂

## The reality of the wavefunction

The wavefunction is, obviously, a mathematical construct: a *description *of reality using a very specific language. What language? Mathematics, of course! Math may not be universal (aliens might not be able to decipher our mathematical models) but it’s pretty good as a *global *tool of communication, at least.

The *real *question is: is the description *accurate*? Does it match reality and, if it does, how *good *is the match? For example, the wavefunction for an electron in a hydrogen atom looks as follows:

ψ(* r*,

*t*) =

*e*

^{−i·(E/ħ)·t}·

*f*(

*)*

**r**As I explained in previous posts (see, for example, my recent post on reality and perception), the *f*(* r*) function basically provides some envelope for the two-dimensional

*e*

^{−i·θ}=

*e*

^{−i·(E/ħ)·t}=

*cos*θ +

*i*·

*sin*θ oscillation, with

*= (*

**r***x*,

*y*,

*z*), θ = (E/ħ)·

*t*= ω·

*t*and ω = E/ħ. So it presumes the duration of each oscillation is some constant. Why? Well… Look at the formula: this thing has a constant frequency in time. It’s only the amplitude that is varying as a function of the

*= (x, y, z) coordinates. 🙂 So… Well… If each oscillation is to*

**r***always*pack 6.626070040(81)×10

^{−34}

*joule*, but the amplitude of the oscillation varies from point to point, then… Well… We’ve got a problem. The wavefunction above is likely to be an approximation of reality only. 🙂 The associated energy is the same, but… Well… Reality is probably

*not*the nice geometrical shape we associate with those wavefunctions.

In addition, we should think of the Uncertainty Principle: there *must *be some uncertainty in the energy of the photons when our hydrogen atom makes a transition from one energy level to another. But then… Well… If our photon packs something like 16 million oscillations, and the order of magnitude of the uncertainty is only of the order of *h* (or *ħ =* *h*/2π) which, as mentioned above, is the (average) energy of *one *oscillation only, then we don’t have much of a problem here, do we? 🙂

**Post scriptum**: In previous posts, we offered some analogies – or metaphors – to a two-dimensional oscillation (remember the V-2 engine?). Perhaps it’s all relatively simple. If we have some tiny little ball of mass – and its center of mass has to stay where it is – then any rotation – around any axis – will be some combination of a rotation around *our *x- and z-axis – as shown below. Two axes only. So we may want to think of a two-dimensional oscillation as an oscillation of the polar and azimuthal angle. 🙂

# Thinking again…

One of the comments on my other blog made me think I should, perhaps, write something on waves again. The animation below shows the *elementary *wavefunction ψ = *a*·*e*^{−iθ }= ψ = *a*·*e*^{−i·θ } = *a*·*e*^{−i}^{(}^{ω·t−k·x}^{)} = *a*·*e*^{−}^{(}^{i}^{/ħ)·}^{(E·}^{t−p·x}^{)} .We know this elementary wavefunction cannot* *represent a real-life particle. Indeed, the *a*·*e*^{−i·θ }function implies the probability of finding the particle – an electron, a photon, or whatever – would be equal to P(x, t) = |ψ(x, t)|^{2} = |*a*·*e*^{−}^{(}^{i}^{/ħ)·}^{(E·}^{t−p·x}^{)}|^{2} = |*a*|^{2}·|*e*^{−}^{(}^{i}^{/ħ)·}^{(E·}^{t−p·x}^{)}|^{2} = |*a*|^{2}·1^{2}= *a*^{2} *everywhere*. Hence, the particle would be everywhere – and, therefore, *nowhere* really. We need to *localize* the wave – or build a wave *packet*. We can do so by introducing uncertainty: we then *add* a potentially infinite number of these elementary wavefunctions with slightly different values for E and p, and various amplitudes *a*. Each of these amplitudes will then reflect the *contribution *to the composite wave, which – in three-dimensional space – we can write as:

ψ(* r*,

*t*) =

*e*

^{−i·(E/ħ)·t}·

*f*(

*)*

**r**As I explained in previous posts (see, for example, my recent post on reality and perception), the *f*(* r*) function basically provides some envelope for the two-dimensional

*e*

^{−i·θ}=

*e*

^{−i·(E/ħ)·t}=

*cos*θ +

*i*·

*sin*θ oscillation, with

*= (*

**r***x*,

*y*,

*z*), θ = (E/ħ)·

*t*= ω·

*t*and ω = E/ħ.

