# Schrödinger’s equation and the two de Broglie relations

Post scriptum note added on 11 July 2016: This is one of the more speculative posts which led to my e-publication analyzing the wavefunction as an energy propagation. With the benefit of hindsight, I would recommend you to immediately the more recent exposé on the matter that is being presented here, which you can find by clicking on the provided link. In fact, I actually made some (small) mistakes when writing the post below.

Original post:

I’ve re-visited the de Broglie equations a couple of times already. In this post, however, I want to relate them to Schrödinger’s equation. Let’s start with the de Broglie equations first. Equations. Plural. Indeed, most popularizing books on quantum physics will give you only one of the two de Broglie equations—the one that associates a wavelength (λ) with the momentum (p) of a matter-particle:

λ = h/p

In fact, even the Wikipedia article on the ‘matter wave’ starts off like that and is, therefore, very confusing, because, for a good understanding of quantum physics, one needs to realize that the λ = h/p equality is just one of a pair of two ‘matter wave’ equations:

1. λ = h/p
2. f = E/h

These two equations give you the spatial and temporal frequency of the wavefunction respectively. Now, those two frequencies are related – and I’ll show you how in a minute – but they are not the same. It’s like space and time: they are related, but they are definitely not the same. Now, because any wavefunction is periodic, the argument of the wavefunction – which we’ll introduce shortly – will be some angle and, hence, we’ll want to express it in radians (or – if you’re really old-fashioned – degrees). So we’ll want to express the frequency as an angular frequency (i.e. in radians per second, rather than in cycles per second), and the wavelength as a wave number (i.e. in radians per meter). Hence, you’ll usually see the two de Broglie equations written as:

1. k = p/ħ
2. ω = E/ħ

It’s the same: ω = 2π∙f and f = 1/T (T is the period of the oscillation), and k = 2π/λ and then ħ = h/2π, of course! [Just to remove all ambiguities: stop thinking about degrees. They’re a Babylonian legacy, who thought the numbers 6, 12, and 60 had particular religious significance. So that’s why we have twelve-hour nights and twelve-hour days, with each hour divided into sixty minutes and each minute divided into sixty seconds, and – particularly relevant in this context – why ‘once around’ is divided into 6×60 = 360 degrees. Radians are the unit in which we should measure angles because… Well… Google it. They measure an angle in distance units. That makes things easier—a lot easier! Indeed, when studying physics, the last thing you want is artificial units, like degrees.]

So… Where were we? Oh… Yes. The de Broglie relation. Popular textbooks usually commit two sins. One is that they forget to say we have two de Broglie relations, and the other one is that the E = h∙f relationship is presented as the twin of the Planck-Einstein relation for photons, which relates the energy (E) of a photon to its frequency (ν): E = h∙ν = ħ∙ω. The former is criminal neglect, I feel. As for the latter… Well… It’s true and not true: it’s incomplete, I’d say, and, therefore, also very confusing.

Why? Because both things lead one to try to relate the two equations, as momentum and energy are obviously related. In fact, I’ve wasted days, if not weeks, on this. How are they related? What formula should we use? To answer that question, we need to answer another one: what energy concept should we use? Potential energy? Kinetic energy? Should we include the equivalent energy of the rest mass?

One quickly gets into trouble here. For example, one can try the kinetic energy, K.E. = m∙v2/2, and use the definition of momentum (p = m∙v), to write E = p2/(2m), and then we could relate the frequency f to the wavelength λ using the general rule that the traveling speed of a wave is equal to the product of its wavelength and its frequency (v = λ∙f). But if E = p2/(2m) and f = v/λ, we get:

p2/(2m) = h∙v/λ ⇔  λ = 2∙h/p

So that is almost right, but not quite: that factor 2 should not be there. In fact, it’s easy to see that we’d get de Broglie’s λ = h/p equation from his E = h∙f equation if we’d use E = m∙v2 rather than E = m∙v2/2. In fact, the E = m∙v2 relation comes out of them if we just multiply the two and, yes, use that v = λ relation once again:

1. f·λ = (E/h)·(h/p) = E/p
2. v = λ ⇒ f·λ = v = E/p ⇔ E = v·p = v·(m·v) ⇒ E = m·v2

But… Well… E = m∙v2? How could we possibly justify the use of that formula?

The answer is simple: our v = f·λ equation is wrong. It’s just something one shouldn’t apply to the complex-valued wavefunction. The ‘correct’ velocity formula for the complex-valued wavefunction should have that 1/2 factor, so we’d write 2·f·λ = v to make things come out alright. But where would this formula come from?

Well… Now it’s time to introduce the wavefunction.

#### The wavefunction

You know the elementary wavefunction:

ψ = ψ(x, t) = ei(ωt − kx) = ei(kx − ωt) = cos(kx−ωt) + i∙sin(kx−ωt)

As for terminology, note that the term ‘wavefunction’ refers to what I write above, while the term ‘wave equation’ usually refers to Schrödinger’s equation, which I’ll introduce in a minute. Also note the use of boldface indicates we’re talking vectors, so we’re multiplying the wavenumber vector k with the position vector x = (x, y, z) here, although we’ll often simplify and assume one-dimensional space. In any case…

So the question is: why can’t we use the v = f·λ formula for this wave? The period of cosθ + isinθ is the same as that of the sine and cosine function considered separately: cos(θ+2π) + isin(θ+2π) = cosθ + isinθ, so T = 2π and f = 1/T = 1/2π do not change. So the f, T and λ should be the same, no?

No. We’ve got two oscillations for the price of one here: one ‘real’ and one ‘imaginary’—but both are equally essential and, hence, equally ‘real’. So we’re actually combining two waves. So it’s just like adding other waves: when adding waves, one gets a composite wave that has (a) a phase velocity and (b) a group velocity.

Huh? Yes. It’s quite interesting. When adding waves, we usually have a different ω and k for each of the component waves, and the phase and group velocity will depend on the relation between those ω’s and k’s. That relation is referred to as the dispersion relation. To be precise, if you’re adding waves, then the phase velocity of the composite wave will be equal to vp = ω/k, and its group velocity will be equal to vg = dω/dk. We’ll usually be interested in the group velocity, and so to calculate that derivative, we need to express ω as a function of k, of course, so we write ω as some function of k, i.e. ω = ω(k). There are number of possibilities then:

1. ω and k may be directly proportional, so we can write ω as ω = a∙k: in that case, we find that vp = vg = a.
2. ω and k are not directly proportional but have a linear relationship, so we can write write ω as ω = a∙k + b. In that case, we find that vg = a and… Well… We’ve got a problem calculating vp, because we don’t know what k to use!
3. ω and k may be non-linearly related, in which case… Well… One does has to do the calculation and see what comes out. 🙂

Let’s now look back at our ei(kx − ωt) = cos(kx−ωt) + i∙sin(kx−ωt) function. You’ll say that we’ve got only one ω and one k here, so we’re not adding waves with different ω’s and k’s. So… Well… What?

That’s where the de Broglie equations come in. Look: k = p/ħ, and ω = E/ħ. If we now use the correct energy formula, i.e. the kinetic energy formula E = m·v2/2 (rather than that nonsensical E = m·v2 equation) – which we can also write as E = m·v·v/2 = p·v/2 = p·p/2m = p2/2m, with v = p/m the classical velocity of the elementary particle that Louis de Broglie was thinking of – then we can calculate the group velocity of our ei(kx − ωt) = cos(kx−ωt) + i∙sin(kx−ωt) as:

vg = dω/dk = d[E/ħ]/d[p/ħ] = dE/dp = d[p2/2m]/dp = 2p/2m = p/m = v

However, the phase velocity of our ei(kx − ωt) is:

vp = ω/k = (E/ħ)/(p/ħ) = E/p = (p2/2m)/p = p/2m = v/2

So that factor 1/2 only appears for the phase velocity. Weird, isn’t it? We find that the group velocity (vg) of the ei(kx − ωt) function is equal to the classical velocity of our particle (i.e. v), but that its phase velocity (vp) is equal to v divided by 2.

