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Tag Archives: interpretation of the wavefunction
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 isâŚ WellâŚ Less frivolous. Unfortunately, the Copenhagen interpretation itself seems to be subject to interpretation.
One such interpretation may be referred to as radical skepticism â or radical empiricism[1]: we can only say something meaningful about SchrĂśdingerâs 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 donât make the observation. All we can do is describe its reality by a superposition of the two possible states: dead or alive. Thatâs Hilbertâs 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 â who may well be the most eminent representative of mainstream physics â thinks this epistemological position is nonsensical, and I fully agree with him:
âA 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.â (Feynmanâs Lectures, III-2-6)
So what is the mainstream physicistâs interpretation of the Copenhagen interpretation of quantum mechanics then? To fully answer that question, I should encourage the reader to read all of Feynmanâs Lectures on quantum mechanics. But then you are reading this because you donât want to do that, so let me quote from his introductory Lecture on the Uncertainty Principle: âMaking 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.â (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 toÂ hear it, he wraps up the philosophical discussion as follows: âWe 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.â In short, I think we can all agree that the cat is dead or alive, or that the tree is standing or not standingÂžregardless of the observer. Itâs a binary situation. Not something in-between. The box obscures our view. Thatâs all. There is nothing more to it.
Of course, in quantum physics, we donât study cats but look at the behavior of photons and electrons (we limit our analysis to quantum electrodynamics â so we wonât 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 â or the photon â before we observe it, or before we make any measurement. Think of the Stein-Gerlach experiment, which tells us that weâll always measure the angular momentum of an electron â along any axis we choose â as either +Ä§/2 or, else, as -Ä§/2. So whatâs 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. Itâs 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 â whizzing around in its own space â 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 â or torque â to change it. But in quantum physics, we have the Uncertainty Principle: some energy (force over a distance, remember) can be borrowed â so to speak â as long as itâs swiftly being returned â within the quantitative limits set by the Uncertainty Principle: ÎEÂˇÎt = Ä§/2.
Mainstream physicists â including Feynman â do not try to think about this. For them, the Stern-Gerlach apparatus is just like SchrĂśdingerâs box: it obscures the view. The cat is dead or alive, and each of the two states has some probability â but they must add up to one â 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 youâve seen this before:
|ĎâŞ = C_{up}|upâŞ + C_{down}|downâŞ
Itâs the so-called Dirac or bra-ket notation. C_{up} is the amplitude for the electron spin to be equal to +Ä§/2 along the chosen direction â 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 â and, likewise, C_{up} is the amplitude for the electron spin to be equal to -Ä§/2 (along the same direction, obviously). C_{up} and C_{up} will be functions, and the associated probabilities will vary sinusoidally â 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 â if thatâs what it is â does not feel like a model of a real electron. Itâs like reducing the cat in our box to the mentioned fuzzy state of being alive and dead at the same time. Letâs try to come up with something more exciting. đ
[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 Hilbertâs 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 youâre 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}Âˇexp(-iÎ¸_{i}), so they have an amplitude A_{i}Â and an argument Î¸_{i} = (E_{i}/Ä§)Âˇt â (p_{i}/Ä§)Âˇx. Letâs forget about uncertainty, so we can drop the index (i) and think of a geometric interpretation of AÂˇexp(-iÎ¸) = AÂˇe^{–i}^{Î¸}.
Here we have a weird thing: physicists think the minus sign in the exponent (-iÎ¸) should always be there: the convention is that we get the imaginary unit (i) by a 90Â° rotation of the real unit (1) â 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: AÂˇe^{–i}^{Î¸} and AÂˇe^{+i}^{Î¸}. 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 AÂˇe^{–i}^{Î¸} and AÂˇe^{+i}^{Î¸} as two-dimensional harmonic oscillators:
AÂˇe^{–i}^{Î¸} = cos(-Î¸) + iÂˇsin(-Î¸) = cosÎ¸ – iÂˇsinÎ¸
AÂˇe^{+i}^{Î¸} = cosÎ¸ + iÂˇsinÎ¸
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 electronâs charge may or may not explain all the mass of the electron (most physicists think it doesnât) but letâs assume it does for the sake of the model that weâre 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 weâll 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 wonât repeat ourselves here.
Figure 2: A perpetuum mobile?
Previously unrelated but structurally similar formulas may be related here:
- The energy of an oscillator: E =Â (1/2)ÂˇmÂˇa^{2}Ď^{2}
- Kinetic energy: E =Â (1/2)ÂˇmÂˇv^{2}
- The rotational (kinetic) energy thatâs stored in a flywheel: E = (1/2)ÂˇIÂˇĎ^{2} = (1/2)ÂˇmÂˇr^{2}ÂˇĎ^{2}
- Einsteinâs energy-mass equivalence relation: E =Â mÂˇc^{2}
Of course, we are mixing relativistic and non-relativistic formulas here, and thereâs the 1/2 factor â but these are minor issues. For example, we were talking not one but two oscillators, so we should add their energies: (1/2)ÂˇmÂˇa^{2}ÂˇĎ^{2} + (1/2)ÂˇmÂˇa^{2}ÂˇĎ^{2} = mÂˇa^{2}ÂˇĎ^{2}. Also, one can show that the classical formula for kinetic energy (i.e. E =Â (1/2)ÂˇmÂˇv^{2}) morphs into E =Â mÂˇc^{2} when we use the relativistically correct force equation for an oscillator. So, yes, our metaphor â or our suggested physical interpretation of the wavefunction, I should say â makes sense.
If you know something about physics, then you know the concept of the electromagnetic mass â its mathematical derivation, that is â gives us the classical electron radius, aka as the Thomson radius. Itâs 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 thatâs the radius of an electron orbital, roughly speaking â or the size of a hydrogen atom, I should say.]
Now, if we combine the E =Â mÂˇa^{2}ÂˇĎ^{2} and E =Â mÂˇc^{2} equations, then aÂˇĎ 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 = Ä§/(mÂˇc), and Ď (the (angular) frequency of the oscillation) by using the Planck relation (Ď = E/Ä§): Â Â Â Â
aÂˇĎ = [Ä§/(mÂˇc)]Âˇ[E/Ä§] = E/(mÂˇc) = mÂˇc^{2}/(mÂˇc) = 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 â which makes sense â and the radial velocity of our spinning charge is the speed of light â which may or may not make sense. Of course, we need an explanation of why this spinning charge doesnât radiate its energy away â but then we donât have such explanation anyway. All we can say is that the electron charge seems to be spinning in its own space â that itâs racing along a geodesic. Itâs just like mass creates its own space here: according to Einsteinâs general relativity theory, gravity becomes a pseudo-forceâliterally: no real force. How? I am not sure: the model here assumes the medium â empty space â is, somehow, perfectly elastic: the electron constantly borrows energy from one direction and then returns it to the other â so to speak. A crazy model, yes â 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, letâs think about an interpretation of the results of the Stern-Gerlach experiment.
Precession
A spinning charge is a tiny magnet â and so itâs got a magnetic moment, which we need to explain the Stern-Gerlach experiment. But it doesnât explain the discrete nature of the electronâs angular momentum: itâs either +Ä§/2 or -Ä§/2, nothing in-between, and thatâs 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 donât.
Indeed, the corollary of the above-mentioned binary value of the angular momentum is that the angular momentum â or the electronâs spin â 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 Feynmanâs 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 â 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 âupâ and âdownâ, âleftâ and ârightâ, and âin frontâ and âbehindâ 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 45Â°. Now, if the value of its z-component (i.e. its projection on the z-axis) is to be equal to Ä§/2, then the magnitude of J itself should be larger. To be precise, it should be equal to Ä§/â2Â â 0.7ÂˇÄ§ (just apply Pythagorasâ Theorem). Is that value compatible with our flywheel model?
Maybe. Letâs see. The classical formula for the magnetic moment is Îź = IÂˇA, 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, butâŚ WellâŚ Letâs 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 (2ĎÂˇa) divided by the tangential velocity (v_{t}). Now, we suggest v_{t} = rÂˇĎ = aÂˇĎ = c, and the circumference of the loop is 2ĎÂˇa. For a, we still use the Compton radius a = Ä§/(mÂˇc). Now, the formula for the area is A = ĎÂˇa^{2}, so we get:
Îź = IÂˇA = [q_{e}/T]ÂˇĎÂˇa^{2} = [q_{e}Âˇc/(2ĎÂˇa)]Âˇ[ĎÂˇa^{2}] = [(q_{e}Âˇc)/2]Âˇa = [(q_{e}Âˇc)/2]Âˇ[Ä§/(mÂˇc)] = [q_{e}/(2m)]ÂˇÄ§
In a classical analysis, we have the following relation between angular momentum and magnetic moment:
ÎźÂ = (q_{e}/2m)ÂˇJ
Hence, we find that the angular momentum J is equal to Ä§, so thatâs twice the measured value. Weâve got a problem. We would have hoped to find Ä§/2 or Ä§/â2. Perhaps itâsÂ because a = Ä§/(mÂˇc) is the so-called reduced Compton scattering radius…
WellâŚ No.
Maybe weâll find the solution one day. I think itâs already quite nice we have a model thatâs accurate up to a factor of 1/2 or 1/â2. đ
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 âi would do just as an imaginary unit as i, because theÂ definitionÂ of the imaginary unit says the only requirement is that its square has to be equal toÂ â1, and (âi)^{2}Â is also equal toÂ â1.
So we actually haveÂ two imaginary units: i and âi. However, physicists stubbornly think there is only one direction for measuring angles, and that is counter-clockwise.Â That’s a mathematical convention, Professor: it’s something in your head only.Â It is not real.Â Nature doesn’t care about our conventions and, therefore, I feel the spin ‘up’ versus spin ‘down’ should correspond to the twoÂ mathematicalÂ 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. đ
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. đ It’s… Well… I admit it: it’s just a rant. đ [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. đ ] 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:
âAtomic behaviorÂ appears peculiar and mysterious to everyoneâboth to the novice and to the experienced physicist.Â Even the experts do not understand it the way they would like to.â
SoâŚ WellâŚ If youâd 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 donât really understand it themselves. đ
Take the example of aÂ physical state, which is represented by aÂ state vector, which we can combine and re-combine using the properties of an abstractÂ Hilbert space.Â 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, theÂ baseÂ of our reference frame doesn’t matter: because we’re usingÂ realÂ vectors (such as the electric of magnetic field vectors E and B), our orientation vis-ĂĄ-vis the object – theÂ line of sight, so to speak – doesn’t matter.
In contrast, in quantum mechanics, it does: SchrĂśdinger’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 anyÂ geometricÂ 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 ifÂ bothÂ 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 720Â° 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?Â Yes.Â TheÂ wavefunction is a ‘flat’ concept: it has two dimensions only, unlike theÂ realÂ 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 positionÂ vis-ĂĄ-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’sÂ ourÂ reality. However, because that reality changes as per our line of sight, physicists keep saying the wavefunction (orÂ das Ding an sichÂ itself) is, somehow, not real.
Frankly,Â 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:Â âThese philosophersÂ areÂ alwaysÂ with us, struggling in the periphery toÂ tryÂ toÂ tell us something, but they never really understand the subtleties and depth of the problem.â (Feynman’s Lectures, Vol. I, Chapter 16)
Now, IÂ loveÂ Feynman’s Lectures but…Â 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-jumboÂ for the poor uninitiated. Consistent mumbo-jumbo, for sure, but mumbo-jumbo nevertheless. đ So, yes, I do think we need to re-invent quantum math. đ 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 (SchrĂśdinger published his equation in 1926), then… Well… Then the physicists have a problem. Not the philosophers. đ 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. đ Perhaps quaternions are the answer but… Well… I don’t think so either – for reasons I’ll explain later. đ
Post scriptum: There are many nice videos on Dirac’s belt trick or, more generally, on 720Â° symmetries, but this links to one I particularly like. It clearly shows that the 720Â° 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 360Â°!Â That’s a very different thing than just looking at it, walking around it, and then looking at it again. That explains the 720Â° 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.Â That’s why I think the quantum-mechanical description is defective.
Should we reinvent wavefunction math?
Preliminary note: This post may cause brain damage. đ 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. đŚ But if you have… Then this should be very interesting. Let’s go. đ
If you know one or two things about quantum math – SchrĂśdinger’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 somethingÂ realÂ 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 ofÂ SchrĂśdinger’s equationÂ (Lectures, III-16-1):
âWe can think of SchrĂśdingerâs equation as describing the diffusion of the probability amplitude from one point to the next. [âŚ] 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 SchrĂśdingerâs equation are complex waves.â
Feynman further formalizes this in his Lecture on Superconductivity (Feynman, III-21-2), in which he refers to SchrĂśdingerâs equation as the âequation for continuity of probabilitiesâ. His analysis there is centered on the local conservation of energy, which makes me think SchrĂśdingerâs equation might be an energy diffusion equation. I’ve written about thisÂ ad nauseamÂ 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 SchrĂśdinger’s equation in its non-relativistic form, which is an approximation that is good enough)Â isÂ 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.Â HowÂ exactly? Well… You know we get all of those complicated electron orbitals (i.e. the various wave functionsÂ that satisfy the equation) out of SchrĂśdinger’s differential equation. We can think of these solutions as (complex)Â standing waves. They basically represent someÂ equilibriumÂ situation, and the main characteristic of each is theirÂ 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 whatÂ IÂ want to do!) – the real and imaginary component of a wavefunction will be perpendicular to each other. Let me copy the animation for theÂ elementaryÂ wavefunction Ď(Î¸) =Â aÂˇe^{âiâÎ¸}Â =Â aÂˇe^{âiâ(E/Ä§)Âˇt}Â = aÂˇcos[(E/Ä§)ât]Â âÂ iÂˇaÂˇsin[(E/Ä§)ât] once more:
So… Well… That 90Â° 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 â with no external force fields acting on it â there is no potential (U = 0) and, therefore, the VĎ term – which is just the equivalent of the theÂ sinkÂ or sourceÂ term S in the diffusion equation – disappears. Therefore, SchrĂśdingerâs equation reduces to:
âĎ(x, t)/ât =Â iÂˇ(1/2)Âˇ(Ä§/m_{eff})Âˇâ^{2}Ď(x, t)
Now, the key difference with the diffusion equation – let me write it for you once again: âĎ(x, t)/ât = DÂˇâ^{2}Ď(x, t) – is thatÂ SchrĂśdingerâs equation gives usÂ twoÂ equations for the price of one. Indeed, because Ď is a complex-valued function, with aÂ realÂ and anÂ imaginaryÂ part, we get the following equations:
- Re(âĎ/ât) = â(1/2)Âˇ(Ä§/m_{eff})ÂˇIm(â^{2}Ď)
- Im(âĎ/ât) = (1/2)Âˇ(Ä§/m_{eff})ÂˇRe(â^{2}Ď)
Huh?Â Yes. These equations are easily derived from noting that two complex numbers a +Â iâb and c +Â iâd are equal if, and only if, their real and imaginary parts are the same. Now, the âĎ/ât =Â iâ(Ä§/m_{eff})ââ^{2}Ď equation amounts to writing something like this: a +Â iâb =Â iâ(c +Â iâd). Now, remembering thatÂ i^{2}Â = â1, you can easily figure out thatÂ iâ(c +Â iâd) =Â iâc +Â i^{2}âd = â d +Â iâc. [Now that we’re getting a bit technical, let me note that theÂ 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.] đ OK.Â Onwards !Â đ
The equations above make me think of the equations for an electromagnetic wave in free space (no stationary charges or currents):
- âB/ât = ââĂE
- âE/ât =Â c^{2}âĂB
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. đ Something like that. đ I will let you look up how it goes,Â exactly. The principle is clear enough.Â 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 asÂ vectors, just like E and B? I mean… Representing them as vectorsÂ (I meanÂ realÂ vectors here – something with a magnitude and a direction in aÂ realÂ space – as opposed to state vectors from the Hilbert space) wouldÂ showÂ they are real, and there would be no need for cumbersome transformations when going from one representationalÂ baseÂ 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 theÂ realÂ thing, aren’t they? đ
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:
- EÂ is measured inÂ newton per coulombÂ (N/C).
- BÂ is measured in newton per coulomb divided by m/s, so that’s (N/C)/(m/s).
The weird dimension of BÂ 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(vĂB)
Of course, it is onlyÂ oneÂ force (one and the same physical reality), as evidenced by the fact that we can write B as the following vector cross-product: BÂ = (1/c)âe_{x}ĂE, withÂ e_{x}Â the unit vector pointing in the 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)âe_{x}Ă operator, which amounts to a rotation by 90 degrees, with the s/m dimension. Now, multiplication by i also amounts to a rotation by 90Â° degrees. Hence, if we can agree on a suitable convention for the directionÂ of rotation here,Â we may boldly write:
BÂ = (1/c)âe_{x}ĂE = (1/c)âiâE
This is, in fact, what triggered my geometric interpretation of SchrĂśdingerâs 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 anÂ electromagnetic wave, the magnitudes of E and B reach their maximum, minimum and zero point simultaneouslyÂ (as shown below). So theirÂ phaseÂ 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. đ The question here is the following: could we possibly think of a new formulation of SchrĂśdinger’s equation – usingÂ vectors (again,Â realÂ vectors – not these weirdÂ state vectors)Â rather than complex algebra?
I think we can, but then I wonder why theÂ inventorsÂ of the wavefunction – Heisenberg, Born, Dirac, and SchrĂśdinger himself, of course – never thought of that. đ
Hmm… I need to do some research here. đ
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? đ
P.S. 2: I also published this post on my new blog:Â https://readingeinstein.blog/. However, I thought the followers of this blog should get it first. đ
Wavefunctions, perspectives, reference frames, representations and symmetries
Ouff ! This title is quite a mouthful, isn’t it? đ 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 (orÂ physical) interpretations of the wavefunction is the fact that a wavefunction does not seem to obey the classical 360Â° 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 (orÂ representation, tout court)Â 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 frameâand so that explains the weird stuff (like that 720Â° 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 paperÂ first.
The reality of directions
Huh? TheÂ realityÂ of directions? Yes. I warned you. This post may cause brain damage. đÂ The whole argument revolves around a thoughtÂ experimentâbut one whose results have been verified in zillions of experiments in university student labs so… Well… We do notÂ doubt the results and, therefore, we do not doubt the basic mathematical results: we just want to try to understandÂ 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 âimprovedâ Stern-Gerlach apparatuses. The apparatus basically splits and then re-combines the two new beams along theÂ z-axis. It is also possible to block one of the beams, so we filter out only particles with their spinÂ upÂ or, alternatively, with their spinÂ down. Spin (or angular momentum or the magnetic moment) as measured along theÂ z-axis, of courseâI should immediately add: we’re talking theÂ z-axis of the apparatus here.
The two situations involve a different relative orientation of the apparatuses: in (a), the angle is 0Â°, while in (b) we have a (right-handed) rotation of 90Â° about the z-axis. He then provesâusing geometry and logic onlyâthat the probabilities and, therefore, the magnitudes of the amplitudes (denoted byÂ C_{+} and C_{â} and Câ_{+} and Câ_{â} in the S and T representation respectively) must be the same, but the amplitudes must have different phases, notingâin his typical style, mixing academic and colloquial languageâthat âthere must be some way for a particle to tell that it has turned a corner in (b).â
The various interpretations of what actually happens here may shed some light on the heated discussions on the reality of the wavefunctionâand 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 wavefunctionâwhich captures a continuum of possible states, so to speakâis introduced only later. However, we may look at the amplitude for a particle to be in theÂ up– or down-state as a wavefunction and, therefore (but do note that’s my humble opinion once more), the analysis is actuallyÂ notÂ 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 (90Â°), we know that the amplitudes are given by Câ_{+} = e^{i}^{âĎ/2}âC_{+} = e^{ i}^{âĎ/4}âC_{+} and Câ_{â} = e^{âiâĎ/2}âC_{+} = e^{â iâĎ/4}âC_{â} respectively (the amplitude to go from the down to the up state, or vice versa, is zero). Hence, yes, weânotÂ the particle, Mr. Feynman!âknowÂ that, in (b), the electron has, effectively, turned a corner.
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 â(a) and (b) are differentâ, âthe probabilities are the sameâ. 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 turnâso it is just going in some other direction. Thatâs 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 mightâor mustâhave 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 âmightâ and âmustâ 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…Â Wait a minute…Â Nothing is obvious in quantum mechanics, right? How can weÂ experimentally confirmÂ thatÂ they are different?
Huh?Â I must be joking, right? You canÂ seeÂ they are different, right? No.Â I am not joking. In physics, two things are different if we get differentÂ measurementÂ 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 thingâsame probabilities, remember?âwhy are they different? Think of this: if we look at the two beam splitters as one singleÂ tube (anÂ 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 sameÂ even when it takes a turn, we could say the tube is still the same, despite us having wrenched it over a 90Â° corner.
Now, I am sure you think I’ve just gone nuts, but just tryÂ to stick with me a little bit longer. Feynman actually acknowledges the same: we need to experimentallyÂ proveÂ (a) and (b) are different. He does so by getting aÂ thirdÂ apparatus in (U), as shown below, whose relative orientation to T is the same in both (a) and (b), so there is no difference there.
Now, the axis ofÂ UÂ is not theÂ z-axis: it is theÂ x-axis in (a), and theÂ y-axis in (b). So what? Well… I will quote Feynman hereânot (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 ofÂ SÂ which produces a pure +xÂ state. Such particles would be split into +z andÂ âz intoÂ beams inÂ S,Â but the two beams would be recombined to give aÂ +xÂ state again at P_{1}âthe exit ofÂ S.Â The same thing happens again inÂ T.Â If we followÂ TÂ by a third apparatusÂ U,Â whose axis is in the +xÂ direction and, as shown in (a), all the particles would go into the +Â beam ofÂ U.Â Now imagine what happens ifÂ TÂ and UÂ are swung aroundÂ togetherÂ by 90Â°Â to the positions shown in (b).Â Again, theÂ TÂ apparatus puts out just what it takes in, so the particles that enterÂ UÂ are in a +xÂ stateÂ with respect toÂ S,Â which is different. By symmetry, we would now expect only one-half of the particles to get through.”
I should note that (b) shows theÂ UÂ apparatus wide open so… Well… I must assume that’s a mistake (and should alert the current editors of the LecturesÂ to it): Feynman’s narrative tells us we should also imagine it with theÂ minus channel shut. InÂ thatÂ case, it should, effectively, filter approximately half of the particles out, while they all get through in (a). So that’s aÂ measurementÂ result which shows the direction, as weÂ seeÂ it, makes a difference.
