Re-visiting uncertainty…

I re-visited the Uncertainty Principle a couple of times already, but here I really want to get at the bottom of the thing? What’s uncertain? The energy? The time? The wavefunction itself? These questions are not easily answered, and I need to warn you: you won’t get too much wiser when you’re finished reading this. I just felt like freewheeling a bit. [Note that the first part of this post repeats what you’ll find on the Occam page, or my post on Occam’s Razor. But these post do not analyze uncertainty, which is what I will be trying to do here.]

Let’s first think about the wavefunction itself. It’s tempting to think it actually is the particle, somehow. But it isn’t. So what is it then? Well… Nobody knows. In my previous post, I said I like to think it travels with the particle, but then doesn’t make much sense either. It’s like a fundamental property of the particle. Like the color of an apple. But where is that color? In the apple, in the light it reflects, in the retina of our eye, or is it in our brain? If you know a thing or two about how perception actually works, you’ll tend to agree the quality of color is not in the apple. When everything is said and done, the wavefunction is a mental construct: when learning physics, we start to think of a particle as a wavefunction, but they are two separate things: the particle is reality, the wavefunction is imaginary.

But that’s not what I want to talk about here. It’s about that uncertainty. Where is the uncertainty? You’ll say: you just said it was in our brain. No. I didn’t say that. It’s not that simple. Let’s look at the basic assumptions of quantum physics:

  1. Quantum physics assumes there’s always some randomness in Nature and, hence, we can measure probabilities only. We’ve got randomness in classical mechanics too, but this is different. This is an assumption about how Nature works: we don’t really know what’s happening. We don’t know the internal wheels and gears, so to speak, or the ‘hidden variables’, as one interpretation of quantum mechanics would say. In fact, the most commonly accepted interpretation of quantum mechanics says there are no ‘hidden variables’.
  2. However, as Shakespeare has one of his characters say: there is a method in the madness, and the pioneers– I mean Werner Heisenberg, Louis de Broglie, Niels Bohr, Paul Dirac, etcetera – discovered that method: all probabilities can be found by taking the square of the absolute value of a complex-valued wavefunction (often denoted by Ψ), whose argument, or phase (θ), is given by the de Broglie relations ω = E/ħ and k = p/ħ. The generic functional form of that wavefunction is:

Ψ = Ψ(x, t) = a·eiθ = a·ei(ω·t − k ∙x) = a·ei·[(E/ħ)·t − (p/ħ)∙x]

That should be obvious by now, as I’ve written more than a dozens of posts on this. 🙂 I still have trouble interpreting this, however—and I am not ashamed, because the Great Ones I just mentioned have trouble with that too. It’s not that complex exponential. That eiφ is a very simple periodic function, consisting of two sine waves rather than just one, as illustrated below. [It’s a sine and a cosine, but they’re the same function: there’s just a phase difference of 90 degrees.] sine

No. To understand the wavefunction, we need to understand those de Broglie relations, ω = E/ħ and k = p/ħ, and then, as mentioned, we need to understand the Uncertainty Principle. We need to understand where it comes from. Let’s try to go as far as we can by making a few remarks:

  • Adding or subtracting two terms in math, (E/ħ)·t − (p/ħ)∙x, implies the two terms should have the same dimension: we can only add apples to apples, and oranges to oranges. We shouldn’t mix them. Now, the (E/ħ)·t and (p/ħ)·x terms are actually dimensionless: they are pure numbers. So that’s even better. Just check it: energy is expressed in newton·meter (energy, or work, is force over distance, remember?) or electronvolts (1 eV = 1.6×10−19 J = 1.6×10−19 N·m); Planck’s constant, as the quantum of action, is expressed in J·s or eV·s; and the unit of (linear) momentum is 1 N·s = 1 kg·m/s = 1 N·s. E/ħ gives a number expressed per second, and p/ħ a number expressed per meter. Therefore, multiplying E/ħ and p/ħ by t and x respectively gives us a dimensionless number indeed.
  • It’s also an invariant number, which means we’ll always get the same value for it, regardless of our frame of reference. As mentioned above, that’s because the four-vector product pμxμ = E·t − px is invariant: it doesn’t change when analyzing a phenomenon in one reference frame (e.g. our inertial reference frame) or another (i.e. in a moving frame).
  • Now, Planck’s quantum of action h, or ħ – h and ħ only differ in their dimension: h is measured in cycles per second, while ħ is measured in radians per second: both assume we can at least measure one cycle – is the quantum of energy really. Indeed, if “energy is the currency of the Universe”, and it’s real and/or virtual photons who are exchanging it, then it’s good to know the currency unit is h, i.e. the energy that’s associated with one cycle of a photon. [In case you want to see the logic of this, see my post on the physical constants c, h and α.]
  • It’s not only time and space that are related, as evidenced by the fact that t − x itself is an invariant four-vector, E and p are related too, of course! They are related through the classical velocity of the particle that we’re looking at: E/p = c2/v and, therefore, we can write: E·β = p·c, with β = v/c, i.e. the relative velocity of our particle, as measured as a ratio of the speed of light. Now, I should add that the t − x four-vector is invariant only if we measure time and space in equivalent units. Otherwise, we have to write c·t − x. If we do that, so our unit of distance becomes meter, rather than one meter, or our unit of time becomes the time that is needed for light to travel one meter, then = 1, and the E·β = p·c becomes E·β = p, which we also write as β = p/E: the ratio of the energy and the momentum of our particle is its (relative) velocity.

Combining all of the above, we may want to assume that we are measuring energy and momentum in terms of the Planck constant, i.e. the ‘natural’ unit for both. In addition, we may also want to assume that we’re measuring time and distance in equivalent units. Then the equation for the phase of our wavefunctions reduces to:

θ = (ω·t − k ∙x) = E·t − p·x

Now, θ is the argument of a wavefunction, and we can always re-scale such argument by multiplying or dividing it by some constant. It’s just like writing the argument of a wavefunction as v·t–x or (v·t–x)/v = t –x/v  with the velocity of the waveform that we happen to be looking at. [In case you have trouble following this argument, please check the post I did for my kids on waves and wavefunctions.] Now, the energy conservation principle tells us the energy of a free particle won’t change. [Just to remind you, a ‘free particle’ means it’s in a ‘field-free’ space, so our particle is in a region of uniform potential.] So we can, in this case, treat E as a constant, and divide E·t − p·x by E, so we get a re-scaled phase for our wavefunction, which I’ll write as:

φ = (E·t − p·x)/E = t − (p/E)·x = t − β·x

Alternatively, we could also look at p as some constant, as there is no variation in potential energy that will cause a change in momentum, and the related kinetic energy. We’d then divide by p and we’d get (E·t − p·x)/p = (E/p)·t − x) = t/β − x, which amounts to the same, as we can always re-scale by multiplying it with β, which would again yield the same t − β·x argument.

The point is, if we measure energy and momentum in terms of the Planck unit (I mean: in terms of the Planck constant, i.e. the quantum of energy), and if we measure time and distance in ‘natural’ units too, i.e. we take the speed of light to be unity, then our Platonic wavefunction becomes as simple as:

Φ(φ) = a·eiφ = a·ei(t − β·x)

This is a wonderful formula, but let me first answer your most likely question: why would we use a relative velocity?Well… Just think of it: when everything is said and done, the whole theory of relativity and, hence, the whole of physics, is based on one fundamental and experimentally verified fact: the speed of light is absolute. In whatever reference frame, we will always measure it as 299,792,458 m/s. That’s obvious, you’ll say, but it’s actually the weirdest thing ever if you start thinking about it, and it explains why those Lorentz transformations look so damn complicated. In any case, this fact legitimately establishes as some kind of absolute measure against which all speeds can be measured. Therefore, it is only natural indeed to express a velocity as some number between 0 and 1. Now that amounts to expressing it as the β = v/c ratio.

Let’s now go back to that Φ(φ) = a·eiφ = a·ei(t − β·x) wavefunction. Its temporal frequency ω is equal to one, and its spatial frequency k is equal to β = v/c. It couldn’t be simpler but, of course, we’ve got this remarkably simple result because we re-scaled the argument of our wavefunction using the energy and momentum itself as the scale factor. So, yes, we can re-write the wavefunction of our particle in a particular elegant and simple form using the only information that we have when looking at quantum-mechanical stuff: energy and momentum, because that’s what everything reduces to at that level.

