Thinking again…

One of the comments on my other blog made me think I should, perhaps, write something on waves again. The animation below shows the elementary wavefunction ψ = a·eiθ = ψ = a·ei·θ  = a·ei(ω·t−k·x) = a·e(i/ħ)·(E·t−p·x) .AnimationWe know this elementary wavefunction cannot represent a real-life particle. Indeed, the a·ei·θ function implies the probability of finding the particle – an electron, a photon, or whatever – would be equal to P(x, t) = |ψ(x, t)|2 = |a·e(i/ħ)·(E·t−p·x)|2 = |a|2·|e(i/ħ)·(E·t−p·x)|2 = |a|2·12= a2 everywhere. Hence, the particle would be everywhere – and, therefore, nowhere really. We need to localize the wave – or build a wave packet. We can do so by introducing uncertainty: we then add a potentially infinite number of these elementary wavefunctions with slightly different values for E and p, and various amplitudes a. Each of these amplitudes will then reflect the contribution to the composite wave, which – in three-dimensional space – we can write as:

ψ(r, t) = ei·(E/ħ)·t·f(r)

As I explained in previous posts (see, for example, my recent post on reality and perception), the f(r) function basically provides some envelope for the two-dimensional ei·θ = ei·(E/ħ)·t = cosθ + i·sinθ oscillation, with r = (x, y, z), θ = (E/ħ)·t = ω·t and ω = E/ħ.

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

Now, the axis which points upwards is usually referred to as the z-axis, and the third and final axis – which points towards us – would then be the y-axis, obviously. Unfortunately, this definition would violate the so-called right-hand rule for defining a proper reference frame: the figures below shows the two possibilities – a left-handed and a right-handed reference frame – and it’s the right-handed reference (i.e. the illustration on the right) which we have to use in order to correctly define all directions, including the direction of rotation of the argument of the wavefunction.400px-Cartesian_coordinate_system_handednessHence, if we don’t change the direction of the y– and z-axes – so we keep defining the z-axis as the axis pointing upwards, and the y-axis as the axis pointing towards us – then the positive direction of the x-axis would actually be the direction from right to left, and we should say that the elementary wavefunction in the animation above seems to propagate in the negative x-direction. [Note that this left- or right-hand rule is quite astonishing: simply swapping the direction of one axis of a left-handed frame makes it right-handed, and vice versa.]

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

wave_opposite-group-phase-velocity

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

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

OK. That’s obvious. Let me now return to my interpretation of Einstein’s E = m·c2 formula (see my previous posts on this). I noted that, in the reference frame of the particle itself (see my basics page), the elementary wavefunction a·e(i/ħ)·(E·t−p·x) reduces to a·e(i/ħ)·(E’·t’): the origin of the reference frame then coincides with (the center of) our particle itself, and the wavefunction only varies with the time in the inertial reference frame (i.e. the proper time t’), with the rest energy of the object (E’) as the time scale factor. How should we interpret this?

Well… Energy is force times distance, and force is defined as that what causes some mass to accelerate. To be precise, the newton – as the unit of force – is defined as the magnitude of a force which would cause a mass of one kg to accelerate with one meter per second per second. Per second per second. This is not a typo: 1 N corresponds to 1 kg times 1 m/s per second, i.e. 1 kg·m/s2. So… Because energy is force times distance, the unit of energy may be expressed in units of kg·m/s2·m, or kg·m2/s2, i.e. the unit of mass times the unit of velocity squared. To sum it all up:

1 J = 1 N·m = 1 kg·(m/s)2

This reflects the physical dimensions on both sides of the E = m·c2 formula again but… Well… How should we interpret this? Look at the animation below once more, and imagine the green dot is some tiny mass moving around the origin, in an equally tiny circle. We’ve got two oscillations here: each packing half of the total energy of… Well… Whatever it is that our elementary wavefunction might represent in reality – which we don’t know, of course.

circle_cos_sin

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

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

My answer is: I don’t know. I do think that, somehow, energy should be thought of as some two-dimensional oscillation of something – something which we refer to as mass, but we didn’t define mass very clearly either. It also, somehow, combines linear and rotational motion. Each of the two dimensions packs half of the energy of the particle that is being represented by our wavefunction. It is, therefore, only logical that the physical unit of both is to be expressed as a force over some distance – which is, effectively, the physical dimension of energy – or the rotational equivalent of them: torque over some angle. Indeed, the analogy between linear and angular movement is obvious: the kinetic energy of a rotating object is equal to K.E. = (1/2)·I·ω2. In this formula, I is the rotational inertia – i.e. the rotational equivalent of mass – and ω is the angular velocity – i.e. the rotational equivalent of linear velocity. Noting that the (average) kinetic energy in any system must be equal to the (average) potential energy in the system, we can add both, so we get a formula which is structurally similar to the E = m·c2 formula. But is it the same? Is the effective mass of some object the sum of an almost infinite number of quanta that incorporate some kind of rotational motion? And – if we use the right units – is the angular velocity of these infinitesimally small rotations effectively equal to the speed of light?

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

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

If you know, please let me know. 🙂

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

An introduction to virtual particles (2)

When reading quantum mechanics, it often feels like the more you know, the less you understand. My reading of the Yukawa theory of force, as an exchange of virtual particles (see my previous post), must have left you with many questions. Questions I can’t answer because… Well… I feel as much as a fool as you do when thinking about it all. Yukawa first talks about some potential – which we usually think of as being some scalar function – and then suddenly this potential becomes a wavefunction. Does that make sense? And think of the mass of that ‘virtual’ particle: the rest mass of a neutral pion is about 135 MeV. That’s an awful lot – at the (sub-)atomic scale that is: it’s equivalent to the rest mass of some 265 electrons!

But… Well… Think of it: the use of a static potential when solving Schrödinger’s equation for the electron orbitals around a hydrogen nucleus (a proton, basically) also raises lots of questions: if we think of our electron as a point-like particle being first here and then there, then that’s also not very consistent with a static (scalar) potential either!

One of the weirdest aspects of the Yukawa theory is that these emissions and absorptions of virtual particles violate the energy conservation principle. Look at the animation once again (below): it sort of assumes a rather heavy particle – consisting of a d- or u-quark and its antiparticle – is emitted – out of nothing, it seems – to then vanish as the antiparticle is destroyed when absorbed. What about the energy balance here: are we talking six quarks (the proton and the neutron), or six plus two?Nuclear_Force_anim_smallerNow that we’re talking mass, note a neutral pion (π0) may either be a uū or a dđ combination, and that the mass of a u-quark and a d-quark is only 2.4 and 4.8 MeV – so the binding energy of the constituent parts of this πparticle is enormous: it accounts for most of its mass.

The thing is… While we’ve presented the πparticle as a virtual particle here, you should also note we find πparticles in cosmic rays. Cosmic rays are particle rays, really: beams of highly energetic particles. Quite a bunch of them are just protons that are being ejected by our Sun. [The Sun also ejects electrons – as you might imagine – but let’s think about the protons here first.] When these protons hit an atom or a molecule in our atmosphere, they usually break up in various particles, including our πparticle, as shown below. 850px-Atmospheric_Collision

 

So… Well… How can we relate these things? What is going on, really, inside of that nucleus?

Well… I am not sure. Aitchison and Hey do their utmost to try to explain the pion – as a virtual particle, that is – in terms of energy fluctuations that obey the Uncertainty Principle for energy and time: ΔE·Δt ≥ ħ/2. Now, I find such explanations difficult to follow. Such explanations usually assume any measurement instrument – measuring energy, time, momentum of distance – measures those variables on some discrete scale, which implies some uncertainty indeed. But that uncertainty is more like an imprecision, in my view. Not something fundamental. Let me quote Aitchison and Hey:

“Suppose a device is set up capable of checking to see whether energy is, in fact, conserved while the pion crosses over.. The crossing time Δt must be at least r/c, where r is the distance apart of the nucleons. Hence, the device must be capable of operating on a time scale smaller than Δt to be able to detect the pion, but it need not be very much less than this. Thus the energy uncertainty in the reading by the device will be of the order ΔE ∼ ħ/Δt) = ħ·(c/r).”

As said, I find such explanations really difficult, although I can sort of sense some of the implicit assumptions. As I mentioned a couple of times already, the E = m·c2 equation tells us energy is mass in motion, somehow: some weird two-dimensional oscillation in spacetime. So, yes, we can appreciate we need some time unit to count the oscillations – or, equally important, to measure their amplitude.

[…] But… Well… This falls short of a more fundamental explanation of what’s going on. I like to think of Uncertainty in terms of Planck’s constant itself: ħ or h or – as you’ll usually see it – as half of that value: ħ/2. [The Stern-Gerlach experiment implies it’s ħ/2, rather than h/2 or ħ or h itself.] The physical dimension of Planck’s constant is action: newton times distance times time. I also like to think action can express itself in two ways: as (1) some amount of energy (ΔE: some force of some distance) over some time (Δt) or, else, as (2) some momentum (Δp: some force during some time) over some distance (Δs). Now, if we equate ΔE with the energy of the pion (135 MeV), then we may calculate the order of magnitude of Δt from ΔE·Δt ≥ ħ/2 as follows:

 Δt = (ħ/2)/(135 MeV) ≈ (3.291×10−16 eV·s)/(134.977×10eV) ≈ 0.02438×10−22 s

Now, that’s an unimaginably small time unit – but much and much larger than the Planck time (the Planck time unit is about 5.39 × 10−44 s). The corresponding distance is equal to = Δt·c = (0.02438×10−22 s)·(2.998×10m/s) ≈ 0.0731×10−14 m = 0.731 fm. So… Well… Yes. We got the answer we wanted… So… Well… We should be happy about that but…

Well… I am not. I don’t like this indeterminacy. This randomness in the approach. For starters, I am very puzzled by the fact that the lifetime of the actual πparticle we see in the debris of proton collisions with other particles as cosmic rays enter the atmosphere is like 8.4×10−17 seconds, so that’s like 35 million times longer than the Δt = 0.02438×10−22 s we calculated above.

Something doesn’t feel right. I just can’t see the logic here. Sorry. I’ll be back. :-/

An introduction to virtual particles

We are going to venture beyond quantum mechanics as it is usually understood – covering electromagnetic interactions only. Indeed, all of my posts so far – a bit less than 200, I think 🙂 – were all centered around electromagnetic interactions – with the model of the hydrogen atom as our most precious gem, so to speak.

In this post, we’ll be talking the strong force – perhaps not for the first time but surely for the first time at this level of detail. It’s an entirely different world – as I mentioned in one of my very first posts in this blog. Let me quote what I wrote there:

“The math describing the ‘reality’ of electrons and photons (i.e. quantum mechanics and quantum electrodynamics), as complicated as it is, becomes even more complicated – and, important to note, also much less accurate – when it is used to try to describe the behavior of  quarks. Quantum chromodynamics (QCD) is a different world. […] Of course, that should not surprise us, because we’re talking very different order of magnitudes here: femtometers (10–15 m), in the case of electrons, as opposed to attometers (10–18 m) or even zeptometers (10–21 m) when we’re talking quarks.”

In fact, the femtometer scale is used to measure the radius of both protons as well as electrons and, hence, is much smaller than the atomic scale, which is measured in nanometer (1 nm = 10−9 m). The so-called Bohr radius for example, which is a measure for the size of an atom, is measured in nanometer indeed, so that’s a scale that is a million times larger than the femtometer scale. This gap in the scale effectively separates entirely different worlds. In fact, the gap is probably as large a gap as the gap between our macroscopic world and the strange reality of quantum mechanics. What happens at the femtometer scale, really?

The honest answer is: we don’t know, but we do have models to describe what happens. Moreover, for want of better models, physicists sort of believe these models are credible. To be precise, we assume there’s a force down there which we refer to as the strong force. In addition, there’s also a weak force. Now, you probably know these forces are modeled as interactions involving an exchange of virtual particles. This may be related to what Aitchison and Hey refer to as the physicist’s “distaste for action-at-a-distance.” To put it simply: if one particle – through some force – influences some other particle, then something must be going on between the two of them.

Of course, now you’ll say that something is effectively going on: there’s the electromagnetic field, right? Yes. But what’s the field? You’ll say: waves. But then you know electromagnetic waves also have a particle aspect. So we’re stuck with this weird theoretical framework: the conceptual distinction between particles and forces, or between particle and field, are not so clear. So that’s what the more advanced theories we’ll be looking at – like quantum field theory – try to bring together.

Note that we’ve been using a lot of confusing and/or ambiguous terms here: according to at least one leading physicist, for example, virtual particles should not be thought of as particles! But we’re putting the cart before the horse here. Let’s go step by step. To better understand the ‘mechanics’ of how the strong and weak interactions are being modeled in physics, most textbooks – including Aitchison and Hey, which we’ll follow here – start by explaining the original ideas as developed by the Japanese physicist Hideki Yukawa, who received a Nobel Prize for his work in 1949.

So what is it all about? As said, the ideas – or the model as such, so to speak – are more important than Yukawa’s original application, which was to model the force between a proton and a neutron. Indeed, we now explain such force as a force between quarks, and the force carrier is the gluon, which carries the so-called color charge. To be precise, the force between protons and neutrons – i.e. the so-called nuclear force – is now considered to be a rather minor residual force: it’s just what’s left of the actual strong force that binds quarks together. The Wikipedia article on this has some good text and a really nice animation on this. But… Well… Again, note that we are only interested in the model right now. So how does that look like?

First, we’ve got the equivalent of the electric charge: the nucleon is supposed to have some ‘strong’ charge, which we’ll write as gs. Now you know the formulas for the potential energy – because of the gravitational force – between two masses, or the potential energy between two charges – because of the electrostatic force. Let me jot them down once again:

  1. U(r) = –G·M·m/r
  2. U(r) = (1/4πε0)·q1·q2/r

The two formulas are exactly the same. They both assume U = 0 for → ∞. Therefore, U(r) is always negative. [Just think of q1 and q2 as opposite charges, so the minus sign is not explicit – but it is also there!] We know that U(r) curve will look like the one below: some work (force times distance) is needed to move the two charges some distance away from each other – from point 1 to point 2, for example. [The distance r is x here – but you got that, right?]potential energy

Now, physics textbooks – or other articles you might find, like on Wikipedia – will sometimes mention that the strong force is non-linear, but that’s very confusing because… Well… The electromagnetic force – or the gravitational force – aren’t linear either: their strength is inversely proportional to the square of the distance and – as you can see from the formulas for the potential energy – that 1/r factor isn’t linear either. So that isn’t very helpful. In order to further the discussion, I should now write down Yukawa’s hypothetical formula for the potential energy between a neutron and a proton, which we’ll refer to, logically, as the n-p potential:n-p potentialThe −gs2 factor is, obviously, the equivalent of the q1·q2 product: think of the proton and the neutron having equal but opposite ‘strong’ charges. The 1/4π factor reminds us of the Coulomb constant: k= 1/4πε0. Note this constant ensures the physical dimensions of both sides of the equation make sense: the dimension of ε0 is N·m2/C2, so U(r) is – as we’d expect – expressed in newton·meter, or joule. We’ll leave the question of the units for gs open – for the time being, that is. [As for the 1/4π factor, I am not sure why Yukawa put it there. My best guess is that he wanted to remind us some constant should be there to ensure the units come out alright.]

So, when everything is said and done, the big new thing is the er/a/factor, which replaces the usual 1/r dependency on distance. Needless to say, e is Euler’s number here – not the electric charge. The two green curves below show what the er/a factor does to the classical 1/r function for = 1 and = 0.1 respectively: smaller values for a ensure the curve approaches zero more rapidly. In fact, for = 1, er/a/is equal to 0.368 for = 1, and remains significant for values that are greater than 1 too. In contrast, for = 0.1, er/a/is equal to 0.004579 (more or less, that is) for = 4 and rapidly goes to zero for all values greater than that.

graph 1graph 2Aitchison and Hey call a, therefore, a range parameter: it effectively defines the range in which the n-p potential has a significant value: outside of the range, its value is, for all practical purposes, (close to) zero. Experimentally, this range was established as being more or less equal to ≤ 2 fm. Needless to say, while this range factor may do its job, it’s obvious Yukawa’s formula for the n-p potential comes across as being somewhat random: what’s the theory behind? There’s none, really. It makes one think of the logistic function: the logistic function fits many statistical patterns, but it is (usually) not obvious why.

Next in Yukawa’s argument is the establishment of an equivalent, for the nuclear force, of the Poisson equation in electrostatics: using the E = –Φ formula, we can re-write Maxwell’s ∇•E = ρ/ε0 equation (aka Gauss’ Law) as ∇•E = –∇•∇Φ = –2Φ ⇔ 2Φ= –ρ/ε0 indeed. The divergence operator the • operator gives us the volume density of the flux of E out of an infinitesimal volume around a given point. [You may want to check one of my post on this. The formula becomes somewhat more obvious if we re-write it as ∇•E·dV = –(ρ·dV)/ε0: ∇•E·dV is then, quite simply, the flux of E out of the infinitesimally small volume dV, and the right-hand side of the equation says this is given by the product of the charge inside (ρ·dV) and 1/ε0, which accounts for the permittivity of the medium (which is the vacuum in this case).] Of course, you will also remember the Φ notation: is just the gradient (or vector derivative) of the (scalar) potential Φ, i.e. the electric (or electrostatic) potential in a space around that infinitesimally small volume with charge density ρ. So… Well… The Poisson equation is probably not so obvious as it seems at first (again, check my post on it on it for more detail) and, yes, that • operator – the divergence operator – is a pretty impressive mathematical beast. However, I must assume you master this topic and move on. So… Well… I must now give you the equivalent of Poisson’s equation for the nuclear force. It’s written like this:Poisson nuclearWhat the heck? Relax. To derive this equation, we’d need to take a pretty complicated détour, which we won’t do. [See Appendix G of Aitchison and Grey if you’d want the details.] Let me just point out the basics:

1. The Laplace operator (∇2) is replaced by one that’s nearly the same: ∇2 − 1/a2. And it operates on the same concept: a potential, which is a (scalar) function of the position r. Hence, U(r) is just the equivalent of Φ.

2. The right-hand side of the equation involves Dirac’s delta function. Now that’s a weird mathematical beast. Its definition seems to defy what I refer to as the ‘continuum assumption’ in math.  I wrote a few things about it in one of my posts on Schrödinger’s equation – and I could give you its formula – but that won’t help you very much. It’s just a weird thing. As Aitchison and Grey write, you should just think of the whole expression as a finite range analogue of Poisson’s equation in electrostatics. So it’s only for extremely small that the whole equation makes sense. Outside of the range defined by our range parameter a, the whole equation just reduces to 0 = 0 – for all practical purposes, at least.

Now, of course, you know that the neutron and the proton are not supposed to just sit there. They’re also in these sort of intricate dance which – for the electron case – is described by some wavefunction, which we derive as a solution from Schrödinger’s equation. So U(r) is going to vary not only in space but also in time and we should, therefore, write it as U(r, t). Now, we will, of course, assume it’s going to vary in space and time as some wave and we may, therefore, suggest some wave equation for it. To appreciate this point, you should review some of the posts I did on waves. More in particular, you may want to review the post I did on traveling fields, in which I showed you the following: if we see an equation like:f8then the function ψ(x, t) must have the following general functional form:solutionAny function ψ like that will work – so it will be a solution to the differential equation – and we’ll refer to it as a wavefunction. Now, the equation (and the function) is for a wave traveling in one dimension only (x) but the same post shows we can easily generalize to waves traveling in three dimensions. In addition, we may generalize the analyse to include complex-valued functions as well. Now, you will still be shocked by Yukawa’s field equation for U(r, t) but, hopefully, somewhat less so after the above reminder on how wave equations generally look like:Yukawa wave equationAs said, you can look up the nitty-gritty in Aitchison and Grey (or in its appendices) but, up to this point, you should be able to sort of appreciate what’s going on without getting lost in it all. Yukawa’s next step – and all that follows – is much more baffling. We’d think U, the nuclear potential, is just some scalar-valued wave, right? It varies in space and in time, but… Well… That’s what classical waves, like water or sound waves, for example do too. So far, so good. However, Yukawa’s next step is to associate a de Broglie-type wavefunction with it. Hence, Yukawa imposes solutions of the type:potential as particleWhat? Yes. It’s a big thing to swallow, and it doesn’t help most physicists refer to U as a force field. A force and the potential that results from it are two different things. To put it simply: the force on an object is not the same as the work you need to move it from here to there. Force and potential are related but different concepts. Having said that, it sort of make sense now, doesn’t it? If potential is energy, and if it behaves like some wave, then we must be able to associate it with a de Broglie-type particle. This U-quantum, as it is referred to, comes in two varieties, which are associated with the ongoing absorption-emission process that is supposed to take place inside of the nucleus (depicted below):

p + U → n and n + U+ → p

absorption emission

It’s easy to see that the U and U+ particles are just each other’s anti-particle. When thinking about this, I can’t help remembering Feynman, when he enigmatically wrote – somewhere in his Strange Theory of Light and Matter – that an anti-particle might just be the same particle traveling back in time. In fact, the exchange here is supposed to happen within a time window that is so short it allows for the brief violation of the energy conservation principle.

Let’s be more precise and try to find the properties of that mysterious U-quantum. You’ll need to refresh what you know about operators to understand how substituting Yukawa’s de Broglie wavefunction in the complicated-looking differential equation (the wave equation) gives us the following relation between the energy and the momentum of our new particle:mass 1Now, it doesn’t take too many gimmicks to compare this against the relativistically correct energy-momentum relation:energy-momentum relationCombining both gives us the associated (rest) mass of the U-quantum:rest massFor ≈ 2 fm, mU is about 100 MeV. Of course, it’s always to check the dimensions and calculate stuff yourself. Note the physical dimension of ħ/(a·c) is N·s2/m = kg (just think of the F = m·a formula). Also note that N·s2/m = kg = (N·m)·s2/m= J/(m2/s2), so that’s the [E]/[c2] dimension. The calculation – and interpretation – is somewhat tricky though: if you do it, you’ll find that:

ħ/(a·c) ≈ (1.0545718×10−34 N·m·s)/[(2×10−15 m)·(2.997924583×108 m/s)] ≈ 0.176×10−27 kg

Now, most physics handbooks continue that terrible habit of writing particle weights in eV, rather than using the correct eV/c2 unit. So when they write: mU is about 100 MeV, they actually mean to say that it’s 100 MeV/c2. In addition, the eV is not an SI unit. Hence, to get that number, we should first write 0.176×10−27 kg as some value expressed in J/c2, and then convert the joule (J) into electronvolt (eV). Let’s do that. First, note that c2 ≈ 9×1016 m2/s2, so 0.176×10−27 kg ≈ 1.584×10−11 J/c2. Now we do the conversion from joule to electronvolt. We get: (1.584×10−11 J/c2)·(6.24215×1018 eV/J) ≈ 9.9×107 eV/c2 = 99 MeV/c2Bingo! So that was Yukawa’s prediction for the nuclear force quantum.

Of course, Yukawa was wrong but, as mentioned above, his ideas are now generally accepted. First note the mass of the U-quantum is quite considerable: 100 MeV/c2 is a bit more than 10% of the individual proton or neutron mass (about 938-939 MeV/c2). While the binding energy causes the mass of an atom to be less than the mass of their constituent parts (protons, neutrons and electrons), it’s quite remarkably that the deuterium atom – a hydrogen atom with an extra neutron – has an excess mass of about 13.1 MeV/c2, and a binding energy with an equivalent mass of only 2.2 MeV/c2. So… Well… There’s something there.

As said, this post only wanted to introduce some basic ideas. The current model of nuclear physics is represented by the animation below, which I took from the Wikipedia article on it. The U-quantum appears as the pion here – and it does not really turn the proton into a neutron and vice versa. Those particles are assumed to be stable. In contrast, it is the quarks that change color by exchanging gluons between each other. And we know look at the exchange particle – which we refer to as the pion – between the proton and the neutron as consisting of two quarks in its own right: a quark and a anti-quark. So… Yes… All weird. QCD is just a different world. We’ll explore it more in the coming days and/or weeks. 🙂Nuclear_Force_anim_smallerAn alternative – and simpler – way of representing this exchange of a virtual particle (a neutral pion in this case) is obtained by drawing a so-called Feynman diagram:Pn_scatter_pi0OK. That’s it for today. More tomorrow. 🙂

Reality and perception

It’s quite easy to get lost in all of the math when talking quantum mechanics. In this post, I’d like to freewheel a bit. I’ll basically try to relate the wavefunction we’ve derived for the electron orbitals to the more speculative posts I wrote on how to interpret the wavefunction. So… Well… Let’s go. 🙂

If there is one thing you should remember from all of the stuff I wrote in my previous posts, then it’s that the wavefunction for an electron orbital – ψ(x, t), so that’s a complex-valued function in two variables (position and time) – can be written as the product of two functions in one variable:

ψ(x, t) = ei·(E/ħ)·t·f(x)

In fact, we wrote f(x) as ψ(x), but I told you how confusing that is: the ψ(x) and ψ(x, t) functions are, obviously, very different. To be precise, the f(x) = ψ(x) function basically provides some envelope for the two-dimensional eiθ = ei·(E/ħ)·t = cosθ + i·sinθ oscillation – as depicted below (θ = −(E/ħ)·t = ω·t with ω = −E/ħ).Circle_cos_sinWhen analyzing this animation – look at the movement of the green, red and blue dots respectively – one cannot miss the equivalence between this oscillation and the movement of a mass on a spring – as depicted below.spiral_sThe ei·(E/ħ)·t function just gives us two springs for the price of one. 🙂 Now, you may want to imagine some kind of elastic medium – Feynman’s famous drum-head, perhaps 🙂 – and you may also want to think of all of this in terms of superimposed waves but… Well… I’d need to review if that’s really relevant to what we’re discussing here, so I’d rather not make things too complicated and stick to basics.

First note that the amplitude of the two linear oscillations above is normalized: the maximum displacement of the object from equilibrium, in the positive or negative direction, which we may denote by x = ±A, is equal to one. Hence, the energy formula is just the sum of the potential and kinetic energy: T + U = (1/2)·A2·m·ω2 = (1/2)·m·ω2. But so we have two springs and, therefore, the energy in this two-dimensional oscillation is equal to E = 2·(1/2)·m·ω2 = m·ω2.

This formula is structurally similar to Einstein’s E = m·c2 formula. Hence, one may want to assume that the energy of some particle (an electron, in our case, because we’re discussing electron orbitals here) is just the two-dimensional motion of its mass. To put it differently, we might also want to think that the oscillating real and imaginary component of our wavefunction each store one half of the total energy of our particle.

However, the interpretation of this rather bold statement is not so straightforward. First, you should note that the ω in the E = m·ω2 formula is an angular velocity, as opposed to the in the E = m·c2 formula, which is a linear velocity. Angular velocities are expressed in radians per second, while linear velocities are expressed in meter per second. However, while the radian measures an angle, we know it does so by measuring a length. Hence, if our distance unit is 1 m, an angle of 2π rad will correspond to a length of 2π meter, i.e. the circumference of the unit circle. So… Well… The two velocities may not be so different after all.

There are other questions here. In fact, the other questions are probably more relevant. First, we should note that the ω in the E = m·ω2 can take on any value. For a mechanical spring, ω will be a function of (1) the stiffness of the spring (which we usually denote by k, and which is typically measured in newton (N) per meter) and (2) the mass (m) on the spring. To be precise, we write: ω2 = k/m – or, what amounts to the same, ω = √(k/m). Both k and m are variables and, therefore, ω can really be anything. In contrast, we know that c is a constant: equals 299,792,458 meter per second, to be precise. So we have this rather remarkable expression: c = √(E/m), and it is valid for any particle – our electron, or the proton at the center, or our hydrogen atom as a whole. It is also valid for more complicated atoms, of course. In fact, it is valid for any system.

Hence, we need to take another look at the energy concept that is used in our ψ(x, t) = ei·(E/ħ)·t·f(x) wavefunction. You’ll remember (if not, you should) that the E here is equal to En = −13.6 eV, −3.4 eV, −1.5 eV and so on, for = 1, 2, 3, etc. Hence, this energy concept is rather particular. As Feynman puts it: “The energies are negative because we picked our zero point as the energy of an electron located far from the proton. When it is close to the proton, its energy is less, so somewhat below zero. The energy is lowest (most negative) for n = 1, and increases toward zero with increasing n.”

Now, this is the one and only issue I have with the standard physics story. I mentioned it in one of my previous posts and, just for clarity, let me copy what I wrote at the time:

Feynman gives us a rather casual explanation [on choosing a zero point for measuring energy] in one of his very first Lectures on quantum mechanics, where he writes the following: “If we have a “condition” which is a mixture of two different states with different energies, then the amplitude for each of the two states will vary with time according to an equation like a·eiωt, with ħ·ω = E = m·c2. Hence, we can write the amplitude for the two states, for example as:

ei(E1/ħ)·t and ei(E2/ħ)·t

And if we have some combination of the two, we will have an interference. But notice that if we added a constant to both energies, it wouldn’t make any difference. If somebody else were to use a different scale of energy in which all the energies were increased (or decreased) by a constant amount—say, by the amount A—then the amplitudes in the two states would, from his point of view, be:

ei(E1+A)·t/ħ and ei(E2+A)·t/ħ

All of his amplitudes would be multiplied by the same factor ei(A/ħ)·t, and all linear combinations, or interferences, would have the same factor. When we take the absolute squares to find the probabilities, all the answers would be the same. The choice of an origin for our energy scale makes no difference; we can measure energy from any zero we want. For relativistic purposes it is nice to measure the energy so that the rest mass is included, but for many purposes that aren’t relativistic it is often nice to subtract some standard amount from all energies that appear. For instance, in the case of an atom, it is usually convenient to subtract the energy Ms·c2, where Ms is the mass of all the separate pieces—the nucleus and the electrons—which is, of course, different from the mass of the atom. For other problems, it may be useful to subtract from all energies the amount Mg·c2, where Mg is the mass of the whole atom in the ground state; then the energy that appears is just the excitation energy of the atom. So, sometimes we may shift our zero of energy by some very large constant, but it doesn’t make any difference, provided we shift all the energies in a particular calculation by the same constant.”

It’s a rather long quotation, but it’s important. The key phrase here is, obviously, the following: “For other problems, it may be useful to subtract from all energies the amount Mg·c2, where Mg is the mass of the whole atom in the ground state; then the energy that appears is just the excitation energy of the atom.” So that’s what he’s doing when solving Schrödinger’s equation. However, I should make the following point here: if we shift the origin of our energy scale, it does not make any difference in regard to the probabilities we calculate, but it obviously does make a difference in terms of our wavefunction itself. To be precise, its density in time will be very different. Hence, if we’d want to give the wavefunction some physical meaning – which is what I’ve been trying to do all along – it does make a huge difference. When we leave the rest mass of all of the pieces in our system out, we can no longer pretend we capture their energy.

So… Well… There you go. If we’d want to try to interpret our ψ(x, t) = ei·(En/ħ)·t·f(x) function as a two-dimensional oscillation of the mass of our electron, the energy concept in it – so that’s the Ein it – should include all pieces. Most notably, it should also include the electron’s rest energy, i.e. its energy when it is not in a bound state. This rest energy is equal to 0.511 MeV. […] Read this again: 0.511 mega-electronvolt (106 eV), so that’s huge as compared to the tiny energy values we mentioned so far (−13.6 eV, −3.4 eV, −1.5 eV,…).