Now, Feynman would be very angry with meâbecause, as mentioned, he hates philosophersâbut I’d say: this experiment proves that a direction is something real. Of course, the next philosophical question then is: whatÂ isÂ a direction? I could answer this by pointing to the experiment above: a direction is something that alters the probabilities between the STU tube as set up in (a) versus the STU tube in (b). In factâbut, I admit, that would be pretty ridiculousâwe could use the varying probabilities as we wrench this tube over varying angles toÂ 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. đ 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 720Â° 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 frameâorÂ representation, as it’s referred to in quantum mechanicsâto another?
Well… I should immediately correct myself here: a reference frame and a representation are two different things. They areÂ relatedÂ but… Well… Different… Quite different. Not same-same but different. đ I’ll explain why later. Let’s go for it.
Before talking representations, let us first think about what we reallyÂ 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 ourÂ 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 theÂ shape, if that sounds easier to you) of the measurement apparatusÂ (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 usâ230 years agoâthat 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.Â The apparatus gives us two directions:
(1) TheÂ upÂ direction, whichÂ weÂ associate with theÂ positive direction of theÂ z-axis, and
(2) the direction of travel of our particle, whichÂ we associateÂ with the positive direction of theÂ y-axis.
Now, if we have two axes, then the third axis (theÂ x-axis) will be given by the right-hand rule, right? So we may say the apparatus gives us the reference frame. Full stop.Â So… Well… Everything is relative? Is this reference frame relative? Are directions relative? That’s what you’ve been told, but think about this:Â relativeÂ to what?Â Here is where the object meets the subject. What’s relative? What’s absolute?Â Frankly, I’ve started to think that, in this particular situation, we should, perhaps, not use these two terms. I am notÂ saying thatÂ 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 theÂ realityÂ that we are observing, right? And the third one… Well… That’s given by our perspectiveâby our right-hand rule, which is… Well… OurÂ right-hand rule.
Of course, now you’ll say: if you think that ârelativeâ and âabsoluteâ are ambiguous terms and that we, therefore, may want to avoid them a bit more, then ârealâ and its opposite (unreal?) are ambiguous terms too, right? WellâŚ Maybe. What language would youÂ suggest? đ Just stick to the story for a while. I am not done yet. So… Yes… WhatÂ isÂ theirÂ reality?Â 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, asymmetries in space), I found myself walking around the table with some asymmetrical everyday objects and papers with arrows and clocks and other stuff on itâeffectively analyzing what right-hand screw, thumb or grip rules actuallyÂ mean. đ
So… Well… I want you to distinguishâjust for a whileâbetween the notion of a reference frame (think of the x–y–z reference frame that comes with the apparatus) and yourÂ 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 sideâwhich, if you think about it, you can only defineÂ in terms of the various positive and negative directions of the various axes. đÂ If you think this is getting ridiculous… Well… Don’t. Feynman himselfÂ 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 theÂ axesÂ (i.e. the reference frame) that comes with it. You don’t believe me? This is theÂ very first illustration of hisÂ LectureÂ 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 positiveÂ y-directionâso thatâs the direction in which our particle is movingâthen we might imagine how it would look like whenÂ weÂ would make a 180Â°Â 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 comingÂ towardsÂ us, out of the apparatus.
What happensâbut that’s not scientific language, of courseâis that left becomes right, and right becomes left. Top still is top, and bottom is bottom. We are looking now in theÂ negativeÂ y-direction, and the positive direction of the x-axisâwhich pointed right when we were looking in the positiveÂ y-directionânow points left. I see you nodding your head nowâbecause you’ve heard about parity inversions, mirror symmetries and what have youâand 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 theÂ 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 toÂ this story, which is about transformations of amplitudes (or wavefunctions). [If you really want to knowâbut I know this sounds counterintuitiveâthe 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 mentallyÂ 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 onlyÂ apparent. It’s a common misconception, and… Well… I’ll let you figure this out yourself. I need to move on.]Â Just note the following:
- TheÂ xyzÂ 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. OurÂ perspectiveÂ 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)âa cosine and sine function respectivelyâthen our change in perspectiveÂ might, effectively, mess up our convention for measuring angles.
I am not saying itÂ 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 counterclockwise, right? As shown below… But… Well… If the thing below would be some funny clock going backwardsâyou’ve surely seen them in a bar or so, right?âthen… Well… If they’d be transparent, and you’d go around them, you’d see them as going… Yes… Clockwise. đ [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, ifÂ we wouldÂ assume this clock represents something realâand, of course, I am thinking of theÂ elementary wavefunctionÂ e^{i}^{Î¸}Â =Â cosÎ¸ +Â iÂˇsinÎ¸ nowâthen… 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…Â Think! What’s your answer? Give me the formula!Â đ
[…]
We’d see it asÂ e^{âi}^{Î¸}Â =Â cos(âÎ¸) +Â iÂˇsin(âÎ¸) =Â cosÎ¸ âÂ iÂˇsinÎ¸, right? The hand of our clock now goes clockwise, so that’s theÂ oppositeÂ direction of our convention for measuring angles. Hence, instead ofÂ e^{i}^{Î¸}, we writeÂ e^{âi}^{Î¸}, right? So that’s the complex conjugate. So we’ve got a differentÂ imageÂ of the same thing here. Not good. Not good at all.
You’ll say: so what? We can fix this thing easily, right?Â YouÂ don’t need the convention for measuring angles or for the imaginary unit (i) here.Â 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).Â To hell with the counterclockwise convention for measuring angles!
You are right. WeÂ couldÂ use the right-hand rule more consistently. We could, in fact, use it as anÂ alternativeÂ convention for measuring angles: we could, effectively, measure them clockwise or counterclockwise depending on the direction of our particle.Â But… Well… The fact is:Â we don’t. We do not use that alternative convention when we talk about the wavefunction. Physicists do use theÂ counterclockwiseÂ convention all of the time and just jot down these complex exponential functions and don’t realize that,Â if they are to represent something real, ourÂ perspectiveÂ on the reference frame matters. To put it differently, theÂ directionÂ in which we are looking at things matters! Hence, the direction is not…Â Well… I am tempted to say… NotÂ relative at all but then… Well… We wanted to avoid that term, right? đ
[…]
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 symmetryâorÂ asymmetry, 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 forceÂ 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 massÂ 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, âwhen you do find the electron some place, the entire charge is there.â Of course, we cannot prove anything here, but our elegant yet simple derivation of the Compton radius of an electron is… Well… Just nice. đ
- 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 compriseÂ the direction of its linear motion. Hence, unlike an electromagnetic wave, theÂ planeÂ of the two-dimensional oscillation (so that’s the polarization plane, really) cannotÂ 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.Â The direction of the angular momentum (and the magnetic moment) of an electronâor, to be precise, its component as measured in the direction of the (inhomogeneous) magnetic field through which our electron is travelingâcannotÂ be parallel to the direction of motion. On the contrary, it must be perpendicularÂ 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. theÂ planeÂ of the polarization)Â has toÂ compriseÂ 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 explanationâcombined with the quantization hypothesisâgoes a long way in explaining this: an object with an angular momentumÂ JÂ and a magnetic momentÂ ÎźÂ that is not exactly parallel to some magnetic fieldÂ B, willÂ notÂ line up: it willÂ precessâand, as mentioned, the quantization of angular momentum may well explain the rest.Â [Well… Maybe… We haveÂ detailed our attempts in this regard in various posts on this (just search for spinÂ orÂ angular momentumÂ 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 theÂ up-direction only, then it will be spinningÂ clockwise if its angular momentum is down (so itsÂ magnetic moment isÂ up). Conversely, it will be spinningÂ counterclockwise if its angular momentum isÂ up. Let us take theÂ 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. đ 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 asâI am not ashamed to admit thisâI 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 someÂ positiveÂ angle.Â Walking around itâin whatever direction, clockwise or counterclockwiseâdoesn’t change the counterclockwise direction of our… Well… That weird object that mightâjust mightârepresent 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 wrenchedÂ e^{i}^{Î¸}Â =Â cosÎ¸ +Â iÂˇsinÎ¸ function, right? The x- andÂ y-axesÂ 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 180Â° angle so we’re looking in theÂ negativeÂ 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 theÂ 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 theÂ y-axis to point away, rather than towards us. In short, we have a transformation of the reference frame here:Â z’ =Â z,Â y’ = âÂ y, andÂ x’ =Â âÂ x. Mind you, this is still a regular right-handed reference frame. [That’s the difference with aÂ mirrorÂ image: aÂ mirroredÂ right-hand reference frame is no longer right-handed.]Â So, in our new reference frame, that we choose to coincide with ourÂ perspective,Â we will now describe the same thing as someÂ âcosÎ¸ âÂ iÂˇsinÎ¸ =Â âe^{i}^{Î¸}Â function. Of course,Â âcosÎ¸ =Â cos(Î¸ +Â Ď) andÂ âsinÎ¸ =Â sin(Î¸ +Â Ď) so we can write this as:
âcosÎ¸ âÂ iÂˇsinÎ¸ =Â cos(Î¸ +Â Ď) +Â iÂˇsinÎ¸ =Â e^{i}^{Âˇ(}^{Î¸+Ď)}Â =Â e^{i}^{Ď}Âˇe^{i}^{Î¸}Â = âe^{i}^{Î¸}.
Sweet ! But… Well… First note this isÂ notÂ the complex conjugate:Â e^{âi}^{Î¸}Â =Â cosÎ¸ âÂ iÂˇsinÎ¸Â â Â âcosÎ¸ âÂ iÂˇsinÎ¸ =Â âe^{i}^{Î¸}. 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. đ And, yes, let me lighten up the discussion with that painting here. đ We need to haveÂ someÂ fun while torturing our brain, right?
So, because we assume the plane of polarization is different, we get anÂ âe^{i}^{Î¸}Â function instead of aÂ e^{âi}^{Î¸}Â function.
Let us now think about the e^{i}^{Âˇ(}^{Î¸+Ď)}Â function. It’s the same asÂ âe^{i}^{Î¸}Â but… Well… We walked around theÂ z-axis taking a full 180Â° turn, right? So that’s Ď in radians. So that’s the phase shiftÂ here. Hey!Â Try the following now. Go back and walk around the apparatus once more, but letÂ 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 theÂ z-axis (which points out of this page, toward you, as you are looking at this), let’s say by some small angleÂ Îą. So we rotate the reference frame about theÂ z-axis byÂ Îą and… Well… Of course, ourÂ e^{i}^{Âˇ}^{Î¸}Â now becomes anÂ ourÂ e^{i}^{Âˇ(}^{Î¸+Îą)}Â function, right? We’ve just derived the transformation coefficient for a rotation about theÂ z-axis, didn’t we? It’s equal toÂ e^{i}^{Âˇ}^{Îą}, right? We get the transformed wavefunction in the new reference frame by multiplying the old one byÂ e^{i}^{Âˇ}^{Îą}, right? It’s equal toÂ e^{i}^{Âˇ}^{Îą}Âˇe^{i}^{Âˇ}^{Î¸}Â =Â e^{i}^{Âˇ(}^{Î¸+Îą)}, right?
Well…
[…]
No. The answer is: no. TheÂ transformation coefficient is notÂ e^{i}^{Âˇ}^{Îą}Â butÂ e^{i}^{Âˇ}^{Îą/2}. So we get an additional 1/2 factor in theÂ phase shift.
Huh?Â Yes.Â That’s what it is: when we change the representation, by rotating our apparatus over some angle Îą 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 onlyÂ half ofÂ the rotation angle only.
Huh?Â Yes. It’s even weirder than that. For a spin downÂ electron, the transformation coefficient is e^{âiÂˇ}^{Îą/2}, so we get an additional minus sign in the argument.
Huh?Â 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…Â Hey! Wait a minute! That’s it, right?Â
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:
OurÂ e^{i}^{Âˇ}^{Îą}Â coefficient describes a rotation of the reference frame. In contrast, theÂ e^{i}^{Âˇ}^{Îą/2}Â andÂ e^{âiÂˇ}^{Îą/2}Â coefficients describe what happens when we rotate the T apparatus! Now thatÂ is a very different proposition.Â
Right! You got it! RepresentationsÂ and reference frames are different things.Â QuiteÂ different, I’d say: representations areÂ real, reference frames aren’tâbut then you don’t like philosophical language, do you? đÂ But think of it. When we just go about theÂ z-axis, a full 180Â°, but we don’t touch thatÂ T-apparatus, we don’t changeÂ 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 theÂ z-axis, a full 180Â°, 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 frameâfrom xyz to x’y’z’ to be precise: we doÂ not changeÂ the representation.
In contrast, when we rotate theÂ TÂ apparatus over a full 180Â°, 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 throughÂ S, and now it goes in the opposite directionârelative to the direction it was going in S, that is.
So what happens,Â really, when weÂ change the representation, rather than the reference frame? Well… Let’s think about that. đ
Quantum-mechanical weirdness?
The transformation matrix for the amplitude of a system to be in anÂ upÂ orÂ downÂ state (and, hence, presumably, for a wavefunction) for a rotation about theÂ z-axis is the following one:
Feynman derives this matrix in a rather remarkable intellectualÂ tour de forceÂ in the 6th of hisÂ 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 howÂ IÂ 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? đ So… Well… Because heÂ knowsâfrom experimentâthat the coefficient isÂ e^{i}^{Âˇ}^{Îą/2}Â instead of e^{i}^{Âˇ}^{Îą}, he just says the phase shiftâwhich he denotes by Îťâmust be someÂ proportionalÂ to the angle of rotationâwhich he denotes byÂ Ď rather than Îą (so as to avoid confusion with the EulerÂ angleÂ Îą). So he writes:
Îť =Â mÂˇĎ
Initially, he also tries the obvious thing: m should be one, right? SoÂ Îť = Ď, 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:
“SupposeÂ TÂ is rotated byÂ 360Â°; then, clearly, it is right back at zero degrees, and we should haveÂ Câ_{+} = C_{+}Â andÂ Câ_{â} =Â C_{â}Â or,Â what is the same thing,Â e^{i}^{ÂˇmÂˇ2Ď}Â = 1. We get m =Â 1. [But no!]Â This argument is wrong!Â To see that it is, consider thatÂ TÂ is rotated byÂ 180Â°. If mÂ were equal to 1, we would have Câ_{+} =Â e^{i}^{ÂˇĎ}C_{+}Â = âC_{+}Â and Câ_{â} =Â e^{â}^{i}^{ÂˇĎ}C_{â}Â =Â âC_{â}. [Feynman works with statesÂ here, instead of the wavefunction of the particle as a whole. I’ll come back to this.] However, this is just theÂ originalÂ state all over again.Â BothÂ amplitudes are just multiplied byÂ â1Â which gives back the original physical system. (It is again a case of a common phase change.) This means that if the angle betweenÂ TÂ andÂ SÂ is increased to 180Â°, the system would be indistinguishable from the zero-degree situation, and the particles would again go through the (+)Â state of theÂ UÂ apparatus. AtÂ 180Â°, though, the (+)Â state of theÂ UÂ apparatus is theÂ (âx)Â state of the originalÂ SÂ apparatus. So a (+x)Â state would become aÂ (âx)Â state. But we have done nothing toÂ changeÂ the original state; the answer is wrong. We cannot haveÂ m = 1.Â We must have the situation that a rotation byÂ 360Â°, andÂ no smaller angleÂ reproduces the same physical state. This will happen ifÂ m = 1/2.”
The result, of course, is this weird 720Â° symmetry. While we get the same physics after a 360Â° rotation of the T apparatus, we doÂ notÂ get the same amplitudes. We get the opposite (complex) number:Â Câ_{+} =Â e^{i}^{Âˇ2Ď/2}C_{+}Â = âC_{+}Â and Câ_{â} =Â e^{â}^{i}^{Âˇ2Ď/2}C_{â}Â =Â âC_{â}. That’s OK, because… Well… It’s aÂ commonÂ phase shift, so it’s just like changing the origin of time. Nothing more. Nothing less. Same physics. Same reality. But… Well…Â Câ_{+} â Â âC_{+}Â andÂ Câ_{â} â Â âC_{â}, right? We only get our original amplitudes back if we rotate theÂ T apparatus two times, so that’s by a full 720 degreesâas opposed to the 360Â° we’d expect.
Now, space is isotropic, right? So this 720Â° 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 orÂ âx, or t and ât.Â I’ve said this a couple of times already again, and I’ll keep saying it many times more:Â NatureÂ surely can’t be bothered by how we measure stuff, right? In the positive or the negative directionâthat’s just our choice, right?Â OurÂ convention. So… Well… It’s just like thatÂ âe^{i}^{Î¸}Â function we got when looking at theÂ same experimental set-up from the other side: ourÂ e^{i}^{Î¸}Â and âe^{i}^{Î¸}Â functions didÂ notÂ describe a different reality. We just changed our perspective. TheÂ reference frame. As such, the reference frame isn’tÂ real. The experimental set-up is. AndâI know I will anger mainstream physicists with thisâtheÂ representationÂ is. Yes. Let me say it loud and clear here:
A different representation describes a different reality.
In contrast, a different perspectiveâor a different reference frameâdoes not.
Conventions
While you might have had a lot of trouble going through all of the weird stuff above, the point is: it isÂ notÂ all that weird. WeÂ canÂ 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 formulaâe^{i}^{Î¸}Â =Â cosÎ¸ +Â iÂˇsinÎ¸âwould, one day, be used to representÂ something real: an electron, or any elementary particle, really. If he wouldÂ have known, I am sure he would have noted what I am noting here:Â NatureÂ can’t be bothered by our conventions. Hence, ifÂ e^{i}^{Î¸}Â represents something real, thenÂ e^{âi}^{Î¸}Â 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 notedâand, if he would have known about circularly polarized waves, probably agreed toâthatÂ alternative convention for measuring angles: we could, effectively, measure angles clockwise or counterclockwise depending on the direction of our particleâas opposed to Euler’s ‘one-size-fits-all’ counterclockwise convention. But so we didÂ notÂ adopt that alternative convention because… Well… We want to keep honoring Euler, I guess. đ
So… Well… If we’re going to keep honoring Euler by sticking to that ‘one-size-fits-all’ counterclockwise convention, then I doÂ believe thatÂ e^{i}^{Î¸}Â and e^{âi}^{Î¸}Â represent twoÂ differentÂ 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 seeÂ something different when they go through a Stern-Gerlach apparatus. So it’s not just some convention toÂ countÂ things like 0, 1, 2, etcetera versus 0,Â â1,Â â2 etcetera. It’s the same story again: different but relatedÂ mathematicalÂ notions are (often) related to different but relatedÂ physicalÂ possibilities. So… Well… I think that’s what we’ve got here.Â Think of it. Mainstream quantum math treats all wavefunctions as right-handed but… Well…Â A particle with up spin is a different particle than one withÂ downÂ spin, right? And, again,Â NatureÂ surely cannotÂ be bothered about our convention of measuring phase angles clockwise or counterclockwise, right? So… Well… Kinda obvious, right? đ
Let me spell out my conclusions here:
1. The angular momentum can be positive or, alternatively, negative: J = +Ä§/2 orÂ âÄ§/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 particleâthink of an electron, or whatever other particle you’d think ofâcomes in twoÂ variants:Â right-handed and left-handed. They will, therefore,Â either consist of (elementary) right-handed waves or,Â else, (elementary) left-handed waves. An elementary right-handed wave would be written as: Ď(Î¸_{i})Â = e^{i}^{Î¸i}Â = a_{i}Âˇ(cosÎ¸_{i} + iÂˇsinÎ¸_{i}). In contrast,Â an elementary left-handed wave would be written as: Ď(Î¸_{i})Â =Â e^{âi}^{Î¸i}Â = a_{i}Âˇ(cosÎ¸_{i} â iÂˇsinÎ¸_{i}).Â So that’s the complex conjugate.
So… Well… Yes, I think complex conjugates are not just someÂ mathematicalÂ notion: I believe they represent something real. It’s the usual thing:Â NatureÂ has shown us that (most) mathematical possibilities correspond to realÂ physical situations so… Well… Here you go. It is reallyÂ 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 differencesâdifferent polarization plane and dimensions and what have youâI’ve already summed those up, so I won’t repeat myself here.]Â The point is: ifÂ we have two differentÂ physicalÂ situations, we’ll want to have two different functions to describe it. Think of it like this: why would we haveÂ twoâyes, I admit, two relatedâamplitudes to describe the upÂ or downÂ state of the same system, but only one wavefunction for it?Â 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 (orÂ physical) interpretation of the wavefunction might be, it won’t be compatible with theÂ isotropyÂ of space. You cannot imagineÂ an object with a 720Â° symmetry. That’sÂ geometrically impossible.”
Well… Almost three years ago, I wrote the following on this blog: “As strange as it sounds, aÂ spin-1/2 particle needsÂ twoÂ full rotations (2Ă360Â°=720Â°) until it is again in the same state. Now, in regard to that particularity, youâll often read something like: âThere isÂ nothingÂ in our macroscopic world which has a symmetry like that.â Or, worse, âCommon sense tells us that something like that cannot exist, that it simply is impossible.â [I wonât quote the site from which I took this quotes, because it is, in fact, the site of a very respectable Â research center!]Â Bollocks!Â TheÂ Wikipedia article on spinÂ has this wonderful animation: look at how the spirals flip between clockwise and counterclockwise orientations, and note that itâs only after spinning a full 720 degrees that this âpointâ 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, weÂ canÂ 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 goodÂ image, I should say, because… Well… A representation is something real, remember? đ
Post scriptum (10 December 2017):Â Our flywheel model of an electron makes sense, but also leaves many unanswered questions. The most obvious one question, perhaps, is: why theÂ upÂ andÂ downÂ state only?