Hmm… What to say? Well… Nothing much—except that it makes sense, and very much so, because it’s the group velocity of the wavefunction that’s associated with the classical velocity of a particle, not the phase velocity. In fact, if we include the rest mass in our energy formula, so if we’d use the relativistic E = γm0c2 and p = γm0v formulas (with γ the Lorentz factor), then we find that vp = ω/k = E/p = (γm0c2)/(γm0v) = c2/v, and so that’s a superluminal velocity, because v is always smaller than c!

What? That’s even weirder! If we take the kinetic energy only, we find a phase velocity equal to v/2, but if we include the rest energy, then we get a superluminal phase velocity. It must be one or the other, no? Yep! You’re right! So that makes us wonder: is E = m·v2/2 really the right energy concept to use? The answer is unambiguous: no! It isn’t! And, just for the record, our young nobleman didn’t use the kinetic energy formula when he postulated his equations in his now famous PhD thesis.

So what did he use then? Where did he get his equations?

I am not sure. 🙂 A stroke of genius, it seems. According to Feynman, that’s how Schrödinger got his equation too: intuition, brilliance. In short, a stroke of genius. 🙂 Let’s relate these these two gems.

#### Schrödinger’s equation and the two de Broglie relations

Erwin Schrödinger and Louis de Broglie published their equations in 1924 and 1926 respectively. Can they be related? The answer is: yes—of course! Let’s first look at de Broglie‘s energy concept, however. Louis de Broglie was very familiar with Einsteins’ work and, hence, he knew that the energy of a particle consisted of three parts:

1. The particle’s rest energy m0c2, which de Broglie referred to as internal energy (Eint): this ‘internal energy’ includes the rest mass of the ‘internal pieces’, as he put it (now we call those ‘internal pieces’ quarks), as well as their binding energy (i.e. the quarks’interaction energy);
2. Any potential energy it may have because of some field (so de Broglie was not assuming the particle was traveling in free space), which we’ll denote by V: the field(s) can be anything—gravitational, electromagnetic—you name it: whatever changes the energy because of the position of the particle;
3. The particle’s kinetic energy, which we wrote in terms of its momentum p: K.E. = m·v2/2 = m2·v2/(2m) = (m·v)2/(2m) = p2/(2m).

Indeed, in my previous posts, I would write the wavefunction as de Broglie wrote it, which is as follows:

ψ(θ) = ψ(x, t) = a·eiθ = a·e−i[(Eint + p2/(2m) + V)·t − p∙x]/ħ

In those post – such as my post on virtual particles – I’d also note how a change in potential energy plays out: a change in potential energy, when moving from one place to another, would change the wavefunction, but through the momentum only—so it would impact the spatial frequency only. So the change in potential would not change the temporal frequencies ω= Eint + p12/(2m) + V1 and ω= Eint + p22/(2m) + V2. Why? Or why not, I should say? Because of the energy conservation principle—or its equivalent in quantum mechanics. The temporal frequency f or ω, i.e. the time-rate of change of the phase of the wavefunction, does not change: all of the change in potential, and the corresponding change in kinetic energy, goes into changing the spatial frequency, i.e. the wave number k or the wavelength λ, as potential energy becomes kinetic or vice versa.

So is that consistent with what we wrote above, that E = m·v2? Maybe. Let’s think about it. Let’s first look at Schrödinger’s equation in free space (i.e. a space with zero potential) once again:

If we insert our ψ = ei(kx − ωt) formula in Schrödinger’s free-space equation, we get the following nice result. [To keep things simple, we’re just assuming one-dimensional space for the calculations, so ∇2ψ = ∂2ψ/∂x2. But the result can easily be generalized.] The time derivative on the left-hand side is ∂ψ/∂t = −iω·ei(kx − ωt). The second-order derivative on the right-hand side is ∂2ψ/∂x2 = (ik)·(ik)·ei(kx − ωt) = −k2·ei(kx − ωt) . The ei(kx − ωt) factor on both sides cancels out and, hence, equating both sides gives us the following condition:

iω = −(iħ/2m)·k2 ⇔ ω = (ħ/2m)·k2

Substituting ω = E/ħ and k = p/ħ yields:

E/ħ = (ħ/2m)·p22 = m2·v2/(2m·ħ) = m·v2/(2ħ) ⇔ E = m·v2/2

Bingo! We get that kinetic energy formula! But now… What if we’d not be considering free space? In other words: what if there is some potential? Well… We’d use the complete Schrödinger equation, which is:

Huh? Why is there a minus sign now? Look carefully: I moved the iħ factor on the left-hand side to the other when writing the free space version. If we’d do that for the complete equation, we’d get:

I like that representation a lot more—if only because it makes it a lot easier to interpret the equation—but, for some reason I don’t quite understand, you won’t find it like that in textbooks. Now how does it work when using the complete equation, so we add the −(i/ħ)·V·ψ term? It’s simple: the ei(kx − ωt) factor also cancels out, and so we get:

iω = −(iħ/2m)·k2−(i/ħ)·V ⇔ ω = (ħ/2m)·k+ V/ħ

Substituting ω = E/ħ and k = p/ħ once more now yields:

E/ħ = (ħ/2m)·p22 + V/ħ = m2·v2/(2m·ħ) + V/ħ = m·v2/(2ħ) + V/ħ ⇔ E = m·v2/2 + V

Bingo once more!

The only thing that’s missing now is the particle’s rest energy m0c2, which de Broglie referred to as internal energy (Eint). That includes everything, i.e. not only the rest mass of the ‘internal pieces’ (as said, now we call those ‘internal pieces’ quarks) but also their binding energy (i.e. the quarks’interaction energy). So how do we get that energy concept out of Schrödinger’s equation? There’s only one answer to that: that energy is just like V. We can, quite simply, just add it.

That brings us to the last and final question: what about our vg = result if we do not use the kinetic energy concept, but the E = m·v2/2 + V + Eint concept? The answer is simple: nothing. We still get the same, because we’re taking a derivative and the V and Eint just appear as constants, and so their derivative with respect to p is zero. Check it:

vg = dω/dk = d[E/ħ]/d[p/ħ] = dE/dp = d[p2/2m + V + Eint ]/dp = 2p/2m = p/m = v

It’s now pretty clear how this thing works. To localize our particle, we just superimpose a zillion of these ei(ωt − kx) equations. The only condition is that we’ve got that fixed vg = dω/dk = v relationhip, but so we do have such fixed relationship—as you can see above. In fact, the Wikipedia article on the dispersion relation mentions that the de Broglie equations imply the following relation between ω and k: ω = ħk2/2m. As you can see, that’s not entirely correct: the author conveniently forgets the potential (V) and the rest energy (Eint) in the energy formula here!

What about the phase velocity? That’s a different story altogether. You can think about that for yourself. 🙂

I should make one final point here. As said, in order to localize a particle (or, to be precise, its wavefunction), we’re going to add a zillion elementary wavefunctions, each of which will make its own contribution to the composite wave. That contribution is captured by some coefficient ai in front of every eiθi function, so we’ll have a zillion aieiθi functions, really. [Yep. Bit confusing: I use here as subscript, as well as imaginary unit.] In case you wonder how that works out with Schrödinger’s equation, the answer is – once again – very simple: both the time derivative (which is just a first-order derivative) and the Laplacian are linear operators, so Schrödinger’s equation, for a composite wave, can just be re-written as the sum of a zillion ‘elementary’ wave equations.

So… Well… We’re all set now to effectively use Schrödinger’s equation to calculate the orbitals for a hydrogen atom, which is what we’ll do in our next post.