Note that it *looks like* the wave *propagates *from left to right – in the *positive *direction of an axis which we may refer to as the *x*-axis. Also note this perception results from the fact that, naturally, we’d associate time with the *rotation *of that arrow at the center – i.e. with the *motion* in the illustration, while the spatial dimensions are just what they are: linear spatial dimensions. [This point is, perhaps, somewhat less self-evident than you may think at first.]

Now, the axis which points upwards is usually referred to as the *z*-axis, and the third and final axis – which points *towards *us – would then be the *y-*axis, obviously. Unfortunately, this definition would violate the so-called right-hand rule for defining a proper reference frame: the figures below shows the two possibilities – a left-handed and a right-handed reference frame – and it’s the right-handed reference (i.e. the illustration on the right) which we have to use in order to correctly define all directions, including the direction of *rotation *of the argument of the wavefunction.Hence, if we don’t change the direction of the *y*– and *z*-axes – so we keep defining the *z*-axis as the axis pointing upwards, and the y-axis as the axis pointing *towards *us – then the *positive* direction of the *x*-axis would actually be the direction from right to left, and we should say that the elementary wavefunction in the animation above *seems to* propagate in the *negative* *x*-direction. [Note that this left- or right-hand rule is quite astonishing: simply swapping the direction of *one *axis of a left-handed frame makes it right-handed, and vice versa.]

Note my language when I talk about the direction of propagation of our wave. I wrote: it *looks like*, or it *seems to *go in this or that direction. And I mean that: there is no* real *traveling* *here. At this point, you may want to review a post I wrote for my son, which explains the basic math behind waves, and in which I also explained the animation below.

Note how the peaks and troughs of this pulse *seem to* move leftwards, but the wave *packet *(or the *group *or the *envelope* of the wave—whatever you want to call it) moves to the right. The point is: **the pulse itself doesn’t travel left or right**. Think of the horizontal axis in the illustration above as an oscillating guitar string: each point on the string just moves up and down. Likewise, if our repeated pulse would represent a physical wave in water, for example, then the water just stays where it is: it just moves up and down. Likewise, if we shake up some rope, the rope is not going anywhere: we just started some *motion *that is traveling down the rope. In other words, *the* *phase velocity is just a mathematical* *concept*. The peaks and troughs that seem to be traveling are just *mathematical points that are ‘traveling’ left or right*. That’s why there’s no limit on the phase velocity: it *can* – and, according to quantum mechanics, actually *will *– exceed the speed of light. In contrast, the *group *velocity – which is the actual speed of the particle that is being represented by the wavefunction – may *approach* – or, in the case of a massless photon, will actually *equal *– the speed of light, but will never *exceed *it, and its *direction *will, obviously, have a *physical *significance as it is, effectively, the direction of travel of our particle – be it an electron, a photon (electromagnetic radiation), or whatever.

Hence, you should not think the *spin *of a particle – integer or half-integer – is somehow related to the direction of rotation of the argument of the elementary wavefunction. It isn’t: Nature doesn’t give a damn about our mathematical conventions, and that’s what the direction of rotation of the argument of that wavefunction is: just some mathematical convention. That’s why we write *a*·*e*^{−i}^{(}^{ω·t−k·x}^{)} rather than *a*·*e*^{i}^{(}^{ω·t+k·x}^{)} or *a*·*e*^{i}^{(}^{ω·t−k·x}^{)}: it’s just because of the right-hand rule for coordinate frames, and also because Euler defined the *counter*-clockwise direction as the positive direction of an angle. There’s nothing more to it.

OK. That’s obvious. Let me now return to my interpretation of Einstein’s E = m·*c*^{2} formula (see my previous posts on this). I noted that, in the reference frame of the particle itself (see my basics page), the elementary wavefunction *a*·*e*^{−}^{(}^{i}^{/ħ)·}^{(E·}^{t−p·x}^{)} reduces to *a*·*e*^{−}^{(}^{i}^{/ħ)·}^{(E’·}^{t’}^{)}: the origin of the reference frame then coincides with (the center of) our particle itself, and the wavefunction only varies with the time in the inertial reference frame (i.e. the *proper *time t’), with the rest energy of the object (E’) as the time scale factor. How should we interpret this?