Of course, this gives us a rather phenomenal order of magnitude for the oscillation that we’re looking at. Let’s quickly calculate it. We need to convert to SI units, of course: 0.511 MeV is about 8.2×10−14 joule (J), and so the associated frequency is equal to ν = E/h = (8.2×10−14 J)/(6.626×10−34 J·s) ≈ 1.23559×1020 cycles per second. Now, I know such number doesn’t say all that much: just note it’s the same order of magnitude as the frequency of gamma rays and… Well… No. I won’t say more. You should try to think about this for yourself. [If you do, think – for starters – about the difference between bosons and fermions: matter-particles are fermions, and photons are bosons. Their nature is very different.]

The corresponding angular frequency is just the same number but multiplied by 2π (one cycle corresponds to 2π radians and, hence, ω = 2π·ν = 7.76344×1020 rad per second. Now, if our green dot would be moving around the origin, along the circumference of our unit circle, then its horizontal and/or vertical velocity would approach the same value. Think of it. We have this eiθ = ei·(E/ħ)·t = ei·ω·t = cos(ω·t) + i·sin(ω·t) function, with ω = E/ħ. So the cos(ω·t) captures the motion along the horizontal axis, while the sin(ω·t) function captures the motion along the vertical axis. Now, the velocity along the horizontal axis as a function of time is given by the following formula:

v(t) = d[x(t)]/dt = d[cos(ω·t)]/dt = −ω·sin(ω·t)

Likewise, the velocity along the vertical axis is given by v(t) = d[sin(ω·t)]/dt = ω·cos(ω·t). These are interesting formulas: they show the velocity (v) along one of the two axes is always less than the angular velocity (ω). To be precise, the velocity approaches – or, in the limit, is equal to – the angular velocity ω when ω·t is equal to ω·= 0, π/2, π or 3π/2. So… Well… 7.76344×1020 meter per second!? That’s like 2.6 trillion times the speed of light. So that’s not possible, of course!

That’s where the amplitude of our wavefunction comes in – our envelope function f(x): the green dot does not move along the unit circle. The circle is much tinier and, hence, the oscillation should not exceed the speed of light. In fact, I should probably try to prove it oscillates at the speed of light, thereby respecting Einstein’s universal formula:

c = √(E/m)

Written like this – rather than as you know it: E = m·c2 – this formula shows the speed of light is just a property of spacetime, just like the ω = √(k/m) formula (or the ω = √(1/LC) formula for a resonant AC circuit) shows that ω, the natural frequency of our oscillator, is a characteristic of the system.

Am I absolutely certain of what I am writing here? No. My level of understanding of physics is still that of an undergrad. But… Well… It all makes a lot of sense, doesn’t it? 🙂

Now, I said there were a few obvious questions, and so far I answered only one. The other obvious question is why energy would appear to us as mass in motion in two dimensions only. Why is it an oscillation in a plane? We might imagine a third spring, so to speak, moving in and out from us, right? Also, energy densities are measured per unit volume, right?

Now that‘s a clever question, and I must admit I can’t answer it right now. However, I do suspect it’s got to do with the fact that the wavefunction depends on the orientation of our reference frame. If we rotate it, it changes. So it’s like we’ve lost one degree of freedom already, so only two are left. Or think of the third direction as the direction of propagation of the wave. 🙂 Also, we should re-read what we wrote about the Poynting vector for the matter wave, or what Feynman wrote about probability currents. Let me give you some appetite for that by noting that we can re-write joule per cubic meter (J/m3) as newton per square meter: J/m3 = N·m/m3 = N/m2. [Remember: the unit of energy is force times distance. In fact, looking at Einstein’s formula, I’d say it’s kg·m2/s2 (mass times a squared velocity), but that simplifies to the same: kg·m2/s2 = [N/(m/s2)]·m2/s2.]

I should probably also remind you that there is no three-dimensional equivalent of Euler’s formula, and the way the kinetic and potential energy of those two oscillations works together is rather unique. Remember I illustrated it with the image of a V-2 engine in previous posts. There is no such thing as a V-3 engine. [Well… There actually is – but not with the third cylinder being positioned sideways.]two-timer-576-px-photo-369911-s-original

But… Then… Well… Perhaps we should think of some weird combination of two V-2 engines. The illustration below shows the superposition of two one-dimensional waves – I think – one traveling east-west and back, and the other one traveling north-south and back. So, yes, we may to think of Feynman’s drum-head again – but combining two-dimensional waves – two waves that both have an imaginary as well as a real dimension

dippArticle-14

Hmm… Not sure. If we go down this path, we’d need to add a third dimension – so w’d have a super-weird V-6 engine! As mentioned above, the wavefunction does depend on our reference frame: we’re looking at stuff from a certain direction and, therefore, we can only see what goes up and down, and what goes left or right. We can’t see what comes near and what goes away from us. Also think of the particularities involved in measuring angular momentum – or the magnetic moment of some particle. We’re measuring that along one direction only! Hence, it’s probably no use to imagine we’re looking at three waves simultaneously!

In any case… I’ll let you think about all of this. I do feel I am on to something. I am convinced that my interpretation of the wavefunction as an energy propagation mechanism, or as energy itself – as a two-dimensional oscillation of mass – makes sense. 🙂

Of course, I haven’t answered one key question here: what is mass? What is that green dot – in reality, that is? At this point, we can only waffle – probably best to just give its standard definition: mass is a measure of inertia. A resistance to acceleration or deceleration, or to changing direction. But that doesn’t say much. I hate to say that – in many ways – all that I’ve learned so far has deepened the mystery, rather than solve it. The more we understand, the less we understand? But… Well… That’s all for today, folks ! Have fun working through it for yourself. 🙂

Post scriptum: I’ve simplified the wavefunction a bit. As I noted in my post on it, the complex exponential is actually equal to ei·[(E/ħ)·− m·φ], so we’ve got a phase shift because of m, the quantum number which denotes the z-component of the angular momentum. But that’s a minor detail that shouldn’t trouble or worry you here.

The periodic table

This post is, in essence, a continuation of my series on electron orbitals. I’ll just further tie up some loose ends and then – hopefully – have some time to show how we get the electron orbitals for other atoms than hydrogen. So we’ll sort of build up the periodic table. Sort of. 🙂

We should first review a bit. The illustration below copies the energy level diagram from Feynman’s Lecture on the hydrogen wave function. Note he uses √E for the energy scale because… Well… I’ve copied the En values for n = 1, 2, 3,… 7 next to it: the value for E(-13.6 eV) is four times the value of E(-3.4 eV).

exponential scale

How do we know those values? We discussed that before – long time back: we have the so-called gross structure of the hydrogen spectrum here. The table below gives the energy values for the first seven levels, and you can calculate an example for yourself: the difference between E2 (-3.4 eV) and E(-0.85 eV) is 2.55 eV, so that’s 4.08555×10−19 J, which corresponds to a frequency equal to = E/h = (4.08555×10−19 J)/(6.626×10−34 J·s) ≈ 0.6165872×1015 Hz. Now that frequency corresponds to a wavelength that’s equal to λ = c/= (299,792,458 m/s)/0.6165872×1015/s) ≈ 486×10−9 m. So that’s the 486 nano-meter line the so-called Balmer series, as shown in the illustration next to the table with the energy values.

So far, so good. An interesting point to note is that we only have one solution for = 1. To be precise, we have one spherical solution only: the 1s solution. Now, for n = 2, we have one 2s solution but also three 2solutions (remember the stands for principal lines). In the simplified model we’re using (we’re not discussing the fine or hyperfine structure here), these three solutions are referred to as ‘degenerate states’: they are different states with the same energy. Now, we know that any linear combination of the solutions for a differential equation must also be a solution. Therefore, any linear combination of the 2solutions will also be a stationary state of the same energy. In fact, a superposition of the 2s and one or more of the 2p states should also be a solution. There is an interesting app which visualizes how such superimposed states look like. I copy three illustrations below, but I recommend you google for stuff like this yourself: it’s really fascinating! You should, once again, pay attention to the symmetries planes and/or symmetry axes.

But we’ve written enough about the orbital of one electron now. What if there are two electrons, or three, or more. In other word, how does it work for heliumlithium, and so on? Feynman gives us a bit of an intuitive explanation here – nothing analytical, really. First, he notes Schrödinger’s equation for two electrons would look as follows:

two electronsSecond, the ψ(x) function in the ψ(x, t) = ei·(E/ħ)·t·ψ(x) function now becomes a function in six variables, which he – curiously enough – now no longer writes as ψ but as f:formulaThe rest of the text speaks for itself, although you might be disappointed by what he writes (the bold-face and/or italics are mine):

“The geometrical dependence is contained in f, which is a function of six variables—the simultaneous positions of the two electrons. No one has found an analytic solution, although solutions for the lowest energy states have been obtained by numerical methods. With 34, or 5 electrons it is hopeless to try to obtain exact solutions, and it is going too far to say that quantum mechanics has given a precise understanding of the periodic tableIt is possible, however, even with a sloppy approximation—and some fixing—to understand, at least qualitatively, many chemical properties which show up in the periodic table.

The chemical properties of atoms are determined primarily by their lowest energy states. We can use the following approximate theory to find these states and their energies. First, we neglect the electron spin, except that we adopt the exclusion principle and say that any particular electronic state can be occupied by only one electron. This means that any particular orbital configuration can have up to two electrons—one with spin up, the other with spin down.

Next we disregard the details of the interactions between the electrons in our first approximation, and say that each electron moves in a central field which is the combined field of the nucleus and all the other electrons. For neon, which has 10 electrons, we say that one electron sees an average potential due to the nucleus plus the other nine electrons. We imagine then that in the Schrödinger equation for each electron we put a V(r) which is a 1/r field modified by a spherically symmetric charge density coming from the other electrons.

In this model each electron acts like an independent particle. The angular dependence of its wave function will be just the same as the ones we had for the hydrogen atom. There will be s-states, p-states, and so on; and they will have the various possible m-values. Since V(r) no longer goes as 1/r, the radial part of the wave functions will be somewhat different, but it will be qualitatively the same, so we will have the same radial quantum numbers, n. The energies of the states will also be somewhat different.”

So that’s rather disappointing, isn’t it? We can only get some approximate – or qualitative – understanding of the periodic table from quantum mechanics – because the math is too complex: only numerical methods can give us those orbitals! Wow! Let me list some of the salient points in Feynman’s treatment of the matter:

  • For helium (He), we have two electrons in the lowest state (i.e. the 1s state): one has its spin ‘up’ and the other is ‘down’. Because the shell is filled, the ionization energy (to remove one electron) has an even larger value than the ionization energy for hydrogen: 24.6 eV! That’s why there is “practically no tendency” for the electron to be attracted by some other atom: helium is chemically inert – which explains it being part of the group of noble or inert gases.
  • For lithium (Li), two electrons will occupy the 1s orbital, and the third should go to an = 2 state. But which one? With = 0, or = 1? A 2s state or a 2p state? In hydrogen, these two = 2 states have the same energy, but in other atoms they don’t. Why not? That’s a complicated story, but the gist of the argument is as follows: a 2s state has some amplitude to be near the nucleus, while the 2p state does not. That means that a 2electron will feel some of the triple electric charge of the Li nucleus, and this extra attraction lowers the energy of the 2state relative to the 2state.

To make a long story short, the energy levels will be roughly as shown in the table below. For example, the energy that’s needed to remove the 2s electron of the lithium – i.e. the ionization energy of lithium – is only 5.4 eV because… Well… As you can see, it has a higher energy (less negative, that is) than the 1s state (−13.6 eV for hydrogen and, as mentioned above, −24.6 eV for helium). So lithium is chemically active – as opposed to helium. energy values more electrons

You should compare the table below with the table above. If you do, you’ll understand how electrons ‘fill up’ those electron shells. Note, for example, that the energy of the 4s state is slightly lower than the energy of the 3d state, so it fills up before the 3shell does. [I know the table is hard to read – just check out the original text if you want to see it better.]

periodic table

This, then, is what you learnt in high school and, of course, there are 94 naturally occurring elements – and another 24 heavier elements that have been produced in labs, so we’d need to go all the way to no. 118. Now, Feynman doesn’t do that, and so I won’t do that either. 🙂

Well… That’s it, folks. We’re done with Feynman. It’s time to move to a physics grad course now! Talk stuff like quantum field theory, for example. Or string theory. 🙂 Stay tuned!

Re-visiting electron orbitals (III)

In my previous post, I mentioned that it was not so obvious (both from a physical as well as from a mathematical point of view) to write the wavefunction for electron orbitals – which we denoted as ψ(x, t), i.e. a function of two variables (or four: one time coordinate and three space coordinates) – as the product of two other functions in one variable only.

[…] OK. The above sentence is difficult to read. Let me write in math. 🙂 It is not so obvious to write ψ(x, t) as:

ψ(x, t) = ei·(E/ħ)·t·ψ(x)

As I mentioned before, the physicists’ use of the same symbol (ψ, psi) for both the ψ(x, t) and ψ(x) function is quite confusing – because the two functions are very different:

  • ψ(x, t) is a complex-valued function of two (real) variables: x and t. Or four, I should say, because x = (x, y, z) – but it’s probably easier to think of x as one vector variable – a vector-valued argument, so to speak. And then t is, of course, just a scalar variable. So… Well… A function of two variables: the position in space (x), and time (t).
  • In contrast, ψ(x) is a real-valued function of one (vector) variable only: x, so that’s the position in space only.

Now you should cry foul, of course: ψ(x) is not necessarily real-valued. It may be complex-valued. You’re right. You know the formula:wavefunctionNote the derivation of this formula involved a switch from Cartesian to polar coordinates here, so from = (x, y, z) to r = (r, θ, φ), and that the function is also a function of the two quantum numbers l and m now, i.e. the orbital angular momentum (l) and its z-component (m) respectively. In my previous post(s), I gave you the formulas for Yl,m(θ, φ) and Fl,m(r) respectively. Fl,m(r) was a real-valued function alright, but the Yl,m(θ, φ) had that ei·m·φ factor in it. So… Yes. You’re right: the Yl,m(θ, φ) function is real-valued if – and only if – m = 0, in which case ei·m·φ = 1. Let me copy the table from Feynman’s treatment of the topic once again:spherical harmonics 2The Plm(cosθ) functions are the so-called (associated) Legendre polynomials, and the formula for these functions is rather horrible:Legendre polynomialDon’t worry about it too much: just note the Plm(cosθ) is a real-valued function. The point is the following:the ψ(x, t) is a complex-valued function because – and only because – we multiply a real-valued envelope function – which depends on position only – with ei·(E/ħ)·t·ei·m·φ = ei·[(E/ħ)·− m·φ].

[…]

Please read the above once again and – more importantly – think about it for a while. 🙂 You’ll have to agree with the following:

  • As mentioned in my previous post, the ei·m·φ factor just gives us phase shift: just a re-set of our zero point for measuring time, so to speak, and the whole ei·[(E/ħ)·− m·φ] factor just disappears when we’re calculating probabilities.
  • The envelope function gives us the basic amplitude – in the classical sense of the word: the maximum displacement from the zero value. And so it’s that ei·[(E/ħ)·− m·φ] that ensures the whole expression somehow captures the energy of the oscillation.

Let’s first look at the envelope function again. Let me copy the illustration for n = 5 and = 2 from Wikimedia Commons article. Note the symmetry planes:

  • Any plane containing the z-axis is a symmetry plane – like a mirror in which we can reflect one half of the shape to get the other half. [Note that I am talking the shape only here. Forget about the colors for a while – as these reflect the complex phase of the wavefunction.]
  • Likewise, the plane containing both the x– and the y-axis is a symmetry plane as well.

n = 5

The first symmetry plane – or symmetry line, really (i.e. the z-axis) – should not surprise us, because the azimuthal angle φ is conspicuously absent in the formula for our envelope function if, as we are doing in this article here, we merge the ei·m·φ factor with the ei·(E/ħ)·t, so it’s just part and parcel of what the author of the illustrations above refers to as the ‘complex phase’ of our wavefunction. OK. Clear enough – I hope. 🙂 But why is the the xy-plane a symmetry plane too? We need to look at that monstrous formula for the Plm(cosθ) function here: just note the cosθ argument in it is being squared before it’s used in all of the other manipulation. Now, we know that cosθ = sin(π/2 − θ). So we can define some new angle – let’s just call it α – which is measured in the way we’re used to measuring angle, which is not from the z-axis but from the xy-plane. So we write: cosθ = sin(π/2 − θ) = sinα. The illustration below may or may not help you to see what we’re doing here.angle 2So… To make a long story short, we can substitute the cosθ argument in the Plm(cosθ) function for sinα = sin(π/2 − θ). Now, if the xy-plane is a symmetry plane, then we must find the same value for Plm(sinα) and Plm[sin(−α)]. Now, that’s not obvious, because sin(−α) = −sinα ≠ sinα. However, because the argument in that Plm(x) function is being squared before any other operation (like subtracting 1 and exponentiating the result), it is OK: [−sinα]2 = [sinα]sin2α. […] OK, I am sure the geeks amongst my readers will be able to explain this more rigorously. In fact, I hope they’ll have a look at it, because there’s also that dl+m/dxl+m operator, and so you should check what happens with the minus sign there. 🙂

[…] Well… By now, you’re probably totally lost, but the fact of the matter is that we’ve got a beautiful result here. Let me highlight the most significant results:

  • definite energy state of a hydrogen atom (or of an electron orbiting around some nucleus, I should say) appears to us as some beautifully shaped orbital – an envelope function in three dimensions, really – which has the z-axis – i.e. the vertical axis – as a symmetry line and the xy-plane as a symmetry plane.
  • The ei·[(E/ħ)·− m·φ] factor gives us the oscillation within the envelope function. As such, it’s this factor that, somehow, captures the energy of the oscillation.

It’s worth thinking about this. Look at the geometry of the situation again – as depicted below. We’re looking at the situation along the x-axis, in the direction of the origin, which is the nucleus of our atom.

spherical

The ei·m·φ factor just gives us phase shift: just a re-set of our zero point for measuring time, so to speak. Interesting, weird – but probably less relevant than the ei·[(E/ħ)·t factor, which gives us the two-dimensional oscillation that captures the energy of the state.

Circle_cos_sin

Now, the obvious question is: the oscillation of what, exactly? I am not quite sure but – as I explained in my Deep Blue page – the real and imaginary part of our wavefunction are really like the electric and magnetic field vector of an oscillating electromagnetic field (think of electromagnetic radiation – if that makes it easier). Hence, just like the electric and magnetic field vector represent some rapidly changing force on a unit charge, the real and imaginary part of our wavefunction must also represent some rapidly changing force on… Well… I am not quite sure on what though. The unit charge is usually defined as the charge of a proton – rather than an electron – but then forces act on some mass, right? And the mass of a proton is hugely different from the mass of an electron. The same electric (or magnetic) force will, therefore, give a hugely different acceleration to both.

So… Well… My guts instinct tells me the real and imaginary part of our wavefunction just represent, somehow, a rapidly changing force on some unit of mass, but then I am not sure how to define that unit right now (it’s probably not the kilogram!).

Now, there is another thing we should note here: we’re actually sort of de-constructing a rotation (look at the illustration above once again) in two linearly oscillating vectors – one along the z-axis and the other along the y-axis. Hence, in essence, we’re actually talking about something that’s spinning. In other words, we’re actually talking some torque around the x-axis. In what direction? I think that shouldn’t matter – that we can write E or −E, in other words, but… Well… I need to explore this further – as should you! 🙂

Let me just add one more note on the ei·m·φ factor. It sort of defines the geometry of the complex phase itself. Look at the illustration below. Click on it to enlarge it if necessary – or, better still, visit the magnificent Wikimedia Commons article from which I get these illustrations. These are the orbitals = 4 and = 3. Look at the red hues in particular – or the blue – whatever: focus on one color only, and see how how – for m = ±1, we’ve got one appearance of that color only. For m = ±1, the same color appears at two ends of the ‘tubes’ – or tori (plural of torus), I should say – just to sound more professional. 🙂 For m = ±2, the torus consists of three parts – or, in mathematical terms, we’d say the order of its rotational symmetry is equal to 3. Check that Wikimedia Commons article for higher values of and l: the shapes become very convoluted, but the observation holds. 🙂

l = 3

Have fun thinking all of this through for yourself – and please do look at those symmetries in particular. 🙂

Post scriptum: You should do some thinking on whether or not these = ±1, ±2,…, ±orbitals are really different. As I mentioned above, a phase difference is just what it is: a re-set of the t = 0 point. Nothing more, nothing less. So… Well… As far as I am concerned, that’s not real difference, is it? 🙂 As with other stuff, I’ll let you think about this for yourself.

Re-visiting electron orbitals (II)

I’ve talked about electron orbitals in a couple of posts already – including a fairly recent one, which is why I put the (II) after the title. However, I just wanted to tie up some loose ends here – and do some more thinking about the concept of a definite energy state. What is it really? We know the wavefunction for a definite energy state can always be written as:

ψ(x, t) = ei·(E/ħ)·t·ψ(x)

Well… In fact, we should probably formally prove that but… Well… Let us just explore this formula in a more intuitive way – for the time being, that is – using those electron orbitals we’ve derived.

First, let me note that ψ(x, t) and ψ(x) are very different functions and, therefore, the choice of the same symbol for both (the Greek psi) is – in my humble opinion – not very fortunate, but then… Well… It is the choice of physicists – as copied in textbooks all over – and so we’ll just have to live with it. Of course, we can appreciate why they choose to use the same symbol – ψ(x) is like a time-independent wavefunction now, so that’s nice – but… Well… You should note that it is not so obvious to write some function as the product of two other functions. To be complete, I’ll be a bit more explicit here: if some function in two variables – say F(x, y) – can be written as the product of two functions in one variable – say f(x) and g(y), so we can write F as F(x, y) = f(x)·g(y) – then we say F is a separable function. For a full overview of what that means, click on this link. And note mathematicians do choose a different symbol for the functions F and g. It would probably be interesting to explore what the conditions for separability actually imply in terms of properties of… Well… The wavefunction and its argument, i.e. the space and time variables. But… Well… That’s stuff for another post. 🙂

Secondly, note that the momentum variable (p) – i.e. the p in our elementary wavefunction a·ei·(p·x−E·t)/ħ has sort of vanished: ψ(x) is a function of the position only. Now, you may think it should be somewhere there – that, perhaps, we can write something like ψ(x) = ψ[x), p(x)]. But… No. The momentum variable has effectively vanished. Look at Feynman’s solutions for the electron orbitals of a hydrogen atom:Grand EquationThe Yl,m(θ, φ) and Fn,l(ρ) functions here are functions of the (polar) coordinates ρ, θ, φ only. So that’s the position only (these coordinates are polar or spherical coordinates, so ρ is the radial distance, θ is the polar angle, and φ is the azimuthal angle). There’s no idea whatsoever of any momentum in one or the other spatial direction here. I find that rather remarkable. Let’s see how it all works with a simple example.

The functions below are the Yl,m(θ, φ) for = 1. Note the symmetry: if we swap θ and φ for -θ and -φ respectively, we get the other function: 2-1/2·sin(-θ)·ei(-φ) = -2-1/2·sinθ·eiφ.

example

To get the probabilities, we need to take the absolute square of the whole thing, including ei·(E/ħ), but we know |ei·δ|2 = 1 for any value of δ. Why? Because the absolute square of any complex number is the product of the number with its complex conjugate, so |ei·δ|2 = ei·δ·ei·δ = ei·0 = 1. So we only have to look at the absolute square of the Yl,m(θ, φ) and Fn,l(ρ) functions here. The Fn,l(ρ) function is a real-valued function, so its absolute square is just what it is: some real number (I gave you the formula for the ak coefficients in my post on it, and you shouldn’t worry about them: they’re real too). In contrast, the Yl,m(θ, φ) functions are complex-valued – most of them are, at least. Unsurprisingly, we find the probabilities are also symmetric:

P = |-2-1/2·sinθ·eiφ|2 = (-2-1/2·sinθ·eiφ)·(-2-1/2·sinθ·eiφ)

= (2-1/2·sinθ·eiφ)·(2-1/2·sinθ·eiφ) =  |2-1/2·sinθ·eiφ|2 = (1/2)·sin2θ

Of course, for = 0, the probability is just cos2θ. The graphs below are the polar graphs for the cos2θ and (1/2)·sin2θ functions respectively.

combinationThese polar graphs are not so easy to interpret, so let me say a few words about them. The points that are plotted combine (a) some radial distance from the center – which I wrote as P because this distance is, effectively, a probability – with (b) the polar angle θ (so that’s one of the  three coordinates). To be precise, the plot gives us, for a given ρ, all of the (θ, P) combinations. It works as follows. To calculate the probability for some ρ and θ (note that φ can be any angle), we must take the absolute square of that ψn,l,m, = Yl,m(θ, φ)·Fn,l(ρ) product. Hence, we must calculate |Yl,m(θ, φ)·Fn,l(ρ)|2 = |Fn,l(ρ)|2·cos2θ for = 0, and (1/2)·|Fn,l(ρ)|2·sin2θ for = ±1. Hence, the value of ρ determines the value of Fn,l(ρ), and that Fn,l(ρ) value then determines the shape of the polar graph. The three graphs below – P = cos2θ, P = (1/2)·cos2θ and P = (1/4)·cos2θ – illustrate the idea. polar comparativeNote that we’re measuring θ from the z-axis here, as we should. So that gives us the right orientation of this volume, as opposed to the other polar graphs above, which measured θ from the x-axis. So… Well… We’re getting there, aren’t we? 🙂

Now you’ll have two or three – or even more – obvious questions. The first one is: where is the third lobe? That’s a good question. Most illustrations will represent the p-orbitals as follows:p orbitalsThree lobes. Well… Frankly, I am not quite sure here, but the equations speak for themselves: the probabilities only depend on ρ and θ. Hence, the azimuthal angle φ can be anything. So you just need to rotate those P = (1/2)·sin2θ and P = cos2θ curves about the the z-axis. In case you wonder how to do that, the illustration below may inspire you.sphericalThe second obvious question is about the size of those lobes. That 1/2 factor must surely matter, right? Well… We still have that Fn,l(ρ) factor, of course, but you’re right: that factor does not depend on the value for m: it’s the same for = 0 or ± 1. So… Well… Those representations above – with the three lobes, all of the same volume – may not be accurate. I found an interesting site – Atom in a Box – with an app that visualizes the atomic orbitals in a fun and exciting way. Unfortunately, it’s for Mac and iPhone only – but this YouTube video shows how it works. I encourage you to explore it. In fact, I need to explore it – but what I’ve seen on that YouTube video (I don’t have a Mac nor an iPhone) suggests the three-lobe illustrations may effectively be wrong: there’s some asymmetry here – which we’d expect, because those p-orbitals are actually supposed to be asymmetric! In fact, the most accurate pictures may well be the ones below. I took them from Wikimedia Commons. The author explains the use of the color codes as follows: “The depicted rigid body is where the probability density exceeds a certain value. The color shows the complex phase of the wavefunction, where blue means real positive, red means imaginary positive, yellow means real negative and green means imaginary negative.” I must assume he refers to the sign of and when writing a complex number as + i·b

The third obvious question is related to the one above: we should get some cloud, right? Not some rigid body or some surface. Well… I think you can answer that question yourself now, based on what the author of the illustration above wrote: if we change the cut-off value for the probability, then we’ll give a different shape. So you can play with that and, yes, it’s some cloud, and that’s what the mentioned app visualizes. 🙂

The fourth question is the most obvious of all. It’s the question I started this post with: what are those definite energy states? We have uncertainty, right? So how does that play out? Now that is a question I’ll try to tackle in my next post. Stay tuned ! 🙂

Post scriptum: Let me add a few remarks here so as to – hopefully – contribute to an even better interpretation of what’s going on here. As mentioned, the key to understanding is, obviously, the following basic functional form:

ψ(r, t) = ei·(E/ħ)·t·ψ(r)

Wikipedia refers to the ei·(E/ħ)·t factor as a time-dependent phase factor which, as you can see, we can separate out because we are looking at a definite energy state here. Note the minus sign in the exponent – which reminds us of the minus sign in the exponent of the elementary  wavefunction, which we wrote as:

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

We know this elementary wavefunction is problematic in terms of interpretation because its absolute square gives us some constant probability P(x, t) = |a·ei·[(E/ħ)·t − (p/ħ)∙x]|= a2. In other words, at any point in time, our electron is equally likely to be anywhere in space. That is not consistent with the idea of our electron being somewhere at some point in time.

The other question is: what reference frame do we use to measure E and p? Indeed, the value of E and p = (px, py, pz) depends on our reference frame: from the electron’s own point of view, it has no momentum whatsoever: p = 0. Fortunately, we do have a point of reference here: the nucleus of our hydrogen atom. And our own position, of course, because you should note, indeed, that both the subject and the object of the observation are necessary to define the Cartesian xx, y, z – or, more relevant in this context – the polar r = ρ, θ, φ coordinates.

This, then, defines some finite or infinite box in space in which the (linear) momentum (p) of our electron vanishes, and then we just need to solve Schrödinger’s diffusion equation to find the solutions for ψ(r). These solutions are more conveniently written in terms of the radial distance ρ, the polar angle θ, and the azimuthal angle φ:Grand Equation

The functions below are the Yl,m(θ, φ) functions for = 1.

example

The interesting thing about these Yl,m(θ, φ) functions is the ei·φ and/or ei·φ factor. Indeed, note the following:

  1. Because the sinθ and cosθ factors are real-valued, they only define some envelope for the ψ(r) function.
  2. In contrast, the ei·φ and/or ei·φ factor define some phase shift.

Let’s have a look at the physicality of the situation, which is depicted below.

spherical

The nucleus of our hydrogen atom is at the center. The polar angle is measured from the z-axis, and we know we only have an amplitude there for = 0, so let’s look at what that cosθ factor does. If θ = 0°, the amplitude is just what it is, but when θ > 0°, then |cosθ| < 1 and, therefore, the probability P = |Fn,l(ρ)|2·cos2θ will diminish. Hence, for the same radial distance (ρ), we are less likely to find the electron at some angle θ > 0° than on the z-axis itself. Now that makes sense, obviously. You can work out the argument for = ± 1 yourself, I hope. [The axis of symmetry will be different, obviously!] angle 2In contrast, the ei·φ and/or ei·φ factor work very differently. These just give us a phase shift, as illustrated below. A re-set of our zero point for measuring time, so to speak, and the ei·φ and/or ei·φ factor effectively disappears when we’re calculating probabilities, which is consistent with the fact that this angle clearly doesn’t influence the magnitude of the amplitude fluctuations.phase shiftSo… Well… That’s it, really. I hope you enjoyed this ! 🙂

Some more on symmetries…

In our previous post, we talked a lot about symmetries in space – in a rather playful way. Let’s try to take it further here by doing some more thinking on symmetries in spacetime. This post will pick up some older stuff – from my posts on states and the related quantum math in November 2015, for example – but that shouldn’t trouble you too much. On the contrary, I actually hope to tie up some loose ends here.