In the meanwhile, you can amuse yourself with reading a nice Wikibook article on the Laplacian, which gives you a nice feel for what Schrödinger’s equation actually represents—even if I gave you a good feel for that too on my Essentials page. Whatever. You choose. Just let me know what you liked best. 🙂

Oh… One more point: the vg = dω/dk = d[p2/2m]/dp = p/m = calculation obviously assumes we can treat m as a constant. In fact, what we’re actually doing is a rather complicated substitution of variables: you should write it all out—but that’s not the point here. The point is that we’re actually doing a non-relativistic calculation. Now, that does not mean that the wavefunction isn’t consistent with special relativity. It is. In fact, in one of my posts, I show how we can explain relativistic length contraction using the wavefunction. But it does mean that our calculation of the group velocity is not relativistically correct. But that’s a minor point: I’ll leave it for you as an exercise to calculate the relativistically correct formula for the group velocity. Have fun with it! 🙂

Note: Notations are often quite confusing. One should, generally speaking, denote a frequency by ν (nu), rather than by f, so as to not cause confusion with any function f, but then… Well… You create a new problem when you do that, because that Greek letter nu (ν) looks damn similar to the v of velocity, so that’s why I’ll often use f when I should be using nu (ν). As for the units, a frequency is expressed in cycles per second, while the angular frequency ω is expressed in radians per second. One cycle covers 2π radians and, therefore, we can write: ν = ω/2π. Hence, h∙ν = h∙ω/2π = ħ∙ω. Both ν as well as ω measure the time-rate of change of the phase of the wave function, as opposed to k, i.e. the spatial frequency of the wave function, which depends on the speed of the wave. Physicists also often use the symbol v for the speed of a wave, which is also hugely confusing, because it’s also used to denote the classical velocity of the particle. And then there’s two wave velocities, of course: the group versus the phase velocity. In any case… I find the use of that other symbol (c) for the wave velocity even more confusing, because this symbol is also used for the speed of light, and the speed of a wave is not necessarily (read: usually not) equal to the speed of light. In fact, both the group as well as the phase velocity of a particle wave are very different from the speed of light. The speed of a wave and the speed of light only coincide for electromagnetic waves and, even then, it should be noted that photons also have amplitudes to travel faster or slower than the speed of light.

# Refraction and Dispersion of Light

Pre-scriptum (dated 26 June 2020): Some of the relevant illustrations in this post were removed as a result of an attack by the dark force. Too bad, because I liked this post. In any case, despite the removal of the illustrations, I think you will still be able to reconstruct the main story line.

Original post:

In this post, we go right at the heart of classical physics. It’s going to be a very long post – and a very difficult one – but it will really give you a good ‘feel’ of what classical physics is all about. To understand classical physics – in order to compare it, later, with quantum mechanics – it’s essential, indeed, to try to follow the math in order to get a good feel for what ‘fields’ and ‘charges’ and ‘atomic oscillators’ actually represent.

As for the topic of this post itself, we’re going to look at refraction again: light gets dispersed as it travels from one medium to another, as illustrated below.

Dispersion literally means “distribution over a wide area”, and so that’s what happens as the light travels through the prism: the various frequencies (i.e. the various colors that make up natural ‘white’ light) are being separated out over slightly different angles. In physics jargon, we say that the index of refraction depends on the frequency of the wave – but so we could also say that the breaking angle depends on the color. But that sounds less scientific, of course. In any case, it’s good to get the terminology right. Generally speaking, the term refraction (as opposed to dispersion) is used to refer to the bending (or ‘breaking’) of light of a specific frequency only, i.e. monochromatic light, as shown in the photograph below. […] OK. We’re all set now.

It is interesting to note that the photograph above shows how the monochromatic light is actually being obtained: if you look carefully, you’ll see two secondary beams on the left-hand side (with an intensity that is much less than the central beam – barely visible in fact). That suggests that the original light source was sent through a diffraction grating designed to filter only one frequency out of the original light beam. That beam is then sent through a bloc of transparent material (plastic in this case) and comes out again, but displaced parallel to itself. So the block of plastics ‘offsets’ the beam. So how do we explain that in classical physics?

The index of refraction and the dispersion equation

As I mentioned in my previous post, the Greeks had already found out, experimentally, what the index of refraction was. To be more precise, they had measured the θ1 and θ2 – depicted below – for light going from air to water. For example, if the angle in air (θ1) is 20°, then the angle in the water (θ2) will be 15°. It the angle in air is 70°, then the angle in the water will be 45°.

Of course, it should be noted that a lot of the light will also be reflected from the water surface (yes, imagine the romance of the image of the moon reflected on the surface of glacial lake while you’re feeling damn cold) – but so that’s a phenomenon which is better  explained by introducing probability amplitudes, and looking at light as a bundle of photons, which we will not do here. I did that in previous posts, and so here, we will just acknowledge that there is a reflected beam but not say anything about it.

In any case, we should go step by step, and I am not doing that right now. Let’s first define the index of refraction. It is a number n which relates the angles above through the following relationship, which is referred to as Snell’s Law:

sinθ1 = n sinθ2

Using the numbers given above, we get: sin(20°) = n sin(15°), and sin(70°) = n sin(45°), so n must be equal to n = sin(20°)/sin(15°)  = sin(70°)/sin(45°) ≈ 1.33. Just for the record, Willibrord Snell was a medieval Dutch astronomer but, according to Wikipedia, some smart Persian, Ibn Sahl, had already jotted this down in a treatise – “On Burning Mirrors and Lenses” – while he was serving the Abbasid court of Baghdad, back in 984, i.e. more than a thousand years ago! What to say? It was obviously a time when the Sunni-Shia divide did not matter, and Arabs and ‘Persians’ were leading civilization. I guess I should just salute the Islamic Golden Age here, regret the time lost during Europe’s Dark Ages and, most importantly, regret where Baghdad is right now ! And, as for the ‘burning’ adjective, it just refers to the fact that large convex lenses can concentrate the sun’s rays to a very small area indeed, thereby causing ignition. [It seems that story about Archimedes burning Roman ships with a ‘death ray’ using mirrors – in all likelihood: something that did not happen – fascinated them as well.]

But let’s get back at it. Where were we? Oh – yes – the refraction index. It’s (usually) a positive number written as n = 1 + some other number which may be positive or negative, and which depends on the properties of the material. To be more specific, it depends on the resonant frequencies of the atoms (or, to be precise, I should say: the resonant frequencies of the electrons bound by the atom, because it’s the charges that generate the radiation). Plus a whole bunch of natural constants that we have encountered already, most of which are related to electrons. Let me jot down the formula – and please don’t be scared away now (you can stop a bit later, but not now 🙂 please):

N is just the number of charges (electrons) per unit volume of the material (e.g. the water, or that block of plastic), and qe and m are just the charge and mass of the electron. And then you have that electric constant once again, ε0, and… Well, that’s it ! That’s not too terrible, is it? So the only variables on the right-hand side are ω0 and ω, so that’s (i) the resonant frequency of the material (or the atoms – well, the electrons bound to the nucleus, to be precise, but then you know what I mean and so I hope you’ll allow me to use somewhat less precise language from time to time) and (ii) the frequency of the incoming light.

The equation above is referred to as the dispersion relation. It’s easy to see why: it relates the frequency of the incoming light to the index of refraction which, in turn, determinates that angle θ. So the formula does indeed determine how light gets dispersed, as a function of the frequencies in it, by some medium indeed (glass, air, water,…).