Well… Energy is force times distance, and force is defined as that what causes some *mass *to *accelerate*. To be precise, the *newton *– as the unit of force – is defined as the *magnitude *of a force which would cause a mass of one kg to accelerate with one meter per second *per second*. Per second per second. This is not a typo: 1 N corresponds to 1 kg times 1 m/s *per second*, i.e. 1 kg·m/s^{2}. So… Because energy is force times distance, the unit of *energy *may be expressed in units of kg·m/s^{2}·m, or kg·m^{2}/s^{2}, i.e. the unit of mass times the unit of velocity *squared*. To sum it all up:

1 J = 1 N·m = 1 kg·(m/s)^{2}

This reflects the *physical dimensions *on both sides of the E = m·*c*^{2} formula again but… Well… How should we *interpret *this? Look at the animation below once more, and imagine the green dot is some tiny *mass *moving around the origin, in an equally tiny circle. We’ve got *two *oscillations here: each packing *half *of the total energy of… Well… Whatever it is that our elementary wavefunction might represent *in reality* – which we don’t know, of course.

Now, the blue and the red dot – i.e. the horizontal and vertical *projection *of the green dot – accelerate up and down. If we look carefully, we see these dots accelerate *towards *the zero point and, once they’ve crossed it, they *decelerate*, so as to allow for a *reversal of direction*: the blue dot goes up, and then down. Likewise, the red dot does the same. The *interplay* between the two oscillations, because of the 90° phase difference, is interesting: if the blue dot is at maximum speed (near or at the origin), the red dot reverses speed (its speed is, therefore, (almost) nil), and vice versa. The metaphor of our frictionless V-2 engine, our *perpetuum mobile*, comes to mind once more.

The question is: **what’s going on, really?**

My answer is: I don’t know. I do think that, somehow, energy should be thought of as some two-dimensional oscillation of *something* – something which we refer to as *mass*, but we didn’t define mass very clearly either. It also, *somehow*, combines linear and rotational motion. Each of the two dimensions packs half of the energy of the particle that is being represented by our wavefunction. It is, therefore, only logical that the physical unit of both is to be expressed as a force over some distance – which is, effectively, the physical dimension of energy – or the rotational equivalent of them: *torque* over some *angle*. Indeed, the analogy between linear and angular movement is obvious: the *kinetic *energy of a rotating object is equal to K.E. = (1/2)·I·ω^{2}. In this formula, I is the *rotational inertia* – i.e. the rotational equivalent of mass – and ω is the angular velocity – i.e. the rotational equivalent of *linear *velocity. Noting that the (average) kinetic energy in any system must be equal to the (average) potential energy in the system, we can add both, so we get a formula which is *structurally *similar to the E = m·*c*^{2} formula. But *is *it the same? Is the effective mass of some object the sum of an almost infinite number of *quanta* that incorporate some kind of *rotational *motion? And – if we use the right units – is the angular velocity of these infinitesimally small rotations effectively equal to the speed of light?

I am not sure. *Not at all*, really. But, so far, I can’t think of any explanation of the wavefunction that would make more sense than this one. I just need to keep trying to find better ways to *articulate *or *imagine *what might be going on. 🙂 In this regard, I’d like to add a point – which may or may not be relevant. When I talked about that guitar string, or the water wave, and wrote that each point on the string – or each water drop – just moves up and down, we should think of the physicality of the situation: when the string oscillates, its *length *increases. So it’s only because our string is flexible that it can vibrate between the fixed points at its ends. For a rope that’s *not *flexible, the end points would need to move in and out with the oscillation. Look at the illustration below, for example: the two kids who are holding rope must come closer to each other, so as to provide the necessary space inside of the oscillation for the other kid. 🙂The next illustration – of how water waves actually propagate – is, perhaps, more relevant. Just think of a two-dimensional equivalent – and of the two oscillations as being *transverse *waves, as opposed to longitudinal. See how string theory starts making sense? 🙂

The most fundamental question remains the same: what *is* it, *exactly*, that is oscillating here? What is the *field*? It’s always some force on some charge – but *what* charge, *exactly*? Mass? What *is* it? Well… I don’t have the answer to that. It’s the same as asking: what is *electric *charge, *really*? So the question is: what’s the *reality *of mass, of electric charge, or whatever other charge that causes a force to *act *on it?

If *you *know, please let *me *know. 🙂

**Post scriptum**: The fact that we’re talking some *two*-dimensional oscillation here – think of a surface now – explains the probability formula: we need to *square *the absolute value of the amplitude to get it. And normalize, of course. Also note that, when normalizing, we’d expect to get some factor involving π somewhere, because we’re talking some *circular *surface – as opposed to a rectangular one. But I’ll let *you *figure that out. 🙂