So… Well… We’ve pretty much explained what quantum physics is all about here. You just need to get used to that complex exponential: eiφ = cos(−φ) + i·sin(−φ) = cos(φ) −i·sin(φ). It would have been nice if Nature would have given us a simple sine or cosine function. [Remember the sine and cosine function are actually the same, except for a phase difference of 90 degrees: sin(φ) = cos(π/2−φ) = cos(φ+π/2). So we can go always from one to the other by shifting the origin of our axis.] But… Well… As we’ve shown so many times already, a real-valued wavefunction doesn’t explain the interference we observe, be it interference of electrons or whatever other particles or, for that matter, the interference of electromagnetic waves itself, which, as you know, we also need to look at as a stream of photons , i.e. light quanta, rather than as some kind of infinitely flexible aether that’s undulating, like water or air.

However, the analysis above does not include uncertainty. That’s as fundamental to quantum physics as de Broglie‘s equations, so let’s think about that now.

Introducing uncertainty

Our information on the energy and the momentum of our particle will be incomplete: we’ll write E = E± σE, and p = p± σp. Huh? No ΔE or ΔE? Well… It’s the same, really, but I am a bit tired of using the Δ symbol, so I am using the σ symbol here, which denotes a standard deviation of some density function. It underlines the probabilistic, or statistical, nature of our approach.

The simplest model is that of a two-state system, because it involves two energy levels only: E = E± A, with A some constant. Large or small, it doesn’t matter. All is relative anyway. 🙂 We explained the basics of the two-state system using the example of an ammonia molecule, i.e. an NHmolecule, so it consists on one nitrogen and three hydrogen atoms. We had two base states in this system: ‘up’ or ‘down’, which we denoted as base state | 1 〉 and base state | 2 〉 respectively. This ‘up’ and ‘down’ had nothing to do with the classical or quantum-mechanical notion of spin, which is related to the magnetic moment. No. It’s much simpler than that: the nitrogen atom could be either beneath or, else, above the plane of the hydrogens, as shown below, with ‘beneath’ and ‘above’ being defined in regard to the molecule’s direction of rotation around its axis of symmetry.

Capture

In any case, for the details, I’ll refer you to the post(s) on it. Here I just want to mention the result. We wrote the amplitude to find the molecule in either one of these two states as:

  • C= 〈 1 | ψ 〉 = (1/2)·e(i/ħ)·(E− A)·t + (1/2)·e(i/ħ)·(E+ A)·t
  • C= 〈 2 | ψ 〉 = (1/2)·e(i/ħ)·(E− A)·t – (1/2)·e(i/ħ)·(E+ A)·t

That gave us the following probabilities:

graph

If our molecule can be in two states only, and it starts off in one, then the probability that it will remain in that state will gradually decline, while the probability that it flips into the other state will gradually increase.

Now, the point you should note is that we get these time-dependent probabilities only because we’re introducing two different energy levels: E+ A and E− A. [Note they separated by an amount equal to 2·A, as I’ll use that information later.] If we’d have one energy level only – which amounts to saying that we know it, and that it’s something definite then we’d just have one wavefunction, which we’d write as:

a·eiθ = a·e−(i/ħ)·(E0·t − p·x) = a·e−(i/ħ)·(E0·t)·e(i/ħ)·(p·x)

Note that we can always split our wavefunction in a ‘time’ and a ‘space’ part, which is quite convenient. In fact, because our ammonia molecule stays where it is, it has no momentum: p = 0. Therefore, its wavefunction reduces to:

a·eiθ = a·e−(i/ħ)·(E0·t)

As simple as it can be. 🙂 The point is that a wavefunction like this, i.e. a wavefunction that’s defined by a definite energy, will always yield a constant and equal probability, both in time as well as in space. That’s just the math of it: |a·eiθ|= a2. Always! If you want to know why, you should think of Euler’s formula and Pythagoras’ Theorem: cos2θ +sin2θ = 1. Always! 🙂

That constant probability is annoying, because our nitrogen atom never ‘flips’, and we know it actually does, thereby overcoming a energy barrier: it’s a phenomenon that’s referred to as ‘tunneling’, and it’s real! The probabilities in that graph above are real! Also, if our wavefunction would represent some moving particle, it would imply that the probability to find it somewhere in space is the same all over space, which implies our particle is everywhere and nowhere at the same time, really.

So, in quantum physics, this problem is solved by introducing uncertainty. Introducing some uncertainty about the energy, or about the momentum, is mathematically equivalent to saying that we’re actually looking at a composite wave, i.e. the sum of a finite or potentially infinite set of component waves. So we have the same ω = E/ħ and k = p/ħ relations, but we apply them to energy levels, or to some continuous range of energy levels ΔE. It amounts to saying that our wave function doesn’t have a specific frequency: it now has n frequencies, or a range of frequencies Δω = ΔE/ħ. In our two-state system, n = 2, obviously! So we’ve two energy levels only and so our composite wave consists of two component waves only.

We know what that does: it ensures our wavefunction is being ‘contained’ in some ‘envelope’. It becomes a wavetrain, or a kind of beat note, as illustrated below:

File-Wave_group

[The animation comes from Wikipedia, and shows the difference between the group and phase velocity: the green dot shows the group velocity, while the red dot travels at the phase velocity.]

So… OK. That should be clear enough. Let’s now apply these thoughts to our ‘reduced’ wavefunction

Φ(φ) = a·eiφ = a·ei(t − β·x)

Thinking about uncertainty

Frankly, I tried to fool you above. If the functional form of the wavefunction is a·e−(i/ħ)·(E·t − p·x), then we can measure E and p in whatever unit we want, including h or ħ, but we cannot re-scale the argument of the function, i.e. the phase θ, without changing the functional form itself. I explained that in that post for my kids on wavefunctions:, in which I explained we may represent the same electromagnetic wave by two different functional forms:

 F(ct−x) = G(t−x/c)

So F and G represent the same wave, but they are different wavefunctions. In this regard, you should note that the argument of F is expressed in distance units, as we multiply t with the speed of light (so it’s like our time unit is 299,792,458 m now), while the argument of G is expressed in time units, as we divide x by the distance traveled in one second). But F and G are different functional forms. Just do an example and take a simple sine function: you’ll agree that sin(θ) ≠ sin(θ/c) for all values of θ, except 0. Re-scaling changes the frequency, or the wavelength, and it does so quite drastically in this case. 🙂 Likewise, you can see that a·ei(φ/E) = [a·eiφ]1/E, so that’s a very different function. In short, we were a bit too adventurous above. Now, while we can drop the 1/ħ in the a·e−(i/ħ)·(E·t − p·x) function when measuring energy and momentum in units that are numerically equal to ħ, we’ll just revert to our original wavefunction for the time being, which equals

Ψ(θ) = a·eiθ = a·ei·[(E/ħ)·t − (p/ħ)·x]

Let’s now introduce uncertainty once again. The simplest situation is that we have two closely spaced energy levels. In theory, the difference between the two can be as small as ħ, so we’d write: E = E± ħ/2. [Remember what I said about the ± A: it means the difference is 2A.] However, we can generalize this and write: E = E± n·ħ/2, with n = 1, 2, 3,… This does not imply any greater uncertainty – we still have two states only – but just a larger difference between the two energy levels.

Let’s also simplify by looking at the ‘time part’ of our equation only, i.e. a·ei·(E/ħ)·t. It doesn’t mean we don’t care about the ‘space part’: it just means that we’re only looking at how our function varies in time and so we just ‘fix’ or ‘freeze’ x. Now, the uncertainty is in the energy really but, from a mathematical point of view, we’ve got an uncertainty in the argument of our wavefunction, really. This uncertainty in the argument is, obviously, equal to:

(E/ħ)·t = [(E± n·ħ/2)/ħ]·t = (E0/ħ ± n/2)·t = (E0/ħ)·t ± (n/2)·t

So we can write:

a·ei·(E/ħ)·t = a·ei·[(E0/ħ)·t ± (1/2)·t] = a·ei·[(E0/ħ)·t]·ei·[±(n/2)·t]

This is valid for any value of t. What the expression says is that, from a mathematical point of view, introducing uncertainty about the energy is equivalent to introducing uncertainty about the wavefunction itself. It may be equal to a·ei·[(E0/ħ)·t]·ei·(n/2)·t, but it may also be equal to a·ei·[(E0/ħ)·t]·ei·(n/2)·t. The phases of the ei·t/2 and ei·t/2 factors are separated by a distance equal to t.