Let’s first review some obvious ideas. Think about the direction of time. On a time axis, time goes from left to right. It will usually be measured from some zero point – like when we started our experiment or something 🙂 – to some +point but we may also think of some point in time before our zero point, so the minus (−t) points – the left side of the axis – make sense as well. So the direction of time is clear and intuitive. Now, what does it mean to reverse the direction of time? We need to distinguish two things here: the convention, and… Well… Reality. If we would suddenly decide to reverse the direction in which we measure time, then that’s just another convention. We don’t change reality: trees and kids would still grow the way they always did. 🙂 We would just have to change the numbers on our clocks or, alternatively, the direction of rotation of the hand(s) of our clock, as shown below. [I only showed the hour hand because… Well… I don’t want to complicate things by introducing two time units. But adding the minute hand doesn’t make any difference.]

clock problemNow, imagine you’re the dictator who decided to change our time measuring convention. How would you go about it? Would you change the numbers on the clock or the direction of rotation? Personally, I’d be in favor of changing the direction of rotation. Why? Well… First, we wouldn’t have to change expressions such as: “If you are looking north right now, then west is in the 9 o’clock direction, so go there.” 🙂 More importantly, it would align our clocks with the way we’re measuring angles. On the other hand, it would not align our clocks with the way the argument (θ) of our elementary wavefunction ψ = a·eiθ = ei·(E·t – p·x)/ħ is measured, because that’s… Well… Clockwise.

So… What are the implications here? We would need to change t for −t in our wavefunction as well, right? Yep. Good point. So that’s another convention that would change: we should write our elementary wavefunction now as ψ = a·ei·(E·t – p·x)/ħ. So we would have to re-define θ as θ = –E·t + p·x = p·x –E·t. So… Well… Done!

So… Well… What’s next? Nothing. Note that we’re not changing reality here. We’re just adapting our formulas to a new dictatorial convention according to which we should count time from positive to negative – like 2, 1, 0, -1, -2 etcetera, as shown below. Fortunately, we can fix all of our laws and formulas in physics by swapping for -t. So that’s great. No sweat. time reversal

Is that all? Yes. We don’t need to do anything else. We’ll still measure the argument of our wavefunction as an angle, so that’s… Well… After changing our convention, it’s now clockwise. 🙂 Whatever you want to call it: it’s still the same direction. Our dictator can’t change physical reality 🙂

Hmm… But so we are obviously interested in changing physical reality. I mean… Anyone can become a dictator, right? In contrast, we – enlightened scientists – want to really change the world, don’t we? 🙂 So what’s a time reversal in reality? Well… I don’t know… You tell me. 🙂 We may imagine some movie being played backwards, or trees and kids shrinking instead of growing, or some bird flying backwards – and I am not talking the hummingbird here. 🙂

Hey! The latter illustration – that bird flying backwards – is probably the better one: if we reverse the direction of time – in reality, that is – then we should also reverse all directions in space. But… Well… What does that mean, really? We need to think in terms of force fields here. A stone that’d be falling must now go back up. Two opposite charges that were going towards each other, should now move away from each other. But… My God! Such world cannot exist, can it?

No. It cannot. And we don’t need to invoke the second law of thermodynamics for that. 🙂 None of what happens in a movie that’s played backwards makes sense: a heavy stone does not suddenly fly up and decelerate upwards. So it is not like the anti-matter world we described in our previous post. No. We can effectively imagine some world in which all charges have been replaced by their opposite: we’d have positive electrons (positrons) around negatively charged nuclei consisting of antiprotons and antineutrons and, somehow, negative masses. But Coulomb’s law would still tell us two opposite charges – q1 and –q2 , for example – don’t repel but attract each other, with a force that’s proportional to the product of their charges, i.e. q1·(-q2) = –q1·q2. Likewise, Newton’s law of gravitation would still tell us that two masses m1 and m2 – negative or positive – will attract each other with a force that’s proportional to the product of their masses, i.e. m1·m= (-m1)·(-m2). If you’d make a movie in the antimatter world, it would look just like any other movie. It would definitely not look like a movie being played backwards.

In fact, the latter formula – m1·m= (-m1)·(-m2) – tells us why: we’re not changing anything by putting a minus sign in front of all of our variables, which are time (t), position (x), mass (m) and charge (q). [Did I forget one? I don’t think so.] Hence, the famous CPT Theorem – which tells us that a world in which (1) time is reversed, (2) all charges have been conjugated (i.e. all particles have been replaced by their antiparticles), and (3) all spatial coordinates now have the opposite sign, is entirely possible (because it would obey the same Laws of Nature that we, in our world, have discovered over the past few hundred years) – is actually nothing but a tautology. Now, I mean that literally: a tautology is a statement that is true by necessity or by virtue of its logical form. Well… That’s the case here: if we flip the signs of all of our variables, we basically just agreed to count or measure everything from positive to negative. That’s it. Full stop. Such exotic convention is… Well… Exotic, but it cannot change the real world. Full stop.

Of course, this leaves the more intriguing questions entirely open. Partial symmetries. Like time reversal only. 🙂 Or charge conjugation only. 🙂 So let’s think about that.

We know that the world that we see in a mirror must be made of anti-matter but, apart from that particularity, that world makes sense: if we drop a stone in front of the mirror, the stone in the mirror will drop down too. Two like charges will be seen as repelling each other in the mirror too, and concepts such as kinetic or potential energy look just the same. So time just seems to tick away in both worlds – no time reversal here! – and… Well… We’ve got two CP-symmetrical worlds here, don’t we? We only flipped the sign of the coordinate frame and of the charges. Both are possible, right? And what’s possible must exist, right? Well… Maybe. That’s the next step. Let’s first see if both are possible. 🙂

Now, when you’ve read my previous post, you’ll note that I did not flip the z-coordinate when reflecting my world in the mirror. That’s true. But… Well… That’s entirely beside the point. We could flip the z-axis too and so then we’d have a full parity inversion. [Or parity transformation – sounds more serious, doesn’t it? But it’s only a simple inversion, really.] It really doesn’t matter. The point is: axial vectors have the opposite sign in the mirror world, and so it’s not only about whether or not an antimatter world is possible (it should be, right?): it’s about whether or not the sign reversal of all of those axial vectors makes sense in each and every situation. The illustration below, for example, shows how a left-handed neutrino should be a right-handed antineutrino in the mirror world.right-handed antineutrinoI hope you understand the left- versus right-handed thing. Think, for example, of how the left-circularly polarized wavefunction below would look like in the mirror. Just apply the customary right-hand rule to determine the direction of the angular momentum vector. You’ll agree it will be right-circularly polarized in the mirror, right? That’s why we need the charge conjugation: think of the magnetic moment of a circulating charge! So… Well… I can’t dwell on this too much but – if Maxwell’s equations are to hold – then that world in the mirror must be made of antimatter.animation

Now, we know that some processes – in our world – are not entirely CP-symmetrical. I wrote about this at length in previous posts, so I won’t dwell on these experiments here. The point is: these experiments – which are not easy to understand – lead physicists, philosophers, bloggers and what have you to solemnly state that the world in the mirror cannot really exist. And… Well… They’re right. However, I think their observations are beside the point. Literally.

So… Well… I would just like to make a very fundamental philosophical remark about all those discussions. My point is quite simple:

We should realize that the mirror world and our world are effectively separated by the mirror. So we should not be looking at stuff in the mirror from our perspective, because that perspective is well… Outside of the mirror. A different world. 🙂 In my humble opinion, the valid point of reference would be the observer in the mirror, like the photographer in the image below. Now note the following: if the real photographer, on this side of the mirror, would have a left-circularly polarized beam in front of him, then the imaginary photographer, on the other side of the mirror, would see the mirror image of this left-circularly polarized beam as a left-circularly polarized beam too. 🙂 I know that sounds complicated but re-read it a couple of times and – I hope – you’ll see the point. If you don’t… Well… Let me try to rephrase it: the point is that the observer in the mirror would be seeing our world – just the same laws and what have you, all makes sense! – but he would see our world in his world, so he’d see it in the mirror world. 🙂

Mirror

Capito? If you would actually be living in the mirror world, then all the things you would see in the mirror world would make perfectly sense. But you would be living in the mirror world. You would not look at it from outside, i.e. from the other side of the mirror. In short, I actually think the mirror world does exist – but in the mirror only. 🙂 […] I am, obviously, joking here. Let me be explicit: our world is our world, and I think those CP violations in Nature are telling us that it’s the only real world. The other worlds exist in our mind only – or in some mirror. 🙂

Post scriptum: I know the Die Hard philosophers among you will now have an immediate rapid-backfire question. [Hey – I just invented a new word, didn’t I? A rapid-backfire question. Neat.] How would the photographer in the mirror look at our world? The answer to that question is simple: symmetry! He (or she) would think it’s a mirror world only. His world and our world would be separated by the same mirror. So… What are the implications here?

Well… That mirror is only a piece of glass with a coating. We made it. Or… Well… Some man-made company made it. 🙂 So… Well… If you think that observer in the mirror – I am talking about that image of the photographer in that picture above now – would actually exist, then… Well… Then you need to be aware of the consequences: the corollary of his existence is that you do not exist. 🙂 And… Well… No. I won’t say more. If you’re reading stuff like this, then you’re smart enough to figure it out for yourself. We live in one world. Quantum mechanics tells us the perspective on that world matters very much – amplitudes are different in different reference frames – but… Well… Quantum mechanics – or physics in general – does not give us many degrees of freedoms. None, really. It basically tells us the world we live in is the only world that’s possible, really. But… Then… Well… That’s just because physics… Well… When everything is said and done, it’s just mankind’s drive to ensure our perception of the Universe lines up with… Well… What we perceive it to be. 😦 or 🙂 Whatever your appreciation of it. Those Great Minds did an incredible job. 🙂

Symmetries and transformations

In my previous post, I promised to do something on symmetries. Something simple but then… Well… You know how it goes: one question always triggers another one. 🙂

Look at the situation in the illustration on the left below. We suppose we have something real going on there: something is moving from left to right (so that’s in the 3 o’clock direction), and then something else is going around clockwise (so that’s not the direction in which we measure angles (which also include the argument θ of our wavefunction), because that’s always counter-clockwise, as I note at the bottom of the illustration). To be precise, we should note that the angular momentum here is all about the y-axis, so the angular momentum vector L points in the (positive) y-direction. We get that direction from the familiar right-hand rule, which is illustrated in the top right corner.

mirrorNow, suppose someone else is looking at this from the other side – or just think of yourself going around a full 180° to look at the same thing from the back side. You’ll agree you’ll see the same thing going from right to left (so that’s in the 9 o’clock direction now – or, if our clock is transparent, the 3 o’clock direction of our reversed clock). Likewise, the thing that’s turning around will now go counter-clockwise.

Note that both observers – so that’s me and that other person (or myself after my walk around this whole thing) – use a regular coordinate system, which implies the following:

  1. We’ve got regular 90° degree angles between our coordinates axes.
  2. Our x-axis goes from negative to positive from left to right, and our y-axis does the same going away from us.
  3. We also both define our z-axis using, once again, the ubiquitous right-hand rule, so our z-axis points upwards.

So we have two observers looking at the same reality – some linear as well as some angular momentum – but from opposite sides. And so we’ve got a reversal of both the linear as well as the angular momentum. Not in reality, of course, because we’re looking at the same thing. But we measure it differently. Indeed, if we use the subscripts 1 and 2 to denote the measurements in the two coordinate systems, we find that p2 = –p1. Likewise, we also find that L2 = –L1.

Now, when you see these two equations, you will probably not worry about that p2 = –p1 equation – although you should, because it’s actually only valid for this rather particular orientation of the linear momentum (I’ll come back to that in a moment). It’s the L2 = –L1 equation which should surprise you most. Why? Because you’ve always been told there is a big difference between (1) real vectors (aka polar vectors), like the momentum p, or the velocity v, or the force F, and (2) pseudo-vectors (aka axial vectors), like the angular momentum L. You may also remember how to distinguish between the two: if you change the direction of the axes of your reference frame, polar vectors will change sign too, as opposed to axial vectors: axial vectors do not swap sign if we swap the coordinate signs.

So… Well… How does that work here? In fact, what we should ask ourselves is: why does that not work here? Well… It’s simple, really. We’re not changing the direction of the axes here. Or… Well… Let me be more precise: we’re only swapping the sign of the x– and y-axis. We did not flip the z-axis. So we turned things around, but we didn’t turn them upside down. It makes a huge difference. Note, for example, that if all of the linear momentum would have been in the z-direction only (so our p vector would have been pointing in the z-direction, and in the z-direction only), it would not swap sign. The illustration below shows what really happens with the coordinates of some vector when we’re doing a rotation. It’s, effectively, only the x– and y-coordinates that flip sign.reflection symmetry

It’s easy to see that this rotation about the z-axis here preserves our deep sense of ‘up’ versus ‘down’, but that it swaps ‘left’ for ‘right’, and vice versa. Note that this is not a reflection. We are not looking at some mirror world here. The difference between a reflection (a mirror world) and a rotation (the real world seen from another angle) is illustrated below. It’s quite confusing but, unlike what you might think, a reflection does not swap left for right. It does turn things inside out, but that’s what a rotation does as well: near becomes far, and far becomes near.difference between reflection and rotation

Before we move on, let me say a few things about the mirror world and, more in particular, about the obvious question: could it possibly exist? Well… What do you think? Your first reaction might well be: “Of course! What nonsense question! We just walk around whatever it is that we’re seeing – or, what amounts to the same, we just turn it around – and there it is: that’s the mirror world, right? So of course it exists!” Well… No. That’s not the mirror world. That’s just the real world seen from the opposite direction, and that world… Well… That’s just the real world. 🙂 The mirror world is, literally, the world in the mirror – like the photographer in the illustration below. We don’t swap left for right here: some object going from left to right in the real world is still going from left to right in the mirror world!MirrorOf course, you may now involve the photographer in the picture above and observe – note that you’re now an observer of the observer of the mirror 🙂 – that, if he would move his left arm in the real world, the photographer in the mirror world would be moving his right arm. But… Well… No. You’re saying that because you’re now imaging that you’re the photographer in the mirror world yourself now, who’s looking at the real world from inside, so to speak. So you’ve rotated the perspective in your mind and you’re saying it’s his right arm because you imagine yourself to be the photographer in the mirror. We usually do that because… Well… Because we look in a mirror every day, right? So we’re used to seeing ourselves that way and we always think it’s us we’re seeing. 🙂 However, the illustration above is correct: the mirror world only swaps near for far, and far for near, so it only swaps the sign of the y-axis.

So the question is relevant: could the mirror world actually exist? What we’re really asking here is the following: can we swap the sign of one coordinate axis only in all of our physical laws and equations and… Well… Do we then still get the same laws and equations? Do we get the same Universe – because that’s what those laws and equations describe? If so, our mirror world can exist. If not, then not.

Now, I’ve done a post on that, in which I explain that mirror world can only exist if it would consist of anti-matter. So if our real world and the mirror world would actually meet, they would annihilate each other. 🙂 But that post is quite technical. Here I want to keep it very simple: I basically only want to show what the rotation operation implies for the wavefunction. There is no doubt whatsoever that the rotated world exists. In fact, the rotated world is just our world. We walk around some object, or we turn it around, but so we’re still watching the same object. So we’re not thinking about the mirror world here. We just want to know how things look like when adopting some other perspective.

So, back to the starting point: we just have two observers here, who look at the same thing but from opposite directions. Mathematically, this corresponds to a rotation of our reference frame about the z-axis of 180°. Let me spell out – somewhat more precisely – what happens to the linear and angular momentum here:

  1. The direction of the linear momentum in the xy-plane swaps direction.
  2. The angular momentum about the y-axis, as well as about the x-axis, swaps direction too.

Note that the illustration only shows angular momentum about the y-axis, but you can easily verify the statement about the angular momentum about the x-axis. In fact, the angular momentum about any line in the xy-plane will swap direction.

Of course, the x-, y-, z-axes in the other reference frame are different than mine, and so I should give them a subscript, right? Or, at the very least, write something like x’, y’, z’, so we have a primed reference frame here, right? Well… Maybe. Maybe not. Think about it. 🙂 A coordinate system is just a mathematical thing… Only the momentum is real… Linear or angular… Equally real… And then Nature doesn’t care about our position, does it? So… Well… No subscript needed, right? Or… Well… What do you think? 🙂

It’s just funny, isn’t it? It looks like we can’t really separate reality and perception here. Indeed, note how our p2 = –pand L2 = –L1 equations already mix reality with how we perceive it. It’s the same thing in reality but the coordinates of p1 and L1 are positive, while the coordinates of p2 and Lare negative. To be precise, these coordinates will look like this:

  1. p1 = (p, 0, 0) and L1 = (0, L, 0)
  2. p2 = (−p, 0, 0) and L1 = (0, −L, 0)

So are they two different things or are they not? 🙂 Think about it. I’ll move on in the meanwhile. 🙂

Now, you probably know a thing or two about parity symmetry, or P-symmetry: if if we flip the sign of all coordinates, then we’ll still find the same physical laws, like F = m·a and what have you. [It works for all physical laws, including quantum-mechanical laws – except those involving the weak force (read: radioactive decay processes).] But so here we are talking rotational symmetry. That’s not the same as P-symmetry. If we flip the signs of all coordinates, we’re also swapping ‘up’ for ‘down’, so we’re not only turning around, but we’re also getting upside down. The difference between rotational symmetry and P-symmetry is shown below.up and down swap

As mentioned, we’ve talked about P-symmetry at length in other posts, and you can easily google a lot more on that. The question we want to examine here – just as a fun exercise – is the following:

How does that rotational symmetry work for a wavefunction?

The very first illustration in this post gave you the functional form of the elementary wavefunction  eiθ = ei·(E·t p·x)/ħ. We should actually use a bold type x = (x, y, z) in this formula but we’ll assume we’re talking something similar to that p vector: something moving in the x-direction only – or in the xy-plane only. The z-component doesn’t change. Now, you know that we can reduce all actual wavefunctions to some linear combination of such elementary wavefunctions by doing a Fourier decomposition, so it’s fine to look at the elementary wavefunction only – so we don’t make it too complicated here. Now think of the following.

The energy E in the eiθ = ei·(E·t – p·x)/ħ function is a scalar, so it doesn’t have any direction and we’ll measure it the same from both sides – as kinetic or potential energy or, more likely, by adding both. But… Well… Writing ei·(E·t – p·x)/ħ or ei·(E·t + p·x)/ħ is not the same, right? No, it’s not. However, think of it as follows: we won’t be changing the direction of time, right? So it’s OK to not change the sign of E. In fact, we can re-write the two expressions as follows:

  1. ei·(E·t – p·x)/ħ = ei·(E/ħ)·t·ei·(p/ħ)·x
  2. ei·(E·t + p·x)/ħ = ei·(E/ħ)·t·ei·(p/ħ)·x

The first wavefunction describes some particle going in the positive x-direction, while the second wavefunction describes some particle going in the negative x-direction, so… Well… That’s exactly what we see in those two reference frames, so there is no issue whatsoever. 🙂 It’s just… Well… I just wanted to show the wavefunction does look different too when looking at something from another angle.

So why am I writing about this? Why am I being fussy? Well.. It’s just to show you that those transformations are actually quite natural – just as natural as it is to see some particle go in one direction in one reference frame and see it go in the other in the other. 🙂 It also illustrates another point that I’ve been trying to make: the wavefunction is something real. It’s not just a figment of our imagination. The real and imaginary part of our wavefunction have a precise geometrical meaning – and I explained what that might be in my more speculative posts, which I’ve brought together in the Deep Blue page of this blog. But… Well… I can’t dwell on that here because… Well… You should read that page. 🙂

The point to note is the following: we do have different wavefunctions in different reference frames, but these wavefunctions describe the same physical reality, and they also do respect the symmetries we’d expect them to respect, except… Well… The laws describing the weak force don’t, but I wrote about that a very long time ago, and it was not in the context of trying to explain the relatively simple basic laws of quantum mechanics. 🙂 If you’re interested, you should check out my post(s) on that or, else, just google a bit. It’s really exciting stuff, but not something that will help you much to understand the basics, which is what we’re trying to do here. 🙂

The second point to note is that those transformations of the wavefunction – or of quantum-mechanical states – which we go through when rotating our reference frame, for example – are really quite natural. There’s nothing special about them. We had such transformations in classical mechanics too! But… Well… Yes, I admit they do look complicated. But then that’s why you’re so fascinated and why you’re reading this blog, isn’t it? 🙂

Post scriptum: It’s probably useful to be somewhat more precise on all of this. You’ll remember we visualized the wavefunction in some of our posts using the animation below. It uses a left-handed coordinate system, which is rather unusual but then it may have been made with a software which uses a left-handed coordinate system (like RenderMan, for example). Now the rotating arrow at the center moves with time and gives us the polarization of our wave. Applying our customary right-hand rule,you can see this beam is left-circularly polarized. [I know… It’s quite confusing, but just go through the motions here and be consistent.]AnimationNow, you know that ei·(p/ħ)·x and ei·(p/ħ)·x are each other’s complex conjugate:

  1. ei·k·x cos(k·x) + i·sin(k·x)
  2. ei·k·x cos(-k·x) + i·sin(-k·x) = cos(k·x) − i·sin(k·x)

Their real part – the cosine function – is the same, but the imaginary part – the sine function – has the opposite sign. So, assuming the direction of propagation is, effectively, the x-direction, then what’s the polarization of the mirror image? Well… The wave will now go from right to left, and its polarization… Hmm… Well… What? 

Well… If you can’t figure it out, then just forget about those signs and just imagine you’re effectively looking at the same thing from the backside. In fact, if you have a laptop, you can push the screen down and go around your computer. 🙂 There’s no shame in that. In fact, I did that just to make sure I am not talking nonsense here. 🙂 If you look at this beam from the backside, you’ll effectively see it go from right to left – instead of from what you see on this side, which is a left-to-right direction. And as for its polarization… Well… The angular momentum vector swaps direction too but the beam is still left-circularly polarized. So… Well… That’s consistent with what we wrote above. 🙂 The real world is real, and axial vectors are as real as polar vectors. This real beam will only appear to be right-circularly polarized in a mirror. Now, as mentioned above, that mirror world is not our world. If it would exist – in some other Universe – then it would be made up of anti-matter. 🙂

So… Well… Might it actually exist? Is there some other world made of anti-matter out there? I don’t know. We need to think about that reversal of ‘near’ and ‘far’ too: as mentioned, a mirror turns things inside out, so to speak. So what’s the implication of that? When we walk around something – or do a rotation – then the reversal between ‘near’ and ‘far’ is something physical: we go near to what was far, and we go away from what was near. But so how would we get into our mirror world, so to speak? We may say that this anti-matter world in the mirror is entirely possible, but then how would we get there? We’d need to turn ourselves, literally, inside out – like short of shrink to the zero point and then come back out of it to do that parity inversion along our line of sight. So… Well… I don’t see that happen, which is why I am a fan of the One World hypothesis. 🙂 So think the mirror world is just what it is: the mirror world. Nothing real. But… Then… Well… What do you think? 🙂

Quantum-mechanical magnitudes

As I was writing about those rotations in my previous post (on electron orbitals), I suddenly felt I should do some more thinking on (1) symmetries and (2) the concept of quantum-mechanical magnitudes of vectors. I’ll write about the first topic (symmetries) in some other post. Let’s first tackle the latter concept. Oh… And for those I frightened with my last post… Well… This should really be an easy read. More of a short philosophical reflection about quantum mechanics. Not a technical thing. Something intuitive. At least I hope it will come out that way. 🙂

First, you should note that the fundamental idea that quantities like energy, or momentum, may be quantized is a very natural one. In fact, it’s what the early Greek philosophers thought about Nature. Of course, while the idea of quantization comes naturally to us (I think it’s easier to understand than, say, the idea of infinity), it is, perhaps, not so easy to deal with it mathematically. Indeed, most mathematical ideas – like functions and derivatives – are based on what I’ll loosely refer to as continuum theory. So… Yes, quantization does yield some surprising results, like that formula for the magnitude of some vector J:Magnitude formulasThe J·J in the classical formula above is, of course, the equally classical vector dot product, and the formula itself is nothing but Pythagoras’ Theorem in three dimensions. Easy. I just put a + sign in front of the square roots so as to remind you we actually always have two square roots and that we should take the positive one. 🙂

I will now show you how we get that quantum-mechanical formula. The logic behind it is fairly straightforward but, at the same time… Well… You’ll see. 🙂 We know that a quantum-mechanical variable – like the spin of an electron, or the angular momentum of an atom – is not continuous but discrete: it will have some value = jj-1, j-2, …, -(j-2), -(j-1), –j. Our here is the maximum value of the magnitude of the component of our vector (J) in the direction of measurement, which – as you know – is usually written as Jz. Why? Because we will usually choose our coordinate system such that our z-axis is aligned accordingly. 🙂 Those values jj-1, j-2, …, -(j-2), -(j-1), –j are separated by one unit. That unit would be Planck’s quantum of action ħ ≈ 1.0545718×10−34 N·m·s – by the way, isn’t it amazing we can actually measure such tiny stuff in some experiment? 🙂 – if J would happen to be the angular momentum, but the approach here is more general – action can express itself in various ways 🙂 – so the unit doesn’t matter: it’s just the unit, so that’s just one. 🙂 It’s easy to see that this separation implies must be some integer or half-integer. [Of course, now you might think the values of a series like 2.4, 1.4, 0.4, -0.6, -1.6 are also separated by one unit, but… Well… That would violate the most basic symmetry requirement so… Well… No. Our has to be an integer or a half-integer. Please also note that the number of possible values for is equal to 2j+1, as we’ll use that in a moment.]

OK. You’re familiar with this by now and so I should not repeat the obvious. To make things somewhat more real, let’s assume = 3/2, so =  3/2, 1/2, -1/2 or +3/2. Now, we don’t know anything about the system and, therefore, these four values are all equally likely. Now, you may not agree with this assumption but… Well… You’ll have to agree that, at this point, you can’t come up with anything else that would make sense, right? It’s just like a classical situation: J might point in any direction, so we have to give all angles an equal probability. [In fact, I’ll show you – in a minute or so – that you actually have a point here: we should think some more about this assumption – but so that’s for later. I am asking you to just go along with this story as for now.]

So the expected value of Jz is E[Jz] is equal to E[Jz] = (1/4)·(3/2)+(1/4)·(1/2)+(1/4)·(-1/2)+(1/4)·(-3/2) = 0. Nothing new here. We just multiply probabilities with all of the possible values to get an expected value. So we get zero here because our values are distributed symmetrically around the zero point. No surprise. Now, to calculate a magnitude, we don’t need Jbut Jz2. In case you wonder, that’s what this squaring business is all about: we’re abstracting away from the direction and so we’re going to square both positive as well as negative values to then add it all up and take a square root. Now, the expected value of Jz2 is equal to E[Jz] = (1/4)·(3/2)2+(1/4)·(1/2)2+(1/4)·(-1/2)2+(1/4)·(-3/2)2 = 5/4 = 1.25. Some positive value.

You may note that it’s a bit larger than the average of the absolute value of our variable, which is equal to (|3/2|+|1/2|+|-1/2|+|-3/2|)/4 = 1, but that’s just because the squaring favors larger values 🙂 Also note that, of course, we’d also get some positive value if Jwould be a continuous variable over the [-3/2, +3/2] interval, but I’ll let you think about what positive value we’d get for Jzassuming Jz is uniform distributed over the [-3/2, +3/2] interval, because that calculation is actually not so straightforward as it may seem at first. In any case, these considerations are not very relevant to our story here, so let’s move on.

Of course, our z-direction was random, and so we get the same thing for whatever direction. More in particular, we’ll also get it for the x– and y-directions: E[Jx] = E[Jy] = E[Jz] = 5/4. Now, at this point it’s probably good to give you a more generalized formula for these quantities. I think you’ll easily agree to the following one:magnitude squared formulaSo now we can apply our classical J·J = JxJyJzformula to these quantities by calculating the expected value of JJ·J, which is equal to:

E[J·J] = E[Jx2] + E[Jy2] + E[Jz2] = 3·E[Jx2] = 3·E[Jy2] = 3·E[Jz2]

You should note we’re making use of the E[X Y] = E[X]+ E[Y] property here: the expected value of the sum of two variables is equal to the sum of the expected values of the variables, and you should also note this is true even if the individual variables would happen to be correlated – which might or might not be the case. [What do you think is the case here?]

For = 3/2, it’s easy to see we get E[J·J] = 3·E[Jx] = 3·5/4 = (3/2)·(3/2+1) = j·(j+1). We should now generalize this formula for other values of j,  which is not so easy… Hmm… It obviously involves some formula for a series, and I am not good at that… So… Well… I just checked if it was true for = 1/2 and = 1 (please check that at least for yourself too!) and then I just believe the authorities on this for all other values of j. 🙂

Now, in a classical situation, we know that J·J product will be the same for whatever direction J would happen to have, and so its expected value will be equal to its constant value J·J. So we can write: E[J·J] = J·J. So… Well… That’s why we write what we wrote above:Magnitude formulas

Makes sense, no? E[J·J] = E[Jx2+Jy2+Jz2] = E[Jx2]+E[Jy2]+E[Jz2] = j·(j+1) = J·J = J2, so = +√[j(j+1)], right?

Hold your horses, man! Think! What are we doing here, really? We didn’t calculate all that much above. We only found that E[Jx2]+E[Jy2]+E[Jz2] = E[Jx2+Jy2+Jz2] =  j·(j+1). So what? Well… That’s not a proof that the J vector actually exists.

Huh? 

Yes. That J vector might just be some theoretical concept. When everything is said and done, all we’ve been doing – or at least, we imagined we did – is those repeated measurements of JxJy and Jz here – or whatever subscript you’d want to use, like Jθ,φ, for example (the example is not random, of course) – and so, of course, it’s only natural that we assume these things are the magnitude of the component (in the direction of measurement) of some real vector that is out there, but then… Well… Who knows? Think of what we wrote about the angular momentum in our previous post on electron orbitals. We imagine – or do like to think – that there’s some angular momentum vector J out there, which we think of as being “cocked” at some angle, so its projection onto the z-axis gives us those discrete values for m which, for = 2, for example, are equal to 0, 1 or 2 (and -1 and -2, of course) – like in the illustration below. 🙂cocked angle 2But… Well… Note those weird angles: we get something close to 24.1° and then another value close to 54.7°. No symmetry here. 😦 The table below gives some more values for larger j. They’re easy to calculate – it’s, once again, just Pythagoras’ Theorem – but… Well… No symmetries here. Just weird values. [I am not saying the formula for these angles is not straightforward. That formula is easy enough: θ = sin-1(m/√[j(j+1)]). It’s just… Well… No symmetry. You’ll see why that matters in a moment.]CaptureI skipped the half-integer values for in the table above so you might think they might make it easier to come up with some kind of sensible explanation for the angles. Well… No. They don’t. For example, for = 1/2 and m = ± 1/2, the angles are ±35.2644° – more or less, that is. 🙂 As you can see, these angles do not nicely cut up our circle in equal pieces, which triggers the obvious question: are these angles really equally likely? Equal angles do not correspond to equal distances on the z-axis (in case you don’t appreciate the point, look at the illustration below).  angles distance

So… Well… Let me summarize the issue on hand as follows: the idea of the angle of the vector being randomly distributed is not compatible with the idea of those Jz values being equally spaced and equally likely. The latter idea – equally spaced and equally likely Jz values – relates to different possible states of the system being equally likely, so… Well… It’s just a different idea. 😦

Now there is another thing which we should mention here. The maximum value of the z-component of our J vector is always smaller than that quantum-mechanical magnitude, and quite significantly so for small j, as shown in the table below. It is only for larger values of that the ratio of the two starts to converge to 1. For example, for = 25, it is about 1.02, so that’s only 2% off. convergenceThat’s why physicists tell us that, in quantum mechanics, the angular momentum is never “completely along the z-direction.” It is obvious that this actually challenges the idea of a very precise direction in quantum mechanics, but then that shouldn’t surprise us, does it? After, isn’t this what the Uncertainty Principle is all about?