So the objective of this post is to show how we can derive that dispersion relation using classical physics only. As usual, I’ll follow Feynman – arguably the best physics teacher ever. 🙂 Let me warn you though: it is not a simple thing to do. However, as mentioned above, it goes to the heart of the “classical world view” in physics and so I do think it’s worth the trouble. Before we get going, however, let’s look at the properties of that formula above, and relate it some experimental facts, in order to make sure we more or less understand what it is that we are trying to understand. 🙂

First, we should note that the index of refraction has nothing to do with transparency. In fact, throughout this post, we’ll assume that we’re looking at very transparent materials only, i.e. materials that do not absorb the electromagnetic radiation that tries to go through them, or only absorb it a tiny little bit. In reality, we will have, of course, some – or, in the case of opaque (i.e. non-transparent) materials, a lot – of absorption going on, but so we will deal with that later. So, let me repeat: the index of refraction has nothing to do with transparency. A material can have a (very) high index of refraction but be fully transparent. In fact, diamond is a case in point: it has one of the highest indexes of refraction (2.42) of any material that’s naturally available, but it’s – obviously – perfectly transparent. [In case you’re interested in jewellery, the refraction index of its most popular substitute, cubic zirconia, comes very close (2.15-2.18) and, moreover, zirconia actually works better as a prism, so its disperses light better than diamond, which is why it reflects more colors. Hence, real diamond actually sparkles less than zirconia! So don’t be fooled! :-)]

Second, it’s obvious that the index of refraction depends on two variables indeed: the natural, or resonant frequency, ω0, and the frequency ω, which is the frequency of the incoming light. For most of the ordinary gases, including those that make up air (i.e. nitrogen (78%) and oxygen (21%), plus some vapor (averaging 1%) and the so-called noble gas argon (0.93%) – noble because, just like helium and neon, it’s colorless, odorless and doesn’t react easily), the natural frequencies of the electron oscillators are close to the frequency of ultraviolet light. [The greenhouse gases are a different story – which is why we’re in trouble on this planet. Anyway…] So that’s why air absorbs most of the UV, especially the cancer-causing ultraviolet-C light (UVC), which is formally classified as a carcinogen by the World Health Organization. The wavelength of UVC light is 100 to 300 nanometer – as opposed to visible light, which has a wavelength ranging from 400 to 700 nm – and, hence, the frequency of UV light is in the 1000 to 3000 Teraherz range (1 THz = 1012 oscillations per second) – as opposed to visible light, which has a frequency in the range of 400 to 800 THz. So, because we’re squaring those frequencies in the formula, ω2 can then be disregarded in comparison with ω02: for example, 15002 = 2,250,000 and that’s not very different from 15002 – 5002 = 2,000,000. Hence, if we leave the ω2 out, we are still dividing by a very large number. That’s why n is very close to one for visible light entering the atmosphere from space (i.e. the vacuum). Its value is, in fact, around 1.000292 for incoming light with a wavelength of 589.3 nm (the odd value is the mean of so-called sodium D light, a pretty common yellow-orange light (street lights!), so that’s why it’s used as a reference value – however, don’t worry about it).

That being said, while the n of air is close to one for all visible light, the index is still slightly higher for blue light as compared to red light, and that’s why the sky is blue, except in the morning and evening, when it’s reddish. Indeed, the illustration below is a bit silly, but it gives you the idea. [I took this from http://mathdept.ucr.edu/ so I’ll refer you to that for the full narrative on that. :-)]

Where are we in this story? Oh… Yes. Two frequencies. So we should also note that – because we have two frequency variables – it also makes sense to talk about, for instance, the index of refraction of graphite (i.e. carbon in its most natural occurrence, like in coal) for x-rays. Indeed, coal is definitely not transparent to visible light (that has to do with the absorption phenomenon, which we’ll discuss later) but it is very ‘transparent’ to x-rays. Hence, we can talk about how graphite bends x-rays, for example. In fact, the frequency of x-rays is much higher than the natural frequency of the carbon atoms and, hence, in this case we can neglect the w02 factor, so we get a denominator that is negative (because only the -w2 remains relevant), so we get a refraction index that is (a bit) smaller than 1. [Of course, our body is transparent to x-rays too – to a large extent – but in different degrees, and that’s why we can take x-ray photographs of, for example, a broken rib or leg.]

OK. […] So that’s just to note that we can have a refraction index that is smaller than one and that’s not ‘anomalous’ – even if that’s a historical term that has survived.

Finally, last but not least as they say, you may have heard that scientists and engineers have managed to construct so-called negative index metamaterials. That matter is (much) more complicated than you might think, however, and so I’ll refer you to the Web if you want to find out more about that.

Light going through a glass plate: the classical idea

OK. We’re now ready to crack the nut. We’ll closely follow my ‘Great Teacher’ Feynman (Lectures, Vol. I-31) as he derives that formula above. Let me warn you again: the narrative below is quite complicated, but really worth the trouble – I think. The key to it all is the illustration below. The idea is that we have some electromagnetic radiation emanating from a far-away source hitting a glass plate – or whatever other transparent material. [Of course, nothing is to scale here: it’s just to make sure you get the theoretical set-up.]

So, as I explained in my previous post, the source creates an oscillating electromagnetic field which will shake the electrons up and down in the glass plate, and then these shaking electrons will generate their own waves. So we look at the glass as an assembly of little “optical-frequency radio stations” indeed, that are all driven with a given phase. It creates two new waves: one reflecting back, and one modifying the original field.

Let’s be more precise. What do we have here? First, we have the field that’s generated by the source, which is denoted by Es above. Then we have the “reflected” wave (or field – not much difference in practice), so that’s Eb. As mentioned above, this is the classical theory, not the quantum-electrodynamical one, so we won’t say anything about this reflection really: just note that the classical theory acknowledges that some of the light is effectively being reflected.

OK. Now we go to the other side of the glass. What do we expect to see there? If we would not have the glass plate in-between, we’d have the same Es field obviously, but so we don’t: there is a glass plate. 🙂 Hence, the “transmitted” wave, or the field that’s arriving at point P let’s say, will be different than Es. Feynman writes it as Es + Ea.

Hmm… OK. So what can we say about that? Not easy…

The index of refraction and the apparent speed of light in a medium

Snell’s Law – or Ibn Sahl’s Law – was re-formulated, by a 17th century French lawyer with an interesting in math and physics, Pierre de Fermat, as the Principle of Least Time. It is a way of looking at things really – but it’s very confusing actually. Fermat assumed that light traveling through a medium (water or glass, for instance) would travel slower, by a certain factor n, which – indeed – turns out to be the index of refraction. But let’s not run before we can walk. The Principle is illustrated below. If light has to travel from point S (the source) to point D (the detector), then the fastest way is not the straight line from S to D, but the broken S-L-D line. Now, I won’t go into the geometry of this but, with a bit of trial and error, you can verify for yourself that it turns out that the factor n will indeed be the same factor n as the one which was ‘discovered’ by Ibn Sahl: sinθ1 = n sinθ2.

What we have then, is that the apparent speed of the wave in the glass plate that we’re considering here will be equal to v = c/n. The apparent speed? So does that mean it is not the real speed? Hmm… That’s actually the crux of the matter. The answer is: yes and no. What? An ambiguous answer in physics? Yes. It’s ambiguous indeed. What’s the speed of a wave? We mentioned above that n could be smaller than one. Hence, in that case, we’d have a wave traveling faster than the speed of light. How can we make sense of that?

We can make sense of that by noting that the wave crests or nodes may be traveling faster than c, but that the wave itself – as a signal – cannot travel faster than light. It’s related to what we said about the difference between the group and phase velocity of a wave. The phase velocity – i.e. the nodes, which are mathematical points only – can travel faster than light, but the signal as such, i.e. the wave envelope in the illustration below, cannot.

What is happening really is the following. A wave will hit one of these electron oscillators and start a so-called transient, i.e. a temporary response preceding the ‘steady state’ solution (which is not steady but dynamic – confusing language once again – so sorry!). So the transient settles down after a while and then we have an equilibrium (or steady state) oscillation which is likely to be out of phase with the driving field. That’s because there is damping: the electron oscillators resist before they go along with the driving force (and they continue to put up resistance, so the oscillation will die out when the driving force stops!). The illustration below shows how it works for the various cases:

In case (b), the phase of the transmitted wave will appear to be delayed, which results in the wave appearing to travel slower, because the distance between the wave crests, i.e. the wavelength λ, is being shortened. In case (c), it’s the other way around: the phase appears to be advanced, which translated into a bigger distance between wave crests, or a lengthening of the wavelength, which translates into an apparent higher speed of the transmitted wave.

