Some thoughts on the nature of reality

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

Physical dimensions and Uncertainty

The physical dimension of the quantum of action (h or聽魔 = h/2蟺) is force (expressed in newton)聽times distance (expressed in meter)聽times time (expressed in seconds): N路m路s. This is also the unit in which angular momentum is expressed. Of course, a force of one newton will give a mass of 1 kg an acceleration of 1 m/s per second. Therefore, 1 N = 1 kg路m/s2聽and the physical dimension of h, or the unit of angular momentum, may also be written as 1 N路m路s = 1 (kg路m/s2)路m路s = 1 kg路m2/s.

The newton is a聽derived聽unit in the metric system, as opposed to the units of mass, length and time (kg, m, s). Nevertheless, I like to think of the quantum of action as representing the three fundamental physical dimensions: (1)聽force, (2)聽time and (3) distance – or space. We may then look at energy and (linear) momentum as physical quantities combining (1) force and distance and (2) force and time respectively.

  1. Force times length (think of force that is聽acting on some object over some distance) is energy: 1 joule聽(J) =聽1 newtonmeter (N). Hence, we may think of the concept of energy as a projection聽of action in space only: we make abstraction of time. The physical dimension of the quantum of action should then be written as [h] = [E]路[t]
  2. Conversely, the magnitude of linear momentum (p = m路v) is expressed in newtonseconds: 1 kg路m/s = 1 (kg路m/s2)路s = 1 N路s. Hence, we may think of (linear) momentum as a projection of action in time only: we make abstraction of its spatial dimension. Think of a force that is acting on some object聽during some time.聽The physical dimension of the quantum of action should then be written as [h] = [p]路[x]

Of course, a force that is acting on some object during some time, will usually also act on the same object over some distance but… Well… Just try to make abstraction of one of the two dimensions here: time聽or聽distance. It is a difficult thing to do because, when everything is said and done, we don’t live in space or in time alone, but in spacetime and, hence, such abstractions are not easy. Also, the聽principle of least action聽in physics tells us it’s action that matters:

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

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

  • 螖E路螖t =
  • 螖p路螖x =

It is best to think of the uncertainty relations as a聽pair聽of equations, if only because you should also think of the concept of energy and momentum as representing different aspects聽of the same reality, as evidenced by the (relativistic) energy-momentum relation (E2聽= p2c2聽鈥 m02c4). Also, as illustrated below, the actual path – or, to be more precise, what we might associate with the concept of the actual path – is likely to be some mix of 螖x and 螖t. If 螖t is very small, then 螖x will be very large. In order to move over such distance, our particle will require a larger energy, so 螖E will be large. Likewise, if 螖t is very large, then 螖x will be very small and, therefore, 螖E will be very small. You can also reason in terms of 螖x, and talk about momentum rather than energy. You will arrive at the same conclusions: the 螖E路螖t = h and 螖p路螖x = h聽relations represent two aspects of the same reality – or, at the very least, what we might聽think聽of as reality.

Uncertainty

We will not further dwell on this here. We want to do some more thinking about those physical dimensions. The idea of a force implies the idea of some object – of some mass on which the force is acting. Hence, let’s think about the concept of mass now.

Note: The actual uncertainty relations have a factor 1/2 in them. This may be explained by thinking of both negative as well as positive variations in space and in time.

Action, energy and mass

Let’s look at the concept of energy once more. What is聽energy, really? In聽real life, we are usually not interested in the energy of a system as such, but by the energy it can deliver, or absorb, per second. This is referred to as the聽power聽of a system, and it’s expressed in J/s. However, in physics, we always talk energy, so what is the energy of a system?

We should – and will – obviously think of the kinetic energy of its parts, their potential energy, their rest聽energy, and – for an atomic system – we may add some internal energy, which may be binding energy, or excitation energy (think of a hydrogen atom in an excited state, for example). Einstein’s mass-equivalence formula comes to mind here: E = m路c2. [The m here refers to mass – not to meter, obviously.] But then… Well… What is it, really?