Different states, rather than different directions… And then Uncertainty because… Well… Because of discrete variables that won’t split in the middle. Hmm… 😦

Perhaps. Perhaps I should just accept all of this and go along with it… But… Well… I am really not satisfied here, despite Feynman’s assurance that that’s OK: “Understanding of these matters comes very slowly, if at all. Of course, one does get better able to know what is going to happen in a quantum-mechanical situation—if that is what understanding means—but one never gets a comfortable feeling that these quantum-mechanical rules are ‘natural’.”

I do want to get that comfortable feeling – on some sunny day, at least. 🙂 And so I’ll keep playing with this, until… Well… Until I give up. 🙂 In the meanwhile, if you’d feel you’ve got some better or some more intuitive explanation for all of this, please do let me know. I’d be very grateful to you. 🙂

Post scriptum: Of course, we would all want to believe that J somehow exists because… Well… We want to explain those states somehow, right? I, for one, am not happy with being told to just accept things and shut up. So let me add some remarks here. First, you may think that the narrative above should distinguish between polar and axial vectors. You’ll remember polar vectors are the real vectors, like a radius vector r, or a force F, or velocity or (linear) momentum. Axial vectors (also known as pseudo-vectors) are vectors like the angular momentum vector: we sort of construct them from… Well… From real vectors. The angular momentum L, for example, is the vector cross product of the radius vector r and the linear momentum vector p: we write L = r×p. In that sense, they’re a figment of our imagination. But then… What’s real and unreal? The magnitude of L, for example, does correspond to something real, doesn’t it? And its direction does give us the direction of circulation, right? You’re right. Hence, I think polar and axial vectors are both real – in whatever sense you’d want to define real. Their reality is just different, and that’s reflected in their mathematical behavior: if you change the direction of the axes of your reference frame, polar vectors will change sign too, as opposed to axial vectors: they don’t swap sign. They do something else, which I’ll explain in my next post, where I’ll be talking symmetries.

But let us, for the sake of argument, assume whatever I wrote about those angles applies to axial vectors only. Let’s be even more specific, and say it applies to the angular momentum vector only. If that’s the case, we may want to think of a classical equivalent for the mentioned lack of a precise direction: free nutation. It’s a complicated thing – even more complicated than the phenomenon of precession, which we should be familiar with by now. Look at the illustration below (which I took from an article of a physics professor from Saint Petersburg), which shows both precession as well as nutation. Think of the movement of a spinning top when you release it: its axis will, at first, nutate around the axis of precession, before it settles in a more steady precession.nutationThe nutation is caused by the gravitational force field, and the nutation movement usually dies out quickly because of dampening forces (read: friction). Now, we don’t think of gravitational fields when analyzing angular momentum in quantum mechanics, and we shouldn’t. But there is something else we may want to think of. There is also a phenomenon which is referred to as free nutation, i.e. a nutation that is not caused by an external force field. The Earth, for example, nutates slowly because of a gravitational pull from the Sun and the other planets – so that’s not a free nutation – but, in addition to this, there’s an even smaller wobble – which is an example of free nutation – because the Earth is not exactly spherical. In fact, the Great Mathematician, Leonhard Euler, had already predicted this, back in 1765, but it took another 125 years or so before an astronomist, Seth Chandler, could finally experimentally confirm and measure it. So they named this wobble the Chandler wobble (Euler already has too many things named after him). 🙂

Now I don’t have much backup here – none, actually 🙂 – but why wouldn’t we imagine our electron would also sort of nutate freely because of… Well… Some symmetric asymmetry – something like the slightly elliptical shape of our Earth. 🙂 We may then effectively imagine the angular momentum vector as continually changing direction between a minimum and a maximum angle – something like what’s shown below, perhaps, between 0 and 40 degrees. Think of it as a rotation within a rotation, or an oscillation within an oscillation – or a standing wave within a standing wave. 🙂wobblingI am not sure if this approach would solve the problem of our angles and distances – the issue of whether we should think in equally likely angles or equally likely distances along the z-axis, really – but… Well… I’ll let you play with this. Please do send me some feedback if you think you’ve found something. 🙂

Whatever your solution is, it is likely to involve the equipartition theorem and harmonics, right? Perhaps we can, indeed, imagine standing waves within standing waves, and then standing waves within standing waves. How far can we go? 🙂

Post scriptum 2: When re-reading this post, I was thinking I should probably do something with the following idea. If we’ve got a sphere, and we’re thinking of some vector pointing to some point on the surface of that sphere, then we’re doing something which is referred to as point picking on the surface of a sphere, and the probability distributions – as a function of the polar and azimuthal angles θ and φ – are quite particular. See the article on the Wolfram site on this, for example. I am not sure if it’s going to lead to some easy explanation of the ‘angle problem’ we’ve laid out here but… Well… It’s surely an element in the explanation. The key idea here is shown in the illustration below: if the direction of our momentum in three-dimensional space is really random, there may still be more of a chance of an orientation towards the equator, rather than towards the pole. So… Well… We need to study the math of this. 🙂 But that’s for later.density

Re-visiting electron orbitals

One of the pieces I barely gave a glance when reading Feynman’s Lectures over the past few years, was the derivation of the non-spherical electron orbitals for the hydrogen atom. It just looked like a boring piece of math – and I thought the derivation of the s-orbitals – the spherically symmetrical ones – was interesting enough already. To some extent, it is – but there is so much more to it. When I read it now, the derivation of those p-, d-, f– etc. orbitals brings all of the weirdness of quantum mechanics together and, while doing so, also provides for a deeper understanding of all of the ideas and concepts we’re trying to get used to. In addition, Feynman’s treatment of the matter is actually much shorter than what you’ll find in other textbooks, because… Well… As he puts it, he takes a shortcut. So let’s try to follow the bright mind of our Master as he walks us through it.

You’ll remember – if not, check it out again – that we found the spherically symmetric solutions for Schrödinger’s equation for our hydrogen atom. Just to be make sure, Schrödinger’s equation is a differential equation – a condition we impose on the wavefunction for our electron – and so we need to find the functional form for the wavefunctions that describe the electron orbitals. [Quantum math is so confusing that it’s often good to regularly think of what it is that we’re actually trying to do. :-)] In fact, that functional form gives us a whole bunch of solutions – or wavefunctions – which are defined by three quantum numbers: n, l, and m. The parameter n corresponds to an energy level (En), l is the orbital (quantum) number, and m is the z-component of the angular momentum. But that doesn’t say much. Let’s go step by step.

First, we derived those spherically symmetric solutions – which are referred to as s-states – assuming this was a state with zero (orbital) angular momentum, which we write as = 0. [As you know, Feynman does not incorporate the spin of the electron in his analysis, which is, therefore, approximative only.] Now what exactly is a state with zero angular momentum? When everything is said and done, we are effectively trying to describe some electron orbital here, right? So that’s an amplitude for the electron to be somewhere, but then we also know it always moves. So, when everything is said and done, the electron is some circulating negative charge, right? So there is always some angular momentum and, therefore, some magnetic moment, right?

Well… If you google this question on Physics Stack Exchange, you’ll get a lot of mumbo jumbo telling you that you shouldn’t think of the electron actually orbiting around. But… Then… Well… A lot of that mumbo jumbo is contradictory. For example, one of the academics writing there does note that, while we shouldn’t think of an electron as some particle, the orbital is still a distribution which gives you the probability of actually finding the electron at some point (x,y,z). So… Well… It is some kind of circulating charge – as a point, as a cloud or as whatever. The only reasonable answer – in my humble opinion – is that = 0 probably means there is no net circulating charge, so the movement in this or that direction must balance the movement in the other. One may note, in this regard, that the phenomenon of electron capture in nuclear reactions suggests electrons do travel through the nucleus for at least part of the time, which is entirely coherent with the wavefunctions for s-states – shown below – which tell us that the most probable (x, y, z) position for the electron is right at the center – so that’s where the nucleus is. There is also a non-zero probability for the electron to be at the center for the other orbitals (pd, etcetera).s-statesIn fact, now that I’ve shown this graph, I should quickly explain it. The three graphs are the spherically symmetric wavefunctions for the first three energy levels. For the first energy level – which is conventionally written as n = 1, not as n = 0 – the amplitude approaches zero rather quickly. For n = 2 and n = 3, there are zero-crossings: the curve passes the r-axis. Feynman calls these zero-crossing radial nodes. To be precise, the number of zero-crossings for these s-states is n − 1, so there’s none for = 1, one for = 2, two for = 3, etcetera.

Now, why is the amplitude – apparently – some real-valued function here? That’s because we’re actually not looking at ψ(r, t) here but at the ψ(r) function which appears in the following break-up of the actual wavefunction ψ(r, t):

ψ(r, t) = ei·(E/ħ)·t·ψ(r)

So ψ(r) is more of an envelope function for the actual wavefunction, which varies both in space as well as in time. It’s good to remember that: I would have used another symbol, because ψ(r, t) and ψ(r) are two different beasts, really – but then physicists want you to think, right? And Mr. Feynman would surely want you to do that, so why not inject some confusing notation from time to time? 🙂 So for = 3, for example, ψ(r) goes from positive to negative and then to positive, and these areas are separated by radial nodes. Feynman put it on the blackboard like this:radial nodesI am just inserting it to compare this concept of radial nodes with the concept of a nodal plane, which we’ll encounter when discussing p-states in a moment, but I can already tell you what they are now: those p-states are symmetrical in one direction only, as shown below, and so we have a nodal plane instead of a radial node. But so I am getting ahead of myself here… 🙂nodal planesBefore going back to where I was, I just need to add one more thing. 🙂 Of course, you know that we’ll take the square of the absolute value of our amplitude to calculate a probability (or the absolute square – as we abbreviate it), so you may wonder why the sign is relevant at all. Well… I am not quite sure either but there’s this concept of orbital parity which you may have heard of.  The orbital parity tells us what will happen to the sign if we calculate the value for ψ for −r rather than for r. If ψ(−r) = ψ(r), then we have an even function – or even orbital parity. Likewise, if ψ(−r) = −ψ(r), then we’ll the function odd – and so we’ll have an odd orbital parity. The orbital parity is always equal to (-1)l = ±1. The exponent is that angular quantum number, and +1, or + tout court, means even, and -1 or just − means odd. The angular quantum number for those p-states is = 1, so that works with the illustration of the nodal plane. 🙂 As said, it’s not hugely important but I might as well mention in passing – especially because we’ll re-visit the topic of symmetries a few posts from now. 🙂

OK. I said I would talk about states with some angular momentum (so ≠ 0) and so it’s about time I start doing that. As you know, our orbital angular momentum is measured in units of ħ (just like the total angular momentum J, which we’ve discussed ad nauseam already). We also know that if we’d measure its component along any direction – any direction really, but physicists will usually make sure that the z-axis of their reference frame coincides with, so we call it the z-axis 🙂 – then we will find that it can only have one of a discrete set of values m·ħ l·ħ, (l-1)·ħ, …, -(l-1)·ħ, –l·ħ. Hence, just takes the role of our good old quantum number here, and m is just Jz. Likewise, I’d like to introduce l as the equivalent of J, so we can easily talk about the angular momentum vector. And now that we’re here, why not write in bold type too, and say that m is the z-component itself – i.e. the whole vector quantity, so that’s the direction and the magnitude.

Now, we do need to note one crucial difference between and l, or between J and l: our j could be an integer or a half-integer. In contrast, must be some integer. Why? Well… If can be zero, and the values of must be separated by a full unit, then l must be 1, 2, 3 etcetera. 🙂 If this simple answer doesn’t satisfy you, I’ll refer you to Feynman’s, which is also short but more elegant than mine. 🙂 Now, you may or may not remember that the quantum-mechanical equivalent of the magnitude of a vector quantity such as l is to be calculated as √[l·(l+1)]·ħ, so if = 1, that magnitude will be √2·ħ ≈ 1.4142·ħ, so that’s – as expected – larger than the maximum value for m, which is +1. As you know, that leads us to think of that z-component m as a projection of l. Paraphrasing Feynman, the limited set of values for m imply that the angular momentum is always “cocked” at some angle. For = 1, that angle is either +45° or, else, −45°, as shown below.cocked angleWhat if l = 2? The magnitude of is then equal to √[2·(2+1)]·ħ = √6·ħ ≈ 2.4495·ħ. How do we relate that to those “cocked” angles? The values of now range from -2 to +2, with a unit distance in-between. The illustration below shows the angles. [I didn’t mention ħ any more in that illustration because, by now, we should know it’s our unit of measurement – always.]

cocked angle 2Note we’ve got a bigger circle here (the radius is about 2.45 here, as opposed to a bit more than 1.4 for m = 0). Also note that it’s not a nice cake with perfectly equal pieces. From the graph, it’s obvious that the formula for the angle is the following:angle formulaIt’s simple but intriguing. Needless to say, the sin −1 function is the inverse sine, also known as the arcsine. I’ve calculated the values for all for l = 1, 2, 3, 4 and 5 below. The most interesting values are the angles for = 1 and l. As the graphs underneath show, for = 1, the values start approaching the zero angle for very large l, so there’s not much difference any more between = ±1 and = 1 for large values of l. What about the l case? Well… Believe it or not, if becomes really large, then these angles do approach 90°. If you don’t remember how to calculate limits, then just calculate θ for some huge value for and m. For = 1,000,000, for example, you should find that θ = 89.9427…°. 🙂angles

graphIsn’t this fascinating? I’ve actually never seen this in a textbook – so it might be an original contribution. 🙂 OK. I need to get back to the grind: Feynman’s derivation of non-symmetrical electron orbitals. Look carefully at the illustration below. If m is really the projection of some angular momentum that’s “cocked”, either at a zero-degree or, alternatively, at ±45º (for the = 1 situation we show here) – a projection on the z-axis, that is – then the value of m (+1, 0 or -1) does actually correspond to some idea of the orientation of the space in which our electron is circulating. For = 0, that space – think of some torus or whatever other space in which our electron might circulate – would have some alignment with the z-axis. For = ±1, there is no such alignment. m = 0

The interpretation is tricky, however, and the illustration on the right-hand side above is surely too much of a simplification: an orbital is definitely not like a planetary orbit. It doesn’t even look like a torus. In fact, the illustration in the bottom right corner, which shows the probability density, i.e. the space in which we are actually likely to find the electron, is a picture that is much more accurate – and it surely does not resemble a planetary orbit or some torus. However, despite that, the idea that, for = 0, we’d have some alignment of the space in which our electron moves with the z-axis is not wrong. Feynman expresses it as follows:

“Suppose m is zero, then there can be some non-zero amplitude to find the electron on the z-axis at some distance r. We’ll call this amplitude Fl(r).”

You’ll say: so what? And you’ll also say that illustration in the bottom right corner suggests the electron is actually circulating around the z-axis, rather than through it. Well… No. That illustration does not show any circulation. It only shows a probability density. No suggestion of any actual movement or circulation. So the idea is valid: if = 0, then the implication is that, somehow, the space of circulation of current around the direction of the angular momentum vector (J), as per the well-known right-hand rule, will include the z-axis. So the idea of that electron orbiting through the z-axis for = 0 is essentially correct, and the corollary is… Well… I’ll talk about that in a moment.

But… Well… So what? What’s so special about that Fl(r) amplitude? What can we do with that? Well… If we would find a way to calculate Fl(r), then we know everything. Huh? Everything? Yes. The reasoning here is quite complicated, so please bear with me as we go through it.

The first thing you need to accept, is rather weird. The thing we said about the non-zero amplitudes to find the electron somewhere on the z-axis for the m = 0 state – which, using Dirac’s bra-ket notation, we’ll write as |l= 0〉 – has a very categorical corollary:

The amplitude to find an electron whose state m is not equal to zero on the z-axis (at some non-zero distance r) is zero. We can only find an electron on the z-axis unless the z-component of its angular momentum (m) is zero. 

Now, I know this is hard to swallow, especially when looking at those 45° angles for J in our illustrations, because these suggest the actual circulation of current may also include at least part of the z-axis. But… Well… No. Why not? Well… I have no good answer here except for the usual one which, I admit, is quite unsatisfactory: it’s quantum mechanics, not classical mechanics. So we have to look at the m and m vectors, which are pointed along the z-axis itself for m = ±1 and, hence, the circulation we’d associate with those momentum vectors (even if they’re the zcomponent only) is around the z-axis. Not through or on it. I know it’s a really poor argument, but it’s consistent with our picture of the actual electron orbitals – that picture in terms of probability densities, which I copy below. For m = −1, we have the yz-plane as the nodal plane between the two lobes of our distribution, so no amplitude to find the electron on the z-axis (nor would we find it on the y-axis, as you can see). Likewise, for m = +1, we have the xz-plane as the nodal plane. Both nodal planes include the z-axis and, therefore, there’s zero probability on that axis. p orbitals

In addition, you may also want to note the 45° angle we associate with = ±1 does sort of demarcate the lobes of the distribution by defining a three-dimensional cone and… Well… I know these arguments are rather intuitive, and so you may refuse to accept them. In fact, to some extent, refuse to accept them. 🙂 Indeed, let me say this loud and clear: I really want to understand this in a better way! 

But… Then… Well… Such better understanding may never come. Feynman’s warning, just before he starts explaining the Stern-Gerlach experiment and the quantization of angular momentum, rings very true here: “Understanding of these matters comes very slowly, if at all. Of course, one does get better able to know what is going to happen in a quantum-mechanical situation—if that is what understanding means—but one never gets a comfortable feeling that these quantum-mechanical rules are “natural.” Of course they are, but they are not natural to our own experience at an ordinary level.” So… Well… What can I say?

It is now time to pull the rabbit out of the hat. To understand what we’re going to do next, you need to remember that our amplitudes – or wavefunctions – are always expressed with regard to a specific frame of reference, i.e. some specific choice of an x-, y– and z-axis. If we change the reference frame – say, to some new set of x’-, y’– and z’-axes – then we need to re-write our amplitudes (or wavefunctions) in terms of the new reference frame. In order to do so, one should use a set of transformation rules. I’ve written several posts on that – including a very basic one, which you may want to re-read (just click the link here).

Look at the illustration below. We want to calculate the amplitude to find the electron at some point in space. Our reference frame is the x, y, z frame and the polar coordinates (or spherical coordinates, I should say) of our point are the radial distance r, the polar angle θ (theta), and the azimuthal angle φ (phi). [The illustration below – which I copied from Feynman’s exposé – uses a capital letter for phi, but I stick to the more usual or more modern convention here.]

change of reference frame

In case you wonder why we’d use polar coordinates rather than Cartesian coordinates… Well… I need to refer you to my other post on the topic of electron orbitals, i.e. the one in which I explain how we get the spherically symmetric solutions: if you have radial (central) fields, then it’s easier to solve stuff using polar coordinates – although you wouldn’t think so if you think of that monster equation that we’re actually trying to solve here:

new de

It’s really Schrödinger’s equation for the situation on hand (i.e. a hydrogen atom, with a radial or central Coulomb field because of its positively charged nucleus), but re-written in terms of polar coordinates. For the detail, see the mentioned post. Here, you should just remember we got the spherically symmetric solutions assuming the derivatives of the wavefunction with respect to θ and φ – so that’s the ∂ψ/∂θ and ∂ψ/∂φ in the equation abovewere zero. So now we don’t assume these partial derivatives to be zero: we’re looking for states with an angular dependence, as Feynman puts it somewhat enigmatically. […] Yes. I know. This post is becoming very long, and so you are getting impatient. Look at the illustration with the (r, θ, φ) point, and let me quote Feynman on the line of reasoning now:

“Suppose we have the atom in some |lm〉 state, what is the amplitude to find the electron at the angles θ and φ and the distance from the origin? Put a new z-axis, say z’, at that angle (see the illustration above), and ask: what is the amplitude that the electron will be at the distance along the new z’-axis? We know that it cannot be found along z’ unless its z’-component of angular momentum, say m’, is zero. When m’ is zero, however, the amplitude to find the electron along z’ is Fl(r). Therefore, the result is the product of two factors. The first is the amplitude that an atom in the state |lm〉 along the z-axis will be in the state |lm’ = 0〉 with respect to the z’-axis. Multiply that amplitude by Fl(r) and you have the amplitude ψl,m(r) to find the electron at (r, θ, φ) with respect to the original axes.”

So what is he telling us here? Well… He’s going a bit fast here. 🙂 Worse, I think he may actually not have chosen the right words here, so let me try to rephrase it. We’ve introduced the Fl(r) function above: it was the amplitude, for m = 0, to find the electron on the z-axis at some distance r. But so here we’re obviously in the x’, y’, z’ frame and so Fl(r) is the amplitude for m’ = 0,  it’s the amplitude to find the electron on the z-axis at some distance r along the z’-axis. Of course, for this amplitude to be non-zero, we must be in the |lm’ = 0〉 state, but are we? Well… |lm’ = 0〉 actually gives us the amplitude for that. So we’re going to multiply two amplitudes here:

Fl(r)·|lm’ = 0〉

So this amplitude is the product of two amplitudes as measured in the the x’, y’, z’ frame. Note it’s symmetric: we may also write it as |lm’ = 0〉·Fl(r). We now need to sort of translate that into an amplitude as measured in the x, y, frame. To go from x, y, z to x’, y’, z’, we first rotated around the z-axis by the angle φ, and then rotated around the new y’-axis by the angle θ. Now, the order of rotation matters: you can easily check that by taking a non-symmetrical object in your hand and doing those rotations in the two different sequences: check what happens to the orientation of your object. Hence, to go back we should first rotate about the y’-axis by the angle −θ, so our z’-axis folds into the old z-axis, and then rotate about the z-axis by the angle −φ.

Now, we will denote the transformation matrices that correspond to these rotations as Ry’(−θ) and Rz(−φ) respectively. These transformation matrices are complicated beasts. They are surely not the easy rotation matrices that you can use for the coordinates themselves. You can click this link to see how they look like for = 1. For larger l, there are other formulas, which Feynman derives in another chapter of his Lectures on quantum mechanics. But let’s move on. Here’s the grand result:

The amplitude for our wavefunction ψl,m(r) – which denotes the amplitude for (1) the atom to be in the state that’s characterized by the quantum numbers and m and – let’s not forget – (2) find the electron at r – note the bold type: = (x, y, z) – would be equal to:

ψl,m(r) = 〈l, m|Rz(−φ) Ry’(−θ)|lm’ = 0〉·Fl(r)

Well… Hmm… Maybe. […] That’s not how Feynman writes it. He writes it as follows:

ψl,m(r) = 〈l, 0|Ry(θ) Rz(φ)|lm〉·Fl(r)

I am not quite sure what I did wrong. Perhaps the two expressions are equivalent. Or perhaps – is it possible at all? – Feynman made a mistake? I’ll find out. [P.S: I re-visited this point in the meanwhile: see the P.S. to this post. :-)] The point to note is that we have some combined rotation matrix Ry(θ) Rz(φ). The elements of this matrix are algebraic functions of θ and φ, which we will write as Yl,m(θ, φ), so we write:

a·Yl,m(θ, φ) = 〈l, 0|Ry(θ) Rz(φ)|lm

Or a·Yl,m(θ, φ) = 〈l, m|Rz(−φ) Ry’(−θ)|lm’ = 0〉, if Feynman would have it wrong and my line of reasoning above would be correct – which is obviously not so likely. Hence, the ψl,m(r) function is now written as:

ψl,m(r) = a·Yl,m(θ, φ)·Fl(r)

The coefficient is, as usual, a normalization coefficient so as to make sure the surface under the probability density function is 1. As mentioned above, we get these Yl,m(θ, φ) functions from combining those rotation matrices. For = 1, and = -1, 0, +1, they are:spherical harmonics A more complete table is given below:spherical harmonics 2So, yes, we’re done. Those equations above give us those wonderful shapes for the electron orbitals, as illustrated below (credit for the illustration goes to an interesting site of the UC Davis school).electron orbitalsBut… Hey! Wait a moment! We only have these Yl,m(θ, φ) functions here. What about Fl(r)?

You’re right. We’re not quite there yet, because we don’t have a functional form for Fl(r). Not yet, that is. Unfortunately, that derivation is another lengthy development – and that derivation actually is just tedious math only. Hence, I will refer you to Feynman for that. 🙂 Let me just insert one more thing before giving you The Grand Equation, and that’s a explanation of how we get those nice graphs. They are so-called polar graphs. There is a nice and easy article on them on the website of the University of Illinois, but I’ll summarize it for you. Polar graphs use a polar coordinate grid, as opposed to the Cartesian (or rectangular) coordinate grid that we’re used to. It’s shown below. 

The origin is now referred to as the pole – like in North or South Pole indeed. 🙂 The straight lines from the pole (like the diagonals, for example, or the axes themselves, or any line in-between) measure the distance from the pole which, in this case, goes from 0 to 10, and we can connect the equidistant points by a series of circles – as shown in the illustration also. These lines from the pole are defined by some angle – which we’ll write as θ to make things easy 🙂 – which just goes from 0 to 2π = 0 and then round and round and round again. The rest is simple: you’re just going to graph a function, or an equation – just like you’d graph y = ax + b in the Cartesian plane – but it’s going to be a polar equation. Referring back to our p-orbitals, we’ll want to graph the cos2θ = ρ equation, for example, because that’s going to show us the shape of that probability density function for = 1 and = 0. So our graph is going to connect the (θ, ρ) points for which the angle (θ) and the distance from the pole (ρ) satisfies the cos2θ = ρ equation. There is a really nice widget on the WolframAlpha site that produces those graphs for you. I used it to produce the graph below, which shows the 1.1547·cos2θ = ρ graph (the 1.1547 coefficient is the normalization coefficient a). Now, you’ll wonder why this is a curve, or a curved line. That widget even calculates its length: it’s about 6.374743 units long. So why don’t we have a surface or a volume here? We didn’t specify any value for ρ, did we? No, we didn’t. The widget calculates those values from the equation. So… Yes. It’s a valid question: where’s the distribution? We were talking about some electron cloud or something, right?

Right. To get that cloud – those probability densities really – we need that Fl(r) function. Our cos2θ = ρ is, once again, just some kind of envelope function: it marks a space but doesn’t fill it, so to speak. 🙂 In fact, I should now give you the complete description, which has all of the possible states of the hydrogen atom – everything! No separate pieces anymore. Here it is. It also includes n. It’s The Grand Equation:The ak coefficients in the formula for ρFn,l(ρ) are the solutions to the equation below, which I copied from Feynman’s text on it all. I’ll also refer you to the same text to see how you actually get solutions out of it, and what they then actually represent. 🙂We’re done. Finally!

I hope you enjoyed this. Look at what we’ve achieved. We had this differential equation (a simple diffusion equation, really, albeit in the complex space), and then we have a central Coulomb field and the rather simple concept of quantized (i.e. non-continuous or discrete) angular momentum. Now see what magic comes out of it! We literally constructed the atomic structure out of it, and it’s all wonderfully elegant and beautiful.

Now think that’s amazing, and if you’re reading this, then I am sure you’ll find it as amazing as I do.

Note: I did a better job in explaining the intricacies of actually representing those orbitals in a later post. I recommend you have a look at it by clicking the link here.

Post scriptum on the transformation matrices:

You must find the explanation for that 〈l, 0|Ry(θ) Rz(φ)|lm〉·Fl(r) product highly unsatisfactory, and it is. 🙂 I just wanted to make you think – rather than just superficially read through it. First note that Fl(r)·|lm’ = 0〉 is not a product of two amplitudes: it is the product of an amplitude with a state. A state is a vector in a rather special vector space – a Hilbert space (just a nice word to throw around, isn’t it?). The point is: a state vector is written as some linear combination of base states. Something inside of me tells me we may look at the three p-states as base states, but I need to look into that.

Let’s first calculate the Ry(θ) Rmatrix to see if we get those formulas for the angular dependence of the amplitudes. It’s the product of the Ry(θ) and Rmatrices, which I reproduce below.

Note that this product is non-commutative because… Well… Matrix products generally are non-commutative. 🙂 So… Well… There they are: the second row gives us those functions, so am wrong, obviously, and Dr. Feynman is right. Of course, he is. He is always right – especially because his Lectures have gone through so many revised editions that all errors must be out by now. 🙂

However, let me – just for fun – also calculate my Rz(−φ) Ry’(−θ) product. I can do so in two steps: first I calculate Rz(φ) Ry’(θ), and then I substitute the angles φ and θ for –φ and –θ, remembering that cos(–α) = cos(α) and sin(–α) = –sin(α). I might have made a mistake, but I got this:The functions look the same but… Well… No. The eiφ and eiφ are in the wrong place (it’s just one minus sign – but it’s crucially different). And then these functions should not be in a column. That doesn’t make sense when you write it all out. So Feynman’s expression is, of course, fully correct. But so how do we interpret that 〈l, 0|Ry(θ) Rz(φ)|lm〉 expression then? This amplitude probably answers the following question:

Given that our atom is in the |lm〉 state, what is the amplitude for it to be in the 〈l, 0| state in the x’, y’, z’ frame?

That makes sense – because we did start out with the assumption that our atom was in the the |lm〉 state, so… Yes. Think about it some more and you’ll see it all makes sense: we can – and should – multiply this amplitude with the Fl(r) amplitude.

OK. Now we’re really done with this. 🙂

Note: As for the 〈 | and  | 〉 symbols to denote a state, note that there’s not much difference: both are state vectors, but a state vector that’s written as an end state – so that’s like 〈 Φ | – is a 1×3 vector (so that’s a column vector), while a vector written as | Φ 〉 is a 3×1 vector (so that’s a row vector). So that’s why 〈l, 0|Ry(θ) Rz(φ)|lm〉 does give us some number. We’ve got a (1×3)·(3×3)·(3×1) matrix product here – but so it gives us what we want: a 1×1 amplitude. 🙂

The state(s) of a photon

While hurrying to try to understand the things I wanted to understand most – like Schrödinger’s equation and, equally important, its solutions explaining the weird shapes of electron orbitals – I skipped some interesting bits and pieces. Worse, I skipped two or three of Feynman’s Lectures on quantum mechanics entirely. These include Chapter 17 – on symmetry and conservation laws – and Chapter 18 – on angular momentum. With the benefit of hindsight, that was not the right thing to do. If anything, doing all of the Lectures would, at the very least, ensure I would have more than an ephemeral grasp of it all. So… In this and the next post, I want to tidy up and go over everything I skipped so far. 🙂

We’ve written a lot on how quantum mechanics applies to both bosons as well as fermions. For example, we pointed out – in very much detail – that the mathematical structure of the electromagnetic wave – light! 🙂 – is quite similar to that of the ubiquitous wavefunction. Equally fundamental – if not more – is the fact that light also arrives in lumps – little light-particles which we call photons. It’s the photoelectric effect, which Einstein explained in 1905 by… Well… By telling us that light consists of quanta – photons – whose energy must be high enough so as to be able to dislodge an electron. It’s what got him his Nobel Prize. [Einstein never got a Nobel Prize for his relativity theory, which is – arguably – at least as important. There’s a lot of controversy around that but, in any case, that’s history.]