So here we just have a mathematical relationship between the (apparent) speed of a wave and its wavelength. The wavelength is the (apparent) speed of the wave (that’s the speed with which the nodes of the wave travel through space, or the phase velocity) divided by the frequency: λ = vp/f. However, from the illustration above, it is obvious that the signal, i.e. the start of the wave, is not earlier – or later – for either wave (b) and (c). In fact, the start of the wave, in time, is exactly the same for all three cases. Hence, the electromagnetic signal travels at the same speed c, always.

While this may seem obvious, it’s quite confusing, and therefore I’ll insert one more illustration below. What happens when the various wave fronts of the traveling field hit the glass plate (coming from the top-left hand corner), let’s say at time t = t0, as shown below, is that the wave crests will have the same spacing along the surface. That’s obvious because we have a regular wave with a fixed frequency and, hence, a fixed wavelength λ0, here. Now, these wave crests must also travel together as the wave continues its journey through the glass, which is what is shown by the red and green arrows below: they indicate where the wave crest is after one and two periods (T and 2T) respectively.

To understand what’s going on, you should note that the frequency f of the wave that is going through the glass sheet and, hence, its period T, has not changed. Indeed, the driven oscillation, which was illustrated for the two possible cases above (n > 1 and n < 1), after the transient has settled down, has the same frequency (f) as the driving source. It must. Always. That being said, the driven oscillation does have that phase delay (remember: we’re in the (b) case here, but we can make a similar analysis for the (c) case). In practice, that means that the (shortest) distance between the crests of the wave fronts at time t = t0 and the crests at time t0 + T will be smaller. Now, the (shortest) distance between the crests of a wave is, obviously, the wavelength divided by the frequency: λ = vp/f, with vp the speed of propagation, i.e. the phase velocity, of the wave, and f = 1/T. [The frequency f is the reciprocal of the period T – always. When studying physics, I found out it’s useful to keep track of a few relationships that hold always, and so this is one of them. :-)]

Now, the frequency is the same, but so the wavelength is shortened as the wave travels through the various layers of electron oscillators, each causing a delay of phase – and, hence, a shortening of the wavelength, as shown above. But, if f is the same, and the wavelength is shorter, then vp cannot be equal to the speed of the incoming light, so vp ≠ c. The apparent speed of the wave traveling through the glass, and the associated shortening of the wavelength, can be calculated using Snell’s Law. Indeed, knowing that n ≈ 1.33, we can calculate the apparent speed of light through the glass as v = c/n  ≈ 0.75c and, therefore, we can calculate the wavelength of the wave in the glass l as λ = 0.75λ0.

OK. I’ve been way too lengthy here. Let’s sum it all up:

• The field in the glass sheet must have the shape that’s depicted above: there is no other way. So that means the direction of ‘propagation’ has been changed. As mentioned above, however, the direction of propagation is a ‘mathematical’ property of the field: it’s not the speed of the ‘signal’.
• Because the direction of propagation is normal to the wave front, it implies that the bending of light rays comes about because the effective speed of the waves is different in the various materials or, to be even more precise, because the electron oscillators cause a delay of phase.
• While the speed and direction of propagation of the wave, i.e. the phase velocity, accurately describes the behavior of the field, it is not the speed with which the signal is traveling (see above). That is why it can be larger or smaller than c, and so it should not raise any eyebrow. For x-rays in particular, we have a refractive index smaller than one. [It’s only slightly less than one, though, and, hence, x-ray images still have a very good resolution. So don’t worry about your doctor getting a bad image of your broken leg. 🙂 In case you want to know more about this: just Google x-ray optics, and you’ll find loads of information. :-)]

Calculating the field

Are you still there? Probably not. If you are, I am afraid you won’t be there ten or twenty minutes from now. Indeed, you ain’t done nothing yet. All of the above was just setting the stage: we’re now ready for the pièce de résistance, as they say in French. We’re back at that illustration of the glass plate and the various fields in front and behind the plate. So we have electron oscillators in the glass plate. Indeed, as Feynman notes: “As far as problems involving light are concerned, the electrons behave as though they were held by springs. So we shall suppose that the electrons have a linear restoring force which, together with their mass m, makes them behave like little oscillators, with a resonant frequency ω0.”

So here we go:

1. From everything I wrote about oscillators in previous posts, you should remember that the equation for this motion can be written as m[d2x/dt2 + ω02) = F. That’s just Newton’s Law. Now, the driving force F comes from the electric field and will be equal to F = qeEs.

Now, we assume that we can chose the origin of time (i.e. the moment from which we start counting) such that the field Es = E0cos(ωt). To make calculations easier, we look at this as the real part of a complex function Es = E0eiωt. So we get:

m[d2x/dt2 + ω02] = qeE0eiωt

We’ve solved this before: its solution is x = x0eiωt. We can just substitute this in the equation above to find x0 (just substitute and take the first- and then second-order derivative of x indeed): x0 = qeE0/m(ω022). That, then, gives us the first piece in this lengthy derivation:

x = qeE0eiωt/m(ω02 2)

Just to make sure you understand what we’re doing: this piece gives us the motion of the electrons in the plate. That’s all.

2. Now, we need an equation for the field produced by a plane of oscillating charges, because that’s what we’ve got here: a plate or a plane of oscillating charges. That’s a complicated derivation in its own, which I won’t do there. I’ll just refer to another chapter of Feynman’s Lectures (Vol. I-30-7) and give you the solution for it (if I wouldn’t do that, this post would be even longer than it already is):

This formula introduces just one new variable, η, which is the number of charges per unit area of the plate (as opposed to N, which was the number of charges per unit volume in the plate), so that’s quite straightforward. Less straightforward is the formula itself: this formula says that the magnitude of the field is proportional to the velocity of the charges at time t – z/c, with z the shortest distance from P to the plane of charges. That’s a bit odd, actually, but so that’s the way it comes out: “a rather simple formula”, as Feynman puts it.

In any case, let’s use it. Differentiating x to get the velocity of the charges, and plugging it into the formula above yields:

Note that this is only Ea, the additional field generated by the oscillating charges in the glass plate. To get the total electric field at P, we still have to add Es, i.e. the field generated by the source itself. This may seem odd, because you may think that the glass plate sort of ‘shields’ the original field but, no, as Feynman puts it: “The total electric field in any physical circumstance is the sum of the fields from all the charges in the universe.”

3. As mentioned above, z is the distance from P to the plate. Let’s look at the set-up here once again. The transmitted wave, or Eafter the plate as we shall note it, consists of two components: Es and Ea. Es here will be equal to (the real part of) Es = E0eiω(t-z/c). Why t – z/c instead of just t? Well… We’re looking at Es here as measured in P, not at Es at the glass plate itself.

Now, we know that the wave ‘travels slower’ through the glass plate (in the sense that its phase velocity is less, as should be clear from the rather lengthy explanation on phase delay above, or – if n would be greater than one – a phase advance). So if the glass plate is of thickness Δz, and the phase velocity is is v = c/n, then the time it will take to travel through the glass plate will be Δz/(c/n) instead of Δz/c (speed is distance divided by time and, hence, time = distance divided by speed). So the additional time that is needed is Δt = Δz/(c/n) – Δz/c = nΔz/c – Δz/c = (n-1)Δz/c. That, then, implies that Eafter the plate is equal to a rather monstrously looking expression:

Eafter plate = E0eiω[t (n1)Δz/c z/c) = eiω(n1)Δz/c)E0eiω(t z/c)

We get this by just substituting t for t – Δt.

So what? Well… We have a product of two complex numbers here and so we know that this involves adding angles – or substracting angles in this case, rather, because we’ve got a minus sign in the exponent of the first factor. So, all that we are saying here is that the insertion of the glass plate retards the phase of the field with an amount equal to w(n-1)Δz/c. What about that sum Eafter the plate = Es + Ea that we were supposed to get?