I am not so worried about that question, even if I can’t answer it right away because… Well… Our apparatusâthe way weÂ measureÂ realityâis set up to measure the angular momentum (or the magnetic moment, to be precise) in one direction only. If our electron isÂ capturedÂ by someÂ harmonicÂ (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 spinÂ upÂ situation (magnetic moment down) is quite peculiar: if our electron is aÂ mini-magnet, why would itÂ notÂ line up with the magnetic field? We understand the precession of a spinning top in a gravitational field, but…Â Hey… It’s actually not that different. Try to imagine some spinning top on the ceiling. đ I am sure we can work out the math. đ 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. đ […] At least, that’s why my instinct tells me. I admit I need to work out the math to convince you. đ
The second question is more important. If we just rotate the reference frame over 360Â°, we see the same thing: some rotating object which we, vaguely, describe by someÂ e^{+i}^{ÂˇÎ¸}Â functionâto be precise, I should say: by some Fourier sum of such functionsâor, if the rotation is in the other direction, by someÂ e^{âi}^{ÂˇÎ¸}Â function (again, you should read: aÂ FourierÂ 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 sameÂ 360Â°, we get aÂ differentÂ object: ourÂ e^{i}^{ÂˇÎ¸}Â andÂ e^{âi}^{ÂˇÎ¸}Â function (again: think of aÂ FourierÂ sum, so that’s a waveÂ packet, really) becomes aÂ âe^{Âąi}^{ÂˇÎ¸}Â thing. We get aÂ minusÂ sign in front of it.Â 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 fullÂ 360Â°, then you’ll end up where you were when you started and, importantly, you’ll see the same thing.Â ExactlyÂ the same thing: if I was anÂ e^{+i}^{ÂˇÎ¸}Â wave packet, I am still anÂ anÂ e^{+i}^{ÂˇÎ¸}Â wave packet now. OrÂ if I was an e^{âi}^{ÂˇÎ¸}Â wave packet, then I am still anÂ an e^{âi}^{ÂˇÎ¸}Â wave packet now. Easy. Logical. Obvious, right?
But so now we try something different:Â IÂ turn around, over a fullÂ 360Â° turn, and youÂ stay where you are. When I am back where I wasâlooking at you again, so to speakâthen… Well… I am not quite the same any more. Or… Well… Perhaps I am but youÂ seeÂ me differently. If I wasÂ e^{+i}^{ÂˇÎ¸}Â wave packet, then I’ve become aÂ âe^{+i}^{ÂˇÎ¸}Â wave packet now. Not hugely different but… Well… ThatÂ minusÂ sign matters, right? OrÂ If I wasÂ wave packet built up from elementaryÂ aÂˇe^{âi}^{ÂˇÎ¸}Â waves, then I’ve become aÂ âe^{âi}^{ÂˇÎ¸}Â wave packet now. What happened?
It makes me think of the twin paradox in special relativity. We know it’s aÂ paradoxâso that’s anÂ apparentÂ 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 someÂ force.
Can we relate this to the twin paradox? Maybe. Note that aÂ minusÂ sign in front of theÂ e^{âÂąi}^{ÂˇÎ¸}Â 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 180Â°: âcosÎ¸ =Â cos(Î¸ Âą Ď) andÂ âsinÎ¸ =Â sin(Î¸ Âą Ď). Now, adding or subtracting aÂ commonÂ phase factor to/from the argument of the wavefunction amounts toÂ changingÂ the origin of time. So… Well… I do think the twin paradox and this rather weird business of 360Â° and 720Â° symmetries are, effectively, related. đ
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. đ [OK. Poor joke. Acknowledged.]
Just to recap the essentials, I part ways with mainstream physicists in regard to theÂ interpretationÂ of the wavefunction. For mainstream physicists, the wavefunction is just some mathematical construct. NothingÂ real. Of course, I acknowledge mainstream physicists have very good reasons for that, but… Well… I believe that, if there is interference, or diffraction, thenÂ somethingÂ must be interfering, or something must be diffracting. I won’t dwell on this because… Well… I have done that too many times already. MyÂ hypothesisÂ is that the wavefunction is, in effect, aÂ rotatingÂ field vector, so itâs just like the electric field vector of a (circularly polarized) electromagnetic wave (illustrated below).
Of course, it must be different, and it is. First, theÂ (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 forceÂ per unit massÂ (as opposed to the force per unit charge dimension of the electric field vector). Of course, the newton/kg dimension reduces to the dimension of acceleration (m/s^{2}), so thatâs 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 doÂ notÂ involve any mass: theyâre just an oscillatingÂ field. Nothing more. Nothing less. Why would I believe there must still be some pointlike particle involved? Well…Â As Feynman puts it: âWhen you do find the electron some place, the entire charge is there.â (FeynmanâsÂ Lectures, III-21-4) So… Well… That’s why.
The third difference is one that I thought of only recently: theÂ planeÂ of the oscillation cannotÂ be perpendicular to the direction of motion of our electron, because then we canât 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. đ
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 toÂ imagineÂ it. That’s great mental exercise, so… Well… Just try it. đ
Let’s now think about rotating reference frames and transformations. If theÂ z-direction is the direction along which we measure the angular momentum (or the magnetic moment), then theÂ up-direction will be theÂ positiveÂ z-direction. We’ll also assume theÂ y-direction is the direction of travel of our elementary particleâand let’s just consider an electron here so we’re moreÂ real. đ 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 apparatusâwhich I’ll refer to as a beam splitterâillustrates this geometry.
So I think the magnetic momentâor the angular momentum, reallyâcomes from an oscillatory motion in the x– and y-directions. One is theÂ realÂ component (the cosine function) and the other is the imaginary component (the sine function), as illustrated below.Â
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 isÂ notÂ in theÂ xz-plane, but in theÂ 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-directionâso that’s the direction in which our particle is movingâ, then we might imagine how it would look like whenÂ weÂ would make a 180Â°Â 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.Â 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 theÂ z-axis as it is (pointing upwards), and we will also want to define theÂ x– andÂ y-axis using the familiar right-hand rule for defining a coordinate frame. So our newÂ x-axis and our newÂ y-axis will the same as the oldÂ x- andÂ y-axes but with the sign reversed. In short, we’ll have the following mini-transformation: (1)Â z‘ =Â z, (2) x’ = âx, and (3) y’ =Â ây.
So… Well… If we’re effectively looking at somethingÂ realÂ that was moving along theÂ y-axis, then it will now still be moving along the y’-axis, butÂ in theÂ negativeÂ direction. Hence, our elementary wavefunctionÂ e^{i}^{Î¸}Â = cosÎ¸ +Â iÂˇsinÎ¸ willÂ transformÂ intoÂ âcosÎ¸ âÂ iÂˇsinÎ¸ =Â âcosÎ¸ âÂ iÂˇsinÎ¸ =Â cosÎ¸ âÂ iÂˇsinÎ¸.Â 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 theÂ z-axis:
Now, ifÂ Ď is equal to 180Â° (so that’s Ď in radians), then theseÂ e^{i}^{Ď/2}Â andÂ e^{âi}^{Ď/2}/â2Â factors areÂ equal toÂ e^{i}^{Ď/2}Â =Â +iÂ andÂ e^{âi}^{Ď/2}Â = âiÂ respectively. Hence, ourÂ e^{i}^{Î¸}Â = cosÎ¸ +Â iÂˇsinÎ¸ becomes…
Hey ! Wait a minute ! We’re talking about twoÂ veryÂ different things here, right? TheÂ e^{i}^{Î¸}Â = cosÎ¸ +Â iÂˇsinÎ¸ is anÂ elementaryÂ wavefunction which, we presume, describes some real-life particleâwe talked about an electron with its spin in theÂ up-directionâwhile these transformation matrices are to be applied to amplitudes describing… Well… Either anÂ up– or a down-state, right?
Right. But… Well… Is itÂ so different, really? Suppose ourÂ e^{i}^{Î¸}Â = cosÎ¸ +Â iÂˇsinÎ¸ wavefunction describes anÂ up-electron, then we still have to apply thatÂ e^{i}^{Ď/2}Â =Â e^{i}^{Ď/2}Â =Â +iÂ factor, right? So we get a new wavefunction that will be equal toÂ e^{i}^{Ď/2}Âˇe^{i}^{Î¸}Â =Â e^{i}^{Ď/2}Âˇe^{i}^{Î¸}Â =Â +iÂˇe^{i}^{Î¸}Â =Â iÂˇcosÎ¸ +Â i^{2}ÂˇsinÎ¸ =Â sinÎ¸ âÂ iÂˇcosÎ¸, right? So how can we reconcile that with the cosÎ¸ âÂ iÂˇsinÎ¸ 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 180Â°, so it is going in the other direction now !Â 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 theÂ sinÎ¸ âÂ iÂˇcosÎ¸ andÂ cosÎ¸ âÂ iÂˇsinÎ¸ 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: cosÎ¸ =Â sin(Î¸ +Â Ď/2) andÂ âsinÎ¸ =Â cos(Î¸ +Â Ď/2). Let’s see what we can do with that. We can write the following, for example:
sinÎ¸ âÂ iÂˇcosÎ¸ =Â âcos(Î¸ +Â Ď/2) âÂ iÂˇsin(Î¸ +Â Ď/2) =Â â[cos(Î¸ +Â Ď/2) +Â iÂˇsin(Î¸ +Â Ď/2)] =Â âe^{i}^{Âˇ(Î¸ +Â Ď/2)}
Well… I guess that’s something at least ! The e^{i}^{ÂˇÎ¸}Â and âe^{i}^{Âˇ(Î¸ +Â Ď/2)}Â functions differ by a phase shiftÂ andÂ a minus sign so… Well… That’s what it takes to reverse the direction of an electron. đ Let us mull over that in the coming days. As I mentioned, these more philosophical topics are not easily exhausted. đ
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 particleâfor an electron, at least: for the proton, we only got the order of magnitude rightâbut then a proton is not an elementary particle.Â We got lots of other interesting equations as well… But… Well… When everything is said and done, it’s that equivalence between theÂ E =Â mÂˇa^{2}ÂˇĎ^{2}Â andÂ E =Â mÂˇc^{2}Â relations that we… Well… We need to be moreÂ specific about it.
Indeed, I’ve been ambiguous here and thereâoscillatingÂ between various interpretations, so to speak. đ In my own mind, I refer to my unanswered questions, or my ambiguous answers to them, as the form factorÂ problem.Â So… Well… That explains the title of my post. But so… Well… I do want to be somewhat moreÂ conclusiveÂ in this post. So let’s go and see where we end up. đ
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. đ
At first sight, the analogy between our flywheel model of an electron and the V-twin engine seems to be complete: the 90 degree angle of ourÂ V-2 engine makes it possible to perfectly balance the pistons and we may, therefore, think of the flywheel as a (symmetric) rotating mass, whose angular momentum is given by the product of the angular frequency and the moment of inertia: L =Â ĎÂˇI. Of course,Â the moment of inertia (aka the angular mass) will depend on theÂ formÂ (orÂ shape) of our flywheel:
- I = mÂˇa^{2}Â for a rotating pointÂ mass m or, what amounts to the same, for a circular hoop of mass m and radiusÂ rÂ =Â a.
- For a rotating (uniformly solid)Â disk, we must add a 1/2 factor: IÂ =Â mÂˇa^{2}/2.
How can we relate those formulas to the E =Â mÂˇa^{2}ÂˇĎ^{2}Â formula? TheÂ kinetic energy that is being stored in a flywheel is equal E_{kinetic}Â = IÂˇĎ^{2}/2, so that is only halfÂ of theÂ E =Â mÂˇa^{2}ÂˇĎ^{2}Â product if we substitute I forÂ I = mÂˇa^{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, theÂ E =Â mÂˇa^{2}ÂˇĎ^{2}Â formula just adds the (kinetic and potential) energy of two oscillators: we do not really consider the energy in the flywheel itself because… Well… The essence of our flywheel model of an electron is not the flywheel: the flywheel justÂ transfersÂ energy from one oscillator to the other, but so… Well… We don’tÂ includeÂ it in our energy calculations. The essence of our model is thatÂ two-dimensional oscillation whichÂ drivesÂ the electron, and which is reflected in Einstein’sÂ E =Â mÂˇc^{2}Â formula.Â That two-dimensional oscillationâtheÂ a^{2}ÂˇĎ^{2}Â = c^{2}Â equation, reallyâtells us that theÂ resonantÂ (orÂ natural) frequencyÂ of the fabric of spacetime is given by theÂ speed of lightâbut measured in units ofÂ a. [If you don’t quite get this, re-write theÂ a^{2}ÂˇĎ^{2}Â = c^{2}Â equation asÂ Ď = c/a: the radius of our electron appears as a naturalÂ distance unit here.]
Now, we were extremely happy with this interpretation not only because of the key results mentioned above, but also because it has lots of other nice consequences. Think of our probabilities as being proportional to energy densities, for exampleâand all of the other stuff I describe in my published paper on this. But there is even more on the horizon: a follower of this blog (a reader with an actual PhD in physics, for a change) sent me an article analyzing elementary particles as tiny black holes because… Well… If our electron is effectively spinning around, then its tangential velocity is equal toÂ vÂ =Â aÂˇĎÂ =Â c. Now, recent research suggest black holes are also spinning at (nearly) the speed of light. Interesting, right? However, in order to understand what she’s trying to tell me, I’ll first need to get a better grasp of general relativity, so I can relate what I’ve been writing here and in previous posts to the Schwarzschild radiusÂ and other stuff.
Let me get back to the lesson here. In the reference frame of our particle, the wavefunction really looks like the animation below: it has two components, and the amplitude of the two-dimensional oscillation is equal to a, which we calculated asÂ aÂ =Â Ä§Âˇ/(mÂˇc) = 3.8616Ă10^{â13}Â m, so that’s the (reduced) Compton scattering radius of an electron.
In my original article on this, I used a more complicated argument involving the angular momentum formula, but I now prefer a more straightforward calculation:
cÂ = aÂˇĎÂ =Â aÂˇE/Ä§ =Â aÂˇmÂˇc^{2}/Ä§Â Â âÂ aÂ =Â Ä§/(mÂˇc)
The question is: whatÂ is that rotating arrow? I’ve been vague and not so vague on this. The thing is: I can’tÂ proveÂ anything in this regard. But myÂ hypothesisÂ is that it is, in effect, aÂ rotatingÂ field vector, so it’s just like the electric field vector of a (circularly polarized) electromagnetic wave (illustrated below).
There are a number of crucial differences though:
- The (physical) dimension of the field vector of the matter-wave is different: I associate the real and imaginary component of the wavefunction with a force per unit massÂ (as opposed to the force per unit charge dimension of the electric field vector). Of course, the newton/kg dimension reduces to the dimension of acceleration (m/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 doÂ notÂ involve any mass: they’re just an oscillating field. Nothing more. Nothing less. In contrast, as Feynman puts it: “When you do find the electron some place, the entire charge is there.” (Feynman’s Lectures, III-21-4)
- The third difference is one that I thought of only recently: theÂ planeÂ of the oscillation cannotÂ be perpendicular to the direction of motion of our electron, because then we can’t explain the direction of its magnetic moment, which is either up or down when traveling through a Stern-Gerlach apparatus.
I mentioned that in my previous post but, for your convenience, I’ll repeat what I wrote there.Â The basic idea here is illustrated below (credit for this illustration goes toÂ another blogger on physics). As for the Stern-Gerlach experiment itself, let me refer you to aÂ YouTube videoÂ from theÂ Quantum Made SimpleÂ site.
The point is: the direction of the angular momentum (and the magnetic moment) of an electronâor, to be precise, its component as measured in the direction of the (inhomogeneous) magnetic field through which our electron isÂ travelingâcannotÂ be parallel to the direction of motion. On the contrary, it isÂ perpendicularÂ to the direction of motion. In other words, if we imagine our electron as spinning around some center, then the disk it circumscribes will compriseÂ the direction of motion.
However, we need to add an interesting detail here. As you know, we don’t really have a precise direction of angular momentum in quantum physics. [If you don’t know this… Well… Just look at one of my many posts on spin and angular momentum in quantum physics.] Now, we’ve explored a number of hypotheses but, when everything is said and done, a rather classical explanation turns out to be the best: an object with an angular momentum JÂ and a magnetic momentÂ ÎźÂ (I used bold-face because these areÂ vector quantities) that is parallel to some magnetic field B, will notÂ line up, as you’d expect a tiny magnet to do in a magnetic fieldâor not completely, at least: it willÂ precess. I explained that in another post on quantum-mechanical spin, which I advise you to re-read if you want to appreciate the point that I am trying to make here. That post integrates some interesting formulas, and so one of the things on my ‘to do’ list is to prove that these formulas are, effectively, compatible with the electron model we’ve presented in this and previous posts.
Indeed, when one advances a hypothesis like this, it’s not enough to just sort ofÂ showÂ that the general geometry of the situation makes sense: we also need to show the numbers come out alright. So… Well… Whatever weÂ thinkÂ our electronâor its wavefunctionâmight be, it needs to be compatible with stuff like the observedÂ precession frequencyÂ of an electron in a magnetic field.
Our model also needs to be compatible with the transformation formulas for amplitudes. I’ve been talking about this for quite a while now, and so it’s about time I get going on that.
Last but not least, those articles that relate matter-particles to (quantum) gravityâsuch as the one I mentioned aboveâare intriguing too and, hence, whatever hypotheses I advance here, I’d better check them against those more advanced theories too, right? đ Unfortunately, that’s going to take me a few more years of studying… But… Well… I still have many years aheadâI hope. đ
Post scriptum: It’s funny how one’s brain keeps working when sleeping. When I woke up this morning, I thought: “But itÂ isÂ that flywheel that matters, right? That’s the energy storage mechanism and also explains how photons possibly interact with electrons. The oscillatorsÂ driveÂ the flywheel but, without the flywheel, nothing is happening. It is really theÂ transferÂ of energyâthrough the flywheelâwhich explains why our flywheel goes round and round.”
It may or may not be useful to remind ourselves of the math in this regard.Â The motionÂ ofÂ our first oscillator is given by the cos(ĎÂˇt) = cosÎ¸ function (Î¸ = ĎÂˇt), and its kinetic energy will be equal toÂ sin^{2}Î¸. Hence, the (instantaneous)Â changeÂ in kinetic energy at any point in time (as a function of the angle Î¸) isÂ equal to:Â d(sin^{2}Î¸)/dÎ¸ = 2âsinÎ¸âd(sinÎ¸)/dÎ¸ = 2âsinÎ¸âcosÎ¸. Now, the motion of theÂ second oscillator (just look at that second piston going up and down in the V-2 engine) is given by theÂ sinÎ¸ function, which is equal to cos(Î¸ â Ď /2). Hence, its kinetic energy is equal toÂ sin^{2}(Î¸ â Ď /2), and how itÂ changesÂ (as a function of Î¸ again) is equal toÂ 2âsin(Î¸ â Ď /2)âcos(Î¸ â Ď /2) =Â = â2âcosÎ¸âsinÎ¸ = â2âsinÎ¸âcosÎ¸. So here we have our energy transfer: the flywheel organizes the borrowing and returning of energy, so to speak. That’s the crux of the matter.
So… Well… WhatÂ if the relevant energy formula isÂ E =Â mÂˇa^{2}ÂˇĎ^{2}/2 instead ofÂ E =Â mÂˇa^{2}ÂˇĎ^{2}? What are the implications? Well… We get aÂ â2 factor in our formula for the radiusÂ a, as shown below.
Now that isÂ notÂ so nice. For the tangential velocity, we getÂ vÂ =Â aÂˇĎ =Â â2Âˇc. This is alsoÂ notÂ so nice. How can we save our model? I am not sure, but here I am thinking of the mentioned precessionâtheÂ wobbling of our flywheel in a magnetic field. Remember we may think of J_{z}âthe angular momentum or, to be precise, its component in theÂ z-direction (the direction in which weÂ measureÂ itâas the projection of theÂ realÂ angular momentumÂ J. Let me insert Feynman’s illustration here again (Feynman’s Lectures, II-34-3), so you get what I am talking about.
Now, all depends on the angle (Î¸) betweenÂ J_{z}Â andÂ J, of course. We did a rather obscure post on these angles, but the formulas there come in handy now. Just click the link and review it if and when you’d want to understand the following formulas for theÂ magnitudeÂ of theÂ presumedÂ actualÂ momentum:In this particular case (spin-1/2 particles),Â j is equal to 1/2 (in units ofÂ Ä§, of course). Hence,Â JÂ is equal toÂ â0.75Â â 0.866. Elementary geometry then tells us cos(Î¸) =Â (1/2)/â(3/4) =Â = 1/â3. Hence,Â Î¸Â â 54.73561Â°. That’s a big angleâlarger than the 45Â° angle we had secretly expected because… Well… The 45Â° angle has thatÂ â2 factor in it:Â cos(45Â°) =Â sin(45Â°) = 1/â2.
Hmm… As you can see, there is no easy fix here. Those damn 1/2 factors! They pop up everywhere, don’t they? đ We’ll solve the puzzle. One day… But not today, I am afraid. I’ll call it the form factor problem… Because… Well… It sounds better than the 1/2 orÂ â2 problem, right? đ
Note: If you’re into quantum math, you’ll noteÂ aÂ =Â Ä§/(mÂˇc) is theÂ reducedÂ Compton scattering radius. The standard Compton scattering radius is equal toÂ aÂˇ2ĎÂ = (2ĎÂˇÄ§)/(mÂˇc) =Â h/(mÂˇc) = h/(mÂˇc). It doesn’t solve theÂ â2 problem. Sorry. The form factor problem. đ
To be honest, I finished my published paper on all of this with a suggestion that, perhaps, we should think of twoÂ circularÂ oscillations, as opposed to linear ones. Think of a tiny ball, whose center of mass stays where it is, as depicted below. Any rotation â around any axis â will be some combination of a rotation around the two other axes. Hence, we may want to think of our two-dimensionalÂ oscillation as an oscillation of a polar and azimuthal angle. It’s just a thought but… Well… I am sure it’s going to keep me busy for a while. đThey are oscillations, still, so I am not thinking ofÂ twoÂ flywheels that keep going around in the same direction. No. More like a wobbling object on a spring. Something like the movement of a bobblehead on a spring perhaps. đ
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):Â Einsteinâs view on aether theories probably still holds true: âWe 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 â 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.â
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: âIt 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. [âŚ] 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. [âŚ]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.â
I really love this: a relativistic aether. MyÂ interpretation of the wavefunction is veryÂ consistent with that.
A physical explanation for relativistic length contraction?
My last posts were all about a possible physicalÂ 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 aÂ physicalÂ 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 unitÂ chargeÂ (newton perÂ coulomb), while the other gives us a force per unitÂ 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:
Ď =Â aÂˇe^{âi(EÂˇt â pâx)/Ä§} =Â aÂˇcos(pâx/Ä§ – Eât/Ä§) + iÂˇaÂˇsin(pâx/Ä§ – Eât/Ä§)
Nature should not care about our conventions for measuring the phase angle clockwise or counterclockwise and, therefore, the Ď =Â aÂˇe^{i}^{[EÂˇt â pâx]/Ä§} function may also be permitted. We know that cos(Î¸) = cos(–Î¸) and sinÎ¸ = –sin(–Î¸), so we can write:Â Â Â
Ď =Â aÂˇe^{i}^{(EÂˇt â pâx)/Ä§} =Â aÂˇcos(Eât/Ä§ – pâx/Ä§) + iÂˇaÂˇsin(Eât/Ä§ – pâx/Ä§)
= aÂˇcos(pâx/Ä§ – Eât/Ä§) – iÂˇaÂˇsin(pâx/Ä§ – Eât/Ä§)
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 â not about the direction nor the magnitude â 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 pâx/Ä§ reduces to pâx/Ä§. 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(Î¸) = cos(0) = 1 and sin(Î¸) = 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: +Ä§/2 or âÄ§/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 Ď/2: sin(Î¸) = cos(Î¸ â Ď/2).