So it shouldn’t surprise you that there’s an equivalent to the spin of an electron. With spin, we refer to the angular momentum of a quantum-mechanical system – an atom, a nucleus, an electron, whatever – which, as you know, can only be one of a set of discrete values when measured along some direction, which we usually refer to as the z-direction. More formally, we write that the z-component of the angular moment J is equal to

Jz = j·ħ, (j-1)·ħ, (j-2)·ħ, …, -(j-2)·ħ, -(j-1)·ħ, –j·ħ

The in this expression is the so-called spin of the system. For an electron, it’s equal to ±1/2, which we referred to as “up” and “down” states respectively because of obvious reasons: one state points upwards – more or less, that is (we know the angular momentum will actually precess around the direction of the magnetic field) – while the other points downwards.

We also know that the magnetic energy of an electron in a (weak) magnetic field – which, as you know, we conveniently assume to be pointing in the same z-direction, so B= B – will be equal to:

Umag = g·μz·B·= ± 2·μz·B·(1/2) = ± μz·B = ± B·(qe·ħ)/(2m)

In short, the magnetic energy is proportional to the magnetic field, and the constant of proportionality is the so-called Bohr magneton qe·ħ/2m. So far, so good. What’s the analog for a photon?

Well… Let’s first discuss the equivalent of a Stern-Gerlach apparatus for photons. That would be a polarizing material, like a piece of calcite, for example. Now, it is, unfortunately, much more difficult to explain how a polarizing material works than to explain how a Stern-Gerlach apparatus works. [If you thought the workings of that (hypothetical) Stern-Gerlach filter were difficult to understand, think again.] We actually have different types of polarizers – some complicated, some easy. We’ll take the easy ones: linear ones. In addition, the phenomenon of polarization itself is a bit intricate. The phenomenon is well described in Chapter 33 of Feynman’s first Volume of Lectures, out of which I copied the two illustrations below the next paragraph.

Of course, to make sure you think about whatever is that you’re reading, Feynman now chooses the z-direction such that it coincides with the direction of propagation of the electromagnetic radiation. So it’s now the x– and y-direction that we’re looking at. Not the z-direction any more. As usual, we forget about the magnetic field vector B and so we think of the oscillating electric field vector E only. Why can we forget about B? Well… If we have E, we know B. Full stop. As you know, I think B is pretty essential in the analysis too but… Well… You’ll see all textbooks on physics quickly forget about B when describing light. I don’t want to do that, but… Well… I need to move on. [I’ll come back to the matter – sideways – at the end of this post. :-)]

So we know the electric field vector E may oscillate in a plane (so that’s up and down and back again) but – interestingly enough – its direction may also rotate around the z-axis (again, remember the z-axis is the direction of propagation). Why? Well… Because E has an x– and a y-component (no z-component!), and these two components may oscillate in phase or out of phase, and so all of the combinations below are possible.Linear polarizationElliptical polarizationTo make a long story short, light comes in two varieties: linearly polarized and elliptically polarized. Of course, elliptically may be circularly – if you’re lucky! 🙂

Now, a (linear) polarizer has an optical axis, and only light whose E vector is oscillating along that axis will go through. […] OK. That’s not true: the component along the optical axis of some E pointing in some other direction will go through too! I’ll show how that works in a moment. But so all the rest is absorbed, and the absorbed energy just heats up the polarizer (which, of course, then radiates heat back out).

In any case, if the optical axis happens to be our x-axis, then we know that the light that comes through will be x-polarized, so that corresponds to the rather peculiar Ex = 1 and Ey = 0 notation. [This notation refers to coefficients we’ll use later to resolve states into base states – but don’t worry about it now.] Needless to say, you shouldn’t confuse the electric field vector E with the energy of our photon, which we denote as E. No bold letter here. No subscript. 🙂

Pfff… This introduction is becoming way too long. What about our photon? We want to talk about one photon only and we’ve already written over a page and haven’t started yet. 🙂

Well… First, we must note that we’ll assume the light is perfectly monochromatic, so all photons will have an energy that’s equal to E = h·f, so the energy is proportional to the frequency of our light, and the constant of proportionality is Planck’s constant. That’s Einstein’s relation, not a de Broglie relation. Just remember: we’re talking definite energy states here.

Second – and much more importantly – we may define two base states for our photon, |x〉 and |y〉 respectively, which correspond to the classical linear x– and y-polarization. So a photon can be in state |x〉 or |y〉 but, as usual, it is much more likely to be in some state that is some linear combination of these two base states.

OK. Now we can start playing with these ideas. Imagine a polarizer – or polaroid, as Feynman calls it – whose optical axis is tilted – say, it’s at an angle θ from the x-axis, as shown below. Classically, the light that comes through will be polarized in the x’-direction, which we associate with that angle θ. So we say the photons will be in the |x‘〉 state. linear combinationSo far, so good. But what happens if we have two polarizers, set up as shown below, with the optical axis of the first one at an angle θ, which is, say, equal to 30°? Will any light get through?two polarizers

Well? No answer? […] Think about it. What happens classically? […] No answer? Let me tell you. In a classical analysis, we’d say that only the x-component of the light that comes through the first polarizer would get through the second one. Huh? Yes. It is not all or nothing in a classical analysis. This is where the magnitude of E comes in, which we’ll write as E0, so as to not confuse it with the energy E. [I know you’ll confuse it anyway but… Well… I need to move on or I won’t get anywhere with this story.] So if E0 is the (maximum) magnitude (or amplitude – in the classical sense of the word, that is) of E as the light leaves the first polarizer, then its x-component will be equal to E0·cosθ. [I don’t need to make a drawing here, do I?] Of course, you know that the intensity of the light will be proportional to the square of the (maximum) field, which is equal to E02·cos2θ = 0.75·E02 for θ = 30°.

So our classical theory says that only 3/4 of the energy that we were sending in will get through. The rest (1/4) will be absorbed. So how do we model that quantum-mechanically? It’s amazingly simple. We’ve already associated the |x‘〉 state with the photons coming out of the first polaroid, and so now we’ll just say that this |x‘〉 state is equal to the following linear combination of the |x〉 and |y〉 base states:

|x‘〉 = cosθ·|x〉 + sinθ·|y

Huh? Yes. As Feynman puts it, we should think our |x‘〉 beam of photons can, somehow, be resolved into |x〉 and |y〉 beams. Of course, we’re talking amplitudes here, so we’re talking 〈x|x‘〉 and 〈y|x‘〉 amplitudes here, and the absolute square of those amplitudes will give us the probability that a photon in the |x‘〉 state gets into the |x〉 and |y〉 state respectively. So how do we calculate that? Well… If |x‘〉 = cosθ·|x〉 + sinθ·|y〉, then we can obviously write the following:

x|x‘〉 = cosθ·〈x|x〉 + sinθ·〈x|y

Now, we know that 〈x|y〉 = 0, because |x〉 and |y〉 are base states. Because of the same reason, 〈x|x〉 = 1. That’s just an implication of the definition of base states: 〈i|j〉 = δij. So we get:

x|x‘〉 = cosθ

Lo and behold! The absolute square of that is equal to cos2θ, so each of these photons have an (average) probability of 3/4 to get through. So if we were to have like 10 billion photons, then some 7.5 billion of them would get through. As these photons are all associated with a definite energy – and they go through as one whole, of course (no such thing as a 3/4 photon!) – we find that 3/4 of all of the energy goes through. The quantum-mechanical theory gives the same result as the classical theory – as it should, in this case at least!

Now that’s all good for linear polarization. What about elliptical or circular polarization? Hmm… That’s a bit more complicated, but equally feasible. If we denote the state of a photon with a right-hand circular polarization (RHC) as |R〉 and, likewise, the state of a photon with a left-hand circular polarization (LHC) as |L〉, then we can write these as the following linear combinations of our base states |x〉 and |y〉:linear combination RHC and LHCThat’s where those coefficients under illustrations (c) and (g) come in, although I think they’ve got the sign of i (the imaginary unit) wrong. 🙂 So how does it work? Well… That 1/√2 factor is – obviously – just there to make sure everything’s normalized, so all probabilities over all states add up to 1. So that is taken care of and now we just need to explain how and why we’re adding |x〉 and |y〉. For |R〉, the amplitudes must be the same but with a phase difference of 90°. That corresponds to the sine and cosine function, which are the same except for a phase difference of π/2 (90°), indeed: sin(φ + π/2) = cosφ. Now, a phase shift of 90° corresponds to a multiplication with the imaginary unit i. Indeed, ei·π/2 and, therefore, it is obvious that ei·π/2·ei·φ = ei·(φ + π/2).

Of course, if we can write RHC and LHC states as a linear combination of the base states |x〉 and |y〉, then you’ll believe me if I say that we can write any polarization state – including non-circular elliptical ones – as a linear combination of these base states. Now, there are two or three other things I’d like to point out here:

1. The RHC and LHC states can be used as base states themselves – so they satisfy all of the conditions for a set of base states. Indeed, it’s easy to add and then subtract the two equations above to get the following:new base setAs an exercise, you should verify the right and left polarization states effectively satisfy the conditions for a set of base states.

2. We can also rotate the xy-plane around the z-axis (as mentioned, that’s the direction of propagation of our beam) and use the resulting |x‘〉 and |y‘〉 states as base states. In short, we can effectively, as Feynman puts it, “You can resolve light into x– and y– polarizations, or into x’– and y’-polarizations, or into right and left polarizations as a basis.” These pairs are always orthogonal and also satisfy the other conditions we’d impose on a set of base states.

3. The last point I want to make here is much more enigmatic but, as far as I am concerned – by far – the most interesting of all of Feynman’s Lecture on this topic. It’s actually just a footnote, but I am very excited about it. So… Well… What is it?

Well… Feynman does the calculations to show how a circularly polarized photon looks like when we rotate the coordinates around the z-axis, and shows the phase of the right and left polarized states effectively keeps track of the x– and y-axes, so all of our “right-hand” rules don’t get lost somehow. He compares this analysis to an analysis he actually did – in a much earlier Lecture (in Chapter 5) – for spin-one particles. But, of course, here we’ve been analyzing the photon as a two-state system, right?

So… Well… Don’t we have a contradiction here? If photons are spin-one particles, then they’re supposed to be analyzed in terms of three base states, right? Well… I guess so… But then Feynman adds a footnote – with very important remark:

“The photon is a spin-one particle which has, however, no ‘zero’-state.”

Why I am noting that? Because it confirms my theory about photons – force-particles – being different from matter-particles not only because of the different rules for adding amplitudes, but also because we get two wavefunctions for the price of one and, therefore, twice the energy for every oscillation! And so we’ll also have a distance of two Planck units between the equivalent of the “up” and “down” states of the photon, rather than one Planck unit, like what we have for the angular momentum for an electron. 

I described the gist of my argument in my e-book, which you’ll find under another tab of this blog, and so I’ll refer you there. However, in case you’re interested, the summary of the summary is as follows:

  1. We can think of a photon having some energy that’s equal to E = p = m (assuming we choose our time and distance units such that c = 1), but that energy would be split up in an electric and a magnetic wavefunction respectively: ψand ψB.
  2. Now, Schrödinger’s equation would then apply to both wavefunctions, but the E, p and m in those two wavefunctions are the same and not the same: their numerical value is the same (pE =EE = mE = pB =EB = mB), but they’re conceptually different. [They must be: I showed that, if they aren’t, then we get a phase and group velocity for the wave that doesn’t make sense.]

It is then easy to show that – using the B = i·E relation between the magnetic and the electric field vectors – we find a composite wavefunction for our photon which we can write as:

E + B = ψ+ ψ= E + i·E = √2·ei(p·x/2 − E·t/2 + π/4) = √2·ei(π/4)·ei(p·x/2 − E·t/2) = √2·ei(π/4)·E

The whole thing then becomes:

ψ = ψ+ ψ= √2·ei(p·x/2 − E·t/2 + π/4) = √2·ei(π/4)·ei(p·x/2 − E·t/2) 

So we’ve got a √2 factor here in front of our combined wavefunction for our photon which, knowing that the energy is proportional to the square of the amplitude gives us twice the energy we’d associate with a regular amplitude… [With “regular”, I mean the wavefunction for matter-particles – fermions, that is.] So… Well… That little footnote of Feynman seems to confirm I really am on to something. Nice! Very nice, actually! 🙂

Davidson’s function

This post has got nothing to do with quantum mechanics. It’s just… Well… My son – who’s preparing for his entrance examinations for engineering studies – sent me a message yesterday asking me to quickly explain Davidson’s function – as he has to do some presentation on it as part of a class assignment. As I am an economist – and Davidson’s function is used in transport economics – he thought I would be able to help him out quickly, and I was. So I just thought it might be interesting to quickly jot down my explanation as a post in this blog. It won’t help you with quantum mechanics but, if anything, it may help you think about functional forms and some related topics.

In his message, he sent me the function – copied below – and some definitions of the variables which he got from some software package he had seen or used – at least that’s what he told me. 🙂Davidson functionSo… This function tells us that the dependent variable is the travel time t, and that it is seen as a function of some independent variable x and some parameters t0, c and ε. My son defined the variable x as the flow (of vehicles) on the road, and c as the capacity of the road. To be precise, he wrote the formula that was to be used for c as follows:capacityWhat about a formula for x? Well… He said that was the actual flow of vehicles, but he had no formula for it. As for t0, that was the travel time “at free speed.” Finally, he said ε was a “paramètre de sensibilité de congestion.” Sorry for the French, but that’s the language of his school, which is located in some town in southern Belgium. In English, we might translate it as a congestion sensitivity coefficient. And so that’s what he struggled most with – or so he said.

So that got us started. I immediately told him that, if you write something like c − x, then you’d better make sure and have the same physical dimension. The formula above tells us that is the number of vehicles that you can park on that road. Bumper to bumper. So I told him that’s a rather weird definition of capacity. It’s definitely not the dimension of flow: the flow should be some number per second or – much more likely in transport economics – per minute or per hour. So I told him that he should double-check those definitions of x and c, and that I’d get back to him to explain the formula itself after I had googled and read some articles on it. So I did that, and so here’s the full explanation I gave him.

While there’s some pretty awesome theory behind (queuing theory and all that), which transportation gurus take very seriously – see, for example, the papers written by Rahmi Akçelik – a quick look at it all reveals that Davidson’s function is, essentially, just a specific functional form that we impose on some real-life problem. So I’d call it an empirical function: there’s some theory behind, but it’s more based on experience than pure theory. Of course, sound logic is – or should be – applied to both empirical as well as to purely theoretical functions, but… Well… It’s a different approach than, say, modeling the dynamics of quantum-mechanical state changes. 🙂 Just note, for example, that we might just as well have tried something else – some exponential function. Something like this, for example:function alternativeDavidson’s function is, quite simply, just nicer and easier than the one above, because the function above is not linear. It could be quadratic (β = 2), or whatever, but surely not linear. In contrast, Davidson’s function is linear and, therefore, easy to fit onto actual traffic data using the simplest of simple linear regression models – and, speaking from experience, most engineers and economists in a real-life job can barely handle even that! 🙂

So just look at that x/(cx) factor as measuring the congestion or saturation, somehow. We’ll denote it by s. If you can sort of accept that, then you’ll agree that Davidson’s function tells us that the extra time that’s needed to drive from some place a to some place b along our road will be directly proportional to:

  1. That congestion factor x/(cx), about which I’ll write more in a moment;
  2. The free-speed or free-flow travel time t0 – which I’ll call the free-flow travel time from now on, rather than the free-speed travel time, because there’s no such thing as free speed in reality: we have speed limits – or safety limits, or scared moms in the car, whatever – and, a more authoritative argument, the literature on Davidson’s function also talks about free flow rather than free speed;
  3. That epsilon factor (ε), which – of all the stuff I presented so far – mystified my son most.

So the formula for the extra travel time that’s needed is, obviously, equal to:function 1So we have a very simple linear functional form for the extra travel time, and we can easily estimate the actual value of our ε parameter using actual traffic data in a simple linear regression. The data analysis toolkit of MS Excel will do stuff like this – if you have the data, of course – so you don’t need a sophisticated statistical software package here.

So that’s it, really: Davidson’s function is, effectively, just nice and easy to work with. […] Well… […] Of course, we still need to define what x and c actually are. And what’s that so-called free flow (or free speed?) travel time? Well… The free-flow travel time is, obviously, the time you need to go from to at the free-flow speed. But what’s the free-flow speed? My friend’s Maserati is faster than my little Santro. 🙂 And are we allowed to go faster than the maximum authorized speed? Interesting questions.

So that’s where the analysis becomes interesting, and why we need better definitions of x and c. If is some density – what my son’s rather non-sensical formula seems to imply – we may want to express it per unit distance. Per kilometer, for example. So we should probably re-define c more simply: as the number of lanes divided by the average length of the vehicles that are using it. We get that by dividing the c above by the length of the road – so we divide the length of the road by the length of the road, which gives 1. 🙂 You may think that’s weird, because we get something like 3/5 = 0.6… So… What? Well… Yes. 0.6 vehicles per meter, so that’s 600 vehicles per kilometer! Does that sound OK? I think it does. So let’s express that capacity (c) as a maximum density – for the time being, at least.

Now, none of those cars can move, of course: they are all standing still. Bumper to bumper. It’s only when we decrease the density that they’re able to move. In fact, you can – and should – visualize the process: the first car moves and opens a space of, say, one or two meter, and then the second one, and so and so on – till all cars are moving with a few meter in-between them. So the density will obviously decrease and, as a result, we’re getting some flow of vehicles here. If there’s three meter between them, for example, then the density goes down to 3/8 vehicles per meter, so that’s 375 vehicles per kilometer. Still a lot, and you’ll have to agree that – with only 3 meters between them – they’ll probably only move very slowly!

You get the idea. We can now define as a density too – some density that is smaller than the maximum density c. Then that x/(cx) factor – measuring the saturation – obviously makes a lot of sense. The graph below shows how it looks like for c = 5. [The value of 5 is just random, and its order of magnitude doesn’t matter either: we can always re-scale from m to km, or from seconds to minutes and what have you. So don’t worry about it.] Look at this example: when is small – like 1 or 2 only – then x/(5−x) doesn’t increase all that much. So that means we add little to the travel time. Conversely, when x approaches c = 5 – so that’s the limit (as you can see, the = 5 line is a (vertical) asymptote of the function) – then the travel time becomes huge and starts approaching infinity. So… Well… Yes. That’s when all cars are standing still – bumper to bumper. asymptoteBut so what’s the free-flow speed? Is it the maximum speed of my friend’s Maserati -which is like 275 km/h? Well… I don’t think my friend ever drove that fast, so probably not. What else? Think about it. What should we choose here? The obvious choice is the speed limit: 120 km/h, or 90 km/h, or 60 km/h – or whatever. Why? Because you don’t want a ticket, I guess… In any case, let’s analyze that question later. Let’s first look at something else.

Of course, you’ll want to keep some distance between you and the car in front of you when driving at relatively high speeds, and that’s the crux of the analysis really. You may or, more likely, you may not remember that your driving instructor told you to always measure the safety distance between you and the car(s) in front in seconds rather than in meter. In Belgium, we’re told to stay two seconds away from the car in front of us. So when it passes a light pole, we’ll count “twenty-one, twenty-two” and… Well… If we pass that same light pole while we’re still counting those two seconds, then we’d better keep some more distance. It’s got to do with reaction time: when the car in front of you slams the brakes, you need some time to react, and then that car might also have better brakes than yours, so you want to build in some extra safety margin in case you don’t slow down as fast as the car in front of you. So that two-seconds rule is not about the breaking distance really – or not about the breaking distance alone. No. It’s more about the reaction time. In any case, the point is that you’ll want to measure the safety distance in time rather than in meterCapito? OK… Onwards…

Now, 120 km/h amounts to 120,000/3,600 = 33.333 meter per second. So the safety distance here is almost 67 meter! If the maximum authorized velocity is only 90 km/h, then the safety distance shrinks to 2 × (90,000/3,600) = 50 meter. For a maximum authorized velocity of 60 km/h, the safety distance would be equal to 33.333 meter. Both are much larger distances than the average length of the vehicles and, hence, it’s basically the safety distance – not the length of the vehicle – that we need to consider! Let’s quickly calculate the related densities:

  • For a three-lane highway, with all vehicles traveling at 120 km/h and keeping their safety distance, the density will be equal to 3·1,000/66.666… = 45 vehicles per kilometer of highway, so that’s 15 vehicles per lane.
  • If the travel speed is 90 km/h, then the density will be equal to 60 vehicles per km (20 vehicles per lane).
  • Finally, at 60 km/h, the density will be 90 vehicles per km (30 vehicles per lane).

Note that our two-seconds rule implies a linear relation between the safety distance and the maximum authorized speed. You can also see that the relation between the density and the maximum authorized speed is inversely proportional: if we halve the speed, the density doubles.

Now, you can easily come up with some more formulas, and play around a bit. For example, if we denote the security distance by d, and the mentioned two seconds as td – so that’s the time (t) that defines the security distance d – then d is, logically, equal to: d = td∙vmax. But rather than trying to find more formulas and play with them, let’s think about that concept of flow now. If we would want to define the capacity – or the actual flow – in terms of the number of vehicles that are passing along any point along this highway, how should we calculate that?

Well… The flow is the number of vehicles that will pass us in one hour, right? So if vmax is 120 km/h, then – assuming full capacity – all the vehicles on the next 120 km of highway will all pass us, right? So that makes 45 vehicles per km times 120 km = 5,400 vehicles – per hour, of course. Hence, the flow is just the product of the density times the speed.

Now, look at this: if vmax is equal to 90 km/h, then we’ll have 60 vehicles per km times 90 km = … Well… It’s – interestingly enough – the same number: 5,400 vehicles per hour. Let’s calculate for vmax = 60 km/h… Security distance is 33.333 meter, so we can have 90 vehicles on each km of highway which means that, over one hour, 90 times 60 = 5,400 vehicles will pass us! It’s, once again, the same number: 5,400! Now that’s a very interesting conclusion. Let me highlight it:

If we assume the vehicles will keep some fixed time distance between them (e.g. two seconds), then the capacity of our highway – expressed as some number of vehicles passing along it per time unit – does not depend on the velocity.

So the capacity – expressed as a flow rather than as a density – is just a fixed number: vehicles per hour. The density affects only the (average) speed of all those vehicles. Hence, increasing densities are associated with lower speeds, and higher travel times, but they don’t change the capacity.

It’s really a rather remarkable conclusion, even if the relation between the density and the flow is easily understood – both mathematically and, more importantly, intuitively. For example, if the density goes down to 60 vehicles per km of highway, then they will only be able to move at a speed of 90 km/h, but we’ll still have that flow of 5,400 vehicles per hour – which we can look at as the capacity but expressed as some flow rather than as a density. Lower densities allow for even higher speeds: we calculated above that a density of 45 vehicles per km would allow them to drive at a maximum speed of 120 km/h, so travel time would be reduced even more, but we’d still have 5,400 vehicles per hour! So… Well… Yes. It all makes sense.

Now what happens if the density is even lower, so we could – theoretically – drive safely, or not so safely, at some speed that’s way above the speed limit? If we have enough cars – say 30 vehicles per km, but all driving more than 120 km/h, while respecting the two-seconds rule – we’d still have the same flow: 5,400 vehicles per hour. And travel time would go down. But so we can think of lower densities and higher speeds but, again, there’s got to be some limit here – a speed limit, safety considerations, a limit to what our engine or our car can do, and, finally, there’s the speed of light too. 🙂 I am just joking, of course, but I hope you see the point. At some point, it doesn’t matter whether or not the density goes down even further: the travel time should hit some minimum. And it’s that minimum – the lowest possible travel time – that you’d probably like to define as t0.

As mentioned, the minimum travel time is associated with some maximum speed, and – after some consideration of the possible candidates for the maximum speed – you’ll agree the speed limit is a better candidate than the 275 km/h limit of my friends’ Maserati Quattroporte. Likewise, you would probably also like to define x0 as the (maximum) density at the speed limit.

What we’re saying here is that – in theory at least – our t = t(x) function should start with a linear section, between x = 0 and x = x0. That linear section defines a density 0 < x < x0 which is compatible with us driving at the speed limit – say, 120 km/h – and, hence, with us only needing the time t = t0 to arrive at our destination. Only when x becomes larger than x0, we’ve got to reduce speed – below the speed limit (say, 120 km/h) – to keep the flow going while keeping an appropriate safety distance. A reduction of speed implies a increase in travel time, of course. So that’s what’s illustrated in the graph below.

graphTo be specific, if the speed limit is 120 km/h, then – assuming you don’t want to be caught speeding – the minimum travel time will always be equal to 30 seconds per km, even if you’re alone on the highway. Now, as long as the density is less than 45 vehicles per km, you can keep that travel time the same, because you can do your 120 km/h while keeping the safety distance. But if the density increases, above 45 vehicles per km, then stuff starts slowing down because everyone is uncomfortable with the shorter distance between them and the car in front. As the density goes up even more – say, to 60 vehicles per km – we can only do 90 km/h, and so the travel time will then be equal to 40 seconds per km. And then it goes to 90 vehicles per km, so speed slows down to 60 km/h, and so that’s a travel time of 60 seconds per km. Of course, you’re smart – very smart – and so you’ll immediately say this implies that the second section of our graph should be linear too, like this:

graph 2You’re right. But then… Well… That doesn’t work with our limit for x, which is c. As I pointed out, is an absolute maximum density: you just can’t park any more cars on that highway – unless you fold them up or so. 🙂 So what’s the conclusion? Well… We may think of the Davidson function as a primitive combination of both shapes, as shown below.

graph 3

I call it a primitive approximation, because that Davidson function (so that’s the green smooth curve above) is not a precise (linear or non-linear) combination of the two functions we presented (I am talking about the blue broken line and the smooth red curve here). It’s just… Well… Some primitive approximation. 🙂 Now you can write some very complicated papers – as other authors do – to sort of try to explain this shape, but you’ll find yourself fiddling with variable time distance rules and other hypotheses that may or may not make sense. In short, you’re likely to introduce other logical inconsistencies when trying to refine the model. So my advice is to just accept the Davidson’s function as some easy empirical fit to some real-life situation, and think of what the parameters actually do – mathematically speaking, that is. How do they change the shape of our graph?

So we’re now ready to explain that epsilon factor (ε) by looking at what it does, indeed. Please try an online graphing tool with a slider (e.g. https://www.desmos.com/calculator) – just type something like a + bx in the function box, and you’ll see the sliders appear – so you can see how the function changes for different parameter values. The two graphs below, for example, which I made using that graphing tool, show you the function t = 2 + 2∙ε∙x/(10−x) for ε = 0.5 and ε = 10 respectively. As you can see, both functions start at t = 2 and have the same asymptote at x = c = 10. However, you’ll agree that they look very different – and that’s because of the value of the ε parameter. For ε = 0.5, the travel time does not increase all that much – initially at least. Indeed, as you can see, t is equal to 3 if the density is half of the capacity (t = 3 for x = 5 = c/2). In contrast, for ε = 10, we have immediate saturation, so to speak: travel time goes through the roof almost immediately! For example, for = 3, t ≈ 10.6, so while the density is less than a third of the capacity, the associated travel time is already more than five times the free-flow travel time!

Now I have a tricky question for you: does it make sense to allow ε to take on values larger than one? Think about it. 🙂 In any case, now you’ve seen what the ε factor does from a math point of view. So… Well… I’ll conclude here by just noting that it does, indeed, make sense to refer to ε as a “paramètre de sensibilité de congestion”, because that’s what it is: a congestion sensitivity coefficient. Indeed, it’s not the congestion or saturation parameter itself (that’s a term we should reserve term for the x/(cx) factor), but a congestion sensitivity coefficient alright!

Of course, you will still want some theoretical interpretation. Well… To be honest, I can’t give you one. I don’t want to get lost in all of those theoretical excursions on Davidson’s function, because… Well… It’s no use. That ε is just what it is: it’s a proportionality coefficient that we are imposing upon our functional form for our travel-time function. You can sum it up as follows:

If x/(cx) is the congestion parameter (or variable, I should say), then it goes from 0 to ∞ (infinity) when the traffic flow (x) goes x = 0 to x = (full capacity). So, yes, we can call the x/(cx) factor the congestion or saturation variable and write it as s = x/(cx). And then we can refer to ε as the “paramètre de sensibilité de congestion”, because it is a measure not of the congestion itself, but of the sensitivity of the travel time to the congestion.

If you’d absolutely want some mathematical formula for it, then you could use this one, which you get from re-writing Δt = t0·ε·s as Δt/t0 = ε·s:

∂(Δt/t0)/∂s = ε

But… Frankly. You can stare at this formula for a long while – it’s a derivative alright, and you know what derivatives stand for – but you’ll probably learn nothing much from it. [Of course, please write me if you don’t agree, Vincent!] I just looked at those two graphs, and note how their form changes as a function of ε. Perhaps you have some brighter idea about it!

So… Well… I am done. You should now fully understand Davidson’s function. Let me write it down once more:

formula 3

Again, as mentioned, its main advantage is its linearity. Because of its linearity, it is easy to actually estimate the parameters: it’s just a simple linear regression – using actual travel times and actual congestion measurements – and so then we can estimate the value of ε and see if it works. Huh? How do we see if it works? Well… I told you already: when everything is said and done, Davidson’s function is just one of the many models for the actual reality, so it tries to explain how travel time increases because of congestion. There are other models, which come with other functions – but they are more complicated, and so are the functions that come with them (check out that paper from Rahmi Akçelik, for example). Only reality can tell us what model is the best fit to whatever it is that we’re trying to model. So that’s why I call Davidson’s function an empirical function, and so you should check it against reality. That’s when a statistical software package might be handy: it allows you to test the fit of various functional – linear and non-linear – forms against a real-life data set.

So that’s it. I tasked my son to go through this post and correct any errors – only typos, I hope! – I may have made. I hope he’ll enjoy this little exercise. 🙂

Comments on the MIT’s Stern-Gerlach lab experiment

In my previous post, I noted that I’d go through the MIT’s documentation on the Stern-Gerlach experiment that their undergrad students have to do, because we should now – after 175 posts on quantum physics 🙂 – be ready to fully understand what is said in there. So this post is just going to be a list of comments. I’ll organize it section by section.

Theory of atomic beam experiments

The theory is known – and then it isn’t, of course. The key idea is that individual atoms behave like little magnets. Why? In the simplest and most naive of models, it’s because the electrons somehow circle around the nucleus. You’ve seen the illustration below before. Note that current is, by convention, the flow of positive charge, which is, of course, opposite to the electron flow. You can check the direction by applying the right-hand rule: if you curl the fingers of your right hand in the direction of the current in the loop (so that’s opposite to v), your thumb will point in the direction of the magnetic moment (μ).orbital-angular-momentumSo the electron orbit – in whatever way we’d want to visualize it – gives us L, which we refer to as the orbital angular momentum. We know the electron is also supposed to spin about its own axis – even if we know this planetary model of an electron isn’t quite correct. So that gives us a spin angular momentum S. In the so-called vector model of the atom, we simply add the two to get the total angular momentum J = L + S.