As I explained in several posts, it is very tempting to think of energy as some kind of two-dimensional oscillation of mass. A force over some distance will cause a mass to accelerate. This is reflected in the聽dimensional analysis:

[E] = [m]路[c2] = 1 kg路m2/s2聽= 1 kg路m/s2路m = 1 N路m

The kg and m/s2聽factors make this abundantly clear: m/s2聽is the physical dimension of acceleration: (the change in) velocity per time unit.

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

E/f聽= E/(n/s) = E路s/n聽= 6.626070040(81)脳10鈭34聽J路s 鈬 E/n聽= 6.626070040(81)脳10鈭34聽J

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

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

Well… What’s an oscillation? We’re used to聽counting聽them:聽n聽oscillations per second, so that’s聽per time unit. How many do we have in total? We wrote about that in our posts on the shape and size of a photon. We know photons are emitted by atomic oscillators – or, to put it simply, just atoms going from one energy level to another. Feynman calculated the Q of these atomic oscillators: it鈥檚 of the order of 108聽(see his聽Lectures,聽I-33-3: it鈥檚 a wonderfully simple exercise, and one that really shows his greatness as a physics teacher), so… Well… This wave train will last about 10鈥8聽seconds (that鈥檚 the time it takes for the radiation to die out by a factor 1/e). To give a somewhat more precise example,聽for sodium light, which has a frequency of 500 THz (500脳1012聽oscillations per second) and a wavelength of 600 nm (600脳10鈥9聽meter), the radiation will lasts about 3.2脳10鈥8聽seconds. [In fact, that鈥檚 the time it takes for the radiation鈥檚 energy to die out by a factor 1/e, so(i.e. the so-called decay time 蟿), so the wavetrain will actually last聽longer, but so the amplitude becomes quite small after that time.]聽So… Well… That鈥檚 a very short time but… Still, taking into account the rather spectacular frequency (500 THz) of sodium light, that makes for some 16 million oscillations and, taking into the account the rather spectacular speed of light (3脳108聽m/s), that makes for a wave train with a length of, roughly,聽9.6 meter. Huh? 9.6 meter!? But a photon is supposed to be pointlike, isn’it it? It has no length, does it?

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

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

The reality of the wavefunction

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

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

蠄(r, t) = ei路(E/魔)路tf(r)

As I explained in previous posts (see, for example, my recent post聽on reality and perception), the聽f(r) function basically provides some envelope for the two-dimensional ei路胃聽=聽ei路(E/魔)路t聽= cos胃 + isin胃聽oscillation, with r= (x, y, z),聽胃 = (E/魔)路t聽= 蠅路t聽and 蠅 = E/魔. So it presumes the聽duration of each oscillation is some constant. Why? Well… Look at the formula: this thing has a constant frequency in time. It’s only the amplitude that is varying as a function of the r= (x, y, z) coordinates. 馃檪 So… Well… If each oscillation is to always聽pack聽6.626070040(81)脳10鈭34joule, but the amplitude of the oscillation varies from point to point, then… Well… We’ve got a problem. The wavefunction above is likely to be an approximation of reality only. 馃檪 The associated energy is the same, but… Well… Reality is probably聽not聽the nice geometrical shape we associate with those wavefunctions.