Well… We’ll use the formula for a first-order (linear) approximation of an exponential once again: ex ≈ 1 + x. Yes. We can do that because Δz is assumed to be very small, infinitesimally small in fact. [If it is not, then we’ll just have to assume that the plate consists of a lot of very thin plates.] So we can write that eiω(n-1)Δz/c) = 1 – iω(n-1)Δz/c, and then we, finally, get that sum we wanted:

Eafter plate = E0eiω[t z/c) iω(n-1)Δz·E0eiω(t z/c)/c

The first term is the original Es field, and the second term is the Ea field. Geometrically, they can be represented as follows:

Why is Ea perpendicular to Es? Well… Look at the –i = 1/i factor. Multiplication with –i amounts to a clockwise rotation by 90°, and then just note that the magnitude of the vector must be small because of the ω(n-1)Δz/c factor.

4. By now, you’ve either stopped reading (most probably) or, else, you wonder what I am getting at. Well… We have two formulas for Ea now:

and Ea = – iω(n-1)Δz·E0eiω(t – z/c)/c

Equating both yields:

But η, the number of charges per unit area, must be equal to NΔz, with N the number of charges per unit volume. Substituting and then cancelling the Δz finally gives us the formula we wanted, and that’s the classical dispersion relation whose properties we explored above:

Absorption and the absorption index

The model we used to explain the index of refraction had electron oscillators at its center. In the analysis we did, we did not introduce any damping factor. That’s obviously not correct: it means that a glass plate, once it had illuminated, would continue to emit radiation, because the electrons would oscillate forever. When introducing damping, the denominator in our dispersion relation becomes m(ω02 – ω2 + iγω), instead of m(ω02 – ω2). We derived this in our posts on oscillators. What it means is that the oscillator continues to oscillate with the same frequency as the driving force (i.e. not its natural frequency) – so that doesn’t change – but that there is an envelope curve, ensuring the oscillation dies out when the driving force is no longer being applied. The γ factor is the damping factor and, hence, determines how fast the damping happens.

We can see what it means by writing the complex index of refraction as n = n’ – in’’, with n’ and n’’ real numbers, describing the real and imaginary part of n respectively. Putting that complex n in the equation for the electric field behind the plate yields:

Eafter plate = eωn’’Δz/ceiω(n’1)Δz/cE0eiω(t z/c)

This is the same formula that we had derived already, but so we have an extra exponential factor: eωn’’Δz/c. It’s an exponential factor with a real exponent, because there were two i‘s that cancelled. The e-x function has a familiar shape (see below): e-x is 1 for x = 0, and between 0 and 1 for any value in-between. That value will depend on the thickness of the glass sheet. Hence, it is obvious that the glass sheet weakens the wave as it travels through it. Hence, the wave must also come out with less energy (the energy being proportional to the square of the amplitude). That’s no surprise: the damping we put in for the electron oscillators is a friction force and, hence, must cause a loss of energy.

Note that it is the n’’ term – i.e. the imaginary part of the refractive index n – that determines the degree of absorption (or attenuation, if you want). Hence, n’’ is usually referred to as the “absorption index”.

The complete dispersion relation

We need to add one more thing in order to get a fully complete dispersion relation. It’s the last thing: then we have a formula which can really be used to describe real-life phenomena. The one thing we need to add is that atoms have several resonant frequencies – even an atom with only one electron, like hydrogen ! In addition, we’ll usually want to take into account the fact that a ‘material’ actually consists of various chemical substances, so that’s another reason to consider more than one resonant frequency. The formula is easily derived from our first formula (see the previous post), when we assumed there was only one resonant frequency. Indeed, when we have Nk electrons per unit of volume, whose natural frequency is ωk and whose damping factor is γk, then we can just add the contributions of all oscillators and write:

The index described by this formula yields the following curve:

So we have a curve with a positive slope, and a value n > 1, for most frequencies, except for a very small range of ω’s for which the slope is negative, and for which the index of refraction has a value n < 1. As Feynman notes, these ω’s– and the negative slope – is sometimes referred to as ‘anomalous’ dispersion but, in fact, there’s nothing ‘abnormal’ about it.

The interesting thing is the iγkω term in the denominator, i.e. the imaginary component of the index, and how that compares with the (real) “resonance term” ωk2– ω2. If the resonance term becomes very small compared to iγkω, then the index will become almost completely imaginary, which means that the absorption effect becomes dominant. We can see that effect in the spectrum of light that we receive from the sun: there are ‘dark lines’, i.e. frequencies that have been strongly absorbed at the resonant frequencies of the atoms in the Sun and its ‘atmosphere’, and that allows us to actually tell what the Sun’s ‘atmosphere’ (or that of other stars) actually consists of.

So… There we are. I am aware of the fact that this has been the longest post of all I’ve written. I apologize. But so it’s quite complete now. The only piece that’s missing is something on energy and, perhaps, some more detail on these electron oscillators. But I don’t think that’s so essential. It’s time to move on to another topic, I think.

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

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

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

# Re-visiting the matter wave (I)

Pre-scriptum (dated 26 June 2020): This post did not suffer from the DMCA take-down of some material. It is, therefore, still quite readable—even if my views on these  matters have evolved quite a bit as part of my realist interpretation of QM. However, I now think de Broglie’s intuition in regard to particles being waves was correct but that he should have used a circular rather than a linear wave concept. Also, the idea of a particle being some wave packet is erroneous. It leads to the kind of contradictions I already start mentioning here, such as super-luminous velocities and other nonsense. Such critique is summarized in my paper on de Broglie’s wave concept. I also discuss it in the context of analyzing wavefunction math in the context of signal transmission in a crystal lattice.

Original post:

In my previous posts, I introduced a lot of wave formulas. They are essential to understanding waves – both real ones (e.g. electromagnetic waves) as well as probability amplitude functions. Probability amplitude function is quite a mouthful so let me call it a matter wave, or a de Broglie wave. The formulas are necessary to create true understanding – whatever that means to you – because otherwise we just keep on repeating very simplistic but nonsensical things such as ‘matter behaves (sometimes) like light’, ‘light behaves (sometimes) like matter’ or, combining both, ‘light and matter behave like wavicles’. Indeed: what does ‘like‘ mean? Like the same but different? 🙂 So that means it’s different. Let’s therefore re-visit the matter wave (i.e. the de Broglie wave) and point out the differences with light waves.

In fact, this post actually has its origin in a mistake in a post scriptum of a previous post (An Easy Piece: On Quantum Mechanics and the Wave Function), in which I wondered what formula to use for the energy E in the (first) de Broglie relation E = hf (with the frequency of the matter wave and h the Planck constant). Should we use (a) the kinetic energy of the particle, (b) the rest mass (mass is energy, remember?), or (c) its total energy? So let us first re-visit these de Broglie relations which, you’ll remember, relate energy and momentum to frequency (f) and wavelength (λ) respectively with the Planck constant as the factor of proportionality:

E = hf and p = h/λ

The de Broglie wave

I first tried kinetic energy in that E = h equation. However, if you use the kinetic energy formula (K.E. = mv2/2, with the velocity of the particle), then the second de Broglie relation (p = h/λ) does not come out right. The second de Broglie relation has the wavelength λ on the right side, not the frequency f. But it’s easy to go from one to the other: frequency and wavelength are related through the velocity of the wave (v). Indeed, the number of cycles per second (i.e. the frequency f) times the length of one cycle (i.e. the wavelength λ) gives the distance traveled by the wave per second, i.e. its velocity v. So fλ = v. Hence, using that kinetic energy formula and that very obvious fλ = v relation, we can write E = hf as mv2/2 = v/λ and, hence, after moving one of the two v’s in v2 (and the 1/2 factor) on the left side to the right side of this equation, we get mv = 2h/λ. So there we are:

p = mv = 2h/λ.

Well… No. The second de Broglie relation is just p = h/λ. There is no factor 2 in it. So what’s wrong?

A factor of 2 in an equation like this surely doesn’t matter, does it? It does. We are talking tiny wavelengths here but a wavelength of 1 nanometer (1×10–9 m) – this is just an example of the scale we’re talking about here – is not the same as a wavelength of 0.5 nm. There’s another problem too. Let’s go back to our an example of an electron with a mass of 9.1×10–31 kg (that’s very tiny, and so you’ll usually see it expressed in a unit that’s more appropriate to the atomic scale), moving about with a velocity of 2.2×106 m/s (that’s the estimated speed of orbit of an electron around a hydrogen nucleus: it’s fast (2,200 km per second), but still less than 1% of the speed of light), and let’s do the math.