Of course, now you’ll say: only hydrogen has one electron, so how does it work with multiple electrons? Well… Then we have multiple orbital angular momentum li which are to be added to give a total orbital angular momentum L. Likewise, the electrons spins si can also be added to give some total spin angular momentum S. So we write:

J = L + S with L = Σi li and S = Σi si

Really? Well… If you’d google this to double-check – check the Wikipedia article on it, for example – then you’ll find this additivity property is valid only for relatively light atoms (Z ≤ 30) and only if any external magnetic field is weak enough. The way individual orbital and spin angular momenta have to be combined so as to arrive at some total L, S and J is referred to a coupling scheme: the additivity rule above is referred to as LS coupling, but one may also encounter LK coupling, or jj coupling, or other stuff. The US National Institute of Standards and Technology (NIST) has a nice article on all these models – but we need to move on here. Just note that we do assume the LS coupling scheme applies to our potassium beam – because its atomic number (Z) is 19, and the external magnetic field is assumed to be weak enough.

The vector model of the atom describes the atom using angular momentum vectors. Of course, we know that a magnetic field will cause our atomic magnet to precess – rather than line up. At this point, the classical analogy between a spinning top – or a gyroscope – and our atomic magnet becomes quite problematic. First, think about the implications for L and S when assuming,  as we usually do, that J precesses nicely about an axis that is parallel to the magnetic field – as shown in the illustration below, which I took from Feynman’s treatment of the matterprecessionIf J is the sum of two other vectors L and S, then this has rather weird implications for the precession of L and S, as shown in the illustration below – which I took from the Wikipedia article on LS coupling. Think about: if L and S are independent, then the axis of precession for these two vectors should be just the same as for J, right? So their axis of precession should also be parallel to the magnetic field (B), so that’s the direction of the Jz component, which is just the z-axis of our reference frame here.375px-ls_couplingMore importantly, our classical model also gets into trouble when actually measuring the magnitude of Jz: repeated measurements will not yield some randomly distributed continuous variable, as one would classically expect. No. In fact, that’s what this experiment is all about: it shows that Jz will take only certain quantized values. That is what is shown in the illustration below (which once again assumes the magnetic field (B) is along the z-axis). vector-modelI copied the illustration above from the HyperPhysics site, because I found it to be enlightening and intriguing at the same time. First, it also shows this rather weird implication of the vector model: if J continually changes direction because of its precession in a weak magnetic field, then L and S must, obviously, also continually change direction. However, this illustration is even more intriguing than the Wikipedia illustration because it assumes the axis of precession of L and S and L actually the same!

So what’s going on here? To better understand what’s going on, I started to read the whole HyperPhysics article on the vector model, which also includes the illustration below, with the following comments: “When orbital angular momentum L and electron spin S are combined to produce the total angular momentum of an atomic electron, the combination process can be visualized in terms of a vector model. Both the orbital and spin angular momenta are seen as precessing about the direction of the total angular momentum J. This diagram can be seen as describing a single electron, or multiple electrons for which the spin and orbital angular momenta have been combined to produce composite angular momenta S and L respectively. In so doing, one has made assumptions about the coupling of the angular momenta which are described by the LS coupling scheme which is appropriate for light atoms with relatively small external magnetic fields.”vector-model-2Hmm… What about those illustrations on the right-hand side – with the vector sums and those values for and mj? I guess the idea may also be illustrated by the table below: combining different values for l (±1) and (±1/2) gives four possible values, ranging from +3/2 to -1/2, for l + s.tableHaving said that, the illustration raises a very fundamental question: the length of the sum of two vectors is definitely not the same as the sum of the length of the two vectors! So… Well… Hmm… Something doesn’t make sense here! However, I can’t dwell any longer on this. I just wanted to note you should not take all that’s published on those oft-used sites on quantum mechanics for granted. But so I need to move on. Back to the other illustration – copied once more below.vector-modelWe have that very special formula for the magnitude (J) of the angular momentum J:

J│= J = √(J·J) = √[j·(j+1)·ħ2] = √[j·(j+1)]·ħ 

So if = 3/2, then J is equal to √[(3/2)·(3/2+1)]·ħ = √[(15/4)·ħ ≈ 1.9635·ħ, so that’s almost 2ħ. 🙂 At the same time, we know that for = 3/2, the possible values of Jz can only be +3ħ/2, +ħ/2, -ħ/2, and -3ħ/2. So that’s what’s shown in that half-circular diagram: the magnitude of J is larger than its z-component – always!

OK. Next. What’s that 3p3/2 notation? Excellent question! Don’t think this 3p denotes an electron orbital, like 1s or 3d – i.e. the orbitals we got from solving Schrödinger’s equation. No. In fact, the illustration above is somewhat misleading because the correct notation is not 3p3/2 but 3P3/2. So we have a capital P which is preceded by a superscript 3. This is the notation for the so-called term symbol for a nuclear, atomic or molecular (ground) state which – assuming our LS coupling model is valid – because we’ve got other term symbols for other coupling models – we can write, more generally, as:

2S+1LJ

The J, L and S in this thing are the following:

1. The J is the total angular momentum quantum number, so it is – the notation gets even more confusing now – the in the │J│= J = √(J·J) = √[j·(j+1)·ħ2] = √[j·(j+1)]·ħ expression. We know that number is 1/2 for electrons, but it may take on other values for nuclei, atoms or molecules. For example, it is 3/2 for nitrogen, and 2 for oxygen, for which the corresponding terms are 4S3/2 and 3P2 respectively.

2. The S in the term symbol is the total spin quantum number, and 2S+1 itself is referred to as the fine-structure multiplicity. It is not an easy concept. Just remember that the fine structure describes the splitting of the spectral lines of atoms due to electron spin. In contrast, the gross structure energy levels are those we effectively get from solving Schrödinger’s equation assuming our electrons have no spin.  We also have a hyperfine structure, which is due to the existence of a (small) nuclear magnetic moment, which we do not take into consideration here, which is why the 4S3/2 and 3P2 terms are sometimes being referred to as describing electronic ground states. In fact, the MIT lab document, which we are studying here, refers to the ground state of the potassium atoms in the beam as an electronic ground state, which is written up as 2S1/2. So is, effectively, equal to 1/2. [Are you still there? If so, just write it down: 2S+1 = 2 ⇒ = 1/2. That means the following: our potassium atom behaves like an electron: its spin is either ‘up’ or, else, it is ‘down’. There is no in-between.]

3. Finally, the in the term symbol is the total orbital angular momentum quantum number but, rather than using a number, the values of are often represented as S, P, D, F, etcetera. This number is very confusing because – as mentioned above – one would think it represents those s, p, d, f, g,… orbitals. However, that is not the case. The difference may easily be illustrated by observing that a carbon atom, for example, has six electrons, which are distributed over the 1s, 2s and 2p orbitals (one pair each). However, its ground state only gets one number: L = P. Hence, its value is 1. Of course, now you will wonder how we get that number.

Well… I wish I could give you an easy answer, but I can’t. For two electrons – think of our carbon atom once again – we can have = 0, 1 or 2, or S, P and D. They effectively correspond to different energy levels, which are related to the way these two electrons interact with each other. The phenomenon is referred to as angular momentum coupling. In fact, all of the numbers we discussed so far – J, S and L – are numbers resulting from angular momentum coupling. As Wikipedia puts it: “Angular momentum coupling refers to the construction of eigenstates of total angular momentum out of eigenstates of separate angular momentum.” [As you know, each eigenstate corresponds to an energy level, of course.]

Now that should clear some of the confusion on the 2S+1LJ notation: the capital letters J, S and L refer to some total, as opposed to the quantum numbers you are used to, i.e. n, l, m and s, i.e. the so-called principalorbitalmagnetic and spin quantum number respectively. The lowercase letters are quantum numbers that describe an electron in an atom, while those capital letters denote quantum numbers describing the atom – or a molecule – itself.

OK. Onwards. But where were we? 🙂 Oh… Yes. That J = L + S formula gives us some total electronic angular momentum, but we’ll also have some nuclear angular momentum, which our MIT paper denotes as I. Our vector model of our potassium atom allows us, once again, to simply add the two to get the total angular momentum, which is written as F = J + I = L + S + I. This, then, explains why the MIT experiment writes the magnitude of the total angular momentum as:

magnitude-of-f

Of course, here I don’t need to explain – or so I hope – why this quantum-mechanical formula for the calculation of the magnitude is what it is (or, equivalently, why the usual Euclidean metric – i.e. √(x2 + y2 + z2) – is not to be used here. If you do need an explanation, you’ll need to go through the basics once again.

Now, the whole point, of course, is that the z-component of F can have only the discrete values that are specified by the Fz = mf·ħ equation, with mf – i.e. the (total) magnetic quantum number – having an equally discrete value equal to mf = −f, −(f−1), …, +(f+1), f.

For the rest, I probably shouldn’t describe the experiment itself: you know it. But let me just copy the set-up below, so it’s clear what it is that we’re expecting to happen. In addition, you’ll also need the illustration below because I’ll refer to those d1 and d2 distances shown in what follows.set-up

Note the MIT documentation does spell out some additional assumptions. Most notably, it says that the potassium atoms that emerge from the oven (at a temperature of 200°) will be:

(1) almost exclusively in the ground electronic state,

(2) nearly equally distributed among the two (magnetic) sub-states characterized by f, and, finally,

(3) very nearly equally distributed among the hyperfine states, i.e. the states with the same but with different mf.

I am just noting these assumptions because it is interesting to note that – according to the man or woman who wrote this paper – we would actually have states within states here. The paper states that the hyperfine splitting of the two sub-beams we expect to come out of the magnet can only be resolved by very advanced atomic beam techniques, so… Well… That’s not the apparatus that’s being used for this experiment.

However, it’s all a bit weird, because the paper notes that the rules for combining the electronic and nuclear angular momentum – using that F = J + I = L + S + I formula – imply that our quantum number f = i ± j can be eitheror 2. These two values would be associated with the following mf and mforce values:

= 1 ⇒ Fz = mf·ħ = −ħ, 0 or +ħ (so we’d have three beams here)

= 2 ⇒ Fz = mf·ħ = −2ħ, −ħ, 0, +ħ or +2ħ (so we’d have five beams here)

Neither of the two possibilities relates to the situation at hand – which assumes two beams only. In short, I think the man or women who wrote the theoretical introduction – an assistant professor, most likely (no disrespect here: that’s how far progressed in economics – nothing more, nothing less) – might have made a mistake. Or perhaps he or she may have wanted to confuse us.

I’ll look into it over the coming days. As for now, all you need to know – please jot it down! – is that our potassium atom is fully described by 2S1/2. That shorthand notation has all the quantum number we need to know. Most importantly, it tells us is, effectively, equal to 1/2. So… Well… That 2S1/2 notation tells us our potassium atom should behave like an electron: its spin is either ‘up’ or ‘down’. No in-between. 🙂 So we should have two beams. Not three or five. No fine or hyperfine sub-structures! 🙂 In any case, the rest of the paper makes it clear the assumption is, effectively, that the angular momentum number is equal to = 1/2. So… Two beams only. 🙂

How to calculate the expected deflection

We know that the inhomogeneous magnetic field (B), whose direction is the z-axis, will result in a force, which we have calculated a couple of times already as being equal to:f1In case you’d want to check this, you can check one of my posts on this. I just need to make one horrifying remark on notation here: while the same symbol is used, the force Fis, obviously, not to be confused with the z-component of the angular momentum F = J + I = L + S + I that we described above. Frankly, I hope that the MIT guys have corrected that in the meanwhile, because it’s really terribly confusing notation! In any case… Let’s move on.

Now, we assume the deflecting force is constant because of the rather particular design of the magnet pole pieces (see Appendix I of the paper). We can then use Newton’s Second Law (F = m·a) to calculate the velocity in the z-direction, which is denoted by Vz (I am not sure why a capital letter is used here, but that’s not important, of course). That velocity is assumed to go from 0 to its final value Vz while our potassium atom travels between the two magnet poles but – to be clear – at any point in time, Vz will increase linearly – not exponentially – so we can write: Vz = a·t1, with t1 the time that is needed to travel through the magnet. Now, the relevant mass is the mass of the atom, of course, which is denoted by M. Hence, it is easy to see that = Fz/M = Vz/t1. Hence, we find that V= Fz·t1/M.

Now, the vertical distance traveled (z) can be calculated by solving the usual integral: z = ∫0t1 v(t)·dt = ∫0t1 a·t·dt = a·t12/2 = (Vz/t1)·t12/2 = Vz·t1/2. Of course, once our potassium atom comes out of the magnetic field, it will continue to travel upward or downward with the same velocity Vz, which adds Vz·t2 to the total distance traveled along the z-direction. Hence, the formula for the deflection is, effectively, the one that you’ll find in the paper:

z = Vz·t1/2 + Vz·t= Vz·(t1/2 + t2)

Now, the travel times depend on the velocity of our potassium atom along the y-axis, which is approximated by equating it with │V│= V, because the y-component of the velocity is easily the largest – by far! Hence, t1 = d1/V and t2 = d2/V. Some more manipulation will then give you the expression we need, which is a formula for the deflection in terms of variables that we actually know:

z-formula

Statistical mechanics

We now need to combine this with the Maxwell-Boltzmann distribution for the velocities we gave you in our previous post:formula-aThe next step is to use this formula so as to be able to calculate a distribution which would describe the intensity of the beam. Now, it’s easy to understand such intensity will be related to the flux of potassium atoms, and it’s equally easy to get that a flux is defined as the rate of flow per unit area. Hmm… So how does this get us the formula below?density-formula-2The tricky thing – of course – is the use of those normalized velocities because… Well… It’s easy to see that the right-hand side of the equation above – just forget about the d(V/V0 ) bit for a second, as we have it on both sides of the equation and so it cancels out anyway – is just density times velocity. We do have a product of the density of particles and the velocity with which they emerge here – albeit a normalized velocity. But then… Who cares? The normalization is just a division by V– or a multiplication by 1/V0, which is some constant. From a math point of view, it doesn’t make any difference: our variable is V/V0 instead of V. It’s just like using some other unit. No worries here – as long as you use the new variable consistently everywhere. 🙂

Alright. […] What’s next? Well… Nothing much. The only thing that we still need to explain now is that factor 2. It’s easy to see that’s just a normalization factor – just like that 4/√π factor in the first formula. So we get it from imposing the usual condition:densitySo… What’s next… Well… We’re almost there. 🙂 As the MIT paper notes, the f(V) and I(V/V0) functions can be mapped to each other: the related transformation maps a velocity distribution to an intensity distribution – i.e. a distribution of the deflection – and vice versa.

Now, the rest of the paper is just a lot of algebraic manipulations – distinguishing the case of a quantized Fversus a continuous Fz. Here again, I must admit I am a bit shocked by the mix-up of concepts and symbols. The paper talks about a quantized deflecting force – while it’s obvious we should be talking a quantized angular momentum. The two concepts – and their units – are fundamentally different: the unit in which angular momentum is measured is the action unit: newton·meter·second (N·m·s). Force is just force: x newton.

Having said that, the mix-up does trigger an interesting philosophical question: what is quantized really? Force (expressed in N)? Energy (expressed in N·m)? Momentum (expressed in N·s)? Action (expressed in N·m·s, i.e. the unit of angular momentum)? Space? Time? Or space-time – related through the absolute speed of light (c)? Three factors (force, distance and time), six possibilities. What’s your guess?

[…]

What’s my guess? Well… The formulas tell us the only thing that’s quantized is action: Nature itself tells us we have to express it in terms of Planck units. However, because action is a product involving all of these factors, with different dimensions, the quantum-mechanical quantization of action can, obviously, express itself in various ways. 🙂

Statistical mechanics re-visited

Quite a while ago – in June and July 2015, to be precise – I wrote a series of posts on statistical mechanics, which included digressions on thermodynamics, Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac statistics (probability distributions used in quantum mechanics), and so forth. I actually thought I had sort of exhausted the topic. However, when going through the documentation on that Stern-Gerlach experiment that MIT undergrad students need to analyze as part of their courses, I realized I did actually not present some very basic formulas that you’ll definitely need in order to actually understand that experiment.

One of those basic formulas is the one for the distribution of velocities of particles in some volume (like an oven, for instance), or in a particle beam – like the beam of potassium atoms that is used to demonstrate the quantization of the magnetic moment in the Stern-Gerlach experiment. In fact, we’ve got two formulas here, which are subtly – as subtle as the difference between (boldface, so it’s a vector) and v (lightface, so it’s a scalar) 🙂 – but fundamentally different:

velocity-distribution

Both functions are referred to as the Maxwell-Boltzmann density distribution, but the first distribution gives us the density for some v in the velocity space, while the second gives us the distribution density of the absolute value (or modulus) of the velocity, so that is the distribution density of the speed, which is just a scalar – without any direction. As you can see, the second formula includes a 4π·v2 factor.

The question is: how are these formulas related to Boltzmann’s f(E) = C·e−energy/kT Law? The answer is: we can derive all of these formulas – for the distribution of velocities, or of momenta – by clever substitutions. However, as evidenced by the two formulas above, these substitutions are not always straightforward. So let me quickly show you a few things here.

First note the two formulas above already include the e−energy/kT function if we equate the energy E with the kinetic energy: E = K.E. = m·v2/2. Of course, if you’ve read those June-July 2015 posts, you’ll note that we derived Boltzmann’s Law in the context of a force field, like gravity, or an electric potential. For example, we wrote the law for the density (n = N/V) of gas in a gravitational field (like the Earth’s atmosphere) as n = n0·e−P.E./kT. In this formula, we only see the potential energy: P.E. = m·g·h, i.e. the product of the mass (m), the gravitational constant (g), and the height (h). However, when we’re talking the distribution of velocities – or of momenta – then the kinetic energy comes into play.

So that’s a first thing to note: Boltzmann’s Law is actually a whole set of laws. For example, the frequency distribution of particles in a system over various possible states, also involves the same exponential function: F(state) ∝ e−E/kT. E is just the total energy of the state here (which varies from state to state, of course), so we don’t distinguish between potential and kinetic energy here.

So what energy concept should we use in that Stern-Gerlach experiment? Because these potassium atoms in that oven – or when they come out of it in a beam – have kinetic energy only, our E = m·v2/2 substitution does the trick: we can say that the potential energy is taken to be zero, so that all energy is in the form of kinetic energy. So now we understand the e−m·v2/2kT function in those f(v) and f(v) formulas. Now we only need to explain those complicated coefficients. How do we get these?

We get them through clever substitutions using equations such as:

fv(v)·dv  = fp(p)·dp

What are we writing here? We’re basically combining two normalization conditions: if fv(v) and fp(p) are proper probability density functions, then they must give us 1 when integrating over their domain. The domain of these two functions is, obviously, the velocity (v) and momentum (p) space. The velocity and momentum space are the same mathematical space, but they are obviously not the same physical space. But the two physical spaces are closely related: p = m·v, and so it’s easy to do the required transformation of variables. For example, it’s easy to see that, if E = m·v2/2, then E is also equal to E = p2/2m.

However, when doing these substitutions, things get tricky. We already noted that p and v are vectors, unlike E, or p and v – which are scalars, or magnitudes. So we write: p = (px, py, pz) and |p| = p, and v = (vx, vy, v z) and |v| = v. Of course, you also know how we calculate those magnitudes:

magnitude

Note that this also implies the following: p·p = p= px+ py+pz= p2. Trivial, right? Yes. But have a look now at the following differentials:

  • d3p
  • dp
  • dp = d(px, py, pz)
  • dpx·dpy·dpz

Are these the same or not? Now you need to think, right? That d3p and dp are different beasts is obvious: d3p is, obviously, some infinitesimal volume, as opposed to dp, which is, equally obviously, an (infinitesimal) interval. But what volume exactly? Is it the same as that dp = d(px, py, pz) volume, and is that the same as the dpx·dpy·dpz volume?

Fortunately, the volume differentials are, in fact, the same – so you can start breathing again. 🙂 Let’s get going with that d3p notation for the time being, as you will find that’s the notation which is used in the Wikipedia article on the Maxwell-Boltzmann distribution – which I warmly recommend, because – for a change – it is a much easier read than other Wikipedia articles on stuff like this. Among other things, the mentioned article writes the following:

fE(E)·dE = fp(p)·d3p

What is this? Well… It’s just like that fv(v)·dv  = fp(p)·dp equation: it combines the normalization condition for both distributions. However, it’s much more interesting, because, on the left-hand side, we multiply a density with an (infinitesimal) interval (dE), while on the right-hand side we multiply with an (infinitesimal) volume (d3p). Now, the (infinitesimal) energy interval dE must, obviously, correspond with the (infinitesimal) momentum volume d3p. So how does that work?

Well… The mentioned Wikipedia article talks about the “spherical symmetry of the energy-momentum dispersion relation” (that dispersion relation is just E = |p|2/2m, of course), but that doesn’t make us all that wiser, so let’s try a more heuristic approach. You might remember the formula for the volume of a spherical shell, which is simply the difference between the volume of the outer sphere minus the volume of the inner sphere: V = (4π/3)·R− (4π/3)·r= (4π/3)·(R− r3). Now, for a very thin shell of thickness Δr, we can use the following first-order approximation: V = 4π·r2·Δr. In case you wonder, I hereby copy a nice explanation from the Physics Stack Exchange site:

approximation

Perfect. That’s all we need to know. We’ll use that first-order approximation to re-write d3as:

d3= dp = 4π·|p|2·d|p| = 4π·p2·dp

Note that we’ll have the same formula for d3v, of course: d3v = dv = 4π·|v|2·d|v| = 4π·v2·dv, and also note that we get that same 4π·v2 factor which we mentioned when discussing the f(v) and f(v) formulas. That is not a coincidence, of course, but – as I’ll explain in a moment – it is not so easy to immediately relate the formulas. In any case, we’re now ready to relate dE and dp so we can re-write that d3p formula in terms of m, E and dE:

substitution-2

We are now – finally! – sufficiently armed to derive all of the formulas we want – or need. Let me just copy them from the mentioned Wikipedia article:

momenta

energy

velocity

As said, you’ll encounter these formulas regularly – and so it’s good that you know how you can derive them. Indeed, the derivation is very straightforward and is done in the same article: the tips I gave you should allow you to read it in a couple of minutes only. Only the density function for velocities might cause you a bit of trouble – but only for a very short moment: just use the p = m·v equation to write d3p as d3p = 4π·p2·dp = 4π·m2·v2·m·dv = 4π·m3·v2·dv = m3·d3v, and you’re all set. 🙂

Of course, you will recognize the formula for the distribution of velocities: it’s the f(v) we mentioned in the introduction. However, you’re more likely to need the f(v) formula (i.e. the probability density function for the speed) than the f(v) function. So how can we derive get the f(v) – i.e. that formula for the distribution of speeds, with the 4π·v2 factor – from the f(v) formula?

Well… I wish I could give you an easy answer. In fact, the same Wikipedia article suggests it’s easy – but it’s not. It involves a transformation from Cartesian to polar coordinates: the volume element dvx·dvy·dvz is to be written as v2·sinθ·dv·dθ·dφ. And then… Well… Have a look at this link. 🙂 It involves a so-called Jacobian transformation matrix. If you want to know more about it, then I recommend you read some article on how to transform distribution functions: here’s a link to one of those, but you can easily google others. Frankly, as for now, I’d suggest you just accept the formula for f(v) as for now. 🙂 Let me copy it from the same article in a slightly different form:density-formulaNow, the final thing to note is that you’ll often want to use so-called normalized velocities, i.e. velocities that are defined as a v/v0 ratio, with vthe most probable speed, which is equal to √(2kT/m). You get that value by calculating the df(v)/dv derivative, and then finding the value v = v0 for which df(v)/dv = 0. You should now be able to verify the formula that is used in the mentioned MIT version of the Stern-Gerlach experiment:mit-formulaIndeed, when you write it all out – note that π/π3/2 = 1/√π 🙂 – you’ll see the two formulas are effectively equivalent. Of course, by now you are completely formula-ed out, and so you probably don’t even wonder what that f(v)·dv product actually stands for. What does it mean, really? Now you’ll sigh: why would I even want to know that? Well… I want you to understand that MIT experiment. 🙂 And you won’t if you don’t know what f(v)·dv actually represents. So think about it. […]

[…] OK. Let me help you once more. Remember the normalization condition once again: the integral of the whole thing – over the whole range of possible velocities – needs to add up to 1, so f(v)·dv is really the fraction of (potassium) atoms (inside the oven) with a velocity in the (infinitesimally small) dv interval. It’s going to be a tiny fraction, of course: just a tiny bit larger than zero. Surely not larger than 1, obviously. 🙂 Think of integrating the function between two values – say v1 and v2 – that are pretty close to each other.

So… Well… We’re done as for now. So where are we now in terms of understanding the calculations in that description of that MIT experiment? Well… We’ve got the meat. But we need a lot of other ingredients now. We’ll want formulas for the intensity of the beam at some point along the axis measuring its deflection from its main direction. That axis is the z-axis. So we’ll want a formula for some I(z) function.

Deflection? Yes. There are a lot of steps to go through now. Here’s the set-up:set-upFirst, we’ll need some formula measuring the flux of (potassium) atoms coming out of the oven. And then… Well… Just have a look and try to make your way through the whole thing now – which is just what I want to do in the coming days, so I’ll give you some more feedback soon. 🙂 Here I only wanted to introduce those formulas for the distribution of velocities and momenta, because you’ll need them in other contexts too.

So I hope you found this useful. Stuff like this all makes it somewhat more real, doesn’t it? 🙂 Frankly, I think the math is at least as fascinating as the physics. We could have a closer look at those distributions, for example, by noting the following:

1. The probability density function for the momenta is the product of three normal distributions. Which ones? Well…  The distribution of px, py and pz respectively: three normal distributions whose variance is equal to mkT. 🙂

2. The fE(E) function is a chi-squared (χ2) distribution with 3 degrees of freedom. Now, we have the equipartition theorem (which you should know – if you don’t, see my post on it), which tells us that this energy is evenly distributed among all three degrees of freedom. It is then relatively easy to show – if you know something about χ2 distributions at least 🙂 – that the energy per degree of freedom (which we’ll write as ε below) will also be distributed as a chi-squared distribution with one degree of freedom:chi-square-2This holds true for any number of degrees of freedom. For example, a diatomic molecule will have extra degrees of freedom, which are related to its rotational and vibrational motion (I explained that in my June-July 2015 posts too, so please go there if you’d want to know more). So we can really use this stuff in, for example, the theory of the specific heat of gases. 🙂

3. The function for the distribution of the velocities is also a product of three independent normally distributed variables – just like the density function for momenta. In this case, we have the vx, vy and vz variables that are normally distributed, with variance kT/m.

So… Well… I’m done – for the time being, that is. 🙂 Isn’t it a privilege to be alive and to be able to savor all these little wonderful intellectual excursions? I wish you a very nice day and hope you enjoy stuff like this as much as I do. 🙂

The quantization of magnetic moments

You may not have many questions after a first read of Feynman’s Lecture on the Stern-Gerlach experiment and his more general musings on the quantization of the magnetic moment of an elementary particle. [At least I didn’t have all that many after my first reading, which I summarized in a previous post.]

However, a second, third or fourth reading should trigger some, I’d think. My key question is the following: what happens to that magnetic moment of a particle – and its spin [1] – as it travels through a homogeneous or inhomogeneous magnetic field? We know – or, to be precise, we assume – its spin is either “up” (Jz = +ħ/2) or “down” (Jz = −ħ/2) when it enters the Stern-Gerlach apparatus, but then – when it’s moving in the field itself – we would expect that the magnetic field would, somehow, line up the magnetic moment, right?

Feynman says that it doesn’t: from all of the schematic drawings – and the subsequent discussion of Stern-Gerlach filters – it is obvious that the magnetic field – which we denote as B, and which we assume to be inhomogeneous [2] – should not result in a change of the magnetic moment. Feynman states it as follows: “The magnetic field produces a torque. Such a torque you would think is trying to line up the (atomic) magnet with the field, but it only causes its precession.”

[…] OK. That’s too much information already, I guess. Let’s start with the basics. The key to a good understanding of this discussion is the force formula:

f1

We should first explain this formula before discussing the obvious question: over what time – or over what distance – should we expect this force to pull the particle up or down in the magnetic field? Indeed, if the force ends up aligning the moment, then the force will disappear!

So let’s first explain the formula. We start by explaining the energy U. U is the potential energy of our particle, which it gets from its magnetic moment μ and its orientation in the magnetic field B. To be precise, we can write the following:

f2

Of course, μ and B are the magnitudes of μ and B respectively, and θ is the angle between μ and B: if the angle θ is zero, then Umag will be negative. Hence, the total energy of our particle (U) will actually be less than what it would be without the magnetic field: it is the energy when the magnetic moment of our particle is fully lined up with the magnetic field. When the angle is a right angle (θ = ±π/2), then the energy doesn’t change (Umag = 0). Finally, when θ is equal to π or −π, then its energy will be more than what it would be outside of the magnetic field. [Note that the angle θ effectively varies between –π and π – not between 0 and 2π!]anglesOf course, we may already note that, in quantum mechanics, Umag will only take on a very limited set of values. To be precise, for a particle with spin number j = 1/2, the possible values of Umag will be limited to two values only. We will come back to that in a moment. First that force formula.

Energy is force over a distance. To be precise, when a particle is moved from point a to point b, then its change in energy can be written as the following line integral:

f3

Note that the minus sign is there because of the convention that we’re doing work against the force when increasing the (potential) energy of that what we’re moving. Also note that F∙ds product is a vector (dot) product: it is, obviously, equal to Ft times ds, with Ft the magnitude of the tangential component of the force. The equation above gives us that force formula:

f4

Feynman calls it the principle of virtual work, which sounds a bit mysterious – but so you get it by taking the derivative of both sides of the energy formula.

Let me now get back to the real mystery of quantum mechanics, which tells us that the magnetic moment – as measured along our z-axis – will only take one of two possible values. To be precise, we have the following formula for μz:

f5

This is a formula you just have to accept for the moment. It needs a bit of interpretation, and you need to watch out for the sign. The g-factor is the so-called Landé g-factor: it is equal to 1 for a so-called pure orbital moment, 2 for a so-called pure spin moment, and some number in-between in reality, which is always some mixture of the two: both the electron’s orbit around the nucleus as well as the electron’s rotation about its own axis contribute to the total angular momentum and, hence, to the total magnetic moment of our electron. As for the other factors, m and qe are, of course, the mass and the charge of our electron, and Jz is either +ħ/2 or −ħ/2. Hence, if we know g, we can easily calculate the two possible values for μz.

Now, that also means we could – theoretically – calculate the two possible values of that angle θ. For some reason, no handbook in physics ever does that. The reason is probably a good one: electron orbits, and the concept of spin itself, are not like the orbit and the spin of some planet in a planetary system. In fact, we know that we should not think of electrons like that at all: quantum physicists tell us we may only think of it as some kind of weird cloud around a center. That cloud has a density which is to be calculated by taking the absolute square of the quantum-mechanical amplitude of our electron.