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

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

oscillation of a ball

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聽elementary聽wavefunction聽蠄 =聽aei胃聽= 蠄 =聽aei路胃聽聽= aei(蠅路t鈭択路x)聽= ae(i/魔)路(E路t鈭抪路x)聽.AnimationWe know this elementary wavefunction cannotrepresent a real-life聽particle. Indeed, the aei路胃聽function implies the probability of finding the particle – an electron, a photon, or whatever – would be equal to P(x, t) = |蠄(x, t)|2聽= |ae(i/魔)路(E路t鈭抪路x)|2聽= |a|2路|e(i/魔)路(E路t鈭抪路x)|2聽= |a|2路12= a2everywhere. 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聽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聽contribution聽to the composite wave, which – in three-dimensional space – we can write as:

蠄(r, t) = ei路(E/魔)路tf(r)

As I explained in previous posts (see, for example, my recent post聽on reality and perception), the聽f(r) function basically provides some envelope for the two-dimensional ei路胃聽=聽ei路(E/魔)路t聽= cos胃 + isin胃聽oscillation, with r= (x, y, z),聽胃 = (E/魔)路t聽= 蠅路t聽and 蠅 = E/魔.

Note that it looks like the wave propagates聽from left to right – in the聽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聽rotation聽of that arrow at the center – i.e. with the motion in the illustration,聽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聽us –聽would then be the y-axis, obviously.聽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聽rotation聽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聽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 negativex-direction. [Note that this left- or right-hand rule is quite astonishing: simply swapping the direction of聽one聽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聽go in this or that direction. And I mean that: there is no real travelinghere. 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聽group聽or the聽envelope聽of the wave鈥攚hatever you want to call it) moves to the right. The point is: the pulse itself doesn’t聽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聽that is traveling down the rope.聽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 鈥榯raveling鈥 left or right. That鈥檚 why there鈥檚 no limit on the phase velocity: it can聽– and, according to quantum mechanics, actually will聽–聽exceed the speed of light. In contrast, the group聽velocity – which is the actual speed of the particle that is being represented by the wavefunction – may approach聽– or, in the case of a massless photon, will actually equal聽–聽the speed of light, but will never exceed聽it, and its聽direction聽will, obviously, have a聽physical聽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聽spin聽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 aei(蠅路t鈭択路x)聽rather than聽aei(蠅路t+k路x)聽or聽aei(蠅路t鈭択路x): it’s just because of the right-hand rule for coordinate frames, and also because Euler defined the counter-clockwise direction as the聽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路c2聽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 ae(i/魔)路(E路t鈭抪路x)聽reduces to ae(i/魔)路(E’路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聽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聽to聽accelerate. To be precise, the聽newton聽– as the unit of force – is defined as the聽magnitude of a force which would cause a mass of one kg to accelerate with one meter per second聽per second. Per second per second. This is not a typo: 1 N corresponds to 1 kg times 1 m/s聽per second, i.e. 1 kg路m/s2. So… Because energy is force times distance, the unit of energy聽may be expressed in units of kg路m/s2路m, or kg路m2/s2, i.e. the unit of mass times the unit of聽velocity squared. To sum it all up:

1 J = 1 N路m = 1 kg路(m/s)2

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

circle_cos_sin

Now, the blue and the red dot – i.e. the horizontal and vertical projection聽of the green dot –聽accelerate up and down. If we look carefully, we see these dots accelerate聽towards聽the zero point and, once they’ve crossed it, they聽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掳 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,聽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聽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聽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聽over some angle.聽Indeed, the analogy between linear and angular movement is obvious: the聽kinetic聽energy of a rotating object is equal to K.E. = (1/2)路I路蠅2. In this formula, I is the rotational inertia聽– i.e. the rotational equivalent of mass – and 蠅 is the angular velocity – i.e. the rotational equivalent of linear聽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聽similar to the聽E = m路c2聽formula. But is聽it the same? Is the effective mass of some object the sum of an almost infinite number of quanta聽that incorporate some kind of聽rotational聽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聽articulate聽or聽imagine聽what might be going on. 馃檪 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聽length聽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聽not聽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. 馃檪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聽waves, as opposed to longitudinal.聽See how string theory starts making sense? 馃檪

rayleighwaveThe most fundamental question remains the same: what is it,聽exactly, that is oscillating here? What is the聽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聽electric聽charge,聽really? So the question is: what’s the聽reality聽of mass, of electric charge, or whatever other charge that causes a force to聽act聽on it?