[Before I do the math, however, let me quickly insert a line on that ‘other unit’ to measure mass. You will usually see it written down as eV, so that’s electronvolt. Electronvolt is a measure of energy but that’s OK because mass is energy according to Einstein’s mass-energy equation: E = mc2. The point to note is that the actual measure for mass at the atomic scale is eV/c2, so we make the unit even smaller by dividing the eV (which already is a very tiny amount of energy) by c2: 1 eV/ccorresponds to 1.782662×10−36 kg, so the mass of our electron (9.1×10–31 kg) is about 510,000 eV/c2, or 0.510 MeV/c2. I am spelling it out because you will often just see 0.510 MeV in older or more popular publications, but so don’t forget that cfactor. As for the calculations below, I just stick to the kg and m measures because they make the dimensions come out right.]

According to our kinetic energy formula (K.E. = mv2/2), these mass and velocity values correspond to an energy value of 22 ×10−19 Joule (the Joule is the so-called SI unit for energy – don’t worry about it right now). So, from the first de Broglie equation (f = E/h) – and using the right value for Planck’s constant (6.626 J·s), we get a frequency of 3.32×1015 hertz (hertz just means oscillations per second as you know). Now, using v once again, and fλ = v, we see that corresponds to a wavelength of 0.66 nanometer (0.66×10−9 m). [Just take the numbers and do the math.]

However, if we use the second de Broglie relation, which relates wavelength to momentum instead of energy, then we get 0.33 nanometer (0.33×10−9 m), so that’s half of the value we got from the first equation. So what is it: 0.33 or 0.66 nm? It’s that factor 2 again. Something is wrong.

It must be that kinetic energy formula. You’ll say we should include potential energy or something. No. That’s not the issue. First, we’re talking a free particle here: an electron moving in space (a vacuum) with no external forces acting on it, so it’s a field-free space (or a region of constant potential). Second, we could, of course, extend the analysis and include potential energy, and show how it’s converted to kinetic energy (like a stone falling from 100 m to 50 m: potential energy gets converted into kinetic energy) but making our analysis more complicated by including potential energy as well will not solve our problem here: it will only make you even more confused.

Then it must be some relativistic effect you’ll say. No. It’s true the formula for kinetic energy above only holds for relatively low speeds (as compared to light, so ‘relatively’ low can be thousands of km per second) but that’s not the problem here: we are talking electrons moving at non-relativistic speeds indeed, so their mass or energy is not (or hardly) affected by relativistic effects and, hence, we can indeed use the more simple non-relativistic formulas.

The real problem we’re encountering here is not with the equations: it’s the simplistic model of our wave. We are imagining one wave here indeed, with a single frequency, a single wavelength and, hence, one single velocity – which happens to coincide with the velocity of our particle. Such wave cannot possibly represent an actual de Broglie wave: the wave is everywhere and, hence, the particle it represents is nowhere. Indeed, a wave defined by a specific wavelength λ (or a wave number k = 2π/λ if we’re using complex number notation) and a specific frequency f or period T (or angular frequency ω = 2π/T = 2πf) will have a very regular shape – such as Ψ = Aei(ωt-kx) and, hence, the probability of actually locating that particle at some specific point in space will be the same everywhere: |Ψ|= |Aei(ωt-kx)|= A2. [If you are confused about the math here, I am sorry but I cannot re-explain this once again: just remember that our de Broglie wave represents probability amplitudes – so that’s some complex number Ψ = Ψ(x, t) depending on space and time – and that we  need to take the modulus squared of that complex number to get the probability associated with some (real) value x (i.e. the space variable) and some value t (i.e. the time variable).]

So the actual matter wave of a real-life electron will be represented by a wave train, or a wave packet as it is usually referred to. Now, a wave packet is described by (at least) two types of wave velocity:

1. The so-called group velocity: the group velocity of a wave is denoted by vand is the velocity of the wave packet as a whole is traveling. Wikipedia defines it as “the velocity with which the overall shape of the waves’ amplitudes — known as the modulation or envelope of the wave — propagates through space.”
2. The so-called phase velocity: the phase velocity is denoted by vp and is what we usually associate with the velocity of a wave. It is just what it says it is: the rate at which the phase of the (composite) wave travels through space.

The term between brackets above – ‘composite’ – already indicates what it’s all about: a wave packet is to be analyzed as a composite wave: so it’s a wave composed of a finite or infinite number of component waves which all have their own wave number k and their own angular frequency ω. So the mistake we made above is that, naively, we just assumed that (i) there is only one simple wave (and, of course, there is only one wave, but it’s not a simple one: it’s a composite wave), and (ii) that the velocity v of our electron would be equal to the velocity of that wave. Now that we are a little bit more enlightened, we need to answer two questions in regard to point (ii):

1. Why would that be the case?
2. If it’s is the case, then what wave velocity are we talking about: the group velocity or the phase velocity?

To answer both questions, we need to look at wave packets once again, so let’s do that. Just to visualize things, I’ll insert – once more – that illustration you’ve seen in my other posts already:

The de Broglie wave packet

The Wikipedia article on the group velocity of a wave has wonderful animations, which I would advise you to look at in order to make sure you are following me here. There are several possibilities:

1. The phase velocity and the group velocity are the same: that’s a rather unexciting possibility but it’s the easiest thing to work with and, hence, most examples will assume that this is the case.
2. The group and phase velocity are not the same, but our wave packet is ‘stable’, so to say. In other words, the individual peaks and troughs of the wave within the envelope travel at a different speed (the phase velocity vg), but the envelope as a whole (so the wave packet as such) does not get distorted as it travels through space.
3. The wave packet dissipates: in this case, we have a constant group velocity, but the wave packet delocalizes. Its width increases over time and so the wave packet diffuses – as time goes by – over a wider and wider region of space, until it’s actually no longer there. [In case you wonder why it did not group this third possibility under (2): it’s a bit difficult to assign a fixed phase velocity to a wave like this.]

How the wave packet will behave depends on the characteristics of the component waves. To be precise, it will depend on their angular frequency and their wave number and, hence, their individual velocities. First, note the relationship between these three variables: ω = 2πf and k = 2π/λ so ω/k = fλ = v. So these variables are not independent: if you have two values (e.g. v and k), you also have the third one (ω). Secondly, note that the component waves of our wave packet will have different wavelengths and, hence, different wave numbers k.

Now, the de Broglie relation p = ħk (i.e. the same relation as p = h/λ but we replace λ with 2π/k and then ħ is the so-called reduced Planck constant ħ = h/2π) makes it obvious that different wave numbers k correspond to different values p for the momentum of our electron, so allowing for a spread in k (or a spread in λ as illustrates above) amounts to allowing for some spread in p. That’s where the uncertainty principle comes in – which I actually derived from a theoretical wave function in my post on Fourier transforms and conjugate variables. But so that’s not something I want to dwell on here.

We’re interested in the ω’s. What about them? Well… ω can take any value really – from a theoretical point of view that is. Now you’ll surely object to that from a practical point of view, because you know what it implies: different velocities of the component waves. But you can’t object in a theoretical analysis like this. The only thing we could possible impose as a constraint is that our wave packet should not dissipate – so we don’t want it to delocalize and/or vanish after a while because we’re talking about some real-life electron here, and so that’s a particle which just doesn’t vanish like that.

To impose that condition, we need to look at the so-called dispersion relation. We know that we’ll have a whole range of wave numbers k, but so what values should ω take for a wave function to be ‘well-behaved’, i.e. not disperse in our case? Let’s first accept that k is some variable, the independent variable actually, and so then we associate some ω with each of these values k. So ω becomes the dependent variable (dependent on k that is) and that amounts to saying that we have some function ω = ω(k).