As for the wave velocity, and its direction of propagation, we know that the (phase) velocity of any wave F(kx – Ďt) is given by v_{p} =Â Ď/k = (E/Ä§)/(p/Ä§) = 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} =Â Ď/k = –E/p.
The de Broglie relations
E/Ä§ = Ď gives the frequency in time (expressed in radians per second), while p/Ä§ = k gives us the wavenumber, or the frequency in space (expressed in radians per meter). Of course, we may write: f = Ď/2ĎÂ and Îť = 2Ď/k, which gives us the two de Broglie relations:
- E = Ä§âĎ = hâf
- p = Ä§âk = h/Îť
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/Îť relation, the wavelength is inversely proportional to the momentum: Îť = 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}âv = m_{c}âc = mâc (all of the energy is kinetic). Hence, we can write: pâc = mâc^{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:
Îť = h/p = hc/E = h/mc
However, this is a limiting situation â 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 pÂ â 0 then Îť âÂ â. How should we interpret this inverse proportionality between Îť and p? To answer this question, let us first see what this wavelength Îť actually represents.
If we look at the Ď = aÂˇcos(pâx/Ä§ – Eât/Ä§) – iÂˇaÂˇsin(pâx/Ä§ – Eât/Ä§) once more, and if we write pâx/Ä§ as Î, then we can look at pâx/Ä§ as a phase factor, and so we will be interested to know for what x this phase factor Î = pâx/Ä§ will be equal to 2Ď. So we write:
Î =pâx/Ä§ = 2Ď âÂ x = 2ĎâÄ§/p = h/p = Îť
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 Îť actually represents for our one-dimensional elementary wavefunction. Now, the time that is needed for one cycle is equal to T = 1/f = 2ĎÂˇ(Ä§/E). Hence, we can now calculate the wave velocity:
v = Îť/T = (h/p)/[2ĎÂˇ(Ä§/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 = mâc^{2}/mâv_{g }Â = c/Î˛_{g}, in which Î˛_{g} is the relative classical velocity[3] of our particle Î˛_{g} = v_{g}/c) tells us that the phase velocities will effectively be superluminal (Î˛_{g}Â < 1 so 1/ Î˛_{g} > 1), but what if Î˛_{g} approaches zero? The conclusion seems unavoidable: for a particle at rest, we only have a frequency in time, as the wavefunction reduces to:
Ď =Â aÂˇe^{âiÂˇEÂˇt/Ä§} =Â aÂˇcos(Eât/Ä§) – iÂˇaÂˇsin(Eât/Ä§)
How should we interpret this?
A physical interpretation of relativistic length contraction?
In my previous posts,Â 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 Îť decreases. Hence, the physical interpretation of the wavefunction that is offered here may explain relativistic length contraction.
đ
Yep. Think about it. đ
[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 (Îł), we can write the relativistically correct formula for the kinetic energy as KE = EÂ âÂ E_{0}Â =Â m_{v}c^{2}Â â m_{0}c^{2}Â =Â m_{0}Îłc^{2}Â â m_{0}c^{2}Â =Â m_{0}c^{2}(Îł â 1). As v approaches c, Îł 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.Â The elementary wavefunction is written as:
Ď =Â aÂˇe^{âi[EÂˇtÂ â pâx]/Ä§} =Â aÂˇcos(pâx/Ä§Â â Eât/Ä§) + iÂˇaÂˇsin(pâx/Ä§Â â Eât/Ä§)
Of course, NatureÂ (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 Ď =Â aÂˇe^{i}^{[EÂˇt â pâx]/Ä§} function is also permitted. We know that cos(Î¸) = cos(âÎ¸) and sinÎ¸ = âsin(âÎ¸), so we can write:Â Â Â
Ď =Â aÂˇe^{i}^{[EÂˇt â pâx]/Ä§} =Â aÂˇcos(Eât/Ä§Â âÂ pâx/Ä§) + iÂˇaÂˇsin(Eât/Ä§Â âÂ pâx/Ä§)
= aÂˇcos(pâx/Ä§Â â Eât/Ä§) âÂ iÂˇaÂˇsin(pâx/Ä§Â â Eât/Ä§)
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 â not about the direction, and not about the magnitude â then the direction of p can be our x-axis. In this reference frame,Â x = (x, y, z) reduces to (x, 0, 0), and pâx/Ä§ reduces to pâx/Ä§. 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(Î¸) = cos(0) = 1 and sin(Î¸) = 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 (+Ä§/2 or âÄ§/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 Ď/2: sin(Î¸) = cos(Î¸ â Ď/2).
As for the wave velocity, and its direction of propagation, we know that the (phase) velocity of any waveform F(kx â Ďt) is given by v_{p} =Â Ď/k. In our case, we find thatÂ v_{p} =Â Ď/k = (E/Ä§)/(p/Ä§) = E/p. Of course, the momentum might also be in the negative x-direction, in which case k would be equal to âp and, therefore, we would get a negative phase velocity: v_{p} =Â Ď/k = (E/Ä§)/(âp/Ä§) = âE/p.
As you know, E/Ä§ = Ď gives the frequency in time (expressed in radians per second), while p/Ä§ = 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 = Ď/2ĎÂ and Îť = 2Ď/k, which gives us the two de Broglie relations:
- E = Ä§âĎ = hâf
- p = Ä§âk = h/Îť
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: Îť = h/p. Hence, if p goes to zero, then the wavelength becomes infinitely long, so we write:
If pÂ â 0 then Îť âÂ â.
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}âv = m_{c}âc = mâc (all of the energy is kinetic) and, therefore, pâc = mâc^{2} = E, which we may also write as: E/p = c. Hence, for a particle with zero rest mass (m_{0}Â = 0), the wavelength can be written as:
Îť = h/p = hc/E = h/mc
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 neverÂ equalÂ c.[2] The question remains: how should we interpret the inverse proportionality between Îť and p?
Let us first see what this wavelength Îť actually represents. If we look at the Ď = aÂˇcos(pâx/Ä§ â Eât/Ä§) âÂ iÂˇaÂˇsin(pâx/Ä§ – Eât/Ä§) once more, and if we write pâx/Ä§ as Î, then we can look at pâx/Ä§ as a phase factor, and so we will be interested to know for what x this phase factor Î = pâx/Ä§ will be equal to 2Ď. So we write:
Î =pâx/Ä§ = 2Ď âÂ x = 2ĎâÄ§/p = h/p = Îť
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 Îť actually represent for our one-dimensional elementary wavefunction. Now, the time that is needed for one cycle is equal to T = 1/f = 2ĎÂˇ(Ä§/E). Hence, we can now calculate the wave velocity:
v = Îť/T = (h/p)/[2ĎÂˇ(Ä§/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 â or the concept of a precise energy, and a precise momentum â does not make sense. This is where the Uncertainty Principle comes in: stating that p = 0, implies zero uncertainty. Hence, the Ď_{p} factor in the Ď_{p}âĎ_{x}Â â¤ Ä§/2 would be zero and, therefore, Ď_{p}âĎ_{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 Ď_{p} refers to the standard deviation from the mean, as illustrated below. Of course, the distribution may be or may not be normal â we donât know â 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 Ä§/2, so thatâs about 0.726Ă10^{â17} m for the distance unit and 0.726Ă10^{â17} Nâs for the momentum unit, as the product of both gives us Ä§/2. To make this somewhat more real, we may note that 0.726Ă10^{â17} m is the attometer scale (1 am = 1Ă10^{â18} 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 â which should correspond to the classical idea of the velocity of our particle â 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}Â = âĎ_{i}/âk_{i}Â = â(E_{i}/Ä§)/â(p_{i}/Ä§) = â(E_{i})/â(p_{i})
This assumes the existence of a dispersion relation which gives us Ď_{i}Â as a function of k_{i} â what amounts to the same â E_{i}Â as a function of p_{i}. How do we get that?Â Well… There are a few ways to go about it but one interesting way of doing it is to re-write SchrĂśdinger’s equation as the following pairÂ of equations[4]:
- Re(âĎ/ât) = â[Ä§/(2m_{eff})]ÂˇIm(â^{2}Ď)Â âÂ ĎÂˇcos(kx â Ďt) =Â k^{2}Âˇ[Ä§/(2m_{eff})]Âˇcos(kx â Ďt)
- Im(âĎ/ât) = [Ä§/(2m_{eff})]ÂˇRe(â^{2}Ď)Â â ĎÂˇsin(kx â Ďt) = k^{2}Âˇ[Ä§/(2m_{eff})]Âˇsin(kx â Ďt)
These equations imply the following dispersion relation:
Ď =Â Ä§Âˇk^{2}/(2m)
Of course, we need to think about the subscripts now: we have Ď_{i}, k_{i}, but… What about m_{eff} or, dropping the subscript, about m? Do we write it as m_{i}? If so, what is it? Well… It is the equivalent mass of E_{i}Â obviously, and so we get it from the mass-energy equivalence relation: m_{i}Â = E_{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: Ď_{m} = Ď_{E}/c^{2}. We are tempted to do a few substitutions here. Letâs first check what we get when doing the m_{i}Â = E_{i}/c^{2} substitution:
Ď_{i} =Â Ä§Âˇk_{i}^{2}/(2m_{i}) = (1/2)âÄ§Âˇk_{i}^{2}âc^{2}/E_{i} = (1/2)âÄ§Âˇk_{i}^{2}âc^{2}/(Ď_{i}âÄ§)Â = (1/2)âÄ§Âˇk_{i}^{2}âc^{2}/Ď_{i}
â Ď_{i}^{2}/k_{i}^{2} = c^{2}/2Â â Ď_{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. đŚ
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/Ä§ = m_{i} Âˇv_{g}. This gives us the following result:
Ď_{i} =Â Ä§Âˇ(m_{i} Âˇv_{g})^{2}/(2m_{i}) = Ä§Âˇm_{i}Âˇv_{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 weÂ dropÂ it. Now you will object that SchrĂśdinger’s equation gives us the electron orbitals – and many other correct descriptions of quantum-mechanical stuff – so, surely,Â SchrĂśdinger’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 atÂ oneÂ of the two dimensions of the oscillation only and, therefore, it’s onlyÂ halfÂ 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(âĎ/ât) = â(Ä§/m_{eff})ÂˇIm(â^{2}Ď)Â âÂ ĎÂˇcos(kx â Ďt) =Â k^{2}Âˇ(Ä§/m_{eff})Âˇcos(kx â Ďt)
- Im(âĎ/ât) = (Ä§/m_{eff})ÂˇRe(â^{2}Ď)Â â ĎÂˇsin(kx â Ďt) = k^{2}Âˇ(Ä§/m_{eff})Âˇsin(kx â Ďt)
We then get the dispersion relation withoutÂ that 1/2 factor:
Ď_{i} =Â Ä§Âˇk_{i}^{2}/m_{i}
TheÂ m_{i}Â = E_{i}/c^{2} substitution then gives us the result we sort of expected to see:
Ď_{i} =Â Ä§Âˇk_{i}^{2}/m_{i}Â = Ä§Âˇk_{i}^{2}âc^{2}/E_{i} = Ä§Âˇk_{i}^{2}âc^{2}/(Ď_{i}âÄ§)Â â Ď_{i}/k_{i} = v_{p} = c
Likewise, the other calculation also looks more meaningful now:
Ď_{i} =Â Ä§Âˇ(m_{i} Âˇv_{g})^{2}/m_{i}Â = Ä§Âˇm_{i}Âˇv_{g}^{2}
Sweet ! đ
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 â the speed with which those wave crests (or troughs) move â and (2) some kind of circular or tangential velocity â the velocity along the red contour lineÂ above. Weâll need the formula for a tangential velocity: v_{t} = aâĎ.
Now, if Îť is zero, then v_{t} = aâĎ = aâE/Ä§ is just all there is. We may double-check this as follows: the distance traveled in one period will be equal to 2Ďa, and the period of the oscillation is T = 2ĎÂˇ(Ä§/E). Therefore, v_{t} will, effectively, be equal to v_{t} = 2Ďa/(2ĎÄ§/E) = aâE/Ä§.Â However, if Îť is non-zero, then the distance traveled in one period will be equal to 2Ďa + Îť. The period remains the same: T = 2ĎÂˇ(Ä§/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 theÂ real-life +Ä§/2 or âÄ§/2 values of its spin, and we got aÂ numericalÂ value forÂ a. It was the Compton radius: the scattering radius for an electron. Let us write it out:
Using the right numbers, youâll find the numerical value for a: 3.8616Ă10^{â13} m. But let us just substitute the formula itself here:Â
This is fascinating ! And we just calculated that v_{p}Â is equal toÂ c. For the elementary wavefunction, that is. Hence, we get this amazing result:
v_{t} = 2c
ThisÂ tangentialÂ velocity isÂ twiceÂ the linearÂ velocity !
Of course, the question is: what is theÂ physicalÂ significance of this? I need to further look at this. Wave velocities are, essentially, mathematicalÂ concepts only: the wave propagates through space, butÂ nothing elseÂ 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 aÂ propertyÂ of the vacuum – or of the fabric of spacetime itself. đ
[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 (Îł), we can write the relativistically correct formula for the kinetic energy as KE = EÂ âÂ E_{0}Â =Â m_{v}c^{2}Â â m_{0}c^{2}Â =Â m_{0}Îłc^{2}Â â m_{0}c^{2}Â =Â m_{0}c^{2}(Îł â 1). As v approaches c, Îł 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)Ă10^{â35} 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 +Â iâb and c +Â iâd are equal if, and only if, their real and imaginary parts are the same. Now, the âĎ/ât =Â iâ(Ä§/m_{eff})ââ^{2}Ď equation amounts to writing something like this: a +Â iâb =Â iâ(c +Â iâd). Now, remembering thatÂ i^{2}Â = â1, you can easily figure out thatÂ iâ(c +Â iâd) =Â iâc +Â i^{2}âd = â d +Â iâc.
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 theÂ elementary wavefunction by heart:
Ď =Â aÂˇe^{âi[EÂˇt â pâx]/Ä§} =Â aÂˇcos(pâx/Ä§ â Eât/Ä§) + iÂˇaÂˇsin(pâx/Ä§ â Eât/Ä§)
If we assume the momentum p is all in the x-direction, then the p and x vectors will have the same direction, and pâx/Ä§ reduces to pâx/Ä§. 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(Î¸) = cos(0) = 1 and sin(Î¸) = 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: +Ä§/2 or âÄ§/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 Ď/2: sin(Î¸) = cos(Î¸ â Ď/2) Â
Surely, Nature doesn’t care a hoot about our conventions for measuring the phase angle clockwise or counterclockwise and therefore, the Ď =Â aÂˇe^{i}^{[EÂˇt â pâx]/Ä§} function should, effectively, also be permitted. We know that cos(Î¸) = cos(–Î¸) and sinÎ¸ = –sin(–Î¸), so we can write: Â Â Â
Ď =Â aÂˇe^{i}^{[EÂˇt â pâx]/Ä§} =Â aÂˇcos(Eât/Ä§ â pâx/Ä§) + iÂˇaÂˇsin(Eât/Ä§ â pâx/Ä§)
= aÂˇcos(pâx/Ä§ â Eât/Ä§) âÂ iÂˇaÂˇsin(pâx/Ä§ â Eât/Ä§)
E/Ä§ = Ď gives the frequency in time (expressed in radians per second), while p/Ä§ = k gives us the wavenumber, or the frequency in space (expressed in radians per meter). Of course, we may write: f = Ď/2ĎÂ and Îť = 2Ď/k, which gives us the two de Broglie relations:
- E = Ä§âĎ = hâf
- p = Ä§âk = h/Îť
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: Îť = h/p. Hence, if p goes to zero, then the wavelength becomes infinitely long: if pÂ â 0, then Îť âÂ â.Â 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}âv = mâc Â and, therefore, pâc = mâc^{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:
Îť = 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 Îť 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} =Â Ď/k = (E/Ä§)/(p/Ä§) = E/p. Of course, we know that, classically, the momentum will be equal to the group velocity times the mass: p = mÂˇv_{g}. However, when p is zero, we have a division by zero once more: if pÂ â 0, then v_{p} = E/pÂ â â. 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}Â =Â Ď/k = E/p = E/(mÂˇv_{g}) = (mÂˇc^{2})/(mÂˇv_{g}) = c^{2}/v_{g}
We can re-write this as v_{p}Âˇv_{g}Â = c^{2}, which reminds us of the relationship between the electric and magnetic constant (1/Îľ_{0})Âˇ(1/Îź_{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 Ď/k. In contrast, the group velocity is the derivative of Ď with respect to k. So we need to write Ď as a function of k. Can we do that even if we have only one wave? We doÂ notÂ have a wave packet here, right? Just some hypotheticalÂ building blockÂ 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 thisÂ elementaryÂ wavefunction. Let’s first get thatÂ Ď = Ď(k) relation. You’ll remember we can write SchrĂśdingerâs equation – the equation that describes theÂ propagationÂ mechanism for matter-waves – asÂ the following pair of equations:
- Re(âĎ/ât) = â[Ä§/(2m)]ÂˇIm(â^{2}Ď)Â âÂ ĎÂˇcos(kx â Ďt) =Â k^{2}Âˇ[Ä§/(2m)]Âˇcos(kx â Ďt)
- Im(âĎ/ât) = [Ä§/(2m)]ÂˇRe(â^{2}Ď)Â â ĎÂˇsin(kx â Ďt) = k^{2}Âˇ[Ä§/(2m)]Âˇsin(kx â Ďt)
This tells us that Ď =Â Ä§Âˇk^{2}/(2m). Therefore, we can calculate âĎ/âkÂ as:
âĎ/âk = Ä§Âˇk/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 mathematicalÂ formula for the group velocity – but we don’t know what it represents – and a finite wavelength. In fact, using the p = Ä§âk = h/Îť relation, we can write one as a function of the other:
Îť = h/p = h/mv_{g} âÂ v_{g} = h/mÎť
What does this mean? ItÂ resembles the c = h/mÎť relation we had for a particle with zero rest mass. Of course, it does: the Îť = h/mc relation is, once again, a limit for v_{g} going to c. By the way, it is interesting to note that the v_{p}Âˇv_{g}Â = c^{2} relation implies that the phase velocity is always superluminal. That’ easy to see when you re-write the equation in terms ofÂ relativeÂ velocities: (v_{p}/c)Âˇ(v_{g}/c) =Â Î˛_{phase}ÂˇÎ˛_{group}Â = 1. Hence, ifÂ Î˛_{group}Â < 1, thenÂ Î˛_{phase}Â > 1.
So whatÂ isÂ the geometry,Â really? Letâs look at the Ď = aÂˇcos(pâx/Ä§ – Eât/Ä§) – iÂˇaÂˇsin(pâx/Ä§ – Eât/Ä§) formula once more. If we write pâx/Ä§ as Î, then we will be interested to know for what x this phase factor will be equal to 2Ď. So we write:
Î =pâx/Ä§ = 2Ď âÂ x = 2ĎâÄ§/p = h/p = ÎťÂ Â
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.Â 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 lineÂ above). Weâll probably need the formula for the tangential velocity: v = aâĎ. If p = 0 (so we have that weird infinitesimally long wavelength), then we have two velocities:
- The tangential velocity around theÂ aÂˇe^{i}^{ÂˇEÂˇt}Â circle, so to speak, and that will just be equal toÂ v = aâĎ =Â aâE/Ä§.
- The red contour line sort of gets stretched out, like infinitely long, and the velocity becomes… What does it do? Does it go toÂ â , or toÂ 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 theÂ real-life +Ä§/2 or âÄ§/2 values of its spin. And so we got aÂ numericalÂ value forÂ 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:Â aÂ = 3.8616Ă10^{â13} m.Â We did then another weird calculation. We said all of the energy of the electron had to be packed in thisÂ cylinderÂ that might of might not be there. The point was: the energy is finite, so thatÂ elementaryÂ wavefunction cannotÂ have an infinite length in space. Indeed, assuming that the energy was distributed uniformly, we jotted down this formula, which reflects the formula for theÂ volume of a cylinder:
E = ĎÂˇa^{2}Âˇl âÂ lÂ = E/(ĎÂˇa^{2})
Using the value we got for the Compton scattering radius (aÂ =Â 3.8616Ă10^{â13} m), we got an astronomical value forÂ l. Let me write it out:
lÂ =Â (8.19Ă10^{â14})/(ĎÂˇ14.9Ă10^{â26}) â 0.175Ă10^{12}Â m
It is,Â literally, an astronomical value:Â 0.175Ă10^{12}Â m is 175 millionÂ kilometer, so that’s like theÂ 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 forÂ Îť? We’d get that linear velocity, right? Let’s try it. TheÂ periodÂ is equal to T =Â T = 2ĎÂˇ(Ä§/E) = h/E and Îť =Â E/(ĎÂˇa^{2}), so we write:We can write this as a function of m and theÂ cÂ andÂ Ä§ 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 toÂ showÂ the geometry of the wavefunction a bit more in detail.
[1]Â The discussions on the mass of neutrinos are interesting in this regard. Scientists all felt the neutrinoÂ had toÂ haveÂ 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. đ 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. đ
Abstract : This paper explores the implications of associating the components of the wavefunction with a physical dimension: force per unit mass â 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, SchrĂśdingerâs 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 (Ď = aÂˇe^{âiâÎ¸} = aâcosÎ¸ – aâsinÎ¸) differ by 90 degrees (Ď/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 = mÂˇa^{2}ÂˇĎ^{2}/2) and Einsteinâs relativistic energy equation E = mâc^{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. SchrĂśdingerâs 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 Einsteinâs 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 Einsteinâs relativistic energy equation (E = mc^{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 = mc^{2}
- E = mĎ^{2}/2
- E = mv^{2}/2
In these formulas, Ď, v and c all describe some velocity.[8] Of course, there is the 1/2 factor in the E = mĎ^{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 = 2ÂˇmÂˇĎ^{2}/2 = mÂˇĎ^{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 90Â° 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 = aÂˇcos(ĎÂˇt + Î).[12] Needless to say, Î is just a phase factor which defines our t = 0 point, and Ď is the natural angular frequency of our oscillator. Because of the 90Â° angle between the two cylinders, Î would be 0 for one oscillator, and âĎ/2 for the other. Hence, the motion of one piston is given by x = aÂˇcos(ĎÂˇt), while the motion of the other is given by x = aÂˇcos(ĎÂˇtâĎ/2) = aÂˇsin(ĎÂˇt).