If you聽know, please let聽me聽know. 馃檪

Post scriptum: The fact that we’re talking some聽two-dimensional oscillation here – think of a surface now – explains the probability formula: we need to聽square聽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聽蟺 somewhere, because we’re talking some聽circular聽surface – as opposed to a rectangular one. But I’ll let聽you聽figure that out. 馃檪

Re-visiting electron orbitals (III)

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

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

蠄(x, t) = ei路(E/魔)路t路蠄(x)

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

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

Now you should cry foul, of course: 蠄(x) is not聽necessarily聽real-valued. It may聽be complex-valued. You’re right.聽You know the formula:wavefunctionNote the derivation of this formula involved a switch from Cartesian to polar coordinates here, so from x聽= (x, y, z) to r= (r, 胃, 蠁), and that the function is also a function of the two聽quantum numbers聽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(胃, 蠁) and Fl,m(r) respectively. Fl,m(r) was a real-valued function alright, but the Yl,m(胃, 蠁) had that eim路蠁聽factor in it. So… Yes. You’re right: the Yl,m(胃, 蠁) function is real-valued if – and only聽if – m = 0, in which case eim路蠁聽= 1.聽Let me copy the table from Feynman’s treatment of the topic once again:spherical harmonics 2The Plm(cos胃) 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(cos胃)聽is a聽real-valued聽function. The point is the following:the聽蠄(x, t) is a complex-valued聽function because – and聽only聽because – we multiply a real-valued envelope function – which depends on position聽only – with ei路(E/魔)路teim路蠁聽= ei路[(E/魔)路t聽鈭捖m路蠁].

[…]

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

  • As mentioned in my previous post,聽the eim路蠁聽factor just gives us phase shift: just a聽re-set of our zero point for measuring time, so to speak, and the whole ei路[(E/魔)路t聽鈭捖m路蠁]聽factor just disappears when we鈥檙e calculating probabilities.
  • The envelope function gives us the basic amplitude – in the聽classical聽sense of the word:聽the maximum displacement from聽the聽zero聽value. And so it’s that ei路[(E/魔)路t聽鈭捖m路蠁]聽that ensures the whole expression somehow captures the energy聽of the oscillation.

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

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

n = 5

The first symmetry plane – or symmetry聽line, really (i.e. the聽z-axis) – should not surprise us, because the azimuthal angle 蠁 is conspicuously absent in the formula for our envelope function if, as we are doing in this article here, we merge the聽eim路蠁聽factor with the ei路(E/魔)路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.聽OK. Clear enough – I hope. 馃檪 But why is the聽the xy-plane a symmetry plane too? We need to look at that monstrous formula for the Plm(cos胃) function here: just note the cos胃 argument in it is being squared聽before it’s used in all of the other manipulation. Now, we know that cos胃 = sin(蟺/2聽鈭捖犖). So we can define some聽new聽angle – let’s just call it 伪 – which is measured in the way we’re used to measuring angle, which is not聽from the z-axis but from the xy-plane. So we write: cos胃 = sin(蟺/2聽鈭捖犖) = sin伪. 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 cos胃 argument in the Plm(cos胃) function for sin伪 = sin(蟺/2聽鈭捖犖). Now, if the xy-plane is a symmetry plane, then we must find the same value for Plm(sin伪) and Plm[sin(鈭捨)]. Now, that’s not obvious, because sin(鈭捨) = 鈭sin伪 鈮犅sin伪. 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: [鈭sin伪]2聽= [sin伪]2聽=聽sin2伪. […] 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聽operator, and so you should check what happens with the minus sign there. 馃檪

[…] 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:

  • A聽definite聽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聽function in three dimensions, really – which聽has the z-axis – i.e. the vertical axis – as a symmetry line and the xy-plane as a symmetry plane.
  • The ei路[(E/魔)路t聽鈭捖m路蠁]聽factor gives us the oscillation within the envelope function. As such, it’s this factor that, somehow,聽captures the energy聽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 eim路蠁聽factor just gives us phase shift: just a聽re-set of our zero point for measuring time, so to speak. Interesting, weird – but probably less relevant than the ei路[(E/魔)路t聽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聽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聽on a unit charge, the real and imaginary part of our wavefunction must also represent some rapidly changing聽force聽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聽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聽mass, but then I am not sure how to define that unit right now (it’s probably not聽the kilogram!).