So… Well…

[…]

Hmm… I am stuck. How is this going to lead me to the ΔE·Δt = ħ/2 principle? To anyone out there: can you help? 🙂

[…]

The thing is: you won’t get the Uncertainty Principle by staring at that formula above. It’s a bit more complicated. The idea is that we have some distribution of the observables, like energy and momentum, and that implies some distribution of the associated frequencies, i.e. ω for E, and k for p. The Wikipedia article on the Uncertainty Principle gives you a formal derivation of the Uncertainty Principle, using the so-called Kennard formulation of it. You can have a look, but it involves a lot of formalism—which is what I wanted to avoid here!

I hope you get the idea though. It’s like statistics. First, we assume we know the population, and then we describe that population using all kinds of summary statistics. But then we reverse the situation: we don’t know the population but we do have sample information, which we also describe using all kinds of summary statistics. Then, based on what we find for the sample, we calculate the estimated statistics for the population itself, like the mean value and the standard deviation, to name the most important ones. So it’s a bit the same here, except that, in quantum mechanics, there may not be any real value underneath: the mean and the standard deviation represent something fuzzy, rather than something precise.

Hmm… I’ll leave you with these thoughts. We’ll develop them further as we will be digging into all much deeper over the coming weeks. 🙂

Post scriptum: I know you expect something more from me, so… Well… Think about the following. If we have some uncertainty about the energy E, we’ll have some uncertainty about the momentum p according to that β = p/E. [By the way, please think about this relationship: it says, all other things being equal (such as the inertia, i.e. the mass, of our particle), that more energy will all go into more momentum. More specifically, note that ∂p/∂p = β according to this equation. In fact, if we include the mass of our particle, i.e. its inertia, as potential energy, then we might say that (1−β)·E is the potential energy of our particle, as opposed to its kinetic energy.] So let’s try to think about that.

Let’s denote the uncertainty about the energy as ΔE. As should be obvious from the discussion above, it can be anything: it can mean two separate energy levels E = E± A, or a potentially infinite set of values. However, even if the set is infinite, we know the various energy levels need to be separated by ħ, at least. So if the set is infinite, it’s going to be a countable infinite set, like the set of natural numbers, or the set of integers. But let’s stick to our example of two values E = E± A only, with A = ħ so E + ΔE = E± ħ and, therefore, ΔE = ± ħ. That implies Δp = Δ(β·E) = β·ΔE = ± β·ħ.

Hmm… This is a bit fishy, isn’t it? We said we’d measure the momentum in units of ħ, but so here we say the uncertainty in the momentum can actually be a fraction of ħ. […] Well… Yes. Now, the momentum is the product of the mass, as measured by the inertia of our particle to accelerations or decelerations, and its velocity. If we assume the inertia of our particle, or its mass, to be constant – so we say it’s a property of the object that is not subject to uncertainty, which, I admit, is a rather dicey assumption (if all other measurable properties of the particle are subject to uncertainty, then why not its mass?) – then we can also write: Δp = Δ(m·v) = Δ(m·β) = m·Δβ. [Note that we’re not only assuming that the mass is not subject to uncertainty, but also that the velocity is non-relativistic. If not, we couldn’t treat the particle’s mass as a constant.] But let’s be specific here: what we’re saying is that, if ΔE = ± ħ, then Δv = Δβ will be equal to Δβ = Δp/m = ± (β/m)·ħ. The point to note is that we’re no longer sure about the velocity of our particle. Its (relative) velocity is now:

β ± Δβ = β ± (β/m)·ħ

But, because velocity is the ratio of distance over time, this introduces an uncertainty about time and distance. Indeed, if its velocity is β ± (β/m)·ħ, then, over some time T, it will travel some distance X = [β ± (β/m)·ħ]·T. Likewise, it we have some distance X, then our particle will need a time equal to T = X/[β ± (β/m)·ħ].

You’ll wonder what I am trying to say because… Well… If we’d just measure X and T precisely, then all the uncertainty is gone and we know if the energy is E+ ħ or E− ħ. Well… Yes and no. The uncertainty is fundamental – at least that’s what’s quantum physicists believe – so our uncertainty about the time and the distance we’re measuring is equally fundamental: we can have either of the two values X = [β ± (β/m)·ħ] T = X/[β ± (β/m)·ħ], whenever or wherever we measure. So we have a ΔX and ΔT that are equal to ± [(β/m)·ħ]·T and X/[± (β/m)·ħ] respectively. We can relate this to ΔE and Δp:

  • ΔX = (1/m)·T·Δp
  • ΔT = X/[(β/m)·ΔE]

You’ll grumble: this still doesn’t give us the Uncertainty Principle in its canonical form. Not at all, really. I know… I need to do some more thinking here. But I feel I am getting somewhere. 🙂 Let me know if you see where, and if you think you can get any further. 🙂

The thing is: you’ll have to read a bit more about Fourier transforms and why and how variables like time and energy, or position and momentum, are so-called conjugate variables. As you can see, energy and time, and position and momentum, are obviously linked through the E·t and p·products in the E0·t − p·x sum. That says a lot, and it helps us to understand, in a more intuitive way, why the ΔE·Δt and Δp·Δx products should obey the relation they are obeying, i.e. the Uncertainty Principle, which we write as ΔE·Δt ≥ ħ/2 and Δp·Δx ≥ ħ/2. But so proving involves more than just staring at that Ψ(θ) = a·eiθ = a·ei·[(E/ħ)·t − (p/ħ)·x] relation.

Having said, it helps to think about how that E·t − p·x sum works. For example, think about two particles, a and b, with different velocity and mass, but with the same momentum, so p= pb ⇔ ma·v= ma·v⇔ ma/v= mb/va. The spatial frequency of the wavefunction  would be the same for both but the temporal frequency would be different, because their energy incorporates the rest mass and, hence, because m≠ mb, we also know that E≠ Eb. So… It all works out but, yes, I admit it’s all very strange, and it takes a long time and a lot of reflection to advance our understanding.

Re-visiting the Uncertainty Principle

Let me, just like Feynman did in his last lecture on quantum electrodynamics for Alix Mautner, discuss some loose ends. Unlike Feynman, I will not be able to tie them up. However, just describing them might be interesting and perhaps you, my imaginary reader, could actually help me with tying them up ! Let’s first re-visit the wave function for a photon by way of introduction.

The wave function for a photon

Let’s not complicate things from the start and, hence, let’s first analyze a nice Gaussian wave packet, such as the right-hand graph below: Ψ(x, t). It could be a de Broglie wave representing an electron but here we’ll assume the wave packet might actually represent a photon. [Of course, do remember we should actually show both the real as well as the imaginary part of this complex-valued wave function but we don’t want to clutter the illustration and so it’s only one of the two (cosine or sine). The ‘other’ part (sine or cosine) is just the same but with a phase shift. Indeed, remember that a complex number reθ is equal to r(cosθ + isinθ), and the shape of the sine function is the same as the cosine function but it’s shifted to the left with π/2. So if we have one, we have the other. End of digression.] 

example of wave packet

The assumptions associated with this wonderful mathematical shape include the idea that the wave packet is a composite wave consisting of a large number of harmonic waves with wave numbers k1, k2k3,… all lying around some mean value μk. That is what is shown in the left-hand graph. The mean value is actually noted as k-bar in the illustration above but because I can’t find a k-bar symbol among the ‘special characters’ in the text editor tool bar here, I’ll use the statistical symbols μ and σ to represent a mean value (μ) and some spread around it (σ). In any case, we have a pretty normal shape here, resembling the Gaussian distribution illustrated below.

normal probability distributionThese Gaussian distributions (also known as a density function) have outliers, but you will catch 95,4% of the observations within the μ ± 2σ interval, and 99.7% within the μ ± 3σ interval (that’s the so-called two- and three-sigma rule). Now, the shape of the left-hand graph of the first illustration, mapping the relation between k and A(k), is the same as this Gaussian density function, and if you would take a little ruler and measure the spread of k on the horizontal axis, you would find that the values for k are effectively spread over an interval that’s somewhat bigger than k-bar plus or minus 2Δk. So let’s say 95,4% of the values of k lie in the interval [μ– 2Δk, μk + 2Δk]. Hence, for all practical purposes, we can write that μ– 2Δk  < k< μ+ 2Δk. In any case, we do not care too much about the rest because their contribution to the amplitude of the wave packet is minimal anyway, as we can see from that graph. Indeed, note that the A(k) values on the vertical axis of that graph do not represent the density of the k variable: there is only one wave number for each component wave, and so there’s no distribution or density function of k. These A(k) numbers represent the (maximum) amplitude of the component waves of our wave packet Ψ(x, t). In short, they are the values A(k) appearing in the summation formula for our composite wave, i.e. the wave packet:

Wave packet summation

I don’t want to dwell much more on the math here (I’ve done that in my other posts already): I just want you to get a general understanding of that ‘ideal’ wave packet possibly representing a photon above so you can follow the rest of my story. So we have a (theoretical) bunch of (component) waves with different wave numbers kn, and the spread in these wave numbers – i.e. 2Δk, or let’s take 4Δk to make sure we catch (almost) all of them – determines the length of the wave packet Ψ, which is written here as 2Δx, or 4Δx if we’d want to include (most of) the tail ends as well. What else can we say about Ψ? Well… Maybe something about velocities and all that? OK.