The kinetic and potential energy of one oscillator (think of one piston or one spring only) can then be calculated as:
- K.E. = T = mÂˇv^{2}/2 = (1/2)ÂˇmÂˇĎ^{2}Âˇa^{2}Âˇsin^{2}(ĎÂˇt + Î)
- P.E. = U = kÂˇx^{2}/2 = (1/2)ÂˇkÂˇa^{2}Âˇcos^{2}(ĎÂˇt + Î)
The coefficient k in the potential energy formula characterizes the restoring force: F = âkÂˇx. From the dynamics involved, it is obvious that k must be equal to mÂˇĎ^{2}. Hence, the total energy is equal to:
E = T + U = (1/2)Âˇ mÂˇĎ^{2}Âˇa^{2}Âˇ[sin^{2}(ĎÂˇt + Î) + cos^{2}(ĎÂˇt + Î)] = mÂˇa^{2}ÂˇĎ^{2}/2
To facilitate the calculations, we will briefly assume k = mÂˇĎ^{2} and a are equal to 1. The motion of our first oscillator is given by the cos(ĎÂˇt) = cosÎ¸ function (Î¸ = ĎÂˇt), and its kinetic energy will be equal to sin^{2}Î¸. Hence, the (instantaneous) change in kinetic energy at any point in time will be equal to:
d(sin^{2}Î¸)/dÎ¸ = 2âsinÎ¸âd(sinÎ¸)/dÎ¸ = 2âsinÎ¸âcosÎ¸
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 sinÎ¸ function, which is equal to cos(Î¸âĎ /2). Hence, its kinetic energy is equal to sin^{2}(Î¸âĎ /2), and how it changes â as a function of Î¸ â will be equal to:
2âsin(Î¸âĎ /2)âcos(Î¸âĎ /2) = = â2âcosÎ¸âsinÎ¸ = â2âsinÎ¸âcosÎ¸
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 = ma^{2}Ď^{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 = mc^{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:
Ď = aÂˇe^{âi[EÂˇt â pâx]/Ä§} = aÂˇe^{âi[EÂˇt â pâx]/Ä§} = aÂˇcos(pâx/Ä§ – Eât/Ä§) + iÂˇaÂˇsin(pâx/Ä§ – Eât/Ä§)
When considering a particle at rest (p = 0) this reduces to:
Ď = aÂˇe^{âiâEÂˇt/Ä§} = aÂˇcos(–Eât/Ä§) + iÂˇaÂˇsin(–Eât/Ä§) = aÂˇcos(Eât/Ä§) – iÂˇaÂˇsin(Eât/Ä§)
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 (Ď) is counter-clockwise.
Figure 2: Eulerâs 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 pâx/Ä§ reduces to pâx/Ä§. Most illustrations â such as the one below â 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 â the real and imaginary part of our wavefunction â 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 Ď = aÂˇe^{âiâEÂˇt/Ä§}. Hence, the angular velocity of both oscillations, at some point x, is given by Ď = -E/Ä§. Now, the energy of our particle includes all of the energy â kinetic, potential and rest energy â and is, therefore, equal to E = mc^{2}.
Can we, somehow, relate this to the mÂˇa^{2}ÂˇĎ^{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 Âľ a^{2}. We may, therefore, think that the a^{2} factor in the E = mÂˇa^{2}ÂˇĎ^{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 Ď_{i} = -E_{i}/Ä§. Each of these wavefunctions will contribute some energy to the total energy of the wave packet. To calculate the contribution of each wave to the total, both a_{i} as well as E_{i} will matter.
What is E_{i}? E_{i} varies around some average E, which we can associate with some 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}Ä§^{2}: it is about 1Â´10^{–}^{51} N^{2}âm^{4}. Let us also do a dimensional analysis: the physical dimensions of the E = mÂˇa^{2}ÂˇĎ^{2} equation make sense if we express m in kg, a in m, and Ď in rad/s. We then get: [E] = kgâm^{2}/s^{2} = (Nâs^{2}/m)âm^{2}/s^{2} = Nâm = J. The dimensions of the left- and right-hand side of the physical normalization condition is N^{3}âm^{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 Einsteinâs 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} = NÂˇmâs^{2}/m^{2} = NÂˇs^{2}/m = kg
This is not very helpful. It only reminds us of Newtonâs 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. Einsteinâs E = mc^{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 Ď^{2}= C^{–}^{1}/L or Ď^{2} = k/m of harmonic oscillators once again.[13] The key difference is that the Ď^{2}= C^{–}^{1}/L and Ď^{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 â 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 Planckâs constant as the constant of proportionality. Now, for one-dimensional oscillations â think of a guitar string, for example â 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 Einsteinâs 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 (SchrĂśdingerâs equation), as we can now analyze it as an energy diffusion equation.
IV. SchrĂśdingerâs equation as an energy diffusion equation
The interpretation of SchrĂśdingerâs equation as a diffusion equation is straightforward. Feynman (Lectures, III-16-1) briefly summarizes it as follows:
âWe can think of SchrĂśdingerâs equation as describing the diffusion of the probability amplitude from one point to the next. [âŚ] 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 SchrĂśdingerâs equation are complex waves.â[17]
Let us review the basic math. For a particle moving in free space â with no external force fields acting on it â there is no potential (U = 0) and, therefore, the UĎ term disappears. Therefore, SchrĂśdingerâs equation reduces to:
âĎ(x, t)/ât = iÂˇ(1/2)Âˇ(Ä§/m_{eff})Âˇâ^{2}Ď(x, t)
The ubiquitous diffusion equation in physics is:
âĎ(x, t)/ât = DÂˇâ^{2}Ď(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 Ď is a complex-valued function, with a real and an imaginary part, we get the following equations[18]:
- Re(âĎ/ât) = â(1/2)Âˇ(Ä§/m_{eff})ÂˇIm(â^{2}Ď)
- Im(âĎ/ât) = (1/2)Âˇ(Ä§/m_{eff})ÂˇRe(â^{2}Ď)
These equations make us think of the equations for an electromagnetic wave in free space (no stationary charges or currents):
- âB/ât = ââĂE
- âE/ât = c^{2}âĂB
The above equations effectively describe a propagation mechanism in spacetime, as illustrated below.
Figure 4: Propagation mechanisms
The Laplacian operator (â^{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 Ď(x, t), so what is the dimension of our wavefunction Ď(x, t)? To answer that question, we should analyze the diffusion constant in SchrĂśdingerâs equation, i.e. the (1/2)Âˇ(Ä§/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 Ä§/m_{eff} factor is expressed in (NÂˇmÂˇs)/(NÂˇ s^{2}/m) = m^{2}/s. Hence, it does ensure the dimensions on both sides of the equation are, effectively, the same: âĎ/ât is a time derivative and, therefore, its dimension is s^{–}^{1} while, as mentioned above, the dimension of â^{2}Ď 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 SchrĂśdingerâs equation. One may argue, effectively, that its argument, (pâx – Eât)/Ä§, is just a number and, therefore, that the real and imaginary part of Ď is also just some number.
To this, we may object that Ä§ may be looked as a mathematical scaling constant only. If we do that, then the argument of Ď will, effectively, be expressed in action units, i.e. in NÂˇmÂˇs. It then does make sense to also associate a physical dimension with the real and imaginary part of Ď. What could it be?
We may have a closer look at Maxwellâs 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)âe_{x}ĂE, with e_{x} the unit vector pointing in the x-direction (i.e. the direction of propagation of the wave). Hence, we may associate the (1/c)âe_{x}Ă operator, which amounts to a rotation by 90 degrees, with the s/m dimension. Now, multiplication by i also amounts to a rotation by 90Â° degrees. Hence, we may boldly write: B = (1/c)âe_{x}ĂE = (1/c)âiâE. This allows us to also geometrically interpret SchrĂśdingerâs 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 Newtonâs and Coulombâs 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 NÂˇs^{2}/m. Hence, our N/kg dimension becomes:
N/kg = N/(NÂˇs^{2}/m)= m/s^{2}
What is this: m/s^{2}? Is that the dimension of the aÂˇcosÎ¸ term in the aÂˇe^{âiÎ¸ }= aÂˇcosÎ¸ â iÂˇaÂˇsinÎ¸ 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 â matter-particles or particles with zero rest mass (photons) â 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 SchrĂśdingerâs 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 ââ operator is the divergence and, therefore, gives us the magnitude of a (vector) fieldâs 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 [EâE] = [E^{2}] = N^{2}/C^{2}.
- B is measured in (N/C)/(m/s), so we get [BâB] = [B^{2}] = (N^{2}/C^{2})Âˇ(s^{2}/m^{2}). However, the dimension of our c^{2} factor is (m^{2}/s^{2}) and so weâre also left with N^{2}/C^{2}.
- The Ďľ_{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: [Îľ_{0}] = C^{2}/(NÂˇm^{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 Ďľ_{0} for an equivalent constant. We may to give it a try. If the energy densities can be calculated â which are also mass densities, obviously â 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)âiâE or for â(1/c)âiâE 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 aÂˇcosÎ¸ and aÂˇsinÎ¸, 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 !
|Ď|^{2 } = |aÂˇe^{âiâEÂˇt/Ä§}|^{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:
âWhy 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.â (Feynman, Lectures, III-4-1)
The physical interpretation of the wavefunction, as presented here, may provide some better understanding of âthe fundamental principle involvedâ: 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 â or the signal â 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 â 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 SchrĂśdinger’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:
Ď = aÂˇe^{âi[EÂˇt â pâx]/Ä§} = aÂˇe^{âi[EÂˇt â pâx]/Ä§} = aÂˇcos(pâx/Ä§ â Eât/Ä§) + iÂˇaÂˇsin(pâx/Ä§ â Eât/Ä§)
The minus sign in the argument of our sine and cosine function defines the direction of travel: an F(xâvât) wavefunction will always describe some wave that is traveling in the positive x-direction (with c the wave velocity), while an F(x+vât) 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 counterclockwise. 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:
Ď = aÂˇe^{âiâEÂˇt/Ä§} = aÂˇcos(âEât/Ä§) + iÂˇaÂˇsin(âE_{0}ât/Ä§) = aÂˇcos(E_{0}ât/Ä§) â iÂˇaÂˇsin(E_{0}ât/Ä§)
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}/Ä§ pops up as the natural frequency of our matter-particle: (E_{0}/Ä§)ât = Ďât. Remembering the Ď = 2ĎÂˇf = 2Ď/T and T = 1/f formulas, we can associate a period and a frequency with this wave, using the Ď = 2ĎÂˇf = 2Ď/T. Noting that Ä§ = h/2Ď, we find the following:
T = 2ĎÂˇ(Ä§/E_{0}) = h/E_{0} â 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} = ÎťÂˇf = (2Ď/k)Âˇ(Ď/2Ď) = Ď/k. In fact, we’ve got something funny here: the wavenumber k = p/Ä§ 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} = Ď/k = (E/Ä§)/(p/Ä§) = E/p = E/(mÂˇv_{g}) = (mÂˇc^{2})/(mÂˇv_{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 = Î˛_{p} = c/v_{p} = 1/Î˛_{g} = 1/(c/v_{p})
Figure 6: Reciprocal relation between phase and group velocity
We can also write the mentioned relationship as v_{p}Âˇv_{g} = c^{2}, which reminds us of the relationship between the electric and magnetic constant (1/Îľ_{0})Âˇ(1/Îź_{0}) = c^{2}. This is interesting in light of the fact we can re-write this as (cÂˇÎľ_{0})Âˇ(cÂˇÎź_{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/Ä§. 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 ÎxÂˇÎp âĽ Ä§, then neither Îx nor Îp 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 = mÂˇc = mÂˇc^{2}/c = E/c. Using the relationship above, we get:
v_{p} = Ď/k = (E/Ä§)/(p/Ä§) = E/p = c â 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 Ď = aÂˇe^{âi[EÂˇt â pâx]/Ä§} or, for a particle at rest, the Ď = aÂˇe^{âiâEÂˇt/Ä§} function. We must build a wave packet for that: a sum of wavefunctions, each with their own amplitude a_{i}, and their own Ď_{i} = âE_{i}/Ä§. Indeed, in section II, we showed that each of these wavefunctions will 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_{i} as well as E_{i} matter. This may or may not resolve the apparent paradox. Let us look at the group velocity.
To calculate a meaningful group velocity, we must assume the v_{g} = âĎ_{i}/âk_{i} = â(E_{i}/Ä§)/â(p_{i}/Ä§) = â(E_{i})/â(p_{i}) exists. So we must have some dispersion relation. How do we calculate it? We need to calculate Ď_{i} as a function of k_{i}_{ }here, or E_{i} 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 SchrĂśdinger’s equation as we did, i.e. by distinguishing the real and imaginary parts of the âĎ/ât =iÂˇ[Ä§/(2m)]Âˇâ^{2}Ď wave equation and, hence, re-write it as the following pair of two equations:
- Re(âĎ/ât) = â[Ä§/(2m_{eff})]ÂˇIm(â^{2}Ď) â ĎÂˇcos(kx â Ďt) = k^{2}Âˇ[Ä§/(2m_{eff})]Âˇcos(kx â Ďt)
- Im(âĎ/ât) = [Ä§/(2m_{eff})]ÂˇRe(â^{2}Ď) â ĎÂˇsin(kx â Ďt) = k^{2}Âˇ[Ä§/(2m_{eff})]Âˇsin(kx â Ďt)
Both equations imply the following dispersion relation:
Ď = Ä§Âˇk^{2}/(2m_{eff})
Of course, we need to think about the subscripts now: we have Ď_{i}, k_{i}, but… What about m_{eff} or, dropping the subscript, m? Do we write it as m_{i}? If so, what is it? Well… It is the equivalent mass of E_{i} obviously, and so we get it from the mass-energy equivalence relation: m_{i} = E_{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_{i} varies around some average energy E and, therefore, the Uncertainty Principle kicks in.
VII. Explaining spin
The elementary wavefunction vector â i.e. the vector sum of the real and imaginary component â 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 â defined as the vector sum of the electric and magnetic field vectors â 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 Ď = aÂˇe^{âiâEÂˇt/Ä§} as some real vector â as a two-dimensional oscillation of mass, to be precise â 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 = rĂp or, perhaps easier for our purposes here as the product of an angular velocity (Ď) and rotational inertia (I), aka as the moment of inertia or the angular mass. We write:
L = IÂˇĎ
Note we can write L and Ď in boldface here because they are (axial) vectors. If we consider their magnitudes only, we write L = IÂˇĎ (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 = 2ĎÂˇ(Ä§/E_{0}). Hence, the angular velocity must be equal to:
Ď = 2Ď/[2ĎÂˇ(Ä§/E_{0})] = E_{0}/Ä§
We also know the distance r, so that is the magnitude of r in the L = rĂp vector cross-product: it is just a, so that is the magnitude of Ď = aÂˇe^{âiâEÂˇt/Ä§}. Now, the momentum (p) is the product of a linear velocity (v) – in this case, the tangential velocity – and some mass (m): p = mÂˇv. If we switch to scalar instead of vector quantities, then the (tangential) velocity is given by v = rÂˇĎ. So now we only need to think about what we should use for m or, if we want to work with the angular velocity (Ď), 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 â and, therefore, the mass â is distributed uniformly. In that case, we may use the formula for the angular mass of a solid cylinder: I = mÂˇr^{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 = IÂˇĎ = (m_{0}Âˇr^{2}/2)Âˇ(E_{0}/Ä§) = (1/2)Âˇa^{2}Âˇ(E_{0}/c^{2})Âˇ(E_{0}/Ä§) = a^{2}ÂˇE_{0}^{2}/(2ÂˇÄ§Âˇc^{2})
Does it make sense? Maybe. Maybe not. Let us do a dimensional analysis: that wonât check our logic, but it makes sure we made no mistakes when mapping mathematical and physical spaces. We have m^{2}ÂˇJ^{2} = m^{2}ÂˇN^{2}Âˇm^{2} in the numerator and NÂˇmÂˇsÂˇm^{2}/s^{2} in the denominator. Hence, the dimensions work out: we get NÂˇmÂˇs 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 = mc^{2} equation allows us to re-write it as:
L = a^{2}ÂˇE_{0}^{2}/(2ÂˇÄ§Âˇc^{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 NÂˇmÂˇs units, and which can only take on one of two possible values: J = +Ä§/2 and âÄ§/2? It looks like a long shot, right? How do we go from (1/2)Âˇa^{2}Âˇm_{0}^{2}/Ä§ to Âą (1/2)âÄ§? Let us do a numerical example. The energy of an electron is typically 0.510 MeV Âť 8.1871Ă10^{â14} Nâm, and aâŚ 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.Ă10^{â10} Nâm. We get L = a^{2}ÂˇE_{0}^{2}/(2ÂˇÄ§Âˇc^{2}) = 9.9Ă10^{â31} Nâmâs. Now that is about 1.88Ă10^{4} times Ä§/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)Ă10^{â34} joule in energy. So our electron should pack about 1.24Ă10^{â20} oscillations. The angular momentum (L) we get when using the Bohr radius for a and the value of 6.626Ă10^{â34} joule for E_{0} and the Bohr radius is equal to 6.49Ă10^{â59} Nâmâs. So that is the angular momentum per oscillation. When we multiply this with the number of oscillations (1.24Ă10^{â20}), we get about 8.01Ă10^{â51} Nâmâs, so that is a totally different number.
The classical electron radius is about 2.818Ă10^{â15} m. We get an L that is equal to about 2.81Ă10^{â39} Nâmâs, so now it is a tiny fraction of Ä§/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 2.42631Ă10^{â12} m.
This gives us an L of 2.08Ă10^{â33} Nâmâs, which is only 20 times Ä§. 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}ÂˇE_{0}^{2}/(2ÂˇÄ§Âˇc^{2}) = Ä§/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 3.8616Ă10^{â33} m), we get what we should find:
This is a rather spectacular result, and one that would â a priori â 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 Ď = aÂˇe^{âi[EÂˇt â pâx]/Ä§} or, for a particle at rest, the Ď = aÂˇe^{âiâEÂˇt/Ä§} function. We must build a wave packet for that: a sum of wavefunctions, each with their own amplitude a_{i}, and their own Ď_{i} = âE_{i}/Ä§. Each of these wavefunctions will 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 = +Ä§/2 or âÄ§/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:
Ď(Î¸_{i}) = a_{i}Âˇ(cosÎ¸_{i} + iÂˇsinÎ¸_{i})
In contrast, an elementary left-handed wave would be written as:
Ď(Î¸_{i}) = a_{i}Âˇ(cosÎ¸_{i} â iÂˇsinÎ¸_{i})
How does that work out with the E_{0}Âˇt 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 â1, â2, â3, etcetera is just the same. If we count like 1, 2, 3, etcetera, then we write our wavefunction like:
Ď = aÂˇcos(E_{0}ât/Ä§) â iÂˇaÂˇsin(E_{0}ât/Ä§)
If we count time like â1, â2, â3, etcetera then we write it as:
Ď = aÂˇcos(âE_{0}ât/Ä§) â iÂˇaÂˇsin(âE_{0}ât/Ä§)= aÂˇcos(E_{0}ât/Ä§) + iÂˇaÂˇsin(E_{0}ât/Ä§)
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 (+Ä§/2 or âÄ§/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 Feynmanâs Lecture on it (Feynman, III-4), which is confusing and â I would dare to say â 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}Âˇc)Âˇ(v_{group}Âˇc) = 1 â v_{p}Âˇv_{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 â 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 â around any axis â 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 âhookâ 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 SchrĂśdingerâs 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 â aka as the matter-wave equations â one may be tempted to derive the following energy concept:
- E = hÂˇf and p = h/Îť. Therefore, f = E/h and Îť = p/h.
- v = fÂˇÎť = (E/h)â(p/h) = E/p
- p = mÂˇv. Therefore, E = vÂˇp = mÂˇv^{2}
E = mÂˇv^{2}? This resembles the E = mc^{2} equation and, therefore, one may be enthused by the discovery, especially because the mÂˇv^{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 â PE = mÂˇv^{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 â as used in SchrĂśdingerâs equation â is a rather enigmatic concept. To make sure we are making the right analysis here, I should start by noting you will usually see SchrĂśdingerâs 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:
âĎ(x, t)/ât = iÂˇ(1/2)Âˇ(Ä§/m_{eff})Âˇâ^{2}Ď(x, t)
We just moved the iÂˇÄ§ coefficient to the other side, noting that 1/i = –i. Now, in one-dimensional space, and assuming Ď is just the elementary wavefunction (so we substitute aÂˇe^{âiâ[EÂˇt â pâx]/Ä§} for Ď), this implies the following:
âaÂˇiÂˇ(E/Ä§)Âˇe^{â}iâ^{[EÂˇt â pâx]/Ä§} = âiÂˇ(Ä§/2m_{eff})ÂˇaÂˇ(p^{2}/Ä§^{2})Âˇ e^{âiâ[EÂˇt â pâx]/Ä§ }
â E = p^{2}/(2m_{eff}) â m_{eff} = mâ(v/c)^{2}/2 = mâÎ˛^{2}/2
It is an ugly formula: it resembles the kinetic energy formula (K.E. = mâv^{2}/2) but it is, in fact, something completely different. The Î˛^{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} = 2âm_{eff}^{OLD}), as a result of which the formula will look somewhat better:
m_{eff} = mâ(v/c)^{2} = mâÎ˛^{2}
We know Î˛ 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 â which is the equivalent of a source term in a diffusion equation â and that may explain why the above-mentioned m_{eff} = mâ(v/c)^{2} = mâÎ˛^{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: Feynmanâs 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 Ď = aÂˇe^{âiâÎ¸} = aÂˇe^{âi[EÂˇt â pâx]/Ä§} = aÂˇ(cosÎ¸ – iÂˇaÂˇsinÎ¸). We must build a wave packet for that: a sum of wavefunctions, each with its own amplitude a_{k} and its own argument Î¸_{k} = (E_{k}ât – p_{k}âx)/Ä§. 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 Newtonâs force law.
[3] In physics, a two-spring metaphor is more common. Hence, the pistons in the authorâs 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 SchrĂśdingerâs 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: âGod does not play dice.âThe actual quote comes out of one of Einsteinâs 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: Ď is an angular velocity, while v is a linear velocity: Ď 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 = mÂˇa^{2}âĎ^{2}/2. The additional factor (a) is the (maximum) amplitude of the oscillator.