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

Let me just add one more note on the eim路蠁聽factor. It sort of defines the geometry聽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 n聽= 4 and l聽= 3. Look at the red hues in particular – or the blue – whatever: focus on one color only, and see how how – for m= 卤1, we’ve got one appearance of that color only. For m= 卤1, the same color appears at two ends of the ‘tubes’ – or tori聽(plural of torus), I should say – just to sound more professional. 馃檪 For m= 卤2, the torus consists of three parts – or, in mathematical terms, we’d say the order of its rotational symmetry聽is equal to 3.聽Check that Wikimedia Commons article for higher values of聽n聽and聽l: the shapes become very convoluted, but the observation holds. 馃檪

l = 3

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

Post scriptum: You should do some thinking on whether or not these聽m聽=聽卤1, 卤2,…, 卤l聽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聽a聽real聽difference, is it? 馃檪 As with other stuff, I’ll let you think about this for yourself.

Feynman’s Lecture on Superconductivity

The ultimate challenge for students of Feynman’s iconic Lectures聽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聽present the detail of each and every step in the development and, therefore,聽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聽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 a聽to聽b聽in a vector potential (think of a classical magnetic field) is the amplitude for the same particle to go from聽a聽to聽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聽the phase of the amplitude of a particle with an amount equal to:

integral

Hence, if we write 鈱b|a鈱 for A = 0 as 鈱b|aA = 0聽= Cei, then 鈱b|a鈱in Awill, naturally, be equal to 鈱b|a鈱in A = Cei(胃+蠁)聽= Ceiei聽=聽鈱b|aA = 0聽ei, and so that explains it. 馃檪 Alright… Next.

The Schr枚dinger equation in an electromagnetic field

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

schrodinger

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

i路魔路鈭傁/鈭倀 = 鈭(1/2)路(魔2/m)鈭2蠄 + V路蠄

Of course, it is easy to see that we replaced V by q路桅, which makes sense: the potential of a charge in an electric field is the product of the charge (q) and the (electric) potential (桅), because 桅 is, obviously, the potential energy of the unit聽charge. It’s also easy to see we can re-write 鈭捘2路鈭2蠄 as [(魔/i)路鈭嘳路[(魔/i)路鈭嘳蠄 because (1/i)路(1/i) = 1/i2聽= 1/(鈭1) = 鈭1. 馃檪 Alright. So it’s just that 鈭抭路A聽term in the (魔/i)鈭 鈭 q路Aexpression that we need to explain now.

Unfortunately, that explanation is聽not聽so easy. Feynman basically re-derives Schr枚dinger’s equation using his trade-mark聽historical argument – which did not聽include any magnetic field –聽with聽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 魔/i)路鈭 operator聽– which is the momentum operator– ought to be replaced by a new momentum operator:

new-momentum-operator

So… Well… There we are… 馃檪 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聽equation of continuity for probabilities. I find it brilliant, because it confirms聽my聽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 Schr枚dinger 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.聽If energy is disappearing, there must be a flow of energy to correspond. In the same way, we would like to find a 鈥渃urrent鈥 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聽flow聽– or probability聽currentas Feynman refers to it? Well… Here’s the formula:

probability-current-2

Huh?聽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聽he聽explains how聽pairs of electrons聽start behaving when temperatures are low enough to render Boltzmann’s Law irrelevant: the kinetic energy that’s associated with temperature聽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聽on the subject-matter in a, hopefully, somewhat more readable format – we calculated electron orbitals neglecting spin. In Feynman’s words:

“We make another聽approximation by forgetting that the electron has spin. […] The non-relativistic Schr枚dinger equation disregards magnetic effects. [However] Small magnetic effects [do] occur because, from the electron鈥檚 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鈥攚hat is usually called 鈥渙rbital鈥 angular momentum鈥攚ill also be conserved. To an excellent approximation the electron moves in the hydrogen atom like a particle without spin鈥攖he 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聽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聽Lecture), so I must assume the theory is well accepted now. 馃檪

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. 馃檪 And I am really talking the latest聽breakthrough: Science聽just published the findings of this experiment last month! 馃檪 馃檪 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聽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. 馃槮

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

An interpretation of the wavefunction

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

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

That’s obviously not the way we want to understand quantum mechanics. [With we,聽I mean, me, of course, and you, if you’re reading this blog.]聽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聽like聽to understand it. Such statements should not prevent us from trying聽harder. So let’s look for better聽metaphors.聽The animation below shows the two components of the archetypal wavefunction 鈥 a simple sine and cosine. They’re the same function actually, but their phases differ by 90 degrees (蟺/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聽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聽not at the center聽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聽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聽being transferred back and forth between the rotating shaft and the moving pistons… So… Well… Let’s model this: the total聽energy, potential聽and聽kinetic, in each harmonic oscillator is constant. Hence, the piston only delivers or receives聽kinetic聽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聽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鈥攂ut the formulas are basically the same for聽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)聽is the maximum amplitude. Of course, you will also recognize 蠅0聽as the聽natural聽frequency of the oscillator, and聽螖 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 蟺/2 and, hence, 螖 would be 0 for one oscillator, and 鈥撓/2 for the other. [The formula to apply here is sin胃 = cos(胃 鈥 蟺/2).] Also note that we can equate our 胃 argument to 蠅0路t.聽Now, if聽a聽= 1 (which is the case here), then these formulas simplify to:

  1. K.E. = T = m路v2/2 =聽m路蠅02路sin2(胃 + 螖) = m路蠅02路sin2(蠅0路t + 螖)
  2. P.E. = U = k路x2/2 = k路cos2(胃 + 螖)

The coefficient k in the potential energy formula characterizes the force: F = 鈭択路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路蠅02., so that gives us the famous T + U = m路蠅02/2 formula or, including a聽once again, T + U = m路a2路蠅02/2.

Now, if we normalize聽our functions by equating k to one (k = 1), then聽the motion of聽our first oscillator is given by the cos胃 function, and its kinetic energy will be equal to聽sin2胃. Hence, the (instantaneous)聽change聽in kinetic energy at any point in time will be equal to:

d(sin2胃)/d胃 = 2鈭檚in胃鈭檇(sin胃)/dt = 2鈭檚in胃鈭檆os胃

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聽sin胃 function which, as mentioned above, is equal to cos(胃鈭捪 /2). Hence, its kinetic energy is equal to聽sin2(胃鈭捪 /2), and how it聽changes聽– as a function of 胃 – will be equal to:

2鈭檚in(胃鈭捪 /2)鈭檆os(胃鈭捪 /2) =聽= 鈭2鈭檆os胃鈭檚in胃 = 鈭2鈭檚in胃鈭檆os胃

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路c2聽equation, and it may not have any direction (when everything is said and done, it’s a scalar聽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聽square this circle. 馃檪

Schr枚dinger’s equation

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

i路魔鈭欌垈蠄/鈭倀 = 鈥(魔2/2m)鈭欌垏2蠄 + V路蠄

We can do a quick dimensional analysis of both sides:

  • [i路魔鈭欌垈蠄/鈭倀] = N鈭檓鈭檚/s = N鈭檓
  • [鈥(魔2/2m)鈭欌垏2蠄] = N鈭檓3/m2 = N鈭檓
  • [V路蠄] = N鈭檓