In fact, when thinking about the two possible values for θ, we may want to remind ourselves of another peculiar consequence of the fact that the angular momentum – and, hence, the magnetic moment – is not continuous but quantized: the magnitude of the angular momentum J is not  J = √(J·J) = √J2 in quantum mechanics but J = √(J·J) = √[j·(j+1)·ħ2] = √[j·(j+1)]·ħ. For our electron, j = 1/2 and, hence, the magnitude of J is equal to J = √[(1/2)∙(3/2)]∙ ħ = √(3/4)∙ħ ≈ 0.866∙ħ. Hence, the magnitude of the angular momentum is larger than the maximum value of Jz – and not just a little bit, because the maximum value of ħ is ħ/2! That leads to that weird conclusion: in quantum mechanics, we find that the angular momentum is never completely along any one direction [3]! In fact, this conclusion basically undercuts the very idea of the angular momentum – and, hence, the magnetic moment – of having any precise direction at all! [This may sound spectacular, but there is actually a classical equivalent to the idea of the angular momentum having no precisely defined direction: gyroscopes may not only precess, but nutate as well. Nutation refers to a kind of wobbling around the direction of the angular momentum. For more details, see the post I wrote after my first reading of Feynman’s Lecture on the quantization of magnetic moments. :-)] 

Let’s move on. So if, in quantum mechanics, we cannot associate the magnetic moment – or the angular momentum – with some specific direction, then how should we imagine it? Well… I won’t dwell on that here, but you may want to have a look at another post of mine, where I develop a metaphor for the wavefunction which may help you to sort of understand what it might be. The metaphor may help you to think of some oscillation in two directions – rather than in one only – with the two directions separated by a right angle. Hence, the whole thing obviously points in some direction but it’s not very precise. In any case, I need to move on here.

We said that the magnetic moment will take one of two values only, in any direction along which we’d want to measure it. We also said that the (maximum) value along that direction – any direction, really – will be smaller than the magnitude of the moment. [To be precise, we said that for the angular momentum, but the formulas above make it clear the conclusions also hold for the magnetic moment.] So that means that the magnetic moment is, in fact, never fully aligned with the magnetic field. Now, if it is not aligned – and, importantly, if it also does not line up – then it should precess. Now, precession is a difficult enough concept in classical mechanics, so you may think it’s going to be totally abstruse in quantum mechanics. Well… That is true – to some extent. At the same time, it is surely not unintelligible. I will not repeat Feynman’s argument here, but he uses the classical formulas once more to calculate an angular velocity and a precession frequency – although he doesn’t explain what they might actually physically represent. Let me just jot down the formula for the precession frequency:

f7

We get the same factors: g, qe and m. In addition, you should also note that the precession frequency is directly proportional  to the strength of the magnetic field, which makes sense. Now, you may wonder: what is the relevance of this? Can we actually measure any of this?

We can. In fact, you may wonder about the if I inserted above: if we can measure the Landé g-factor… Can we? We can. It’s done in a resonance experiment, which is referred to as the Rabi molecular-beam method – but then it might also be just an atomic beam, of course!

The experiment is interesting, because it shows the precession is – somehow – real. It also illustrates some other principles we have been describing above.

The set-up looks pretty complicated. We have a series of three magnets. The first magnet is just a Stern-Gerlach apparatus: a magnet with a very sharp edge on one of the pole tips so as to produce an inhomogeneous magnetic field. Indeed, a homogeneous magnetic field implies that ∂B/∂z = 0 and, hence, the force along the z-direction would be zero and our atomic magnets would not be displaced.

The second magnet is more complicated. Its magnetic field is uniform, so there are no vertical forces on the atoms and they go straight through. However, the magnet includes an extra set of coils that can produce an alternating horizontal field as well. I’ll come back to that in a moment. Finally, the third magnet is just like the first one, but with the field inverted. Have a look at it:

rabi-apparatus

It may not look very obvious but, after some thinking, you’ll agree that the atoms can only arrive at the detector if they follow the trajectories a and/or b. In fact, these trajectories are the only possible ones because of the slits S1 and S2.

Now what’s the idea of that horizontal field B’ in magnet 2? In a classical situation, we could change the angular momentum – and the magnetic moment – by applying some torque about the z-axis. The idea is shown in Figure (a) and (b) below.

changing-angular-momentum

Figure (a) shows – or tries to show – some rotating field B’ – one that is always at right angles to both the angular momentum as well as to the (uniform) B field. That would be effective. However, Figure (b) shows another arrangement that is almost equally effective: an oscillating field that sort of pulls and pushes at some frequency ω. Classically, such fields would effectively change the angle of our gyroscope with respect to the z-axis. Is it also the case quantum-mechanically?

It turns out it sort of works the same in quantum mechanics. There is a big difference though. Classically, μz would change gradually, but in quantum mechanics it cannot: in quantum mechanics, it must jump suddenly from one value to the other, i.e. from +ħ/2 to −ħ/2, or the other way around. In other words, it must flip up or down. Now, if an atom flips, then it will, of course, no longer follow the (a) or (b) trajectories: it will follow some other path, like a’ or b’, which make it crash into the magnet. Now, it turns out that almost all atoms will flip if we get that frequency ω right. The graph below shows this ‘resonance’ phenomenon: there is a sharp drop in the ’current’ of atoms if ω is close or equal to ωp.

resonance

What’s ωp? It’s that precession frequency for which we gave you that formula above. To make a long story short, from the experiment, we can calculate the Landé g-factor for that particular beam of atoms – say, silver atoms [4]. So… Well… Now we know it all, don’t we?

Maybe. As mentioned when I started this post, when going through all of this material, I always wonder why there is no magnetization effect: why would an atom remain in the same state when it crosses a magnetic field? When it’s already aligned with the magnetic field – to the maximum extent possible, that is – then it shouldn’t flip, but what if its magnetic moment is opposite? It should lower its energy by flipping, right? And it should flip just like that. Why would it need an oscillating B’ field?

In fact, Feynman does describe how the magnetization phenomenon can be analyzed – classically and quantum-mechanically, but he does that for bulk materials: solids, or liquids, or gases – anything that involves lots of atoms that are kicked around because of the thermal motions. So that involves statistical mechanics – which I am sure you’ve skipped so far. 🙂 It is a beautiful argument – which ends with an equally beautiful formula, which tells us the magnetization (M) of a material – which is defined as the net magnetic moment per unit volume – has the same direction as the magnetic field (B) and a magnitude M that is proportional the magnitude of B:

f6The μ in this formula is the magnitude of the magnetic moment of the individual atoms and so… Well… It’s just like the formula for the electric polarization P, which we described in some other post. In fact, the formula for P and M are same-same but different, as they would say in Thailand. 🙂 But this wonderful story doesn’t answer our question. The magnetic moment of an individual particle should not stay what it is: if it doesn’t change because of all the kicking around as a result of thermal motions, then… Well… These little atomic magnets should line up. That means atoms with their spin “up” should go into the “spin-down” state.

I don’t have an answer to my own question as for now. I suspect it’s got to do with the strength of the magnetic field: a Stern-Gerlach apparatus involves a weak magnetic field. If it’s too strong, the atomic magnets must flip. Hence, a more advanced analysis should probably include that flipping effect. When quickly googling – just now – I found an MIT lab exercise on it, which also provides a historical account of the Stern-Gerlach experiment itself. I skimmed through it – and will read all of it in the coming days – but let me just quote this from the historical background section:

“Stern predicted that the effect would be be just barely observable. They had difficulty in raising support in the midst of the post war financial turmoil in Germany. The apparatus, which required extremely precise alignment and a high vacuum, kept breaking down. Finally, after a year of struggle, they obtained an exposure of sufficient length to give promise of an observable silver deposit. At first, when they examined the glass plate they saw nothing. Then, gradually, the deposit became visible, showing a beam separation of 0.2 millimeters! Apparently, Stern could only afford cheap cigars with a high sulfur content. As he breathed on the glass plate, sulfur fumes converted the invisible silver deposit into visible black silver sufide, and the splitting of the beam was discovered.”

Isn’t this funny? And great at the same time? 🙂 But… Well… The point is: the paper for that MIT lab exercise makes me realize Feynman does cut corners when explaining stuff – and some corners are more significant than others. I note, for example, that they talk about interference peaks rather than “two distinct spots on the glass plate.” Hence, the analysis is somewhat more sophisticated than Feynman pretends it to be. So, when everything is said and done, Feynman’s Lectures may indeed be reading for undergraduate students only. Is it time to move on?

[1] The magnetic moment – as measured in a particular coordinate system – is equal to μ = −g·[q/(2m)]·J. The factor J in this expression is the angular momentum, and the coordinate system is chosen such that its z-axis is along the direction of the magnetic field B. The component of J along the z-axis is written as Jz. This z-component of the angular momentum is what is, rather loosely, being referred to as the spin of the particle in this context. In most other contexts, spin refers to the spin number j which appears in the formula for the value of Jz, which is Jz = j∙ħ, (j−1)∙ħ, (j−2)∙ħ,…, (−j+2)∙ħ, (−j+1), −j∙ħ. Note the separation between the possible values of Jz is equal to ħ. Hence, j itself must be an integer (e.g. 1 or 2) or a half-integer (e.g. 1/2). We usually look at electrons, whose spin number j is 1/2.

[2] One of the pole tips of the magnet that is used in the Stern-Gerlach experiment has a sharp edge. Therefore, the magnetic field strength varies with z. We write: ∂B/∂z ≠ 0.

[3] The z-direction can be any direction, really.

[4] The original experiment was effectively done with a beam of silver atoms. The lab exercise which MIT uses to show the effect to physics students involves potassium atoms.

Feynman’s Lecture on Superconductivity

The ultimate challenge for students of Feynman’s iconic Lectures series is, of course, to understand his final one: A Seminar on Superconductivity. As he notes in his introduction to this formidably dense piece, the text does not present the detail of each and every step in the development and, therefore, we’re not supposed to immediately understand everything. As Feynman puts it: we should just believe (more or less) that things would come out if we would be able to go through each and every step. Well… Let’s see. It took me one long maddening day to figure out the first formula:f1It says that the amplitude for a particle to go from to in a vector potential (think of a classical magnetic field) is the amplitude for the same particle to go from to b when there is no field (A = 0) multiplied by the exponential of the line integral of the vector potential times the electric charge divided by Planck’s constant.

Of course, after a couple of hours, I recognized the formula for the magnetic effect on an amplitude, which I described in my previous post, which tells us that a magnetic field will shift the phase of the amplitude of a particle with an amount equal to:

integral

Hence, if we write 〈b|a〉 for A = 0 as 〈b|aA = 0 = C·eiθ, then 〈b|a〉 in A will, naturally, be equal to 〈b|a〉 in A = C·ei(θ+φ) = C·eiθ·eiφ = 〈b|aA = 0 ·eiφ, and so that explains it. 🙂 Alright… Next.

The Schrödinger equation in an electromagnetic field

Feynman then jots down Schrödinger’s equation for the same particle (with charge q) moving in an electromagnetic field that is characterized not only by a vector potential but also by the (scalar) potential Φ:

schrodinger

Now where does that come from? We know the standard formula in an electric field, right? It’s the formula we used to find the energy states of electrons in a hydrogen atom:

i·ħ·∂ψ/∂t = −(1/2)·(ħ2/m)∇2ψ + V·ψ

Of course, it is easy to see that we replaced V by q·Φ, which makes sense: the potential of a charge in an electric field is the product of the charge (q) and the (electric) potential (Φ), because Φ is, obviously, the potential energy of the unit charge. It’s also easy to see we can re-write −ħ2·∇2ψ as [(ħ/i)·∇]·[(ħ/i)·∇]ψ because (1/i)·(1/i) = 1/i2 = 1/(−1) = −1. 🙂 Alright. So it’s just that −q·A term in the (ħ/i)∇ − q·A expression that we need to explain now.

Unfortunately, that explanation is not so easy. Feynman basically re-derives Schrödinger’s equation using his trade-mark historical argument – which did not include any magnetic field – with a vector potential. The re-derivation is rather annoying, and I didn’t have the courage to go through it myself, so you should – just like me – just believe Feynman when he says that, when there’s a vector potential – i.e. when there’s a magnetic field – then that ħ/i)·∇ operator – which is the momentum operator– ought to be replaced by a new momentum operator:

new-momentum-operator

So… Well… There we are… 🙂 So far, so good.

Local conservation of probability

The title of this section in Feynman’s Lecture (yes, still the same Lecture – we’re not switching topics here) is the equation of continuity for probabilities. I find it brilliant, because it confirms my interpretation of the wave function as describing some kind of energy flow. Let me quote Feynman on his endeavor here:

“An important part of the Schrödinger equation for a single particle is the idea that the probability to find the particle at a position is given by the absolute square of the wave function. It is also characteristic of the quantum mechanics that probability is conserved in a local sense. When the probability of finding the electron somewhere decreases, while the probability of the electron being elsewhere increases (keeping the total probability unchanged), something must be going on in between. In other words, the electron has a continuity in the sense that if the probability decreases at one place and builds up at another place, there must be some kind of flow between. If you put a wall, for example, in the way, it will have an influence and the probabilities will not be the same. So the conservation of probability alone is not the complete statement of the conservation law, just as the conservation of energy alone is not as deep and important as the local conservation of energy. If energy is disappearing, there must be a flow of energy to correspond. In the same way, we would like to find a “current” of probability such that if there is any change in the probability density (the probability of being found in a unit volume), it can be considered as coming from an inflow or an outflow due to some current.”

This is it, really ! The wave function does represent some kind of energy flow – between a so-called ‘real’ and a so-called ‘imaginary’ space, which are to be defined in terms of directional versus rotational energy, as I try to point out – admittedly: more by appealing to intuition than to mathematical rigor – in that post of mine on the meaning of the wavefunction.

So what is the flow – or probability current as Feynman refers to it? Well… Here’s the formula:

probability-current-2

Huh? Yes. Don’t worry too much about it right now. The essential point is to understand what this current – denoted by J – actually stands for:

probability-current-1

So what’s next? Well… Nothing. I’ll actually refer you to Feynman now, because I can’t improve on how he explains how pairs of electrons start behaving when temperatures are low enough to render Boltzmann’s Law irrelevant: the kinetic energy that’s associated with temperature can no longer break up electron pairs if temperature comes close to the zero point.

Huh? What? Electron pairs? Electrons are not supposed to form pairs, are they? They carry the same charge and are, therefore, supposed to repel each other. Well… Yes and no. In my post on the electron orbitals in a hydrogen atom – which just presented Feynman’s presentation on the subject-matter in a, hopefully, somewhat more readable format – we calculated electron orbitals neglecting spin. In Feynman’s words:

“We make another approximation by forgetting that the electron has spin. […] The non-relativistic Schrödinger equation disregards magnetic effects. [However] Small magnetic effects [do] occur because, from the electron’s point-of-view, the proton is a circulating charge which produces a magnetic field. In this field the electron will have a different energy with its spin up than with it down. [Hence] The energy of the atom will be shifted a little bit from what we will calculate. We will ignore this small energy shift. Also we will imagine that the electron is just like a gyroscope moving around in space always keeping the same direction of spin. Since we will be considering a free atom in space the total angular momentum will be conserved. In our approximation we will assume that the angular momentum of the electron spin stays constant, so all the rest of the angular momentum of the atom—what is usually called “orbital” angular momentum—will also be conserved. To an excellent approximation the electron moves in the hydrogen atom like a particle without spin—the angular momentum of the motion is a constant.”

To an excellent approximation… But… Well… Electrons in a metal do form pairs, because they can give up energy in that way and, hence, they are more stable that way. Feynman does not go into the details here – I guess because that’s way beyond the undergrad level – but refers to the Bardeen-Coopers-Schrieffer (BCS) theory instead – the authors of which got a Nobel Prize in Physics in 1972 (that’s a decade or so after Feynman wrote this particular Lecture), so I must assume the theory is well accepted now. 🙂

Of course, you’ll shout now: Hey! Hydrogen is not a metal! Well… Think again: the latest breakthrough in physics is making hydrogen behave like a metal. 🙂 And I am really talking the latest breakthrough: Science just published the findings of this experiment last month! 🙂 🙂 In any case, we’re not talking hydrogen here but superconducting materials, to which – as far as we know – the BCS theory does apply.

So… Well… I am done. I just wanted to show you why it’s important to work your way through Feynman’s last Lecture because… Well… Quantum mechanics does explain everything – although the nitty-gritty of it (the Meissner effect, the London equation, flux quantization, etc.) are rather hard bullets to bite. 😦

Don’t give up ! I am struggling with the nitty-gritty too ! 🙂

The Aharonov-Bohm effect

This title sounds very exciting. It is – or was, I should say – one of these things I thought I would never ever understand, until I started studying physics, that is. 🙂

Having said that, there is – incidentally – nothing very special about the Aharonov-Bohm effect. As Feynman puts it: “The theory was known from the beginning of quantum mechanics in 1926. […] The implication was there all the time, but no one paid attention to it.”

To be fair, he also admits the experiment itself – proving the effect – is “very, very difficult”, which is why the first experiment that claimed to confirm the predicted effect was set up in 1960 only. In fact, some claim the results of that experiment were ambiguous, and that it was only in 1986, with the experiment of Akira Tonomura, that the Aharonov-Bohm effect was unambiguously demonstrated. So what is it about?

In essence, it proves the reality of the vector potential—and of the (related) magnetic field. What do we mean with a real field? To put it simply, a real field cannot act on some particle from a distance through some kind of spooky ‘action-at-a-distance’: real fields must be specified at the position of the particle itself and describe what happens there. Now you’ll immediately wonder: so what’s a non-real field? Well… Some field that does act through some kind of spooky ‘action-at-a-distance.’ As for an example… Well… I can’t give you one because we’ve only been discussing real fields so far. 🙂

So it’s about what a magnetic (or an electric) field does in terms influencing motion and/or quantum-mechanical amplitudes. In fact, we discussed this matter  quite a while ago (check my 2015 post on it). Now, I don’t want to re-write that post, but let me just remind you of the essentials. The two equations for the magnetic field (B) in Maxwell’s set of four equations (the two others specify the electric field E) are: (1) B = 0 and (2) c2×B = j0 + ∂E/ ∂t. Now, you can temporarily forget about the second equation, but you should note that the B = 0 equation is always true (unlike the ×E = 0 expression, which is true for electrostatics only, when there are no moving charges). So it says that the divergence of B is zero, always.

Now, from our posts on vector calculus, you may or may not remember that the divergence of the curl of a vector field is always zero. We wrote: div (curl A) = •(×A) = 0, always. Now, there is another theorem that we can now apply, which says the following: if the divergence of a vector field, say D, is zero – so if D = 0, then D will be the curl of some other vector field C, so we can write: D×C. When we now apply this to our B = 0 equation, we can confidently state the following: 

If B = 0, then there is an A such that B×A

We can also write this as follows:·B = ·(×A) = 0 and, hence, B×A. Now, it’s this vector field A that is referred to as the (magnetic) vector potential, and so that’s what we want to talk about here. As a start, it may be good to write out all of the components of our B×A vector:

formula for B

In that 2015 post, I answered the question as to why we’d need this new vector field in a way that wasn’t very truthful: I just said that, in many situations, it would be more convenient – from a mathematical point of view, that is – to first find A, and then calculate the derivatives above to get B.

Now, Feynman says the following about this argument in his Lecture on the topic: “It is true that in many complex problems it is easier to work with A, but it would be hard to argue that this ease of technique would justify making you learn about one more vector field. […] We have introduced A because it does have an important physical significance: it is a real physical field.” Let us follow his argument here.

Quantum-mechanical interference effects

Let us first remind ourselves of the quintessential electron interference experiment illustrated below. [For a much more modern rendering of this experiment, check out the  Tout Est Quantique video on it. It’s much more amusing than my rather dry exposé here, but it doesn’t give you the math.]

interference

We have electrons, all of (nearly) the same energy, which leave the source – one by one – and travel towards a wall with two narrow slits. Beyond the wall is a backstop with a movable detector which measures the rate, which we call I, at which electrons arrive at a small region of the backstop at the distance x from the axis of symmetry. The rate (or intensityI is proportional to the probability that an individual electron that leaves the source will reach that region of the backstop. This probability has the complicated-looking distribution shown in the illustration, which we understand is due to the interference of two amplitudes, one from each slit. So we associate the two trajectories with two amplitudes, which Feynman writes as A1eiΦ1 and A2eiΦ2 respectively.

As usual, Feynman abstracts away from the time variable here because it is, effectively, not relevant: the interference pattern depends on distances and angles only. Having said that, for a good understanding, we should – perhaps – write our two wavefunctions as A1ei(ωt + Φ1and A2ei(ωt + Φ2respectively. The point is: we’ve got two wavefunctions – one for each trajectory – even if it’s only one electron going through the slit: that’s the mystery of quantum mechanics. 🙂 We need to add these waves so as to get the interference effect:

R = A1ei(ωt + Φ1A2ei(ωt + Φ2= [A1eiΦ1 A2eiΦ2eiωt

Now, we know we need to take the absolute square of this thing to get the intensity – or probability (before normalization). The absolute square of a product, is the product of the absolute squares of the factors, and we also know that the absolute square of any complex number is just the product of the same number with its complex conjugate. Hence, the absolute square of the eiωt factor is equal to |eiωt|2 = eiωteiωt = e= 1. So the time-dependent factor doesn’t matter: that’s why we can always abstract away from it. Let us now take the absolute square of the [A1eiΦ1 A2eiΦ2] factor, which we can write as:

|R|= |A1eiΦ1 A2eiΦ2|= (A1eiΦ1 A2eiΦ2)·(A1eiΦ1 A2eiΦ2)

= A1+ A2+ 2·A1·A2·cos(Φ1−Φ2) = A1+ A2+ 2·A1·A2·cosδ with δ = Φ1−Φ2

OK. This is probably going a bit quick, but you should be able to figure it out, especially when remembering that eiΦ eiΦ = 2·cosΦ and cosΦ = cos(−Φ). The point to note is that the intensity is equal to the sum of the intensities of both waves plus a correction factor, which is equal to 2·A1·A2·cos(Φ1−Φ2) and, hence, ranges from −2·A1·A2 to +2·A1·A2. Now, it takes a bit of geometrical wizardry to be able to write the phase difference δ = Φ1−Φas

δ = 2π·a/λ = 2π·(x/L)·d/λ

—but it can be done. 🙂 Well… […] OK. 🙂 Let me quickly help you here by copying another diagram from Feynman – one he uses to derive the formula for the phase difference on arrival between the signals from two oscillators. A1 and A2 are equal here (A1 = A2 = A) so that makes the situation below somewhat simpler to analyze. However, instead, we have the added complication of a phase difference (α) at the origin – which Feynman refers to as an intrinsic relative phasetriangle

When we apply the geometry shown above to our electron passing through the slits, we should, of course, equate α to zero. For the rest, the picture is pretty similar as the two-slit picture. The distance in the two-slit – i.e. the difference in the path lengths for the two trajectories of our electron(s) – is, obviously, equal to the d·sinθ factor in the oscillator picture. Also, because L is huge as compared to x, we may assume that trajectory 1 and 2 are more or less parallel and, importantly, that the triangles in the picture – small and large – are rectangular. Now, trigonometry tells us that sinθ is equal to the ratio of the opposite side of the triangle and the hypotenuse (i.e. the longest side of the rectangular triangle). The opposite side of the triangle is x and, because is very, very small as compared to L, we may approximate the length of the hypotenuse with L. [I know—a lot of approximations here, but… Well… Just go along with it as for now…] Hence, we can equate sinθ to x/L and, therefore, d·x/L. Now we need to calculate the phase difference. How many wavelengths do we have in a? That’s simple: a/λ, i.e. the total distance divided by the wavelength. Now these wavelengths correspond to 2π·aradians (one cycle corresponds to one wavelength which, in turn, corresponds to 2π radians). So we’re done. We’ve got the formula: δ = Φ1−Φ= 2π·a/λ = 2π·(x/L)·d/λ.

Huh? Yes. Just think about it. I need to move on. The point is: when is equal to zero, the two waves are in phase, and the probability will have a maximum. When δ = π, then the waves are out of phase and interfere destructively (cosπ = −1), so the intensity (and, hence, the probability) reaches a minimum. 

So that’s pretty obvious – or should be pretty obvious if you’ve understood some of the basics we presented in this blog. We now move to the non-standard stuff, i.e. the Aharonov-Bohm effect(s).

Interference in the presence of an electromagnetic field

In essence, the Aharonov-Bohm effect is nothing special: it is just a law – two laws, to be precise – that tells us how the phase of our wavefunction changes because of the presence of a magnetic and/or electric field. As such, it is not very different from previous analyses and presentations, such as those showing how amplitudes are affected by a potential − such as an electric potential, or a gravitational field, or a magnetic field − and how they relate to a classical analysis of the situation (see, for example, my November 2015 post on this topic). If anything, it’s just a more systematic approach to the topic and – importantly – an approach centered around the use of the vector potential A (and the electric potential Φ). Let me give you the formulas:

f1

f2

The first formula tells us that the phase of the amplitude for our electron (or whatever charged particle) to arrive at some location via some trajectory is changed by an amount that is equal to the integral of the vector potential along the trajectory times the charge of the particle over Planck’s constant. I know that’s quite a mouthful but just read it a couple of times.

The second formula tells us that, if there’s an electrostatic field, it will produce a phase change given by the negative of the time integral of the (scalar) potential Φ.

These two expressions – taken together – tell us what happens for any electromagnetic field, static or dynamic. In fact, they are really the (two) law(s) replacing the q(v×B) expression in classical mechanics.

So how does it work? Let me further follow Feynman’s treatment of the matter—which analyzes what happens when we’d have some magnetic field in the two-slit experiment (so we assume there’s no electric field: we only look at some magnetic field). We said Φ1 was the phase of the wave along trajectory 1, and Φ2 was the phase of the wave along trajectory 2. Without magnetic field, that is, so B = 0. Now, the (first) formula above tells us that, when the field is switched on, the new phases will be the following:

f3

f4

Hence, the phase difference δ = Φ1−Φwill now be equal to:

f5

Now, we can combine the two integrals into one that goes forward along trajectory 1 and comes back along trajectory 2. We’ll denote this path as 1-2 and write the new integral as follows:

f6

Note that we’re using a notation here which suggests that the 1-2 path is closed, which is… Well… Yet another approximation of the Master. In fact, his assumption that the new 1-2 path is closed proves to be essential in the argument that follows the one we presented above, in which he shows that the inherent arbitrariness in our choice of a vector potential function doesn’t matter, but… Well… I don’t want to get too technical here.

Let me conclude this post by noting we can re-write our grand formula above in terms of the flux of the magnetic field B:

f7

So… Well… That’s it, really. I’ll refer you to Feynman’s Lecture on this matter for a detailed description of the 1960 experiment itself, which involves a magnetized iron whisker that acts like a tiny solenoid—small enough to match the tiny scale of the interference experiment itself. I must warn you though: there is a rather long discussion in that Lecture on the ‘reality’ of the magnetic and the vector potential field which – unlike Feynman’s usual approach to discussions like this – is rather philosophical and partially misinformed, as it assumes there is zero magnetic field outside of a solenoid. That’s true for infinitely long solenoids, but not true for real-life solenoids: if we have some A, then we must also have some B, and vice versa. Hence, if the magnetic field (B) is a real field (in the sense that it cannot act on some particle from a distance through some kind of spooky ‘action-at-a-distance’), then the vector potential A is an equally real field—and vice versa. Feynman admits as much as he concludes his rather lengthy philosophical excursion with the following conclusion (out of which I already quoted one line in my introduction to this post):

“This subject has an interesting history. The theory we have described was known from the beginning of quantum mechanics in 1926. The fact that the vector potential appears in the wave equation of quantum mechanics (called the Schrödinger equation) was obvious from the day it was written. That it cannot be replaced by the magnetic field in any easy way was observed by one man after the other who tried to do so. This is also clear from our example of electrons moving in a region where there is no field and being affected nevertheless. But because in classical mechanics A did not appear to have any direct importance and, furthermore, because it could be changed by adding a gradient, people repeatedly said that the vector potential had no direct physical significance—that only the magnetic and electric fields are “real” even in quantum mechanics. It seems strange in retrospect that no one thought of discussing this experiment until 1956, when Bohm and Aharonov first suggested it and made the whole question crystal clear. The implication was there all the time, but no one paid attention to it. Thus many people were rather shocked when the matter was brought up. That’s why someone thought it would be worthwhile to do the experiment to see if it was really right, even though quantum mechanics, which had been believed for so many years, gave an unequivocal answer. It is interesting that something like this can be around for thirty years but, because of certain prejudices of what is and is not significant, continues to be ignored.”

Well… That’s it, folks! Enough for today! 🙂

An interpretation of the wavefunction

This is my umpteenth post on the same topic. 😦 It is obvious that this search for a sensible interpretation is consuming me. Why? I am not sure. Studying physics is frustrating. As a leading physicist puts it:

“The teaching of quantum mechanics these days usually follows the same dogma: firstly, the student is told about the failure of classical physics at the beginning of the last century; secondly, the heroic confusions of the founding fathers are described and the student is given to understand that no humble undergraduate student could hope to actually understand quantum mechanics for himself; thirdly, a deus ex machina arrives in the form of a set of postulates (the Schrödinger equation, the collapse of the wavefunction, etc); fourthly, a bombardment of experimental verifications is given, so that the student cannot doubt that QM is correct; fifthly, the student learns how to solve the problems that will appear on the exam paper, hopefully with as little thought as possible.”

That’s obviously not the way we want to understand quantum mechanics. [With we, I mean, me, of course, and you, if you’re reading this blog.] Of course, that doesn’t mean I don’t believe Richard Feynman, one of the greatest physicists ever, when he tells us no one, including himself, understands physics quite the way we’d like to understand it. Such statements should not prevent us from trying harder. So let’s look for better metaphors. The animation below shows the two components of the archetypal wavefunction – a simple sine and cosine. They’re the same function actually, but their phases differ by 90 degrees (π/2).

circle_cos_sin

It makes me think of a V-2 engine with the pistons at a 90-degree angle. Look at the illustration below, which I took from a rather simple article on cars and engines that has nothing to do with quantum mechanics. Think of the moving pistons as harmonic oscillators, like springs.

two-timer-576-px-photo-369911-s-original

We will also think of the center of each cylinder as the zero point: think of that point as a point where – if we’re looking at one cylinder alone – the internal and external pressure balance each other, so the piston would not move… Well… If it weren’t for the other piston, because the second piston is not at the center when the first is. In fact, it is easy to verify and compare the following positions of both pistons, as well as the associated dynamics of the situation:

Piston 1

Piston 2

Motion of Piston 1

Motion Piston 2

Top

Center

Compressed air will push piston down

Piston moves down against external pressure

Center

Bottom

Piston moves down against external pressure

External air pressure will push piston up

Bottom

Center

External air pressure will push piston up

Piston moves further up and compresses the air

Center

Top

Piston moves further up and compresses the air

Compressed air will push piston down

When the pistons move, their linear motion will be described by a sinusoidal function: a sine or a cosine. In fact, the 90-degree V-2 configuration ensures that the linear motion of the two pistons will be exactly the same, except for a phase difference of 90 degrees. [Of course, because of the sideways motion of the connecting rods, our sine and cosine function describes the linear motion only approximately, but you can easily imagine the idealized limit situation. If not, check Feynman’s description of the harmonic oscillator.]