[9] We also have a 1/2 factor in the E = mv^{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 (Îł), we can write the relativistically correct formula for the kinetic energy as K.E. = E â E_{0} = m_{v}c^{2} â m_{0}c^{2} = m_{0}Îłc^{2} â m_{0}c^{2} = m_{0}c^{2}(Îł â 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 = mv^{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 Ď. The kinetic energy of a rotating object is then given by K.E. = (1/2)ÂˇIÂˇĎ^{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 Ď^{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 Ď^{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 Îłm and as R = ÎłL 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^{â8 }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 3.2Ă10^{â8 }seconds (this is the so-called decay time Ď). Now, because the frequency of sodium light is some 500 THz (500Ă10^{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 â riding along the wave as it is being emitted, so to speak â 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)ÂˇmÂˇĎ^{2}Âˇa^{2}Âˇsin^{2}(ĎÂˇt + Î) and the P.E. = U = kÂˇx^{2}/2 = (1/2)Âˇ mÂˇĎ^{2}Âˇa^{2}Âˇcos^{2}(ĎÂˇt + Î) formulas for the linear oscillator.
[17] Feynman further formalizes this in his Lecture on Superconductivity (Feynman, III-21-2), in which he refers to SchrĂśdingerâs equation as the âequation for continuity of probabilitiesâ. The analysis is centered on the local conservation of energy, which confirms the interpretation of SchrĂśdingerâs 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 + iâb and c + iâd are equal if, and only if, their real and imaginary parts are the same. Now, the âĎ/ât = iâ(Ä§/m_{eff})ââ^{2}Ď equation amounts to writing something like this: a + iâb = iâ(c + iâd). Now, remembering that i^{2} = â1, you can easily figure out that iâ(c + iâd) = iâc + i^{2}âd = â d + iâc.
[19] The dimension of B is usually written as N/(mâA), using the SI unit for current, i.e. the ampere (A). However, 1 C = 1 Aâs and, hence, 1 N/(mâA) = 1 (N/C)/(m/s).
[20] Of course, multiplication with i amounts to a counterclockwise 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 SchrĂśdingerâs equation in three-dimensional space: if we look at a clock from the back, then its hand will be moving counterclockwise. When writing B = (1/c)âiâE, 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)âiâE. 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}/(NÂˇm^{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: sinÎ¸ = cos(Î¸âĎ /2).
[24] I must thank a physics blogger for re-writing the 1/(Îľ_{0}ÂˇÎź_{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 90Â° 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. đ Let’s go.
Let’s first look at Feynman’s (simplified) formula for the amplitude of a photon to go from point aÂ to point b. If we identify point aÂ by the position vector r_{1}Â and point bÂ by the position vectorÂ r_{2}, and using Dirac’s fancyÂ bra-ketÂ notation, then it’s written as:
So we have a vector dot product here: pâr_{12}Â = |p|â|r_{12}|ÂˇÂ cosÎ¸ = pâr_{12}ÂˇcosÎą. The angle here (Îą) is the angle between theÂ pÂ andÂ r_{12}Â vector. All good. Well… No. We’ve got a problem. When it comes to calculating probabilities, the Îą angle doesn’t matter: |e^{iÂˇÎ¸}/r|^{2}Â = 1/r^{2}. Hence, for the probability, we get: P = |Â âŠr_{2}|r_{1}âŞ |^{2}Â =Â 1/r_{12}^{2}. Always ! Now that’s strange. The Î¸ =Â pâr_{12}/Ä§Â argument gives us a different phase depending on the angle (Îą) between p and r_{12}. But… Well… Think of it:Â cosÎą goes from 1 to 0 when Îą goes from 0 to Âą90Â° and, of course, is negative when p and r_{12}Â have opposite directions but… Well… According to this formula, the probabilitiesÂ doÂ not depend on the direction of the momentum. That’s just weird, I think. Did Feynman, in his iconicÂ 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:
Ď(x, t) = aÂˇe^{âi}^{âÎ¸}Â =Â aÂˇe^{âi}^{â}^{(Eât âÂ pâx)/Ä§}= aÂˇe^{âi}^{â}^{(Eât)/Ä§}Âˇe^{i}^{â}^{(pâx)/Ä§}
The only difference is that the âŠr_{2}|r_{1}âŞ sort of abstracts away from time, so… Well… Let’s get a feel for the quantities. Let’s think of a photon carryingÂ some typical amount of energy. Hence, let’s talk visible light and, therefore, photons of a few eV only – say 5.625 eV = 5.625Ă1.6Ă10^{â19}Â J = 9Ă10^{â19}Â J. Hence, their momentum is equal to p = E/c = (9Ă10^{â19}Â NÂˇm)/(3Ă10^{5}Â m/s) = 3Ă10^{â24}Â NÂˇs. That’s tiny but that’s only becauseÂ newtonsÂ andÂ secondsÂ 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 experimentalÂ 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 ofÂ Î¸ that is equal to 13.6 million. Hence, theÂ densityÂ 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:Â Î¸ is a phase angle) when we go down to the nanometerÂ scale (10^{â9}Â m) or, even better, theÂ angstroms scale ((10^{â9}Â m).Â
So… Well… Again: what can we do with Feynman’s formula? Perhaps he didn’t give us a propagatorÂ 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 đ – I think… Well… Yes. That’s it. Feynman wants us to think about it. đ Are you joking again, Mr. Feynman?Â đÂ So let’s assume the reasonable thing: let’s assume it gives us the amplitude to go from point a toÂ point bÂ by the position vectorÂ along some path r.Â So, then, in line with what we wrote in our previous post, let’s say pÂˇrÂ (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 writeÂ the formula like this:
ĎÂ =Â aÂˇe^{iÂˇÎ¸}Â = (1/r)Âˇe^{iÂˇS/Ä§}Â =Â e^{iÂˇpâr/Ä§}/r
We’ll use an index to denote the various paths: r_{0}Â is the straight-line path and r_{i}Â is any (other) path.Â Now, quantum mechanics tells us we should calculate this amplitudeÂ 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 PlanckÂ units:Â Î¸ =Â S/Ä§.Â
The time interval is given by tÂ = t_{0Â }=Â 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 specificÂ point in space and in timeÂ to some otherÂ specificÂ point in space and in time. Indeed, when everything is said and done, we do think of light as traveling from pointÂ a to pointÂ bÂ at the speed of light (c). In fact, all of the weird stuff here is all about trying to explain howÂ it does that. đ
Now, if we would think of the photon actually traveling along this or that path, then this implies its velocityÂ along any of the nonlinear paths will be largerÂ thanÂ c, which is OK. That’s just the weirdness of quantum mechanics, and you should actuallyÂ notÂ 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. đ
So the action is energy times time, or momentum times distance. Hence, the difference in action between two paths iÂ andÂ jÂ is given by:
Î´SÂ = pÂˇr_{j}Â âÂ pÂˇr_{i}Â = pÂˇ(r_{j}Â â r_{i}) = pÂˇÎr
I’ll explain theÂ Î´S <Â 2ĎÄ§/3 thing in a moment. Let’s first pause and think about theÂ uncertainty and how we’re modeling it. We can effectively think of the variation in SÂ as some uncertaintyÂ in the action: Î´SÂ = ÎS = pÂˇÎr. However, if SÂ is also equal to energy times time (SÂ = EÂˇt), and we insist tÂ is the same for all paths, then we must have some uncertainty in the energy, right? Hence, we can write Î´SÂ as ÎS = ÎEÂˇt. But, of course, E =Â E =Â mÂˇc^{2}Â = pÂˇc, so we will have an uncertainty in the momentum as well. Hence, the variation inÂ SÂ should be written as:
Î´SÂ = ÎSÂ = ÎpÂˇÎr
That’s just logical thinking: if we, somehow, entertain the idea of a photon going from someÂ specificÂ point in spacetime to some otherÂ specificÂ point in spacetime along various paths, then the variation, or uncertainty,Â in the action will effectively combine some uncertainty in the momentum and the distance. We can calculate Îp as ÎE/c, so we get the following:
Î´SÂ = ÎSÂ = ÎpÂˇÎr =Â ÎEÂˇÎr/c = ÎEÂˇÎt with Ît =Â Îr/c
So we have the two expressions for the Uncertainty Principle here: ÎSÂ = ÎpÂˇÎr =Â ÎEÂˇÎt. Just be careful with the interpretation of Ît: it’s just the equivalent of Îr. We just express the uncertainty in distance in secondsÂ using the (absolute) speed of light. We are notÂ changing our spacetime interval: we’re still looking at a photon going fromÂ aÂ toÂ bÂ inÂ tÂ seconds,Â exactly. Let’s now look at theÂ Î´S <Â 2ĎÄ§/3 thing. If we’re addingÂ twoÂ amplitudes (twoÂ arrowsÂ 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 2Ď/3 rad. The illustration below shows how you can figure that out geometrically.Hence, if S_{0}Â is the action for r_{0}, then S_{1}Â = S_{0}Â + Ä§Â and S_{2}Â = S_{0}Â + 2ÂˇÄ§ are still good, but S_{3}Â = S_{0}Â + 3ÂˇÄ§Â isÂ notÂ good. Why? Because the difference in the phase angles is ÎÎ¸Â =Â S_{1}/Ä§Â âÂ S_{0}/Ä§Â = (S_{0}Â + Ä§)/Ä§Â âÂ S_{0}/Ä§ = 1 andÂ ÎÎ¸ =Â S_{2}/Ä§Â âÂ S_{0}/Ä§Â = (S_{0}Â + 2ÂˇÄ§)/Ä§Â âÂ S_{0}/Ä§ = 2 respectively, so that’s 57.3Â°Â and 114.6Â°Â respectively and that’s, effectively,Â lessÂ than 120Â°. In contrast,Â for the next path, we find that ÎÎ¸Â =Â S_{3}/Ä§Â âÂ S_{0}/Ä§Â = (S_{0}Â + 3ÂˇÄ§)/Ä§Â âÂ S_{0}/Ä§ = 3, so that’s 171.9Â°. 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:Â S_{n}Â = S_{0}Â + n. Of course, nÂ = 1, 2,… etcetera, right? Well… Maybe not. We areÂ measuringÂ action in units ofÂ Ä§, butÂ do we actually think actionÂ comesÂ in units ofÂ Ä§?Â I am not sure. It would make sense, intuitively, butâŚ WellâŚ Thereâs uncertainty on the energy (E) and the momentum (p) of our photon, right? And how accurately can we measure the distance? So thereâs some randomness everywhere. đŚ So let’s leave that question open as for now.
We will also assume that the phase angle forÂ S_{0}Â is equal to 0 (or some multiple of 2Ď, if you want). That’s just a matter of choosing the origin of time. This makes it really easy: ÎS_{n}Â =Â S_{n}Â â S_{0}Â = n, and the associated phase angle Î¸_{n}Â = ÎÎ¸_{n}Â is the same. In short, the amplitude for each path reduces to Ď_{n}Â = e^{iÂˇn}/r_{0}. So we need to add these firstÂ andÂ thenÂ 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 rÂˇe^{iÂˇÎ¸}Â = rÂˇ(cosÎ¸ + iÂˇsinÎ¸) = rÂˇcosÎ¸ + iÂˇrÂˇsinÎ¸ formula. Needless to say, |rÂˇe^{iÂˇÎ¸}|^{2}Â = |r|^{2}Âˇ|e^{iÂˇÎ¸}|^{2}Â = |r|^{2}Âˇ(cos^{2}Î¸ + sin^{2}Î¸) = r. Finally, when adding complex numbers, we add the real and imaginary parts respectively, and we’ll denote the Ď_{0}Â + Ď_{1}Â +Ď_{2}Â + … sum as Î¨.
Now, we also need to see how our ÎSÂ = ÎpÂˇÎrÂ 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:Â ÎS_{n}Â = Îp_{n}ÂˇÎr_{n}Â =Â nÂˇÎp_{1}ÂˇÎr_{1}. It also makes sense that you may want Îp_{n}Â and Îr_{n}Â to be proportional to Îp_{1}Â and Îr_{1}Â respectively. Combining both, the assumption would be this:
Îp_{n}Â =Â ânÂˇÎp_{1Â }andÂ Îr_{n}Â =Â ânÂˇÎr_{1}
So now we just need to decide how we will distribute ÎS_{1}Â =Â Ä§Â = 1 over Îp_{1}Â and Îr_{1}Â respectively. For example, if we’d assume Îp_{1}Â = 1, then Îr_{1}Â = Ä§/Îp_{1}Â = 1/1 = 1. These are the calculations. I will let you analyze them. đWell… We get a weird result. It reminds me ofÂ 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. đ Look at the graph more carefully. The peaks sort of oscillate out so… Well… That might make sense… đ
Does it? Are we doingÂ something wrongÂ here? These amplitudes should reflect the ones that are reflected in those nice animations (like this one, for example, which is part of thatâs part of the Wikipedia article on Feynmanâs 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 distanceÂ rÂ varies as a function ofÂ n.Â
If we’d use a model in which the distance wouldÂ increaseÂ linearly or, preferably, exponentially, then we’d get the result we want to get, right?
Well… Maybe. Let’s try it.Â Hmm… We need to think about the geometry here. Look at the triangle below.Â IfÂ bÂ is the straight-line path (r_{0}), thenÂ acÂ could be one of the crooked paths (r_{n}). To simplify, we’ll assume isosceles triangles, soÂ aÂ equalsÂ cÂ and, hence, r_{n}Â = 2Âˇa = 2Âˇc. We will also assume theÂ successive paths are separated by the same vertical distance (h =Â h_{1}) right in the middle, so h_{b}Â =Â h_{n}Â = nÂˇh_{1}.Â It is then easy to show the following:This gives the following graph for r_{n}Â = 10 and h_{1Â }= 0.01.
Is this the right step increase? Not sure. We can vary the values in our spreadsheet. Let’s first build it. TheÂ 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 (nÂ = 1). In order to cover the extraÂ distance Îr_{1}, the velocity c_{1}Â must be equal to (r_{0}Â + Îr_{1})/tÂ = r_{0}/tÂ + Îr_{1}/t =Â cÂ + Îr_{1}/tÂ = c_{0}Â + Îr_{1}/t. We can write c_{1}Â as c_{1}Â =Â c_{0}Â + Îc_{1}, so Îc_{1}Â = Îr_{1}/t.Â Now, theÂ ratioÂ of p_{1}Â and p_{0}Â will be equal to theÂ ratioÂ of c_{1}Â andÂ c_{0}Â because p_{1}/p_{0Â }= (mc_{1})/mc_{0}) = c_{1}/c_{0}. Hence, we have the following formula for p_{1}:
p_{1}Â = p_{0}Âˇc_{1}/c_{0}Â = p_{0}Âˇ(c_{0}Â + Îc_{1})/c_{0}Â = p_{0}Âˇ[1 + Îr_{1}/(c_{0}Âˇt) = p_{0}Âˇ(1 + Îr_{1}/r_{0})
ForÂ p_{n}, the logic is the same, so we write:
p_{n}Â = p_{0}Âˇc_{n}/c_{0}Â = p_{0}Âˇ(c_{0}Â + Îc_{n})/c_{0}Â = p_{0}Âˇ[1 + Îr_{n}/(c_{0}Âˇt) = p_{0}Âˇ(1 + Îr_{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.Â
Pretty interesting. In fact, this looksÂ reallyÂ good. TheÂ probabilityÂ 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 veryÂ meaningful result with this model. Sweet ! đ I’m lovin’ it ! đ And, here you go, this is (part of) the calculation table, so you can see what I am doing. đ
The graphs below look even better: I just changed the h_{1}/r_{0}Â ratio from 1/100 to 1/10. The probability stabilizes almost immediately. đ 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 ! đ
đ This is good stuff… đ
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}Â = ,Â 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 oneÂ arrow for everyÂ Î´Â Â having oneÂ 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 ĎÂˇh_{n}Âˇr_{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 theÂ weightÂ of the paths that are further removed from the center. So… Well… We shouldn’t be doing that, I guess. đ I’ll you look for the right formula, OK? Let me know when you found it. đ
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. đ
Physical dimensions and Uncertainty
The physical dimension of the quantum of action (h orÂ Ä§ = h/2Ď) is force (expressed in newton)Â times distance (expressed in meter)Â times time (expressed in seconds): NÂˇmÂˇs. Now, you may think this NÂˇmÂˇs 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 NÂˇmÂˇs 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] = NÂˇmÂˇs = (NÂˇm)Âˇs = [E]Âˇ[t]
- [h] = NÂˇmÂˇs = (NÂˇs)Âˇm = [p]Âˇ[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]Âˇ[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:Â [h] = [E]Âˇ[t]. What does it mean?
Energy… Hmm… InÂ 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 theÂ powerÂ 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 385Ă10^{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 someÂ definiteÂ amount of time. So, yes: someÂ definite amount of mass or energy (E) for someÂ definiteÂ amount of time (t).
Let’s bring the mass-energy equivalence formula in here: E = mc^{2}. Hence, the physical dimension of action can also be written as [h] = [E]Âˇ[t] = [mc]^{2}Âˇ[t] = (kgÂˇm^{2}/s^{2})Âˇs =Â kgÂˇm^{2}/s.Â What does that say? Not all that much – for the time being, at least. We can get thisÂ [h] = kgÂˇm^{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 kgÂˇm/s^{2}Â and, hence, the physical dimension of h, or the unit of angular momentum, may also be written as 1 NÂˇmÂˇs = 1 (kgÂˇm/s^{2})ÂˇmÂˇs = 1 kgÂˇm^{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, theÂ newton is aÂ derivedÂ 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:Â a force is what causes an object to deviate from its straight trajectory in spacetime. Hence, we may want to think of theÂ quantum of action as representing three fundamental physical dimensions: (1)Â force, (2)Â 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 isÂ acting on some object over some distance) is energy: 1 jouleÂ (J) =Â 1 newtonÂˇmeter (N). Hence, we may think of the concept of energy as a projectionÂ 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]Âˇ[t]. [Note the square brackets tell us we are looking at aÂ dimensionalÂ 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 = mÂˇv) is expressed in newtonÂˇseconds: 1 kgÂˇm/s = 1 (kgÂˇm/s^{2})Âˇs = 1 NÂˇs. 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 objectÂ during some time.Â The physical dimension of the quantum of action should then be written as [h] = [p]Âˇ[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: timeÂ orÂ 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 spaceÂ 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 actionÂ is so essential and deep in physics:
- In classical physics, the path of some object in a force field will minimizeÂ 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^{âiÂˇÎ¸}Â =Â e^{iÂˇ(S/Ä§)}. Because Ä§ is so tiny, even a small change in S will give a completely different phase angle Î¸. Therefore, most amplitudes will cancel each other out as we take the sum of the amplitudes over all possible paths: only the paths that nearlyÂ 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 Ä§.
The paragraph above summarizes, in essence, Feynman’s path integral formulation of quantum mechanics. We may, therefore, think of the quantum of actionÂ expressingÂ 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 magnitudeÂ of the uncertainty – think of writing something like S Âą Ä§, we may re-write our dimensional [Ä§] = [E]Âˇ[t] and [Ä§] = [p]Âˇ[x] equations as the uncertainty equations:
- ÎEÂˇÎt = Ä§Â
- ÎpÂˇÎx = Ä§
You should note here that it is best to think of the uncertainty relations as aÂ pairÂ of equations, if only because you should also think of the concept of energy and momentum as representing different aspectsÂ of the same reality, as evidenced by the (relativistic) energy-momentum relation (E^{2}Â = p^{2}c^{2}Â â 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 Îx and Ît. If Ît is very small, then Îx will be very large. In order to move over such distance, our particle will require a larger energy, so ÎE will be large. Likewise, if Ît is very large, then Îx will be very small and, therefore, ÎE will be very small. You can also reason in terms of Îx, and talk about momentum rather than energy. You will arrive at the same conclusions: the ÎEÂˇÎt = h and ÎpÂˇÎx = hÂ relations represent two aspects of the same reality – or, at the very least, what we mightÂ thinkÂ of as reality.
Also think of the following: ifÂ ÎEÂˇÎt =Â hÂ and ÎpÂˇÎx =Â h, thenÂ ÎEÂˇÎt =Â ÎpÂˇÎx and, therefore,Â ÎE/Îp must be equal to Îx/Ît. Hence, theÂ ratioÂ of the uncertainty about x (the distance) and the uncertainty about t (the time) equals theÂ ratioÂ 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 energyÂ too.
Action, energy and mass
What isÂ energy, really? InÂ 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 theÂ powerÂ 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 BroglieÂ 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 restÂ 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. đ But, obviously, Einstein’s mass-equivalence formula comes to mind here, and summarizes it all:
E = mÂˇc^{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,Â really? What’s thatÂ equivalenceÂ between mass and energy,Â 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 theÂ dimensional analysis:
[E] = [m]Âˇ[c^{2}] = 1 kgÂˇm^{2}/s^{2}Â = 1 kgÂˇm/s^{2}Âˇm = 1 NÂˇm
The kg and m/s^{2Â }factors make this abundantly clear: m/s^{2}Â 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 = hÂˇf = ĎÂˇÄ§. We could also write: E = h/T. Needless to say, T = 1/fÂ is theÂ periodÂ 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)Ă10^{â34}Â JÂˇs. Now, fÂ is the number of oscillationsÂ per second. Let’s write it asÂ fÂ = n/s, so we get:
E/fÂ = E/(n/s) = EÂˇs/nÂ = 6.626070040(81)Ă10^{â34}Â JÂˇs â E/nÂ = 6.626070040(81)Ă10^{â34}Â J
What an amazing result! Our wavicle – be it a photon or a matter-particle – will alwaysÂ packÂ 6.626070040(81)Ă10^{â34}Â jouleÂ inÂ oneÂ oscillation, so that’s the numericalÂ value of Planck’s constant which, of course, depends on our fundamentalÂ units (i.e. kg, meter, second, etcetera in the SI system).
Of course, the obvious question is: what’s oneÂ 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 toÂ countingÂ them:Â nÂ oscillations per second, so that’sÂ 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: itâs of the order of 10^{8Â }(see hisÂ Lectures,Â I-33-3: itâs a wonderfully simple exercise, and one that really shows his greatness as a physics teacher), so… Well… This wave train will last about 10^{â8Â }seconds (thatâs the time it takes for the radiation to die out by a factor 1/e). To give a somewhat more precise example,Â for sodium light, which has a frequency of 500 THz (500Ă10^{12Â }oscillations per second) and a wavelength of 600 nm (600Ă10^{â9Â }meter), the radiation will lasts about 3.2Ă10^{â8Â }seconds. [In fact, thatâs the time it takes for the radiationâs energy to die out by a factor 1/e, so(i.e. the so-called decay time Ď), so the wavetrain will actually lastÂ longer, but so the amplitude becomes quite small after that time.]Â So… Well… Thatâs 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 (3Ă10^{8Â }m/s), that makes for a wave train with a length of, roughly,Â 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 photonÂ – so if we’d be traveling at speed c,â ridingâ with the photon, so to say, as itâs being emitted – then we’d âseeâ the electromagnetic transient as itâs being radiated into space.