What kind of function? Well… It’s called the dispersion relation – for rather obvious reasons: because this function determines how the wave packet will behave: non-dispersive or – what we don’t want here – dispersive. Indeed, there are several possibilities:

1. The speed of all component waves is the same: that means that the ratio ω/k = is the same for all component waves. Now that’s the case only if ω is directly proportional to k, with the factor of proportionality equal to v. That means that we have a very simple dispersion relation: ω = αk with α some constant equal to the velocity of the component waves as well as the group and phase velocity of the composite wave. So all velocities are just the same (vvp = vg = α) and we’re in the first of the three cases explained at the beginning of this section.
2. There is a linear relation between ω and k but no direct proportionality, so we write ω = αk + β, in which β can be anything but not some function of k. So we allow different wave speeds for the component waves. The phase velocity will, once again, be equal to the ratio of the angular frequency and the wave number of the composite wave (whatever that is), but what about the group velocity, i.e. the velocity of our electron in this example? Well… One can show – but I will not do it here because it is quite a bit of work – that the group velocity of the wave packet will be equal to vg = dω/dk, i.e. the (first-order) derivative of ω with respect to k. So, if we want that wave packet to travel at the same speed of our electron (which is what we want of course because, otherwise, the wave packet would obviously not represent our electron), we’ll have to impose that dω/dk (or ∂ω/∂k if you would want to introduce more independent variables) equals v. In short, we have the condition that dω/dk = d(αk + β)/dk = α = k.
3. If the relation between ω and k is non-linear, well… Then we have none of the above. Hence, we then have a wave packet that gets distorted and stretched out and actually vanishes after a while. That case surely does not represent an electron.

Back to the de Broglie wave relations

Indeed, it’s now time to go back to our de Broglie relations – E = hf and p = h/λ and the question that sparked the presentation above: what formula to use for E? Indeed, for p it’s easy: we use p = mv and, if you want to include the case of relativistic speeds, you will write that formula in a more sophisticated way by making it explicit that the mass m is the relativistic mass m = γm0: the rest mass multiplied with a factor referred to as the Lorentz factor which, I am sure, you have seen before: γ = (1 – v2/c2)–1/2. At relativistic speeds (i.e. speeds close to c), this factor makes a difference: it adds mass to the rest mass. So the mass of a particle can be written as m = γm0, with m0 denoting the rest mass. At low speeds (e.g. 1% of the speed of light – as in the case of our electron), m will hardly differ from m0 and then we don’t need this Lorentz factor. It only comes into play at higher speeds.

At this point, I just can’t resist a small digression. It’s just to show that it’s not ‘relativistic effects’ that cause us trouble in finding the right energy equation for our E = hf relation. What’s kinetic energy? Well… There’s a few definitions – such as the energy gathered through converting potential energy – but one very useful definition in the current context is the following: kinetic energy is the excess of a particle over its rest mass energy. So when we’re looking at high-speed or high-energy particles, we will write the kinetic energy as K.E. = mc– m0c= (m – m0)c= γm0c– m0c= m0c2(γ – 1). Before you think I am trying to cheat you: where is the v of our particle? [To make it specific: think about our electron once again but not moving at leisure this time around: imagine it’s speeding at a velocity very close to c in some particle accelerator. Now, v is close to c but not equal to c and so it should not disappear. […]

It’s in the Lorentz factor γ = (1 – v2/c2)–1/2.

Now, we can expand γ into a binomial series (it’s basically an application of the Taylor series – but just check it online if you’re in doubt), so we can write γ as an infinite sum of the following terms: γ = 1 + (1/2)·v2/c+ (3/8)·v4/c+ (3/8)·v4/c+ (5/16)·v6/c+ … etcetera. [The binomial series is an infinite Taylor series, so it’s not to be confused with the (finite) binomial expansion.] Now, when we plug this back into our (relativistic) kinetic energy equation, we can scrap a few things (just do it) to get where I want to get:

K.E. = (1/2)·m0v+ (3/8)·m0v4/c+ (5/16)·m0v6/c+ … etcetera

So what? Well… That’s it – for the digression at least: see how our non-relativistic formula for kinetic energy (K.E. = m0v2/2 is only the first term of this series and, hence, just an approximation: at low speeds, the second, third etcetera terms represent close to nothing (and more close to nothing as you check out the fourth, fifth etcetera terms). OK, OK… You’re getting tired of these games. So what? Should we use this relativistic kinetic energy formula in the de Broglie relation?

No. As mentioned above already, we don’t need any relativistic correction, but the relativistic formula above does come in handy to understand the next bit. What’s the next bit about?

Well… It turns out that we actually do have to use the total energy – including (the energy equivalent to) the rest mass of our electron – in the de Broglie relation E = hf.

WHAT!?

If you think a few seconds about the math of this – so we’d use γm0c2 instead of (1/2)m0v2 (so we use the speed of light instead of the speed of our particle) – you’ll realize we’ll be getting some astronomical frequency (we got that already but so here we are talking some kind of truly fantastic frequency) and, hence, combining that with the wavelength we’d derive from the other de Broglie equation (p = h/λ) we’d surely get some kind of totally unreal speed. Whatever it is, it will surely have nothing to do with our electron, does it?

Let’s go through the math.

The wavelength is just the same as that one given by p = h/λ, so we have λ = 0.33 nanometer. Don’t worry about this. That’s what it is indeed. Check it out online: it’s about a thousand times smaller than the wavelength of (visible) light but that’s OK. We’re talking something real here. That’s why electron microscopes can ‘see’ stuff that light microscopes can’t: their resolution is about a thousand times higher indeed.

But so when we take the first equation once again (E =hf) and calculate the frequency from f = γm0c2/h, we get an frequency in the neighborhood of 12.34×1019 herz. So that gives a velocity of v = fλ = 4.1×1010 meter per second (m/s). But… THAT’S MORE THAN A HUNDRED TIMES THE SPEED OF LIGHT. Surely, we must have got it wrong.

We don’t. The velocity we are calculating here is the phase velocity vp of our matter wave – and IT’S REAL! More in general, it’s easy to show that this phase velocity is equal to vp = fλ = E/p = (γm0c2/h)·(h/γm0v) = c2/v. Just fill in the values for c and v (3×108 and 2.2×106 respectively and you will get the same answer.

But that’s not consistent with relativity, is it? It is: phase velocities can be (and, in fact, usually are – as evidenced by our real-life example) superluminal as they say – i.e. much higher than the speed of light. However, because they carry no information – the wave packet shape is the ‘information’, i.e. the (approximate) location of our electron – such phase velocities do not conflict with relativity theory. It’s like amplitude modulation, like AM radiowaves): the modulation of the amplitude carries the signal, not the carrier wave.

The group velocity, on the other hand, can obviously not be faster than and, in fact, should be equal to the speed of our particle (i.e. the electron). So how do we calculate that? We don’t have any formula ω(k) here, do we? No. But we don’t need one. Indeed, we can write:

v= ∂ω/∂k = ∂(E/ ħ)/∂(p/ ħ) = ∂E/∂p

[Do you see why we prefer the ∂ symbol instead of the d symbol now? ω is a function of k but it’s – first and foremost – a function of E, so a partial derivative sign is quite appropriate.]

So what? Well… Now you can use either the relativistic or non-relativistic relation between E and p to get a value for ∂E/∂p. Let’s take the non-relativistic one first (E = p2/2m) : ∂E/∂p = ∂(p2/2m)/∂p = p/m = v. So we get the velocity of our electron! Just like we wanted. 🙂 As for the relativistic formula (E = (p2c+ m02c4)1/2), well… I’ll let you figure that one out yourself. [You can also find it online in case you’d be desperate.]

Wow! So there we are. That was quite something! I will let you digest this for now. It’s true I promised to ‘point out the differences between matter waves and light waves’ in my introduction but this post has been lengthy enough. I’ll save those ‘differences’ for the next post. In the meanwhile, I hope you enjoyed and – more importantly – understood this one. If you did, you’re a master! A real one! 🙂