The question is: if we’d have a set-up like this, two springs – or two harmonic oscillators – attached to a shaft through a crank, would this really work as a perpetuum mobile? We obviously talk energy being transferred back and forth between the rotating shaft and the moving pistons… So… Well… Let’s model this: the total energy, potential and kinetic, in each harmonic oscillator is constant. Hence, the piston only delivers or receives kinetic energy from the rotating mass of the shaft.

Now, in physics, that’s a bit of an oxymoron: we don’t think of negative or positive kinetic (or potential) energy in the context of oscillators. We don’t think of the direction of energy. But… Well… If we’ve got two oscillators, our picture changes, and so we may have to adjust our thinking here.

Let me start by giving you an authoritative derivation of the various formulas involved here, taking the example of the physical spring as an oscillator—but the formulas are basically the same for any harmonic oscillator.

energy harmonic oscillator

The first formula is a general description of the motion of our oscillator. The coefficient in front of the cosine function (a) is the maximum amplitude. Of course, you will also recognize ω0 as the natural frequency of the oscillator, and Δ as the phase factor, which takes into account our t = 0 point. In our case, for example, we have two oscillators with a phase difference equal to π/2 and, hence, Δ would be 0 for one oscillator, and –π/2 for the other. [The formula to apply here is sinθ = cos(θ – π/2).] Also note that we can equate our θ argument to ω0·t. Now, if = 1 (which is the case here), then these formulas simplify to:

  1. K.E. = T = m·v2/2 = m·ω02·sin2(θ + Δ) = m·ω02·sin20·t + Δ)
  2. P.E. = U = k·x2/2 = k·cos2(θ + Δ)

The coefficient k in the potential energy formula characterizes the force: F = −k·x. The minus sign reminds us our oscillator wants to return to the center point, so the force pulls back. From the dynamics involved, it is obvious that k must be equal to m·ω02., so that gives us the famous T + U = m·ω02/2 formula or, including once again, T + U = m·a2·ω02/2.

Now, if we normalize our functions by equating k to one (k = 1), then the motion of our first oscillator is given by the cosθ function, and its kinetic energy will be equal to sin2θ. Hence, the (instantaneous) change in kinetic energy at any point in time will be equal to:

d(sin2θ)/dθ = 2∙sinθ∙d(sinθ)/dt = 2∙sinθ∙cosθ

Let’s look at the second oscillator now. Just think of the second piston going up and down in our V-twin engine. Its motion is given by the sinθ function which, as mentioned above, is equal to cos(θ−π /2). Hence, its kinetic energy is equal to sin2(θ−π /2), and how it changes – as a function of θ – will be equal to:

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

We have our perpetuum mobile! While transferring kinetic energy from one piston to the other, the rotating shaft moves at constant speed. Linear motion becomes circular motion, and vice versa, in a frictionless Universe. We have the metaphor we were looking for!

Somehow, in this beautiful interplay between linear and circular motion, energy is being borrowed from one place to another, and then returned. From what place to what place? I am not sure. We may call it the real and imaginary energy space respectively, but what does that mean? One thing is for sure, however: the interplay between the real and imaginary part of the wavefunction describes how energy propagates through space!

How exactly? Again, I am not sure. Energy is, obviously, mass in motion – as evidenced by the E = m·c2 equation, and it may not have any direction (when everything is said and done, it’s a scalar quantity without direction), but the energy in a linear motion is surely different from that in a circular motion, and our metaphor suggests we need to think somewhat more along those lines. Perhaps we will, one day, able to square this circle. 🙂

Schrödinger’s equation

Let’s analyze the interplay between the real and imaginary part of the wavefunction through an analysis of Schrödinger’s equation, which we write as:

i·ħ∙∂ψ/∂t = –(ħ2/2m)∙∇2ψ + V·ψ

We can do a quick dimensional analysis of both sides:

  • [i·ħ∙∂ψ/∂t] = N∙m∙s/s = N∙m
  • [–(ħ2/2m)∙∇2ψ] = N∙m3/m2 = N∙m
  • [V·ψ] = N∙m

Note the dimension of the ‘diffusion’ constant ħ2/2m: [ħ2/2m] = N2∙m2∙s2/kg = N2∙m2∙s2/(N·s2/m) = N∙m3. Also note that, in order for the dimensions to come out alright, the dimension of V – the potential – must be that of energy. Hence, Feynman’s description of it as the potential energy – rather than the potential tout court – is somewhat confusing but correct: V must equal the potential energy of the electron. Hence, V is not the conventional (potential) energy of the unit charge (1 coulomb). Instead, the natural unit of charge is used here, i.e. the charge of the electron itself.

Now, Schrödinger’s equation – without the V·ψ term – can be written as the following pair of equations:

  1. Re(∂ψ/∂t) = −(1/2)∙(ħ/m)∙Im(∇2ψ)
  2. Im(∂ψ/∂t) = (1/2)∙(ħ/m)∙Re(∇2ψ)

This closely resembles the propagation mechanism of an electromagnetic wave as described by Maxwell’s equation for free space (i.e. a space with no charges), but E and B are vectors, not scalars. How do we get this result. Well… ψ is a complex function, which we can write as a + i∙b. Likewise, ∂ψ/∂t is a complex function, which we can write as c + i∙d, and ∇2ψ can then be written as e + i∙f. If we temporarily forget about the coefficients (ħ, ħ2/m and V), then Schrödinger’s equation – including V·ψ term – amounts to writing something like this:

i∙(c + i∙d) = –(e + i∙f) + (a + i∙b) ⇔ a + i∙b = i∙c − d + e+ i∙f  ⇔ a = −d + e and b = c + f

Hence, we can now write:

  1. V∙Re(ψ) = −ħ∙Im(∂ψ/∂t) + (1/2)∙( ħ2/m)∙Re(∇2ψ)
  2. V∙Im(ψ) = ħ∙Re(∂ψ/∂t) + (1/2)∙( ħ2/m)∙Im(∇2ψ)

This simplifies to the two equations above for V = 0, i.e. when there is no potential (electron in free space). Now we can bring the Re and Im operators into the brackets to get:

  1. V∙Re(ψ) = −ħ∙∂Im (ψ)/∂t + (1/2)∙( ħ2/m)∙∇2Re(ψ)
  2. V∙Im(ψ) = ħ∙∂Re(ψ)/∂t + (1/2)∙( ħ2/m)∙∇2Im(ψ)

This is very interesting, because we can re-write this using the quantum-mechanical energy operator H = –(ħ2/2m)∙∇2 + V· (note the multiplication sign after the V, which we do not have – for obvious reasons – for the –(ħ2/2m)∙∇2 expression):

  1. H[Re (ψ)] = −ħ∙∂Im(ψ)/∂t
  2. H[Im(ψ)] = ħ∙∂Re(ψ)/∂t

A dimensional analysis shows us both sides are, once again, expressed in N∙m. It’s a beautiful expression because – if we write the real and imaginary part of ψ as r∙cosθ and r∙sinθ, we get:

  1. H[cosθ] = −ħ∙∂sinθ/∂t = E∙cosθ
  2. H[sinθ] = ħ∙∂cosθ/∂t = E∙sinθ

Indeed, θ = (E∙t − px)/ħ and, hence, −ħ∙∂sinθ/∂t = ħ∙cosθ∙E/ħ = E∙cosθ and ħ∙∂cosθ/∂t = ħ∙sinθ∙E/ħ = E∙sinθ.  Now we can combine the two equations in one equation again and write:

H[r∙(cosθ + i∙sinθ)] = r∙(E∙cosθ + i∙sinθ) ⇔ H[ψ] = E∙ψ

The operator H – applied to the wavefunction – gives us the (scalar) product of the energy E and the wavefunction itself. Isn’t this strange?

Hmm… I need to further verify and explain this result… I’ll probably do so in yet another post on the same topic… 🙂

Post scriptum: The symmetry of our V-2 engine – or perpetuum mobile – is interesting: its cross-section has only one axis of symmetry. Hence, we may associate some angle with it, so as to define its orientation in the two-dimensional cross-sectional plane. Of course, the cross-sectional plane itself is at right angles to the crankshaft axis, which we may also associate with some angle in three-dimensional space. Hence, its geometry defines two orthogonal directions which, in turn, define a spherical coordinate system, as shown below.

558px-3d_spherical

We may, therefore, say that three-dimensional space is actually being implied by the geometry of our V-2 engine. Now that is interesting, isn’t it? 🙂

Quantum-mechanical operators

I wrote a post on quantum-mechanical operators some while ago but, when re-reading it now, I am not very happy about it, because it tries to cover too much ground in one go. In essence, I regret my attempt to constantly switch between the matrix representation of quantum physics – with the | state 〉 symbols – and the wavefunction approach, so as to show how the operators work for both cases. But then that’s how Feynman approaches this.

However, let’s admit it: while Heisenberg’s matrix approach is equivalent to Schrödinger’s wavefunction approach – and while it’s the only approach that works well for n-state systems – the wavefunction approach is more intuitive, because:

  1. Most practical examples of quantum-mechanical systems (like the description of the electron orbitals of an atomic system) involve continuous coordinate spaces, so we have an infinite number of states and, hence, we need to describe it using the wavefunction approach.
  2. Most of us are much better-versed in using derivatives and integrals, as opposed to matrix operations.
  3. A more intuitive statement of the same argument above is the following: the idea of one state flowing into another, rather than being transformed through some matrix, is much more appealing. 🙂

So let’s stick to the wavefunction approach here. So, while you need to remember that there’s a ‘matrix equivalent’ for each of the equations we’re going to use in this post, we’re not going to talk about it.

The operator idea

In classical physics – high school physics, really – we would describe a pointlike particle traveling in space by a function relating its position (x) to time (t): x = x(t). Its (instantaneous) velocity is, obviously, v(t) = dx/dt. Simple. Obvious. Let’s complicate matters now by saying that the idea of a velocity operator would sort of generalize the v(t) = dx/dt velocity equation by making abstraction of the specifics of the x = x(t) function.

Huh? Yes. We could define a velocity ‘operator’ as:

velocity operator

Now, you may think that’s a rather ridiculous way to describe what an operator does, but – in essence – it’s correct. We have some function – describing an elementary particle, or a system, or an aspect of the system – and then we have some operator, which we apply to our function, to extract the information from it that we want: its velocity, its momentum, its energy. Whatever. Hence, in quantum physics, we have an energy operator, a position operator, a momentum operator, an angular momentum operator and… Well… I guess I listed the most important ones. 🙂

It’s kinda logical. Our velocity operator looks at one particular aspect of whatever it is that’s going on: the time rate of change of position. We do refer to that as the velocity. Our quantum-mechanical operators do the same: they look at one aspect of what’s being described by the wavefunction. [At this point, you may wonder what the other properties of our classical ‘system’ – i.e. other properties than velocity – because we’re just looking at a pointlike particle here, but… Well… Think of electric charge and forces acting on it, so it accelerates and decelerates in all kinds of ways, and we have kinetic and potential energy and all that. Or momentum. So it’s just the same: the x = x(t) function may cover a lot of complexities, just like the wavefunction does!]

The Wikipedia article on the momentum operator is, for a change (I usually find Wikipedia quite abstruse on these matters), quite simple – and, therefore – quite enlightening here. It applies the following simple logic to the elementary wavefunction ψ = ei·(ω·t − k∙x), with the de Broglie relations telling us that ω = E/ħ and k = p/ħ:

mom op 1

Note we forget about the normalization coefficient a here. It doesn’t matter: we can always stuff it in later. The point to note is that we can sort of forget about ψ (or abstract away from it—as mathematicians and physicists would say) by defining the momentum operator, which we’ll write as:

mom op 2

Its three-dimensional equivalent is calculated in very much the same way:

wiki

So this operator, when operating on a particular wavefunction, gives us the (expected) momentum when we would actually catch our particle there, provided the momentum doesn’t vary in time. [Note that it may – and actually is likely to – vary in space!]

So that’s the basic idea of an operator. However, the comparison goes further. Indeed, a superficial reading of what operators are all about gives you the impression we get all these observables (or properties of the system) just by applying the operator to the (wave)function. That’s not the case. There is the randomness. The uncertainty. Actual wavefunctions are superpositions of several elementary waves with various coefficients representing their amplitudes. So we need averages, or expected values: E[X] Even our velocity operator ∂/∂t – in the classical world – gives us an instantaneous velocity only. To get the average velocity (in quantum mechanics, we’ll be interested in the the average momentum, or the average position, or the average energy – rather than the average velocity), we’re going to have the calculate the total distance traveled. Now, that’s going to involve a line integral:

= ∫ds.

The principle is illustrated below.

line integral

You’ll say: this is kids stuff, and it is. Just note how we write the same integral in terms of the x and t coordinate, and using our new velocity operator:

integral

Kids stuff. Yes. But it’s good to think about what it represents really. For example, the simplest quantum-mechanical operator is the position operator. It’s just for the x-coordinate, for the y-coordinate, and z for the z-coordinate. To get the average position of a stationary particle – represented by the wavefunction ψ(r, t) – in three-dimensional space, we need to calculate the following volume integral:

position operator 3D V2

Simple? Yes and no. The r·|ψ(r)|2 integrand is obvious: we multiply each possible position (r) by its probability (or likelihood), which is equal to P(r) = |ψ(r)|2. However, look at the assumptions: we already omitted the time variable. Hence, the particle we’re describing here must be stationary, indeed! So we’ll need to re-visit the whole subject allowing for averages to change with time. We’ll do that later. I just wanted to show you that those integrals – even with very simple operators, like the position operator – can become very complicated. So you just need to make sure you know what you’re looking at.

One wavefunction—or two? Or more?

There is another reason why, with the immeasurable benefit of hindsight, I now feel that my earlier post is confusing: I kept switching between the position and the momentum wavefunction, which gives the impression we have different wavefunctions describing different aspects of the same thing. That’s just not true. The position and momentum wavefunction describe essentially the same thing: we can go from one to the other, and back again, by a simple mathematical manipulation. So I should have stuck to descriptions in terms of ψ(x, t), instead of switching back and forth between the ψ(x, t) and φ(x, t) representations.

In any case, the damage is done, so let’s move forward. The key idea is that, when we know the wavefunction, we know everything. I tried to convey that by noting that the real and imaginary part of the wavefunction must, somehow, represent the total energy of the particle. The structural similarity between the mass-energy equivalence relation (i.e. Einstein’s formula: E = m·c2) and the energy formulas for oscillators and spinning masses is too obvious:

  1. The energy of any oscillator is given by the E = m·ω02/2. We may want to liken the real and imaginary component of our wavefunction to two oscillators and, hence, add them up. The E = m·ω02 formula we get is then identical to the E = m·c2 formula.
  2. The energy of a spinning mass is given by an equivalent formula: E = I·ω2/2 (I is the moment of inertia in this formula). The same 1/2 factor tells us our particle is, somehow, spinning in two dimensions at the same time (i.e. a ‘real’ as well as an ‘imaginary’ space—but both are equally real, because amplitudes interfere), so we get the E = I·ω2 formula. 

Hence, the formulas tell us we should imagine an electron – or an electron orbital – as a very complicated two-dimensional standing wave. Now, when I write two-dimensional, I refer to the real and imaginary component of our wavefunction, as illustrated below. What I am asking you, however, is to not only imagine these two components oscillating up and down, but also spinning about. Hence, if we think about energy as some oscillating mass – which is what the E = m·c2 formula tells us to do, we should remind ourselves we’re talking very complicated motions here: mass oscillates, swirls and spins, and it does so both in real as well as in imaginary space.  rising_circular

What I like about the illustration above is that it shows us – in a very obvious way – why the wavefunction depends on our reference frame. These oscillations do represent something in absolute space, but how we measure it depends on our orientation in that absolute space. But so I am writing this post to talk about operators, not about my grand theory about the essence of mass and energy. So let’s talk about operators now. 🙂

In that post of mine, I showed how the position, momentum and energy operator would give us the average position, momentum and energy of whatever it was that we were looking at, but I didn’t introduce the angular momentum operator. So let me do that now. However, I’ll first recapitulate what we’ve learnt so far in regard to operators.

The energy, position and momentum operators

The equation below defines the energy operator, and also shows how we would apply it to the wavefunction:

energy operator

To the purists: sorry for not (always) using the hat symbol. [I explained why in that post of mine: it’s just too cumbersome.] The others 🙂 should note the following:

  • Eaverage is also an expected value: Eav = E[E]
  • The * symbol tells us to take the complex conjugate of the wavefunction.
  • As for the integral, it’s an integral over some volume, so that’s what the d3r shows. Many authors use double or triple integral signs (∫∫ or ∫∫∫) to show it’s a surface or a volume integral, but that makes things look very complicated, and so I don’t that. I could also have written the integral as ∫ψ(r)*·H·ψ(r) dV, but then I’d need to explain that the dV stands for dVolume, not for any (differental) potential energy (V).
  • We must normalize our wavefunction for these formulas to work, so all probabilities over the volume add up to 1.

OK. That’s the energy operator. As you can see, it’s a pretty formidable beast, but then it just reflects Schrödinger’s equation which, as I explained a couple of times already, we can interpret as an energy propagation mechanism, or an energy diffusion equation, so it is actually not that difficult to memorize the formula: if you’re able to remember Schrödinger’s equation, then you’ll also have the operator. If not… Well… Then you won’t pass your undergrad physics exam. 🙂

I already mentioned that the position operator is a much simpler beast. That’s because it’s so intimately related to our interpretation of the wavefunction. It’s the one thing you know about quantum mechanics: the absolute square of the wavefunction gives us the probability density function. So, for one-dimensional space, the position operator is just:

position operator

The equivalent operator for three-dimensional space is equally simple:

position operator 3D V2

Note how the operator, for the one- as well as for the three-dimensional case, gets rid of time as a variable. In fact, the idea itself of an average makes abstraction of the temporal aspect. Well… Here, at least—because we’re looking at some box in space, rather than some box in spacetime. We’ll re-visit that rather particular idea of an average, and allow for averages that change with time, in a short while.

Next, we introduced the momentum operator in that post of mine. For one dimension, Feynman shows this operator is given by the following formula:

momentum operator

Now that does not look very simple. You might think that the ∂/∂x operator reflects our velocity operator, but… Well… No: ∂/∂t gives us a time rate of change, while ∂/∂x gives us the spatial variation. So it’s not the same. Also, that ħ/i factor is quite intriguing, isn’t it? We’ll come back to it in the next section of this post. Let me just give you the three-dimensional equivalent which, remembering that 1/i = −i, you’ll understand to be equal to the following vector operator:

momentum vector operator

Now it’s time to define the operator we wanted to talk about, i.e. the angular momentum operator.

The angular momentum operator

The formula for the angular momentum operator is remarkably simple:

angular momentum operator

Why do I call this a simple formula? Because it looks like the familiar formula of classical mechanics for the z-component of the classical angular momentum L = r × p. I must assume you know how to calculate a vector cross product. If not, check one of my many posts on vector analysis. I must also assume you remember the L = r × p formula. If not, the following animation might bring it all back. If that doesn’t help, check my post on gyroscopes. 🙂

torque_animation-1.gif

Now, spin is a complicated phenomenon, and so, to simplify the analysis, we should think of orbital angular momentum only. This is a simplification, because electron spin is some complicated mix of intrinsic and orbital angular momentum. Hence, the angular momentum operator we’re introducing here is only the orbital angular momentum operator. However, let us not get bogged down in all of the nitty-gritty and, hence, let’s just go along with it for the time being.

I am somewhat hesitant to show you how we get that formula for our operator, but I’ll try to show you using an intuitive approach, which uses only bits and pieces of Feynman’s more detailed derivation. It will, hopefully, give you a bit of an idea of how these differential operators work. Think about a rotation of our reference frame over an infinitesimally small angle – which we’ll denote as ε – as illustrated below.

rotation

Now, the whole idea is that, because of that rotation of our reference frame, our wavefunction will look different. It’s nothing fundamental, but… Well… It’s just because we’re using a different coordinate system. Indeed, that’s where all these complicated transformation rules for amplitudes come in.  I’ve spoken about these at length when we were still discussing n-state systems. In contrast, the transformation rules for the coordinates themselves are very simple:

rotation

Now, because ε is an infinitesimally small angle, we may equate cos(θ) = cos(ε) to 1, and cos(θ) = sin(ε) to ε. Hence, x’ and y’ are then written as x’+ εy and y’− εx, while z‘ remains z. Vice versa, we can also write the old coordinates in terms of the new ones: x = x’ − εy, y = y’ + εx, and zThat’s obvious. Now comes the difficult thing: you need to think about the two-dimensional equivalent of the simple illustration below.

izvod

If we have some function y = f(x), then we know that, for small Δx, we have the following approximation formula for f(x + Δx): f(x + Δx) ≈ f(x) + (dy/dx)·Δx. It’s the formula you saw in high school: you would then take a limit (Δ0), and define dy/dx as the Δy/Δx ratio for Δ0. You would this after re-writing the f(x + Δx) ≈ f(x) + (dy/dx)·Δx formula as:

Δy = Δf = f(x + Δx) − f(x) ≈ (dy/dx)·Δx

Now you need to substitute f for ψ, and Δx for ε. There is only one complication here: ψ is a function of two variables: x and y. In fact, it’s a function of three variables – x, y and z – but we keep constant. So think of moving from and to + εy = + Δand to + Δ− εx. Hence, Δ= εy and Δ= −εx. It then makes sense to write Δψ as:

angular momentum operator v2

If you agree with that, you’ll also agree we can write something like this:

formula 2

Now that implies the following formula for Δψ:

repair

This looks great! You can see we get some sort of differential operator here, which is what we want. So the next step should be simple: we just let ε go to zero and then we’re done, right? Well… No. In quantum mechanics, it’s always a bit more complicated. But it’s logical stuff. Think of the following:

1. We will want to re-write the infinitesimally small ε angle as a fraction of i, i.e. the imaginary unit.

Huh? Yes. This little represents many things. In this particular case, we want to look at it as a right angle. In fact, you know multiplication with i amounts to a rotation by 90 degrees. So we should replace ε by ε·i. It’s like measuring ε in natural units. However, we’re not done.

2. We should also note that Nature measures angles clockwise, rather than counter-clockwise, as evidenced by the fact that the argument of our wavefunction rotates clockwise as time goes by. So our ε is, in fact, a −ε. We will just bring the minus sign inside of the brackets to solve this issue.

Huh? Yes. Sorry. I told you this is a rather intuitive approach to getting what we want to get. 🙂

3. The third modification we’d want to make is to express ε·i as a multiple of Planck’s constant.

Huh? Yes. This is a very weird thing, but it should make sense—intuitively: we’re talking angular momentum here, and its dimension is the same as that of physical action: N·m·s. Therefore, Planck’s quantum of action (ħ = h/2π ≈ 1×10−34 J·s ≈ 6.6×10−16 eV·s) naturally appears as… Well… A natural unit, or a scaling factor, I should say.

To make a long story short, we’ll want to re-write ε as −(i/ħ)·ε. However, there is a thing called mathematical consistency, and so, if we want to do such substitutions and prepare for that limit situation (ε → 0), we should re-write that Δψ equation as follows:

final

So now – finally! – we do have the formula we wanted to find for our angular momentum operator:

final 2

The final substitution, which yields the formula we just gave you when commencing this section, just uses the formula for the linear momentum operator in the x– and y-direction respectively. We’re done! 🙂 Finally! 

Well… No. 🙂 The question, of course, is the same as always: what does it all mean, really? That’s always a great question. 🙂 Unfortunately, the answer is rather boring: we can calculate the average angular momentum in the z-direction, using a similar integral as the one we used to get the average energy, or the average linear momentum in some direction. That’s basically it.

To compensate for that very boring answer, however, I will show you something that is far less boring. 🙂

Quantum-mechanical weirdness

I’ll shameless copy from Feynman here. He notes that many classical equations get carried over into a quantum-mechanical form (I’ll copy some of his illustrations later). But then there are some that don’t. As Feynman puts it—rather humorously: “There had better be some that don’t come out right, because if everything did, then there would be nothing different about quantum mechanics. There would be no new physics.” He then looks at the following super-obvious equation in classical mechanics:

x·p− px·x = 0

In fact, this equation is so super-obvious that it’s almost meaningless. Almost. It’s super-obvious because multiplication is commutative (for real as well for complex numbers). However, when we replace x and pby the position and momentum operator, we get an entirely different result. You can verify the following yourself:

strange

This is plain weird! What does it mean? I am not sure. Feynman’s take on it is nice but leaves us in the dark on it:

Feynman quote 2

He adds: “If Planck’s constant were zero, the classical and quantum results would be the same, and there would be no quantum mechanics to learn!” Hmm… What does it mean, really? Not sure. Let me make two remarks here:

1. We should not put any dot (·) between our operators, because they do not amount to multiplying one with another. We just apply operators successively. Hence, commutativity is not what we should expect.

2. Note that Feynman forgot to put the subscript in that quote. When doing the same calculations for the equivalent of the x·p− py·x expression, we do get zero, as shown below:

not strange

These equations – zero or not – are referred to as ‘commutation rules’. [Again, I should not have used any dot between x and py, because there is no multiplication here. It’s just a separation mark.] Let me quote Feynman on it, so the matter is dealt with:

quote

OK. So what do we conclude? What are we talking about?

Conclusions

Some of the stuff above was really intriguing. For example, we found that the linear and angular momentum operators are differential operators in the true sense of the word. The angular momentum operator shows us what happens to the wavefunction if we rotate our reference frame over an infinitesimally small angle ε. That’s what’s captured by the formulas we’ve developed, as summarized below:

angular momentum

Likewise, the linear momentum operator captures what happens to the wavefunction for an infinitesimally small displacement of the reference frame, as shown by the equivalent formulas below:

linear momentum

What’s the interpretation for the position operator, and the energy operator? Here we are not so sure. The integrals above make sense, but these integrals are used to calculate averages values, as opposed to instantaneous values. So… Well… There is not all that much I can say about the position and energy operator right now, except… Well… We now need to explore the question of how averages could possibly change over time. Let’s do that now.

Averages that change with time

I know: you are totally quantum-mechanicked out by now. So am I. But we’re almost there. In fact, this is Feynman’s last Lecture on quantum mechanics and, hence, I think I should let the Master speak here. So just click on the link and read for yourself. It’s a really interesting chapter, as he shows us the equivalent of Newton’s Law in quantum mechanics, as well as the quantum-mechanical equivalent of other standard equations in classical mechanics. However, I need to warn you: Feynman keeps testing the limits of our intellectual absorption capacity by switching back and forth between matrix and wave mechanics. Interesting, but not easy. For example, you’ll need to remind yourself of the fact that the Hamiltonian matrix is equal to its own complex conjugate (or – because it’s a matrix – its own conjugate transpose.

Having said that, it’s all wonderful. The time rate of change of all those average values is denoted by using the over-dot notation. For example, the time rate of change of the average position is denoted by:

p1

Once you ‘get’ that new notation, you will quickly understand the derivations. They are not easy (what derivations are in quantum mechanics?), but we get very interesting results. Nice things to play with, or think about—like this identity:

formula2

It takes a while, but you suddenly realize this is the equivalent of the classical dx/dtv = p/m formula. 🙂

Another sweet result is the following one:

formula3

This is the quantum-mechanical equivalent of Newton’s force law: F = m·a. Huh? Yes. Think of it: the spatial derivative of the (potential) energy is the force. Now just think of the classical dp/dt = d(m·v) = m·dv/dt = m·a formula. […] Can you see it now? Isn’t this just Great Fun?

Note, however, that these formulas also show the limits of our analysis so far, because they treat m as some constant. Hence, we’ll need to relativistically correct them. But that’s complicated, and so we’ll postpone that to another day.

[…]

Well… That’s it, folks! We’re really through! This was the last of the last of Feynman’s Lectures on Physics. So we’re totally done now. Isn’t this great? What an adventure! I hope that, despite the enormous mental energy that’s required to digest all this stuff, you enjoyed it as much as I did. 🙂

Post scriptum 1: I just love Feynman but, frankly, I think he’s sometimes somewhat sloppy with terminology. In regard to what these operators really mean, we should make use of better terminology: an average is something else than an expected value. Our momentum operator, for example, as such returns an expected value – not an average momentum. We need to deepen the analysis here somewhat, but I’ll also leave that for later.

Post scriptum 2: There is something really interesting about that i·ħ or −(i/ħ) scaling factor – or whatever you want to call it – appearing in our formulas. Remember the Schrödinger equation can also be written as:

i·ħ·∂ψ/∂t = −(1/2)·(ħ2/m)∇2ψ + V·ψ = Hψ

This is interesting in light of our interpretation of the Schrödinger equation as an energy propagation mechanism. If we write Schrödinger’s equation like we write it here, then we have the energy on the right-hand side – which is time-independent. How do we interpret the left-hand side now? Well… It’s kinda simple, but we just have the time rate of change of the real and imaginary part of the wavefunction here, and the i·ħ factor then becomes a sort of unit in which we measure the time rate of change. Alternatively, you may think of ‘splitting’ Planck’s constant in two: Planck’s energy, and Planck’s time unit, and then you bring the Planck energy unit to the other side, so we’d express the energy in natural units. Likewise, the time rate of change of the components of our wavefunction would also be measured in natural time units if we’d do that.

I know this is all very abstract but, frankly, it’s crystal clear to me. This formula tells us that the energy of the particle that’s being described by the wavefunction is being carried by the oscillations of the wavefunction. In fact, the oscillations are the energy. You can play with the mass factor, by moving it to the left-hand side too, or by using Einstein’s mass-energy equivalence relation. The interpretation remains consistent.

In fact, there is something really interesting here. You know that we usually separate out the spatial and temporal part of the wavefunction, so we write: ψ(r, t) = ψ(rei·(E/ħ)·t. In fact, it is quite common to refer to ψ(r) – rather than to ψ(r, t) – as the wavefunction, even if, personally, I find that quite confusing and misleading (see my page onSchrödinger’s equation). Now, we may want to think of what happens when we’d apply the energy operator to ψ(r) rather than to ψ(r, t). We may think that we’d get a time-independent value for the energy at that point in space, so energy is some function of position only, not of time. That’s an interesting thought, and we should explore it. For example, we then may think of energy as an average that changes with position—as opposed to the (average) position and momentum, which we like to think of as averages than change with time, as mentioned above. I will come back to this later – but perhaps in another post or so. Not now. The only point I want to mention here is the following: you cannot use ψ(r) in Schrödinger’s equation. Why? Well… Schrödinger’s equation is no longer valid when substituting ψ for ψ(r), because the left-hand side is always zero, as ∂ψ(r)/∂t is zero – for any r.

There is another, related, point to this observation. If you think that Schrödinger’s equation implies that the operators on both sides of Schrödinger’s equation must be equivalent (i.e. the same), you’re wrong:

i·ħ·∂/∂t ≠ H = −(1/2)·(ħ2/m)∇2 + V

It’s a basic thing, really: Schrödinger’s equation is not valid for just any function. Hence, it does not work for ψ(r). Only ψ(r, t) makes it work, because… Well… Schrödinger’s equation gave us ψ(r, t)!