To calculate velocities, we need both ω and k. Indeed, the phase velocity of a wave (vp) is equal to v= ω/k. Now, the wave number k of the wave packet itself – i.e. the wave number of the oscillating ‘carrier wave’ so to say – should be equal to μaccording to the article I took this illustration from. I should check that but, looking at that relationship between A(k) and k, I would not be surprised if the math behind is right. So we have the k for the wave packet itself (as opposed to the k’s of its components). However, I also need the angular frequency ω.

So what is that ω? Well… That will depend on all the ω’s associated with all the k’s, isn’t it? It does. But, as I explained in a previous post, the component waves do not necessarily have to travel all at the same speed and so the relationship between ω and k may not be simple. We would love that, of course, but Nature does what it wants. The only reasonable constraint we can impose on all those ω’s is that they should be some linear function of k. Indeed, if we do not want our wave packet to dissipate (or disperse or, to put it even more plainly, to disappear), then the so-called dispersion relation ω = ω(k) should be linear, so ωn should be equal to ω= ak+ b. What a and b? We don’t know. Random constants. But if the relationship is not linear, then the wave packet will disperse and it cannot possibly represent a particle – be it an electron or a photon.

I won’t go through the math all over again but in my Re-visiting the Matter Wave (I), I used the other de Broglie relationship (E = ħω) to show that – for matter waves that do not disperse – we will find that the phase velocity will equal c/β, with β = v/c, i.e. the ratio of the speed of our particle (v) and the speed of light (c). But, of course, photons travel at the speed of light and, therefore, everything becomes very simple and the phase velocity of the wave packet of our photon would equal the group velocity. In short, we have:

v= ω/k = v= ∂ω/∂k = c

Of course, I should add that the angular frequency of all component waves will also be equal to ω = ck, so all component waves of the wave packet representing a photon are supposed to travel at the speed of light! What an amazingly simple result!

It is. In order to illustrate what we have here – especially the elegance and simplicity of that wave packet for a photon – I’ve uploaded two gif files (see below). The first one could represent our ‘ideal’ photon: group and phase velocity (represented by the speed of the green and red dot respectively) are the same. Of course, our ‘ideal’ photon would only be one wave packet – not a bunch of them like here – but then you may want to think that the ‘beam’ below might represent a number of photons following each other in a regular procession.

Wave - same group and phase velocity

The second animated gif below shows how phase and group velocity can differ. So that would be a (bunch of) wave packets representing a particle not traveling at the speed of light. The phase velocity here is faster than the group velocity (the red dot travels faster than the green dot). [One can actually also have a wave with positive group velocity and negative phase velocity – quite interesting ! – but so that would not represent a particle wave.] Again, a particle would be represented by one wave packet only (so that’s the space between two green dots only) but, again, you may want to think of this as representing electrons following each other in a very regular procession. 

Wave_groupThese illustrations (which I took, once again, from the online encyclopedia Wikipedia) are a wonderful pedagogic tool. I don’t know if it’s by coincidence but the group velocity of the second wave is actually somewhat slower than the first – so the photon versus electron comparison holds (electrons are supposed to move (much) slower). However, as for the phase velocities, they are the same for both waves and that would not reflect the results we found for matter waves. Indeed, you may or may not remember that we calculated superluminal speeds for the phase velocity of matter waves in that post I mentioned above (Re-visiting the Matter Wave): an electron traveling at a speed of 0.01c (1% of the speed of light) would be represented by a wave packet with a group velocity of 0.01c indeed, but its phase velocity would be 100 times the speed of light, i.e. 100c. [That being said, the second illustration may be interpreted as a little bit correct as the red dot does travel faster than the green dot, which – as I explained – is not necessarily always the case when looking at such composite waves (we can have slower or even negative speeds).]

Of course, I should once again repeat that we should not think that a photon or an electron is actually wriggling through space like this: the oscillation only represent the real or imaginary part of the complex-valued probability amplitude associated with our ‘ideal’ photon or our ‘ideal’ electron. That’s all. So this wave is an ‘oscillating complex number’, so to say, whose modulus we have to square to get the probability to actually find the photon (or electron) at some point x and some time t. However, the photon (or the electron) itself are just moving straight from left to right, with a speed matching the group velocity of their wave function.

Are they? 

Well… No. Or, to be more precise: maybe. WHAT? Yes, that’s surely one ‘loose end’ worth mentioning! According to QED, photons also have an amplitude to travel faster or slower than light, and they are not necessarily moving in a straight line either. WHAT? Yes. That’s the complicated business I discussed in my previous post. As for the amplitudes to travel faster or slower than light, Feynman dealt with them very summarily. Indeed, you’ll remember the illustration below, which shows that the contributions of the amplitudes associated with slower or faster speed than light tend to nil because (a) their magnitude (or modulus) is smaller and (b) they point in the ‘wrong’ direction, i.e. not the direction of travel.

Contribution interval

Still, these amplitudes are there and – Shock, horror ! – photons also have an amplitude to not travel in a straight line, especially when they are forced to travel through a narrow slit, or right next to some obstacle. That’s diffraction, described as “the apparent bending of waves around small obstacles and the spreading out of waves past small openings” in Wikipedia. 

Diffraction is one of the many phenomena that Feynman deals with in his 1985 Alix G. Mautner Memorial Lectures. His explanation is easy: “not enough arrows” – read: not enough amplitudes to add. With few arrows, there are also few that cancel out indeed, and so the final arrow for the event is quite random, as shown in the illustrations below.

Many arrows Few arrows

So… Not enough arrows… Feynman adds the following on this: “[For short distances] The nearby, nearly straight paths also make important contributions. So light doesn’t really travel only in a straight line; it “smells” the neighboring paths around it, and uses a small core of nearby space. In the same way, a mirror has to have enough size to reflect normally; if the mirror is too small for the core of neighboring paths, the light scatters in many directions, no matter where you put the mirror.” (QED, 1985, p. 54-56)

Not enough arrows… What does he mean by that? Not enough photons? No. Diffraction for photons works just the same as for electrons: even if the photons would go through the slit one by one, we would have diffraction (see my Revisiting the Matter Wave (II) post for a detailed discussion of the experiment). So even one photon is likely to take some random direction left or right after going through a slit, rather than to go straight. Not enough arrows means not enough amplitudes. But what amplitudes is he talking about?

These amplitudes have nothing to do with the wave function of our ideal photon we were discussing above: that’s the amplitude Ψ(x, t) of a photon to be at point x at point t. The amplitude Feynman is talking about is the amplitude of a photon to go from point A to B along one of the infinitely many possible paths it could take. As I explained in my previous post, we have to add all of these amplitudes to arrive at one big final arrow which, over longer distances, will usually be associated with a rather large probability that the photon will travel in a straight line and at the speed of light – which is why light seems to do at a macro-scale. 🙂

But back to that very succinct statement: not enough arrows. That’s obviously a very relative statement. Not enough as compared to what? What measurement scale are we talking about here? It’s obvious that the ‘scale’ of these arrows for electrons is different than for photons, because the 2012 diffraction experiment with electrons that I referred to used 50 nanometer slits (50×10−9 m), while one of the many experiments demonstrating light diffraction using pretty standard (red) laser light used slits of some 100 micrometer (that 100×10−6 m or – in units you are used to – 0.1 millimeter). 