However, while we can associate some massÂ with the energy of the photon, none of what I wrote above explains what the (rest) mass of a matter-particle could possibly be.Â 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? đŚ 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? đ
The reality of the wavefunction
The wavefunction is, obviously, a mathematical construct: aÂ descriptionÂ 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 globalÂ tool of communication, at least.
The realÂ question is: is the descriptionÂ accurate? Does it match reality and, if it does, howÂ goodÂ is the match? For example, the wavefunction for an electron in a hydrogen atom looks as follows:
Ď(r, t) = e^{âiÂˇ(E/Ä§)Âˇt}Âˇf(r)
As I explained in previous posts (see, for example, my recent postÂ on reality and perception), theÂ f(r) function basically provides some envelope for the two-dimensional e^{âiÂˇÎ¸}Â =Â e^{âiÂˇ(E/Ä§)Âˇt}Â = cosÎ¸ + iÂˇsinÎ¸Â oscillation, with rÂ = (x, y, z),Â Î¸ = (E/Ä§)ÂˇtÂ = ĎÂˇtÂ and Ď = E/Ä§. So it presumes theÂ 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 rÂ = (x, y, z) coordinates. đ So… Well… If each oscillation is to alwaysÂ packÂ 6.626070040(81)Ă10^{â34}Â 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. đ The associated energy is the same, but… Well… Reality is probablyÂ notÂ the nice geometrical shape we associate with those wavefunctions.
In addition, we should think of theÂ Uncertainty Principle: thereÂ mustÂ 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 ofÂ hÂ (or Ä§ = h/2Ď) which, as mentioned above, is the (average) energy of oneÂ oscillation only, then we don’t have much of a problem here, do we? đ
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 ourÂ x- and z-axis – as shown below. Two axes only. So we may want to think of a two-dimensionalÂ oscillation as an oscillation of the polar and azimuthal angle. đ
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 theÂ elementaryÂ wavefunctionÂ Ď =Â aÂˇe^{âiÎ¸Â }= Ď =Â aÂˇe^{âiÂˇÎ¸Â }Â = aÂˇe^{âi}^{(}^{ĎÂˇtâkÂˇx}^{)}Â = aÂˇe^{â}^{(}^{i}^{/Ä§)Âˇ}^{(EÂˇ}^{tâpÂˇx}^{)}Â .We know this elementary wavefunction cannotÂ represent a real-lifeÂ particle. Indeed, the aÂˇe^{âiÂˇÎ¸Â }function implies the probability of finding the particle – an electron, a photon, or whatever – would be equal to P(x, t) = |Ď(x, t)|^{2}Â = |aÂˇe^{â}^{(}^{i}^{/Ä§)Âˇ}^{(EÂˇ}^{tâpÂˇx}^{)}|^{2}Â = |a|^{2}Âˇ|e^{â}^{(}^{i}^{/Ä§)Âˇ}^{(EÂˇ}^{tâpÂˇx}^{)}|^{2}Â = |a|^{2}Âˇ1^{2}= a^{2}Â 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 addÂ 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 theÂ contributionÂ to the composite wave, which – in three-dimensional space – we can write as:
Ď(r, t) = e^{âiÂˇ(E/Ä§)Âˇt}Âˇf(r)
As I explained in previous posts (see, for example, my recent postÂ on reality and perception), theÂ f(r) function basically provides some envelope for the two-dimensional e^{âiÂˇÎ¸}Â =Â e^{âiÂˇ(E/Ä§)Âˇt}Â = cosÎ¸ + iÂˇsinÎ¸Â oscillation, with rÂ = (x, y, z),Â Î¸ = (E/Ä§)ÂˇtÂ = ĎÂˇtÂ and Ď = E/Ä§.
Note that it looks like the wave propagatesÂ from left to right – in theÂ 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 theÂ rotationÂ of that arrow at the center – i.e. with the motion in the illustration,Â 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 towardsÂ us –Â would then be the y-axis, obviously.Â 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 ofÂ rotationÂ 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 towardsÂ 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 negativeÂ x-direction. [Note that this left- or right-hand rule is quite astonishing: simply swapping the direction ofÂ oneÂ 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 toÂ go in this or that direction. And I mean that: there is no real travelingÂ 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 theÂ groupÂ or theÂ envelopeÂ of the waveâwhatever you want to call it) moves to the right. The point is: the pulse itself doesn’tÂ 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 motionÂ that is traveling down the rope.Â 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 âtravelingâ left or right. Thatâs why thereâs no limit on the phase velocity: it canÂ – and, according to quantum mechanics, actually willÂ –Â exceed the speed of light. In contrast, the groupÂ velocity – which is the actual speed of the particle that is being represented by the wavefunction – may approachÂ – or, in the case of a massless photon, will actually equalÂ –Â the speed of light, but will never exceedÂ it, and itsÂ directionÂ will, obviously, have aÂ physicalÂ 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 theÂ spinÂ 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 aÂˇe^{âi}^{(}^{ĎÂˇtâkÂˇx}^{)}Â rather thanÂ aÂˇe^{i}^{(}^{ĎÂˇt+kÂˇx}^{)}Â orÂ aÂˇe^{i}^{(}^{ĎÂˇtâkÂˇx}^{)}: it’s just because of the right-hand rule for coordinate frames, and also because Euler defined the counter-clockwise direction as theÂ 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 = mÂˇc^{2}Â 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 aÂˇe^{â}^{(}^{i}^{/Ä§)Âˇ}^{(EÂˇ}^{tâpÂˇx}^{)}Â reduces to aÂˇe^{â}^{(}^{i}^{/Ä§)Âˇ}^{(E’Âˇ}^{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 properÂ 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 massÂ toÂ accelerate. To be precise, theÂ newtonÂ – as the unit of force – is defined as theÂ magnitude of a force which would cause a mass of one kg to accelerate with one meter per secondÂ per second. Per second per second. This is not a typo: 1 N corresponds to 1 kg times 1 m/sÂ per second, i.e. 1 kgÂˇm/s^{2}. So… Because energy is force times distance, the unit of energyÂ may be expressed in units of kgÂˇm/s^{2}Âˇm, or kgÂˇm^{2}/s^{2}, i.e. the unit of mass times the unit ofÂ velocity squared. To sum it all up:
1 J = 1 NÂˇm = 1 kgÂˇ(m/s)^{2}
This reflects the physical dimensionsÂ on both sides of theÂ E = mÂˇc^{2}Â formula again but… Well… How should weÂ interpretÂ this? Look at the animation below once more, and imagine the green dot is some tinyÂ massÂ moving around the origin, in an equally tiny circle. We’ve gotÂ twoÂ oscillations here: each packingÂ halfÂ of the total energy of… Well… Whatever it is that our elementary wavefunction might represent in realityÂ – which we don’t know, of course.
Now, the blue and the red dot – i.e. the horizontal and vertical projectionÂ of the green dot –Â accelerate up and down. If we look carefully, we see these dots accelerateÂ towardsÂ the zero point and, once they’ve crossed it, theyÂ 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 90Â° 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,Â 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 asÂ 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 unitÂ 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: torqueÂ over some angle.Â Indeed, the analogy between linear and angular movement is obvious: theÂ kineticÂ energy of a rotating object is equal to K.E. = (1/2)ÂˇIÂˇĎ^{2}. In this formula, I is the rotational inertiaÂ – i.e. the rotational equivalent of mass – and Ď is the angular velocity – i.e. the rotational equivalent of linearÂ 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 structurallyÂ similar to theÂ E = mÂˇc^{2}Â formula. But isÂ it the same? Is the effective mass of some object the sum of an almost infinite number of quantaÂ that incorporate some kind ofÂ rotationalÂ 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 toÂ articulateÂ orÂ imagineÂ what might be going on. đ 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, itsÂ lengthÂ 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’sÂ notÂ 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. đThe 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 transverseÂ waves, as opposed to longitudinal.Â See how string theory starts making sense? đ
The most fundamental question remains the same: what is it,Â exactly, that is oscillating here? What is theÂ 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 isÂ electricÂ charge,Â really? So the question is: what’s theÂ realityÂ of mass, of electric charge, or whatever other charge that causes a force toÂ actÂ on it?
If youÂ know, please letÂ meÂ know. đ
Post scriptum: The fact that we’re talking someÂ two-dimensional oscillation here – think of a surface now – explains the probability formula: we need toÂ squareÂ 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 involvingÂ Ď somewhere, because we’re talking someÂ circularÂ surface – as opposed to a rectangular one. But I’ll letÂ youÂ figure that out. đ
Re-visiting electron orbitals (III)
In my previous post, I mentioned that it wasÂ not so obvious (both from a physicalÂ as well as from aÂ mathematicalÂ point of view) to write the wavefunction for electron orbitals – which we denoted as Ď(x, t), i.e. a function of two variables (or four: one time coordinate and three space coordinates) –Â as the product of two other functions in one variable only.
[…] OK. The above sentence is difficult to read. Let me write in math. đ It isÂ notÂ so obvious to write Ď(x, t) as:
Ď(x, t) = e^{âiÂˇ(E/Ä§)Âˇt}ÂˇĎ(x)
As I mentioned before, the physicists’ use of the same symbol (Ď, psi) for both the Ď(x, t) and Ď(x) function is quite confusing – because the two functions areÂ veryÂ different:
- Ď(x, t) is a complex-valued function of twoÂ (real)Â variables: x and t. OrÂ four, I should say, because xÂ = (x, y, z) – but it’s probably easier to think of x as oneÂ vectorÂ variable – aÂ vector-valued argument, so to speak. And then t is, of course, just aÂ scalarÂ variable. So… Well… A function of twoÂ variables: the position in space (x), and time (t).
- In contrast, Ď(x) is a real-valuedÂ function ofÂ oneÂ (vector) variable only: x, so that’s the position in space only.
Now you should cry foul, of course: Ď(x) is notÂ necessarilyÂ real-valued. It mayÂ be complex-valued. You’re right.Â You know the formula:Note the derivation of this formula involved a switch from Cartesian to polar coordinates here, so from xÂ = (x, y, z) to rÂ = (r, Î¸, Ď), and that the function is also a function of the twoÂ quantum numbersÂ l and m now, i.e. the orbital angular momentum (l) and its z-component (m) respectively. In my previous post(s), I gave you the formulas for Y_{l,m}(Î¸, Ď) and F_{l,m}(r) respectively. F_{l,m}(r) was a real-valued function alright, but the Y_{l,m}(Î¸, Ď) had that e^{iÂˇmÂˇĎ}Â factor in it. So… Yes. You’re right: the Y_{l,m}(Î¸, Ď) function is real-valued if – and onlyÂ if – m = 0, in which case e^{iÂˇmÂˇĎ}Â = 1.Â Let me copy the table from Feynman’s treatment of the topic once again:The P_{l}^{m}(cosÎ¸) functions are the so-called (associated) Legendre polynomials, and the formula for these functions is rather horrible:Don’t worry about it too much: just note the P_{l}^{m}(cosÎ¸)Â is aÂ real-valuedÂ function. The point is the following:theÂ Ď(x, t) is a complex-valuedÂ function because – andÂ onlyÂ because – we multiply a real-valued envelope function – which depends on positionÂ only – with e^{âiÂˇ(E/Ä§)Âˇt}Âˇe^{iÂˇmÂˇĎ}Â = e^{âiÂˇ[(E/Ä§)ÂˇtÂ âÂ }^{mÂˇĎ]}.
[…]
Please read the above once again and – more importantly – think about it for a while. đ You’ll have to agree with the following:
- As mentioned in my previous post,Â the e^{iÂˇmÂˇĎ}Â factor just gives us phase shift: just aÂ re-set of our zero point for measuring time, so to speak, and the whole e^{âiÂˇ[(E/Ä§)ÂˇtÂ âÂ }^{mÂˇĎ]}Â factor just disappears when weâre calculating probabilities.
- The envelope function gives us the basic amplitude – in theÂ classicalÂ sense of the word:Â the maximum displacement fromÂ theÂ zeroÂ value. And so it’s that e^{âiÂˇ[(E/Ä§)ÂˇtÂ âÂ }^{mÂˇĎ]}Â that ensures the whole expression somehow captures the energyÂ of the oscillation.
Let’s first look at the envelope function again. Let me copy the illustration forÂ n = 5 and lÂ = 2 from aÂ Wikimedia CommonsÂ article.Â Note the symmetry planes:
- Any plane containing theÂ z-axis is a symmetry plane – like a mirror in which we can reflect one half of theÂ shape to get the other half. [Note that I am talking theÂ shapeÂ only here. Forget about the colors for a while – as these reflect the complex phase of the wavefunction.]
- Likewise, the plane containingÂ bothÂ the x– and the y-axis is a symmetry plane as well.
The first symmetry plane – or symmetryÂ line, really (i.e. theÂ z-axis) – should not surprise us, because the azimuthal angle Ď is conspicuously absent in the formula for our envelope function if, as we are doing in this article here, we merge theÂ e^{iÂˇmÂˇĎ}Â factor with the e^{âiÂˇ(E/Ä§)Âˇt}, so it’s just part and parcel of what the author of the illustrations above refers to as the ‘complex phase’ of our wavefunction.Â OK. Clear enough – I hope. đ But why is theÂ the xy-plane a symmetry plane too? We need to look at that monstrous formula for the P_{l}^{m}(cosÎ¸) function here: just note the cosÎ¸ argument in it is being squaredÂ before it’s used in all of the other manipulation. Now, we know that cosÎ¸ = sin(Ď/2Â âÂ Î¸). So we can define someÂ newÂ angle – let’s just call it Îą – which is measured in the way we’re used to measuring angle, which is notÂ from the z-axis but from the xy-plane. So we write: cosÎ¸ = sin(Ď/2Â âÂ Î¸) = sinÎą. The illustration below may or may not help you to see what we’re doing here.So… To make a long story short, we can substitute the cosÎ¸ argument in the P_{l}^{m}(cosÎ¸) function for sinÎą = sin(Ď/2Â âÂ Î¸). Now, if the xy-plane is a symmetry plane, then we must find the same value for P_{l}^{m}(sinÎą) and P_{l}^{m}[sin(âÎą)]. Now, that’s not obvious, because sin(âÎą) = âsinÎą â Â sinÎą. However, because the argument in that P_{l}^{m}(x) function is being squared before any other operation (like subtracting 1 and exponentiating the result), it is OK: [âsinÎą]^{2}Â = [sinÎą]^{2Â }=Â sin^{2}Îą. […] OK, I am sure the geeks amongst my readers will be able to explain this more rigorously. In fact, I hope they’ll have a look at it, because there’s also that d^{l+m}/dx^{l+m}Â operator, and so you should check what happens with the minus sign there. đ
[…] Well… By now, you’re probably totally lost, but the fact of the matter is that we’ve got a beautiful result here. Let me highlight the most significant results:
- AÂ definiteÂ energy state of a hydrogen atom (or of an electron orbiting around some nucleus, I should say) appears to us as some beautifully shaped orbital – an envelopeÂ function in three dimensions, really – whichÂ has the z-axis – i.e. the vertical axis – as a symmetry line and the xy-plane as a symmetry plane.
- The e^{âiÂˇ[(E/Ä§)ÂˇtÂ âÂ }^{mÂˇĎ]}Â factor gives us the oscillation within the envelope function. As such, it’s this factor that, somehow,Â captures the energyÂ of the oscillation.
It’s worth thinking about this. Look at the geometry of the situation again – as depicted below. We’re looking at the situation along the x-axis, in the direction of the origin, which is the nucleus of our atom.
The e^{iÂˇmÂˇĎ}Â factor just gives us phase shift: just aÂ re-set of our zero point for measuring time, so to speak. Interesting, weird – but probably less relevant than the e^{âiÂˇ[(E/Ä§)Âˇt}Â factor, which gives us the two-dimensional oscillation that captures the energy of the state.
Now, the obvious question is: the oscillation of what, exactly? I am not quite sure but – as I explained in my Deep BlueÂ page – the real and imaginary part of our wavefunction are really like the electric and magnetic field vector of an oscillating electromagnetic field (think of electromagnetic radiation – if that makes it easier). Hence, just like the electric and magnetic field vector represent some rapidly changing forceÂ on a unit charge, the real and imaginary part of our wavefunction must also represent some rapidly changingÂ forceÂ on… Well… I am not quite sure on what though. The unit charge is usually defined as the charge of a proton – rather than an electron – but then forces act on some mass, right? And the massÂ of a proton is hugely different from the mass of an electron. The same electric (or magnetic) force will, therefore, give a hugely different acceleration to both.
So… Well… My guts instinct tells me the real and imaginary part of our wavefunction just represent, somehow, a rapidly changing force on some unit ofÂ mass, but then I am not sure how to define that unit right now (it’s probably notÂ the kilogram!).
Now, there is another thing we should note here: we’re actually sort of de-constructing a rotationÂ (look at the illustration above once again) in two linearly oscillating vectors – one along the z-axis and the other along the y-axis.Â Hence, in essence, we’re actually talking about something that’s spinning.Â In other words, we’re actually talking someÂ torqueÂ around the x-axis. In what direction? I think that shouldn’t matter – that we can write E or âE, in other words, but… Well… I need to explore this further – as should you! đ
Let me just add one more note on the e^{iÂˇmÂˇĎ}Â factor. It sort of defines the geometryÂ of the complex phase itself. Look at the illustration below. Click on it to enlarge it if necessary – or, better still, visit the magnificent Wikimedia Commons article from which I get these illustrations. These are the orbitals nÂ = 4 and lÂ = 3. Look at the red hues in particular – or the blue – whatever: focus on one color only, and see how how – for mÂ = Âą1, we’ve got one appearance of that color only. For mÂ = Âą1, the same color appears at two ends of the ‘tubes’ – or toriÂ (plural of torus), I should say – just to sound more professional. đ For mÂ = Âą2, the torus consists of three parts – or, in mathematical terms, we’d say the order of its rotational symmetryÂ is equal to 3.Â Check that Wikimedia Commons article for higher values ofÂ nÂ andÂ l: the shapes become very convoluted, but the observation holds. đ
Have fun thinking all of this through for yourself – and please do look at those symmetries in particular. đ
Post scriptum: You should do some thinking on whether or not theseÂ mÂ =Â Âą1, Âą2,…, ÂąlÂ orbitals are really different. As I mentioned above, a phase difference is just what it is: a re-set of the t = 0 point. Nothing more, nothing less. So… Well… As far as I am concerned, that’s notÂ aÂ realÂ difference, is it? đ As with other stuff, I’ll let you think about this for yourself.
Feynman’s Seminar on Superconductivity (1)
The ultimate challenge for students of Feynman’s iconic LecturesÂ series is, of course, to understand his final one: A Seminar on Superconductivity. As he notes in his introduction to this formidably dense piece, the text does notÂ present the detail of each and every step in the development and, therefore,Â we’re not supposed to immediately understand everything. As Feynman puts it: we should just believe (more or less) that things would come out if we wouldÂ be able to go through each and every step. Well… Let’s see. Feynman throws a lot of stuff in hereâincluding, I suspect, some stuff that may not beÂ directlyÂ relevant, but that he sort of couldn’t insert into all of his otherÂ Lectures. So where do we start?
It took me one long maddening day to figure out the first formula:It says that the amplitude for a particle to go from aÂ toÂ bÂ in a vector potential (think of a classical magnetic field) is the amplitude for the same particle to go fromÂ aÂ toÂ b when there is no field (A = 0) multiplied by the exponential of the line integral of the vector potential times the electric charge divided by Planck’s constant. I stared at this for quite a while, but thenÂ I recognized the formula for the magnetic effect on an amplitude, which I described in my previous post, which tells us that a magnetic field will shiftÂ the phase of the amplitude of a particle with an amount equal to:
Hence, if we write âŠb|aâŞ for A = 0 as âŠb|aâŞ_{A = 0}Â = CÂˇe^{i}^{Î¸}, then âŠb|aâŞÂ in AÂ will, naturally, be equal to âŠb|aâŞÂ _{in A }= CÂˇe^{i}^{(}^{Î¸+Ď)}Â = CÂˇe^{i}^{Î¸}Âˇe^{i}^{Ď}Â =Â âŠb|aâŞ_{A = 0Â }Âˇe^{i}^{Ď}, and so that explains it. đ Alright… Next. Or… Well… Let us briefly re-examine the concept of the vector potential, because we’ll need it a lot. We introduced it in our post on magnetostatics. Let’s briefly re-cap the development there. In Maxwell’s set of equations, two out of the four equations give us the magnetic field:Â ââ˘BÂ = 0 andÂ c^{2}âĂBÂ =Â j/Îľ_{0}. We noted the following in this regard:
- TheÂ ââ˘BÂ = 0 equation is true,Â always, unlike theÂ âĂEÂ = 0 expression, which is true for electrostatics only (no moving charges). So theÂ ââ˘BÂ = 0 equation says theÂ divergenceÂ ofÂ BÂ is zero,Â always.
- The divergence of the curl of a vector field is always zero. Hence, if AÂ is some vector field, thenÂ div(curlÂ A) =Â ââ˘(âĂA) = 0,Â always.