The key to the ‘scale’ here is the wavelength of these de Broglie waves: the slit needs to be ‘small enough’ as compared to these de Broglie wavelengths. For example, the width of the slit in the laser experiment corresponded to (roughly) 100 times the wavelength of the laser light, and the (de Broglie) wavelength of the electrons in that 2012 diffraction experiment was 50 picometer – that was actually a thousand times the electron wavelength – but it was OK enough to demonstrate diffraction. Much larger slits would not have done the trick. So, when it comes to light, we have diffraction at scales that do not  involve nanotechnology, but when it comes to matter particles, we’re not talking micro but nano: that’s thousand times smaller.

The weird relation between energy and size

Let’s re-visit the Uncertainty Principle, even if Feynman says we don’t need that (we just need to do the amplitude math and we have it all). We wrote the uncertainty principle using the more scientific Kennard formulation: σxσ≥ ħ/2, in which the sigma symbol represents the standard deviation of position x and momentum p respectively. Now that’s confusing, you’ll say, because we were talking wave numbers, not momentum in the introduction above. Well… The wave number k of a de Broglie wave is, of course, related to the momentum p of the particle we’re looking at: p = ħk. Hence, a spread in the wave numbers amounts to a spread in the momentum really and, as I wanted to talk scales, let’s now check the dimensions.

The value for ħ is about 1×10–34 Joule·seconds (J·s) (it’s about 1.054571726(47)×10−34 but let’s go with the gross approximation as for now). One J·s is the same as one kg·m2/s because 1 Joule is a shorthand for km kg·m2/s2. It’s a rather large unit and you probably know that physicists prefer electronVolt·seconds (eV·s) because of that. However, even in expressed in eV·s the value for ħ comes out astronomically small6.58211928(15)×10−16 eV·s. In any case, because the J·s makes dimensions come out right, I’ll stick to it for a while. What does this incredible small factor of proportionality, both in the de Broglie relations as well in that Kennard formulation of the uncertainty principle, imply? How does it work out from a math point of view?

Well… It’s literally a quantum of measurement: even if Feynman says the uncertainty principle should just be seen “in its historical context”, and that “we don’t need it for adding arrows”, it is a consequence of the (related) position-space and momentum-space wave functions for a particle. In case you would doubt that, check it on Wikipedia: the author of the article on the uncertainty principle derives it from these two wave functions, which form a so-called Fourier transform pair. But so what does it say really?

Look at it. First, it says that we cannot know any of the two values exactly (exactly means 100%) because then we have a zero standard deviation for one or the other variable, and then the inequality makes no sense anymore: zero is obviously not greater or equal to 0.527286×10–34 J·s. However, the inequality with the value for  ħ plugged in shows how close to zero we can get with our measurements. Let’s check it out. 

Let’s use the assumption that two times the standard deviation (written as 2Δk or 2Δx on or above the two graphs in the very first illustration of this post) sort of captures the whole ‘range’ of the variable. It’s not a bad assumption: indeed, if Nature would follow normal distributions – and in our macro-world, that seems to be the case – then we’d capture 95.4 of them, so that’s good. Then we can re-write the uncertainty principle as:

Δx·σ≥ ħ or σx·Δp ≥ ħ

So that means we know x within some interval (or ‘range’ if you prefer that term) Δx or, else, we know p within some interval (or ‘range’ if you prefer that term) Δp. But we want to know both within some range, you’ll say. Of course. In that case, the uncertainty principle can be written as:

Δx·Δp ≥ 2ħ

Huh? Why the factor 2? Well… Each of the two Δ ranges corresponds to 2σ (hence, σ= Δx/2 and σ= Δp/2), and so we have (1/2)Δx·(1/2)Δp ≥ ħ/2. Note that if we would equate our Δ with 3σ to get 97.7% of the values, instead of 95.4% only, once again assuming that Nature distributes all relevant properties normally (not sure – especially in this case, because we are talking discrete quanta of action here – so Nature may want to cut off the ‘tail ends’!), then we’d get Δx·Δp ≥ 4.5×ħ: the cost of extra precision soars! Also note that, if we would equate Δ with σ (the one-sigma rule corresponds to 68.3% of a normally distributed range of values), then we get yet another ‘version’ of the uncertainty principle: Δx·Δp ≥ ħ/2. Pick and choose! And if we want to be purists, we should note that ħ is used when we express things in radians (such as the angular frequency for example: E = ħω), so we should actually use h when we are talking distance and (linear) momentum. The equation above then becomes Δx·Δp ≥ h/π.

It doesn’t matter all that much. The point to note is that, if we express x and p in regular distance and momentum units (m and kg·m/s), then the unit for ħ (or h) is 1×10–34. Now, we can sort of choose how to spread the uncertainty over x and p. If we spread it evenly, then we’ll measure both Δx and Δp  in units of 1×10–17  m and 1×10–17 kg·m/s. That’s small… but not that small. In fact, it is (more or less) imaginably small I’d say.

For example, a photon of a blue-violet light (let’s say a wavelength of around 660 nanometer) would have a momentum p = h/λ equal to some 1×10–22 kg·m/s (just work it out using the values for h and λ). You would usually see this value measured in a unit that’s more appropriate to the atomic scale: 6.25 eV/c. [Converting momentum into energy using E = pc, and using the Joule-electronvolt conversion (1 eV ≈ 1.6×10–19 J) will get you there.] Hence, units of 1×10–17  m for momentum are a hundred thousand times the rather average momentum of our light photon. We can’t have that so let’s reduce the uncertainty related to the momentum to that 1×10–22 kg·m/s scale. Then the uncertainty about position will be measured in units of 1×10–12 m. That’s the picometer scale in-between the nanometer (1×10–9 m) and the femtometer (1×10–9 m) scale. You’ll remember that this scale corresponds to the resolution of a (modern) electron microscope (50 pm). So can we see “uncertainty effects” ? Yes. I’ll come back to that.

However, before I discuss these, I need to make a little digression. Despite the sub-title I am using above, the uncertainties in distance and momentum we are discussing here are nowhere near to what is referred to as the Planck scale in physics: the Planck scale is at the other side of that Great Desert I mentioned: the Large Hadron Collider, which smashes particles with  (average) energies of 4 tera-electronvolt (i.e. 4 trillion eV – all packed into one particle !) is probing stuff measuring at a scale of a thousandth of a femtometer (0.001×10–12 m), but we’re obviously at the limits of what’s technically possible, and so that’s where the Great Desert starts. The ‘other side’ of that Great Desert is the Planck scale: 10–35 m. Now, why is that some kind of theoretical limit? Why can’t we just continue to further cut these scales down? Just like Dedekind did when defining irrational numbers? We can surely get infinitely close to zero, can we? Well… No. The reasoning is quite complex (and I am not sure if I actually understand it – the way I should) but it is quite relevant to the topic here (the relation between energy and size), and it goes something like this:

  1. In quantum mechanics, particles are considered to be point-like but they do take space, as evidenced from our discussion on slit widths: light will show diffraction at the micro-scale (10–6 m) but electrons will do that only at the nano-scale (10–9 m), so that’s a thousand times smaller. That’s related to their respective the de Broglie wavelength which, for electrons, is also a thousand times smaller than that of electrons. Now, the de Broglie wavelength is related to the energy and/or the momentum of these particles: E = hf and p = h/λ.
  2. Higher energies correspond to smaller de Broglie wavelengths and, hence, are associated with particles of smaller size. To continue the example, the energy formula to be used in the E = hf relation for an electron – or any particle with rest mass – is the (relativistic) mass-energy equivalence relation: E = γm0c2, with γ the Lorentz factor, which depends on the velocity v of the particle. For example, electrons moving at more or less normal speeds (like in the 2012 experiment, or those used in an electron microscope) have typical energy levels of some 600 eV, and don’t think that’s a lot: the electrons from that cathode ray tube in the back of an old-fashioned TV which lighted up the screen so you could watch it, had energies in the 20,000 eV range. So, for electrons, we are talking energy levels a thousand or a hundred thousand higher than for your typical 2 to 10 eV photon.
  3. Of course, I am not talking X or gamma rays here: hard X rays also have energies of 10 to 100 kilo-electronvolt, and gamma ray energies range from 1 million to 10 million eV (1-10 MeV). In any case, the point to note is ‘small’ particles must have high energies, and I am not only talking massless particles such as photons. Indeed, in my post End of the Road to Reality?, I discussed the scale of a proton and the scale of quarks: 1.7 and 0.7 femtometer respectively, which is smaller than the so-called classical electron radius. So we have (much) heavier particles here that are smaller? Indeed, the rest mass of the u and d quarks that make up a proton (uud) is 2.4 and 4.8 MeV/c2 respectively, while the (theoretical) rest mass of an electron is 0.511 Mev/conly, so that’s almost 20 times more: (2.4+2.4+4.8/0.5). Well… No. The rest mass of a proton is actually 1835 times the rest mass of an electron: the difference between the added rest masses of the quarks that make it up and the rest mass of the proton itself (938 MeV//c2) is the equivalent mass of the strong force that keeps the quarks together.
  4. But let me not complicate things. Just note that there seems to be a strange relationship between the energy and the size of a particle: high-energy particles are supposed to be smaller, and vice versa: smaller particles are associated with higher energy levels. If we accept this as some kind of ‘factual reality’, then we may understand what the Planck scale is all about: : the energy levels associated with theoretical ‘particles’ of the above-mentioned Planck scale (i.e. particles with a size in the 10–35 m range) would have energy levels in the 1019 GeV range. So what? Well… This amount of energy corresponds to an equivalent mass density of a black hole. So any ‘particle’ we’d associate with the Planck length would not make sense as a physical entity: it’s the scale where gravity takes over – everything.

Again: so what? Well… I don’t know. It’s just that this is entirely new territory, and it’s also not the topic of my post here. So let me just quote Wikipedia on this and then move on: “The fundamental limit for a photon’s energy is the Planck energy [that’s the 1019 GeV which I mentioned above: to be precise, that ‘limit energy’ is said to be 1.22 × 1019 GeV], for the reasons cited above [that ‘photon’ would not be ‘photon’ but a black hole, sucking up everything around it]. This makes the Planck scale a fascinating realm for speculation by theoretical physicists from various schools of thought. Is the Planck scale domain a seething mass of virtual black holes? Is it a fabric of unimaginably fine loops or a spin foam network? Is it interpenetrated by innumerable Calabi-Yau manifolds which connect our 3-dimensional universe with a higher-dimensional space? [That’s what’s string theory is about.] Perhaps our 3-D universe is ‘sitting’ on a ‘brane’ which separates it from a 2, 5, or 10-dimensional universe and this accounts for the apparent ‘weakness’ of gravity in ours. These approaches, among several others, are being considered to gain insight into Planck scale dynamics. This would allow physicists to create a unified description of all the fundamental forces. [That’s what’s these Grand Unification Theories (GUTs) are about.]

Hmm… I wish I could find some easy explanation of why higher energy means smaller size. I do note there’s an easy relationship between energy and momentum for massless particles traveling at the velocity of light (like photons): E = p(or p = E/c), but – from what I write above – it is obvious that it’s the spread in momentum (and, therefore, in wave numbers) which determines how short or how long our wave train is, not the energy level as such. I guess I’ll just have to do some more research here and, hopefully, get back to you when I understand things better. 

Re-visiting the Uncertainty Principle

You will probably have read countless accounts of the double-slit experiment, and so you will probably remember that these thought or actual experiments also try to watch the electrons as they pass the slits – with disastrous results: the interference pattern disappears. I copy Feynman’s own drawing from his 1965 Lecture on Quantum Behavior below: a light source is placed behind the ‘wall’, right between the two slits. Now, light (i.e. photons) gets scattered when it hits electrons and so now we should ‘see’ through which slit the electron is coming. Indeed, remember that we sent them through these slits one by one, and we still had interference – suggesting the ‘electron wave’ somehow goes through both slits at the same time, which can’t be true – because an electron is a particle.

Watching the electrons

However, let’s re-examine what happens exactly.

  1. We can only detect all electrons if the light is high intensity, and high intensity does not mean higher energy photons but more photons. Indeed, if the light source is deem, then electrons might get through without being seen. So a high-intensity light source allows us to see all electrons but – as demonstrated not only in thought experiments but also in the laboratory – it destroys the interference pattern.
  2. What if we use lower-energy photons, like infrared light with wavelengths of 10 to 100 microns instead of visible light? We can then use thermal imaging night vision goggles to ‘see’ the electrons. 🙂 And if that doesn’t work, we can use radiowaves (or perhaps radar!). The problem – as Feynman explains it – is that such low frequency light (associated with long wavelengths) only give a ‘big fuzzy flash’ when the light is scattered: “We can no longer tell which hole the electron went through! We just know it went somewhere!” At the same time, “the jolts given to the electron are now small enough so that we begin to see some interference effect again.” Indeed: “For wavelengths much longer than the separation between the two slits (when we have no chance at all of telling where the electron went), we find that the disturbance due to the light gets sufficiently small that we again get the interference curve P12.” [P12 is the curve describing the original interference effect.]

Now, that would suggest that, when push comes to shove, the Uncertainty Principle only describes some indeterminacy in the so-called Compton scattering of a photon by an electron. This Compton scattering is illustrated below: it’s a more or less elastic collision between a photon and electron, in which momentum gets exchanged (especially the direction of the momentum) and – quite important – the wavelength of the scattered light is different from the incident radiation. Hence, the photon loses some energy to the electron and, because it will still travel at speed c, that means its wavelength must increase as prescribed by the λ = h/p de Broglie relation (with p = E/c for a photon). The change in the wavelength is called the Compton shift. and its formula is given in the illustration: it depends on the (rest) mass of the electron obviously and on the change in the direction of the momentum (of the photon – but that change in direction will obviously also be related to the recoil direction of the electron).

Compton scattering

This is a very physical interpretation of the Uncertainty Principle, but it’s the one which the great Richard P. Feynman himself stuck to in 1965, i.e. when he wrote his famous Lectures on Physics at the height of his career. Let me quote his interpretation of the Uncertainty Principle in full indeed:

“It is impossible to design an apparatus to determine which hole the electron passes through, that will not at the same time disturb the electrons enough to destroy the interference pattern. If an apparatus is capable of determining which hole the electron goes through, it cannot be so delicate that it does not disturb the pattern in an essential way. No one has ever found (or even thought of) a way around this. So we must assume that it describes a basic characteristic of nature.”

That’s very mechanistic indeed, and it points to indeterminacy rather than ontological uncertainty. However, there’s weirder stuff than electrons being ‘disturbed’ in some kind of random way by the photons we use to detect them, with the randomness only being related to us not knowing at what time photons leave our light source, and what energy or momentum they have exactly. That’s just ‘indeterminacy’ indeed; not some fundamental ‘uncertainty’ about Nature.

We see such ‘weirder stuff’ in those mega- and now tera-electronvolt experiments in particle accelerators. In 1965, Feynman had access to the results of the high-energy positron-electron collisions being observed in the 3 km long Stanford Linear Accelerator (SLAC), which started working in 1961, but stuff like quarks and all that was discovered only in the late 1960s and early 1970s, so that’s after Feynman’s Lectures on Physics.So let me just mention a rather remarkable example of the Uncertainty Principle at work which Feynman quotes in his 1985 Alix G. Mautner Memorial Lectures on Quantum Electrodynamics.

In the Feynman diagram below, we see a photon disintegrating, at time t = T3into a positron and an electron. The positron (a positron is an electron with positive charge basically: it’s the electron’s anti-matter counterpart) meets another electron that ‘happens’ to be nearby and the annihilation results in (another) high-energy photon being emitted. While, as Feynman underlines, “this is a sequence of events which has been observed in the laboratory”, how is all this possible? We create matter – an electron and a positron both have considerable mass – out of nothing here ! [Well… OK – there’s a photon, so that’s some energy to work with…]

Photon disintegrationFeynman explains this weird observation without reference to the Uncertainty Principle. He just notes that “Every particle in Nature has an amplitude to move backwards in time, and therefore has an anti-particle.” And so that’s what this electron coming from the bottom-left corner does: it emits a photon and then the electron moves backwards in time. So, while we see a (very short-lived) positron moving forward, it’s actually an electron quickly traveling back in time according to Feynman! And, after a short while, it has had enough of going back in time, so then it absorbs a photon and continues in a slightly different direction. Hmm… If this does not sound fishy to you, it does to me.