- We can now apply another theorem: if the divergence of a vector field, sayÂ D, is zeroâso ifÂ ââ˘DÂ = 0âthenÂ DÂ will be theÂ the curl of some other vector fieldÂ C, so we can write:Â DÂ =Â âĂC. Â Applying this toÂ ââ˘BÂ = 0, we can write:Â
IfÂ ââ˘BÂ = 0, then there is anÂ AÂ such thatÂ BÂ =Â âĂA
So, in essence, we’re just re-defining the magnetic field (B) in terms of some other vector field. To be precise, we write it as the curlÂ of some other vector field, which we refer to asÂ the (magnetic)Â vectorÂ potential. The components of the magnetic field vector can then be re-written as:
We need to note an important point here: the equations above suggest that the components of BÂ depend on position only. In other words, we assumeÂ staticÂ magnetic fields, so theyÂ doÂ notÂ change with time. That, in turn, assumes steady currents. We will want to extend the analysis to also includeÂ magnetodynamics. It complicates the analysis but… Well… Quantum mechanicsÂ isÂ complicated. Let us remind ourselves here of Feynman’s re-formulation of Maxwell’s equations as a set ofÂ twoÂ equations (expressed in terms of the magnetic (vector) and the electric potential) only:
These equations are wave equations, as you can see by writing out the second equation:
It is a wave equation in three dimensions. Note that, even in regions where we do no have any charges or currents, we have non-zeroÂ solutions for ĎÂ andÂ A. These non-zero solutions are, effectively, representing the electric and magnetic fields as they travel through free space. As Feynman notes, the advantage of re-writing Maxwell’s equations as we do above, is that the two new equations make it immediately apparent that we’re talking electromagnetic waves, really. As he notes, for many practical purposes, it will still be convenient to use the original equations in terms of E and B, but… Well… Not in quantum mechanics, it turns out. As Feynman puts it: “E and BÂ are on the other side of the mountain we have climbed. Now we are ready to cross over to the other side of the peak. Things will look differentâwe are ready for some new and beautiful views.”
Well… Maybe. Appreciating those views, as part of our study of quantum mechanics, does take time and effort, unfortunately. đŚ
The SchrĂśdinger equation in an electromagnetic field
Feynman then jots down SchrĂśdinger’s equation for the same particle (with chargeÂ q) moving in an electromagnetic field that is characterized not only by the (scalar) potentialÂ ÎŚ but also by a vector potentialÂ A:
Now where doesÂ thatÂ come from? We know the standard formula in anÂ electricÂ field, right? It’s the formula we used to find the energy states of electrons in a hydrogen atom:
iÂˇÄ§ÂˇâĎ/ât = â(1/2)Âˇ(Ä§^{2}/m)â^{2}Ď + VÂˇĎ
Of course, it is easy to see that we replaced V by qÂˇÎŚ, which makes sense: the potential of a charge in an electric field is the product of the charge (q) and the (electric) potential (ÎŚ), because ÎŚ is, obviously, the potential energy of the unitÂ charge. It’s also easy to see we can re-write âÄ§^{2}Âˇâ^{2}Ď as [(Ä§/i)Âˇâ]Âˇ[(Ä§/i)Âˇâ]Ď because (1/i)Âˇ(1/i) = 1/i^{2}Â = 1/(â1) = â1. đ Alright. So it’s just that âqÂˇAÂ term in the (Ä§/i)â â qÂˇAÂ expression that we need to explain now.
Unfortunately, that explanation isÂ notÂ so easy. Feynman basically re-derives SchrĂśdinger’s equation using his trade-markÂ historical argument – which did notÂ include any magnetic field –Â withÂ a vector potential. The re-derivation is rather annoying, and I didn’t have the courage to go through it myself, so you should – just like me – just believe Feynman when he says that, when there’s a vector potential – i.e. when there’s a magnetic field – then that (Ä§/i)Âˇâ operatorÂ – which is the momentum operator– ought to be replaced by a new momentum operator:
So… Well… There we are… đ So far, so good? Well… Maybe.
While, as mentioned, you won’t be interested in the mathematical argument, it is probably worthwhile to reproduce Feynman’s more intuitive explanation of why the operator above is what it is. In other words, let us try to understand thatÂ âqA term. Look at the following situation: we’ve got a solenoid here, and some current I is going through it so there’s a magnetic field B. Think of theÂ dynamicsÂ while weÂ turn onÂ this flux. Maxwell’s second equation (âĂE =Â ââB/ât) tells us theÂ line integral ofÂ EÂ around a loop will be equal to the timeÂ rate of change of the magnetic flux through that loop. TheÂ âĂE =Â ââB/ât equation is aÂ differentialÂ equation, of course, so it doesn’t have the integral, but you get the ideaâI hope.
Now, using the BÂ =Â âĂA equationÂ we can re-write theÂ âĂE =Â ââB/ât asÂ âĂE =Â ââ(âĂA)/ât. This allows us to write the following:
Â âĂE =Â ââ(âĂA)/ât =Â ââĂ(âA/ât)Â â E =Â ââA/ât
This is a remarkable expression. Note its derivation is based on the commutativity of the curl and time derivative operators, which is a property that can easily be explained: if we have a function in two variablesâsay x and tâthen the order of the derivation doesn’t matter: we can first take the derivative with respect toÂ xÂ and then toÂ tÂ or, alternatively, we can first take the time derivative and then do theÂ â/âx operation. So… Well… The curl is, effectively, a derivative with regard to the spatial variables. OK. So what? What’s the point?
Well… If we’d have some charge q, as shown in the illustration above, that would happen to be there as the flux is being switched on, it will experience a force which is equal to F = qE. We can now integrate this over the time interval (t) during which the flux is being built up to get the following:
âŤ_{0}^{t}Â F =Â âŤ_{0}^{t}Â mÂˇa =Â âŤ_{0}^{t}Â mÂˇdv/dt =Â mÂˇv_{t}= âŤ_{0}^{t}Â qÂˇEÂ =Â ââŤ_{0}^{t}Â qÂˇâA/ât =Â âqÂˇA_{t}
AssumingÂ v_{0}Â and A_{0Â }are zero, we may drop the time subscript and simply write:
mÂˇvÂ =Â âqÂˇA
The point is: during the build-up of the magnetic flux, our charge will pick up some (classical)Â momentumÂ that is equal to p =Â mÂˇvÂ =Â âqÂˇA. So… Well… That sort of explains the additional term in our new momentum operator.
Note: For some reason I don’t quite understand, Feynman introduces the weird concept of ‘dynamical momentum’, which he defines as the quantityÂ mÂˇvÂ + qÂˇA, so that quantity must be zero in the analysis above. I quickly googled to see why but didn’t invest too much time in the research here. It’s just… Well… A bit puzzling. I don’t really see the relevance of his point here: I am quite happy to go along with the new operator, as it’s rather obvious that introducingÂ changingÂ magnetic fields must, obviously, also have some impact on our wave equationsâin classical as well as in quantum mechanics.
Local conservation of probability
The title of this section in Feynman’s Lecture (yes, still the same Lecture – we’re not switching topics here) is theÂ equation of continuity for probabilities. I find it brilliant, because it confirmsÂ myÂ interpretation of the wave function as describing some kind of energy flow. Let me quote Feynman on his endeavor here:
“An important part of the SchrĂśdinger equation for a single particle is the idea that the probability to find the particle at a position is given by the absolute square of the wave function. It is also characteristic of the quantum mechanics that probability is conserved in a local sense. When the probability of finding the electron somewhere decreases, while the probability of the electron being elsewhere increases (keeping the total probability unchanged), something must be going on in between. In other words, the electron has a continuity in the sense that if the probability decreases at one place and builds up at another place, there must be some kind of flow between. If you put a wall, for example, in the way, it will have an influence and the probabilities will not be the same. So the conservation of probability alone is not the complete statement of the conservation law, just as the conservation of energy alone is not as deep and important as the local conservation of energy.Â If energy is disappearing, there must be a flow of energy to correspond. In the same way, we would like to find a âcurrentâ of probability such that if there is any change in the probability density (the probability of being found in a unit volume), it can be considered as coming from an inflow or an outflow due to some current.”
This is it, really ! The wave function does represent some kind of energy flow – between a so-called ‘real’ and a so-called ‘imaginary’ space, which are to be defined in terms of directional versus rotational energy, as I try to point out – admittedly: more by appealing to intuition than to mathematical rigor – in that post of mine on the meaning of the wavefunction.
So what is theÂ flowÂ – or probabilityÂ currentÂ as Feynman refers to it? Well… Here’s the formula:
Huh?Â Yes. Don’t worry too much about it right now. The essential point is to understand what this current – denoted by J – actually stands for:
So what’s next? Well… Nothing. I’ll actually refer you to Feynman now, because I can’t improve on howÂ heÂ explains howÂ pairs of electronsÂ start behaving when temperatures are low enough to render Boltzmann’s Law irrelevant: the kinetic energy that’s associated with temperatureÂ can no longer break up electron pairs if temperature comes close to the zero point.
Huh? What? Electron pairs? Electrons are not supposed to form pairs, are they? They carry the same charge and are, therefore, supposed to repel each other. Well… Yes and no. In my post on the electron orbitals in a hydrogen atom – which just presented Feynman’s presentationÂ on the subject-matter in a, hopefully, somewhat more readable format – we calculated electron orbitals neglecting spin. In Feynman’s words:
“We make anotherÂ approximation by forgetting that the electron has spin. […] The non-relativistic SchrĂśdinger equation disregards magnetic effects. [However] Small magnetic effects [do] occur because, from the electronâs point-of-view, the proton is a circulating charge which produces a magnetic field. In this field the electron will have a different energy with its spin up than with it down. [Hence] The energy of the atom will be shifted a little bit from what we will calculate. We will ignore this small energy shift. Also we will imagine that the electron is just like a gyroscope moving around in space always keeping the same direction of spin. Since we will be considering a free atom in space the total angular momentum will be conserved. In our approximation we will assume that the angular momentum of the electron spin stays constant, so all the rest of the angular momentum of the atomâwhat is usually called âorbitalâ angular momentumâwill also be conserved. To an excellent approximation the electron moves in the hydrogen atom like a particle without spinâthe angular momentum of the motion is a constant.”
To an excellent approximation… But… Well… Electrons in a metal do form pairs, because they can give up energy in that way and, hence, they are more stableÂ that way. Feynman does not go into the details here – I guess because that’s way beyond the undergrad level – but refers to the Bardeen-Coopers-Schrieffer (BCS) theory instead – the authors of which got a Nobel Prize in Physics in 1972 (that’s a decade or so after Feynman wrote this particularÂ Lecture), so I must assume the theory is well accepted now. đ
Of course, you’ll shout now: Hey! Hydrogen is not a metal!Â Well… Think again: the latest breakthrough in physics is making hydrogen behave like a metal. đ And I am really talking the latestÂ breakthrough: ScienceÂ just published the findings of this experiment last month! đ đ In any case, we’re not talking hydrogen here but superconducting materials, to which – as far as we know – the BCS theory does apply.
So… Well… I am done. I just wanted to show you why it’s important to work your way through Feynman’s last LectureÂ because… Well… Quantum mechanics does explain everything – although the nitty-gritty of it (the Meissner effect, the London equation, flux quantization, etc.) are rather hard bullets to bite. đŚ
Don’t give up ! I am struggling with the nitty-gritty too ! đ
An interpretation of the wavefunction
This is my umpteenth post on the same topic. đŚ It is obvious that this search for a sensibleÂ interpretation is consuming me. Why? I am not sure. Studying physics is frustrating. As a leading physicist puts it:
“TheÂ teaching of quantum mechanics these days usuallyÂ follows the same dogma: firstly, the student is told about the failure of classical physics atÂ the beginning of the last century; secondly, the heroic confusions of the founding fathersÂ are described and the student is given to understand that no humble undergraduate studentÂ could hope to actually understand quantum mechanics for himself; thirdly, a deus exÂ machina arrives in the form of a set of postulates (the SchrĂśdinger equation, the collapseÂ of the wavefunction, etc); fourthly, a bombardment of experimental verifications is given,Â so that the student cannot doubt that QM is correct; fifthly, the student learns how toÂ solve the problems that will appear on the exam paper, hopefully with as little thought asÂ possible.”
That’s obviously not the way we want to understand quantum mechanics. [With we,Â I mean, me, of course, and you, if you’re reading this blog.]Â Of course, that doesn’t mean I don’t believe Richard Feynman, one of the greatest physicists ever, when he tells us no one, including himself, understands physics quite the way we’dÂ likeÂ to understand it. Such statements should not prevent us from tryingÂ harder. So let’s look for betterÂ metaphors.Â The animation below shows the two components of the archetypal wavefunction â a simple sine and cosine. They’re the same function actually, but their phases differ by 90 degrees (Ď/2).
It makes me think of a V-2 engine with the pistons at a 90-degree angle. Look at the illustration below, which I took from a rather simple article on cars and engines that has nothing to do with quantum mechanics. Think of the moving pistons as harmonic oscillators, like springs.
We will also think of theÂ center of each cylinder as the zero point: think of that point as a point where – if we’re looking at one cylinder alone – the internal and external pressure balance each other, so the piston would not move… Well… If it weren’t for the other piston, because the second piston isÂ not at the centerÂ when the first is. In fact, it is easy to verify and compare the following positions of both pistons, as well as the associated dynamics of the situation:
Piston 1 |
Piston 2 |
Motion of Piston 1 |
Motion Piston 2 |
Top |
Center |
Compressed air will push piston down |
Piston moves down against external pressure |
Center |
Bottom |
Piston moves down against external pressure |
External air pressure will push piston up |
Bottom |
Center |
External air pressure will push piston up |
Piston moves further up and compresses the air |
Center |
Top |
Piston moves further up and compresses the air |
Compressed air will push piston down |
When the pistons move, their linear motion will be described by a sinusoidal function: a sine or a cosine. In fact, the 90-degree V-2 configuration ensures that the linear motion of the two pistons will be exactly the same, except for a phase difference of 90 degrees. [Of course, because of the sideways motion of the connecting rods, our sine and cosine function describes the linear motion only approximately, but you can easily imagine the idealizedÂ limit situation. If not, check Feynman’s description of the harmonic oscillator.]
The question is: if we’d have a set-up like this, two springs – or two harmonic oscillators – attached to a shaft through a crank, would this really work as a perpetuum mobile? We obviously talk energyÂ being transferred back and forth between the rotating shaft and the moving pistons… So… Well… Let’s model this: the totalÂ energy, potentialÂ andÂ kinetic, in each harmonic oscillator is constant. Hence, the piston only delivers or receivesÂ kineticÂ energy from the rotating mass of the shaft.
Now, in physics, that’s a bit of an oxymoron: we don’t think of negative or positive kinetic (or potential) energy in the context of oscillators. We don’t think of the direction of energy. But… Well… If we’ve got twoÂ oscillators, our picture changes, and so we may have to adjust our thinking here.
Let me start by giving you an authoritative derivation of the various formulas involved here, taking the example of the physical spring as an oscillatorâbut the formulas are basically the same forÂ any harmonic oscillator.
The first formula is a general description of the motion of our oscillator. The coefficient in front of the cosine function (a)Â is the maximum amplitude. Of course, you will also recognize Ď_{0}Â as theÂ naturalÂ frequency of the oscillator, andÂ Î as the phase factor, which takes into account our t = 0 point. In our case, for example, we have two oscillators with a phase difference equal to Ď/2 and, hence, Î would be 0 for one oscillator, and âĎ/2 for the other. [The formula to apply here is sinÎ¸ = cos(Î¸ â Ď/2).] Also note that we can equate our Î¸ argument to Ď_{0}Âˇt.Â Now, ifÂ aÂ = 1 (which is the case here), then these formulas simplify to:
- K.E. = T = mÂˇv^{2}/2 =Â mÂˇĎ_{0}^{2}Âˇsin^{2}(Î¸ + Î) = mÂˇĎ_{0}^{2}Âˇsin^{2}(Ď_{0}Âˇt + Î)
- P.E. = U = kÂˇx^{2}/2 = kÂˇcos^{2}(Î¸ + Î)
The coefficient k in the potential energy formula characterizes the force: F = âkÂˇx. The minus sign reminds us our oscillator wants to return to the center point, so the force pulls back. From the dynamics involved, it is obvious that k must be equal to mÂˇĎ_{0}^{2}., so that gives us the famous T + U = mÂˇĎ_{0}^{2}/2 formula or, including aÂ once again, T + U = mÂˇa^{2}ÂˇĎ_{0}^{2}/2.
Now, if we normalizeÂ our functions by equating k to one (k = 1), thenÂ the motion ofÂ our first oscillator is given by the cosÎ¸ function, and its kinetic energy will be equal toÂ sin^{2}Î¸. Hence, the (instantaneous)Â changeÂ in kinetic energy at any point in time will be equal to:
d(sin^{2}Î¸)/dÎ¸ = 2âsinÎ¸âd(sinÎ¸)/dt = 2âsinÎ¸âcosÎ¸
Let’s look at the second oscillator now. Just think of the second piston going up and down in our V-twin engine. Its motion is given by theÂ sinÎ¸ function which, as mentioned above, is equal to cos(Î¸âĎ /2). Hence, its kinetic energy is equal toÂ sin^{2}(Î¸âĎ /2), and how itÂ changesÂ – as a function of Î¸ – will be equal to:
2âsin(Î¸âĎ /2)âcos(Î¸âĎ /2) =Â = â2âcosÎ¸âsinÎ¸ = â2âsinÎ¸âcosÎ¸
We have our perpetuum mobile! While transferring kinetic energy from one piston to the other, the rotating shaft moves at constant speed. Linear motion becomes circular motion, and vice versa, in a frictionless Universe. We have the metaphor we were looking for!
Somehow, in this beautiful interplay between linear and circular motion, energy is being borrowed from one place to another, and then returned. From what place to what place? I am not sure. We may call it the real and imaginary energy space respectively, but what does that mean? One thing is for sure, however: the interplay between the real and imaginary part of the wavefunction describes how energy propagates through space!
How exactly? Again, I am not sure. Energy is, obviously, mass in motion – as evidenced by the E = mÂˇc^{2}Â equation, and it may not have any direction (when everything is said and done, it’s a scalarÂ quantity without direction), but the energy in a linear motion is surely different from that in a circular motion, and our metaphor suggests we need to think somewhat more along those lines. Perhaps we will, one day, able toÂ square this circle. đ
SchrĂśdinger’s equation
Let’s analyze the interplay between the real and imaginary part of the wavefunction through an analysis of SchrĂśdinger’s equation, which we write as:
iÂˇÄ§ââĎ/ât = â(Ä§^{2}/2m)ââ^{2}Ď + VÂˇĎ
We can do a quick dimensional analysis of both sides:
- [iÂˇÄ§ââĎ/ât] = Nâmâs/s = Nâm
- [â(Ä§^{2}/2m)ââ^{2}Ď] = Nâm^{3}/m^{2} = Nâm
- [VÂˇĎ] = Nâm
Note the dimension of the ‘diffusion’ constantÂ Ä§^{2}/2m: [Ä§^{2}/2m] = N^{2}âm^{2}âs^{2}/kg = N^{2}âm^{2}âs^{2}/(NÂˇs^{2}/m) = Nâm^{3}. Also note that, in order for the dimensions to come out alright, the dimension of V â the potential â must be that of energy. Hence, Feynmanâs description of it as the potential energy â rather than the potential tout courtÂ â is somewhat confusing but correct: V must equal the potential energy of the electron. Hence, V is not the conventional (potential) energy of the unit charge (1 coulomb). Instead, the natural unit of charge is used here, i.e. the charge of the electron itself.
Now, SchrĂśdingerâs equation â without the VÂˇĎ term â can be written as the following pair of equations:
- Re(âĎ/ât) = â(1/2)â(Ä§/m)âIm(â^{2}Ď)
- Im(âĎ/ât) = (1/2)â(Ä§/m)âRe(â^{2}Ď)
This closely resembles the propagation mechanism of an electromagnetic wave as described by Maxwell’s equation for free space (i.e. a space with no charges), but E and B are vectors, not scalars. How do we get this result. Well… Ď is a complex function, which we can write as a + iâb. Likewise, âĎ/ât is a complex function, which we can write as c + iâd, and â^{2}Ď can then be written as e + iâf. If we temporarily forget about the coefficients (Ä§, Ä§^{2}/m and V), then SchrĂśdingerâs equation – including VÂˇĎ term – amounts to writing something like this:
iâ(c + iâd) = â(e + iâf) + (a + iâb) â a + iâb = iâc â d + e+ iâf Â â a = âd + e and b = c + f
Hence, we can now write:
- VâRe(Ď) = âÄ§âIm(âĎ/ât) + (1/2)â( Ä§^{2}/m)âRe(â^{2}Ď)
- VâIm(Ď) = Ä§âRe(âĎ/ât) + (1/2)â( Ä§^{2}/m)âIm(â^{2}Ď)
This simplifies to the two equations above for V = 0, i.e. when there is no potential (electron in free space). Now we can bring the Re and Im operators into the brackets to get:
- VâRe(Ď) = âÄ§ââIm (Ď)/ât + (1/2)â( Ä§^{2}/m)ââ^{2}Re(Ď)
- VâIm(Ď) = Ä§ââRe(Ď)/ât + (1/2)â( Ä§^{2}/m)ââ^{2}Im(Ď)
This is very interesting, because we can re-write this using the quantum-mechanical energy operator H = â(Ä§^{2}/2m)ââ^{2} + VÂˇ (note the multiplication sign after the V, which we do not have â for obvious reasons â for the â(Ä§^{2}/2m)ââ^{2} expression):
- H[Re (Ď)] = âÄ§ââIm(Ď)/ât
- H[Im(Ď)] = Ä§ââRe(Ď)/ât
A dimensional analysis shows us both sides are, once again, expressed in Nâm. Itâs a beautiful expression because â if we write the real and imaginary part of Ď as râcosÎ¸ and râsinÎ¸, we get:
- H[cosÎ¸] = âÄ§ââsinÎ¸/ât = EâcosÎ¸
- H[sinÎ¸] = Ä§ââcosÎ¸/ât = EâsinÎ¸
Indeed, Î¸Â = (Eât â pâx)/Ä§ and, hence, âÄ§ââsinÎ¸/ât = Ä§âcosÎ¸âE/Ä§ = EâcosÎ¸ and Ä§ââcosÎ¸/ât = Ä§âsinÎ¸âE/Ä§ = EâsinÎ¸.Â Now we can combine the two equations in one equation again and write:
H[râ(cosÎ¸ + iâsinÎ¸)] = râ(EâcosÎ¸ + iâsinÎ¸) â H[Ď] = EâĎ
The operator H â applied to the wavefunction â gives us the (scalar) product of the energy E and the wavefunction itself. Isn’t this strange?
Hmm… I need to further verify and explain this result… I’ll probably do so in yet another post on the same topic… đ
Post scriptum: The symmetry of our V-2 engine – or perpetuum mobileÂ – is interesting: its cross-section has only one axis of symmetry. Hence, we may associate some angle with it, so as to define its orientation in the two-dimensional cross-sectional plane. Of course, the cross-sectional plane itself is at right angles to the crankshaft axis, which we may also associate with some angle in three-dimensional space. Hence, its geometry defines two orthogonal directions which, in turn, define a spherical coordinate system, as shown below.
We may, therefore, say that three-dimensional space is actually being impliedÂ byÂ the geometry of our V-2 engine. Now thatÂ isÂ interesting, isn’t it? đ