The more standard explanation is in terms of the Uncertainty Principle applied to energy and time. Indeed, I mentioned that we have several pairs of conjugate variables in quantum mechanics: position and momentum are one such pair (related through the de Broglie relation p =ħk), but energy and time are another (related through the other de Broglie relation E = hf = ħω). While the ‘energy-time uncertainty principle’ – ΔE·Δt ≥ ħ/2 – resembles the position-momentum relationship above, it is apparently used for ‘very short-lived products’ produced in high-energy collisions in accelerators only. I must assume the short-lived positron in the Feynman diagram is such an example: there is some kind of borrowing of energy (remember mass is equivalent to energy) against time, and then normalcy soon gets restored. Now THAT is something else than indeterminacy I’d say.

uncertainty principle energy time

But so Feynman would say both interpretations are equivalent, because Nature doesn’t care about our interpretations.

What to say in conclusion? I don’t know. I obviously have some more work to do before I’ll be able to claim to understand the uncertainty principle – or quantum mechanics in general – somewhat. I think the next step is to solve my problem with the summary ‘not enough arrows’ explanation, which is – evidently – linked to the relation between energy and size of particles. That’s the one loose end I really need to tie up I feel ! I’ll keep you posted !

The Uncertainty Principle re-visited: Fourier transforms and conjugate variables

In previous posts, I presented a time-independent wave function for a particle (or wavicle as we should call it – but so that’s not the convention in physics) – let’s say an electron – traveling through space without any external forces (or force fields) acting upon it. So it’s just going in some random direction with some random velocity v and, hence, its momentum is p = mv. Let me be specific – so I’ll work with some numbers here – because I want to introduce some issues related to units for measurement.

So the momentum of this electron is the product of its mass m (about 9.1×10−28 grams) with its velocity v (typically something in the range around 2,200 km/s, which is fast but not even close to the speed of light – and, hence, we don’t need to worry about relativistic effects on its mass here). Hence, the momentum p of this electron would be some 20×10−25 kg·m/s. Huh? Kg·m/s?Well… Yes, kg·m/s or N·s are the usual measures of momentum in classical mechanics: its dimension is [mass][length]/[time] indeed. However, you know that, in atomic physics, we don’t want to work with these enormous units (because we then always have to add these ×10−28 and ×10−25 factors and so that’s a bit of a nuisance indeed). So the momentum p will usually be measured in eV/c, with c representing what it usually represents, i.e. the speed of light. Huh? What’s this strange unit? Electronvolts divided by c? Well… We know that eV is an appropriate unit for measuring energy in atomic physics: we can express eV in Joule and vice versa: 1 eV = 1.6×10−19 Joule, so that’s OK – except for the fact that this Joule is a monstrously large unit at the atomic scale indeed, and so that’s why we prefer electronvolt. But the Joule is a shorthand unit for kg·m2/s2, which is the measure for energy expressed in SI units, so there we are: while the SI dimension for energy is actually [mass][length]2/[time]2, using electronvolts (eV) is fine. Now, just divide the SI dimension for energy, i.e. [mass][length]2/[time]2, by the SI dimension for velocity, i.e. [length]/[time]: we get something expressed in [mass][length]/[time]. So that’s the SI dimension for momentum indeed! In other words, dividing some quantity expressed in some measure for energy (be it Joules or electronvolts or erg or calories or coulomb-volts or BTUs or whatever – there’s quite a lot of ways to measure energy indeed!) by the speed of light (c) will result in some quantity with the right dimensions indeed. So don’t worry about it. Now, 1 eV/c is equivalent to 5.344×10−28 kg·m/s, so the momentum of this electron will be 3.75 eV/c.

Let’s go back to the main story now. Just note that the momentum of this electron that we are looking at is a very tiny amount – as we would expect of course.

Time-independent means that we keep the time variable (t) in the wave function Ψ(x, t) fixed and so we only look at how Ψ(x, t) varies in space, with x as the (real) space variable representing position. So we have a simplified wave function Ψ(x) here: we can always put the time variable back in when we’re finished with the analysis. By now, it should also be clear that we should distinguish between real-valued wave functions and complex-valued wave functions. Real-valued wave functions represent what Feynman calls “real waves”, like a sound wave, or an oscillating electromagnetic field. Complex-valued wave functions describe probability amplitudes. They are… Well… Feynman actually stops short of saying that they are not real. So what are they?

They are, first and foremost complex numbers, so they have a real and a so-called imaginary part (z = a + ib or, if we use polar coordinates, reθ = cosθ + isinθ). Now, you may think – and you’re probably right to some extent – that the distinction between ‘real’ waves and ‘complex’ waves is, perhaps, less of a dichotomy than popular writers – like me 🙂 – suggest. When describing electromagnetic waves, for example, we need to keep track of both the electric field vector E as well as the magnetic field vector B (both are obviously related through Maxwell’s equations). So we have two components as well, so to say, and each of these components has three dimensions in space, and we’ll use the same mathematical tools to describe them (so we will also represent them using complex numbers). That being said, these probability amplitudes usually denoted by Ψ(x), describe something very different. What exactly? Well… By now, it should be clear that that is actually hard to explain: the best thing we can do is to work with them, so they start feeling familiar. The main thing to remember is that we need to square their modulus (or magnitude or absolute value if you find these terms more comprehensible) to get a probability (P). For example, the expression below gives the probability of finding a particle – our electron for example – in in the (space) interval [a, b]:

probability versus amplitude

Of course, we should not be talking intervals but three-dimensional regions in space. However, we’ll keep it simple: just remember that the analysis should be extended to three (space) dimensions (and, of course, include the time dimension as well) when we’re finished (to do that, we’d use so-called four-vectors – another wonderful mathematical invention).

Now, we also used a simple functional form for this wave function, as an example: Ψ(x) could be proportional, we said, to some idealized function eikx. So we can write: Ψ(x) ∝ eikx (∝ is the standard symbol expressing proportionality). In this function, we have a wave number k, which is like the frequency in space of the wave (but then measured in radians because the phase of the wave function has to be expressed in radians). In fact, we actually wrote Ψ(x, t) = (1/x)ei(kx – ωt) (so the magnitude of this amplitude decreases with distance) but, again, let’s keep it simple for the moment: even with this very simple function eikx , things will become complex enough.

We also introduced the de Broglie relation, which gives this wave number k as a function of the momentum p of the particle: k = p/ħ, with ħ the (reduced) Planck constant, i.e. a very tiny number in the neighborhood of 6.582 ×10−16 eV·s. So, using the numbers above, we’d have a value for k equal to 3.75 eV/c divided by 6.582 ×10−16 eV·s. So that’s 0.57×1016 (radians) per… Hey, how do we do it with the units here? We get an incredibly huge number here (57 with 14 zeroes after it) per second? We should get some number per meter because k is expressed in radians per unit distance, right? Right. We forgot c. We are actually measuring distance here, but in light-seconds instead of meter: k is 0.57×1016/s. Indeed, a light-second is the distance traveled by light in one second, so that’s s, and if we want k expressed in radians per meter, then we need to divide this huge number 0.57×1016 (in rad) by 2.998×108 ( in (m/s)·s) and so then we get a much more reasonable value for k, and with the right dimension too: to be precise, k is about 19×106 rad/m in this case. That’s still huge: it corresponds with a wavelength of 0.33 nanometer (1 nm = 10-6 m) but that’s the correct order of magnitude indeed.

[In case you wonder what formula I am using to calculate the wavelength: it’s λ = 2π/k. Note that our electron’s wavelength is more than a thousand times shorter than the wavelength of (visible) light (we humans can see light with wavelengths ranging from 380 to 750 nm) but so that’s what gives the electron its particle-like character! If we would increase their velocity (e.g. by accelerating them in an accelerator, using electromagnetic fields to propel them to speeds closer to and also to contain them in a beam), then we get hard beta rays. Hard beta rays are surely not as harmful as high-energy electromagnetic rays. X-rays and gamma rays consist of photons with wavelengths ranging from 1 to 100 picometer (1 pm = 10–12 m) – so that’s another factor of a thousand down – and thick lead shields are needed to stop them: they are the cause of cancer (Marie Curie’s cause of death), and the hard radiation of a nuclear blast will always end up killing more people than the immediate blast effect. In contrast, hard beta rays will cause skin damage (radiation burns) but they won’t go deeper than that.]

Let’s get back to our wave function Ψ(x) ∝ eikx. When we introduced it in our previous posts, we said it could not accurately describe a particle because this wave function (Ψ(x) = Aeikx) is associated with probabilities |Ψ(x)|2 that are the same everywhere. Indeed,  |Ψ(x)|2 = |Aeikx|2 = A2. Apart from the fact that these probabilities would add up to infinity (so this mathematical shape is unacceptable anyway), it also implies that we cannot locate our electron somewhere in space. It’s everywhere and that’s the same as saying it’s actually nowhere. So, while we can use this wave function to explain and illustrate a lot of stuff (first and foremost the de Broglie relations), we actually need something different if we would want to describe anything real (which, in the end, is what physicists want to do, right?). We already said in our previous posts: real particles will actually be represented by a wave packet, or a wave train. A wave train can be analyzed as a composite wave consisting of a (potentially infinite) number of component waves. So we write:

Composite wave

Note that we do not have one unique wave number k or – what amounts to saying the same – one unique value p for the momentum: we have n values. So we’re introducing a spread in the wavelength here, as illustrated below:

Explanation of uncertainty principle

In fact, the illustration above talks of a continuous distribution of wavelengths and so let’s take the continuum limit of the function above indeed and write what we should be writing:

Composite wave - integral

Now that is an interesting formula. [Note that I didn’t care about normalization issues here, so it’s not quite what you’d see in a more rigorous treatment of the matter. I’ll correct that in the Post Scriptum.] Indeed, it shows how we can get the wave function Ψ(x) from some other function Φ(p). We actually encountered that function already, and we referred to it as the wave function in the momentum space. Indeed, Nature does not care much what we measure: whether it’s position (x) or momentum (p), Nature will not share her secrets with us and, hence, the best we can do – according to quantum mechanics – is to find some wave function associating some (complex) probability amplitude with each and every possible (real) value of x or p. What the equation above shows, then, is these wave functions come as a pair: if we have Φ(p), then we can calculate Ψ(x) – and vice versa. Indeed, the particular relation between Ψ(x) and Φ(p) as established above, makes Ψ(x) and Φ(p) a so-called Fourier transform pair, as we can transform Φ(p) into Ψ(x) using the above Fourier transform (that’s how that  integral is called), and vice versa. More in general, a Fourier transform pair can be written as:

Fourier transform pair

Instead of x and p, and Ψ(x) and Φ(p), we have x and y, and f(x) and g(y), in the formulas above, but so that does not make much of a difference when it comes to the interpretation: x and p (or x and y in the formulas above) are said to be conjugate variables. What it means really is that they are not independent. There are quite a few of such conjugate variables in quantum mechanics such as, for example: (1) time and energy (and time and frequency, of course, in light of the de Broglie relation between both), and (2) angular momentum and angular position (or orientation). There are other pairs too but these involve quantum-mechanical variables which I do not understand as yet and, hence, I won’t mention them here. [To be complete, I should also say something about that 1/2π factor, but so that’s just something that pops up when deriving the Fourier transform from the (discrete) Fourier series on which it is based. We can put it in front of either integral, or split that factor across both. Also note the minus sign in the exponent of the inverse transform.]

When you look at the equations above, you may think that f(x) and g(y) must be real-valued functions. Well… No. The Fourier transform can be used for both real-valued as well as complex-valued functions. However, at this point I’ll have to refer those who want to know each and every detail about these Fourier transforms to a course in complex analysis (such as Brown and Churchill’s Complex Variables and Applications (2004) for instance) or, else, to a proper course on real and complex Fourier transforms (they are used in signal processing – a very popular topic in engineering – and so there’s quite a few of those courses around).

The point to note in this post is that we can derive the Uncertainty Principle from the equations above. Indeed, the (complex-valued) functions Ψ(x) and Φ(p) describe (probability) amplitudes, but the (real-valued) functions |Ψ(x)|2 and |Φ(p)|2 describe probabilities or – to be fully correct – they are probability (density) functions. So it is pretty obvious that, if the functions Ψ(x) and Φ(p) are a Fourier transform pair, then |Ψ(x)|2 and |Φ(p)|2 must be related to. They are. The derivation is a bit lengthy (and, hence, I will not copy it from the Wikipedia article on the Uncertainty Principle) but one can indeed derive the so-called Kennard formulation of the Uncertainty Principle from the above Fourier transforms. This Kennard formulation does not use this rather vague Δx and Δp symbols but clearly states that the product of the standard deviation from the mean of these two probability density functions can never be smaller than ħ/2:

σxσ≥ ħ/2

To be sure: ħ/2 is a rather tiny value, as you should know by now, 🙂 but, so, well… There it is.

As said, it’s a bit lengthy but not that difficult to do that derivation. However, just for once, I think I should try to keep my post somewhat shorter than usual so, to conclude, I’ll just insert one more illustration here (yes, you’ve seen that one before), which should now be very easy to understand: if the wave function Ψ(x) is such that there’s relatively little uncertainty about the position x of our electron, then the uncertainty about its momentum will be huge (see the top graphs). Vice versa (see the bottom graphs), precise information (or a narrow range) on its momentum, implies that its position cannot be known.

2000px-Quantum_mechanics_travelling_wavefunctions_wavelength

Does all this math make it any easier to understand what’s going on? Well… Yes and no, I guess. But then, if even Feynman admits that he himself “does not understand it the way he would like to” (Feynman Lectures, Vol. III, 1-1), who am I? In fact, I should probably not even try to explain it, should I? 🙂

So the best we can do is try to familiarize ourselves with the language used, and so that’s math for all practical purposes. And, then, when everything is said and done, we should probably just contemplate Mario Livio’s question: Is God a mathematician? 🙂

Post scriptum:

I obviously cut corners above, and so you may wonder how that ħ factor can be related to σand σ if it doesn’t appear in the wave functions. Truth be told, it does. Because of (i) the presence of ħ in the exponent in our ei(p/ħ)x function, (ii) normalization issues (remember that probabilities (i.e. Ψ|(x)|2 and |Φ(p)|2) have to add up to 1) and, last but not least, (iii) the 1/2π factor involved in Fourier transforms , Ψ(x) and Φ(p) have to be written as follows:

Position and momentum wave functionNote that we’ve also re-inserted the time variable here, so it’s pretty complete now. One more thing we could do is to substitute x for a proper three-dimensional space vector or, better still, introduce four-vectors, which would allow us to also integrate relativistic effects (most notably the slowing of time with motion – as observed from the stationary reference frame) – which become important when, for instance, we’re looking at electrons being accelerated, which is the rule, rather than the exception, in experiments.

Remember (from a previous post) that we calculated that an electron traveling at its usual speed in orbit (2200 km/s, i.e. less than 1% of the speed of light) had an energy of about 70 eV? Well, the Large Electron-Positron Collider (LEP) did accelerate them to speeds close to light, thereby giving them energy levels topping 104.5 billion eV (or 104.5 GeV as it’s written) so they could hit each other with collision energies topping 209 GeV (they come from opposite directions so it’s two times 104.5 GeV). Now, 209 GeV is tiny when converted to everyday energy units: 209 GeV is 33×10–9 Joule only indeed – and so note the minus sign in the exponent here: we’re talking billionths of a Joule here. Just to put things into perspective: 1 Watt is the energy consumption of an LED (and 1 Watt is 1 Joule per second), so you’d need to combine the energy of billions of these fast-traveling electrons to power just one little LED lamp. But, of course, that’s not the right comparison: 104.5 GeV is more than 200,000 times the electron’s rest mass (0.511 MeV), so that means that – in practical terms – their mass (remember that mass is a measure for inertia) increased by the same factor (204,500 times to be precise). Just to give an idea of the effort that was needed to do this: CERN’s LEP collider was housed in a tunnel with a circumference of 27 km. Was? Yes. The tunnel is still there but it now houses the Large Hadron Collider (LHC) which, as you surely know, is the world’s largest and most powerful particle accelerator: its experiments confirmed the existence of the Higgs particle in 2013, thereby confirming the so-called Standard Model of particle physics. [But I’ll see a few things about that in my next post.]

Oh… And, finally, in case you’d wonder where we get the inequality sign in σxσ≥ ħ/2, that’s because – at some point in the derivation – one has to use the Cauchy-Schwarz inequality (aka as the triangle inequality): |z1+ z1| ≤ |z1|+| z1|. In fact, to be fully complete, the derivation uses the more general formulation of the Cauchy-Schwarz inequality, which also applies to functions as we interpret them as vectors in a function space. But I would end up copying the whole derivation here if I add any more to this – and I said I wouldn’t do that. 🙂 […]