Note the dimension of the ‘diffusion’ constant聽魔2/2m: [魔2/2m] = N2鈭檓2鈭檚2/kg = N2鈭檓2鈭檚2/(N路s2/m) = N鈭檓3. Also note that, in order for the dimensions to come out alright, the dimension of V 鈥 the potential 鈥 must be that of energy. Hence, Feynman鈥檚 description of it as the potential energy 鈥 rather than the potential tout court聽鈥 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, Schr枚dinger鈥檚 equation 鈥 without the V路蠄 term 鈥 can be written as the following pair of equations:

  1. Re(鈭傁/鈭倀) = 鈭(1/2)鈭(魔/m)鈭Im(鈭2蠄)
  2. Im(鈭傁/鈭倀) = (1/2)鈭(魔/m)鈭Re(鈭2蠄)

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… 蠄 is a complex function, which we can write as a + i鈭檅. Likewise, 鈭傁/鈭倀 is a complex function, which we can write as c + i鈭檇, and 鈭2蠄 can then be written as e + i鈭檉. If we temporarily forget about the coefficients (魔, 魔2/m and V), then Schr枚dinger鈥檚 equation – including V路蠄 term – amounts to writing something like this:

i鈭(c + i鈭檇) = 鈥(e + i鈭檉) + (a + i鈭檅) 鈬 a + i鈭檅 = i鈭檆 鈭 d + e+ i鈭檉 聽鈬 a = 鈭抎 + e and b = c + f

Hence, we can now write:

  1. V鈭Re(蠄) = 鈭捘р垯Im(鈭傁/鈭倀) + (1/2)鈭( 魔2/m)鈭Re(鈭2蠄)
  2. V鈭Im(蠄) = 魔鈭Re(鈭傁/鈭倀) + (1/2)鈭( 魔2/m)鈭Im(鈭2蠄)

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鈭Re(蠄) = 鈭捘р垯鈭Im (蠄)/鈭倀 + (1/2)鈭( 魔2/m)鈭欌垏2Re(蠄)
  2. V鈭Im(蠄) = 魔鈭欌垈Re(蠄)/鈭倀 + (1/2)鈭( 魔2/m)鈭欌垏2Im(蠄)

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

  1. H[Re (蠄)] = 鈭捘р垯鈭Im(蠄)/鈭倀
  2. H[Im(蠄)] = 魔鈭欌垈Re(蠄)/鈭倀

A dimensional analysis shows us both sides are, once again, expressed in N鈭檓. It鈥檚 a beautiful expression because 鈥 if we write the real and imaginary part of 蠄 as r鈭檆os胃 and r鈭檚in胃, we get:

  1. H[cos胃] = 鈭捘р垯鈭俿in胃/鈭倀 = E鈭檆os胃
  2. H[sin胃] = 魔鈭欌垈cos胃/鈭倀 = E鈭檚in胃

Indeed, 胃聽= (E鈭檛 鈭 px)/魔 and, hence, 鈭捘р垯鈭俿in胃/鈭倀 = 魔鈭檆os胃鈭橢/魔 = E鈭檆os胃 and 魔鈭欌垈cos胃/鈭倀 = 魔鈭檚in胃鈭橢/魔 = E鈭檚in胃.聽 Now we can combine the two equations in one equation again and write:

H[r鈭(cos胃 + i鈭檚in胃)] = r鈭(E鈭檆os胃 + i鈭檚in胃) 鈬 H[蠄] = E鈭櫹

The operator H 鈥 applied to the wavefunction 鈥 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… 馃檪

Post scriptum: The symmetry of our V-2 engine – or perpetuum mobile聽– 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聽by聽the geometry of our V-2 engine. Now that聽is聽interesting, isn’t it? 馃檪

All what you ever wanted to know about the photon wavefunction…

Post scriptum note added on 11 July 2016: This is one of the more speculative posts which led to my e-publication analyzing the wavefunction as an energy propagation. With the benefit of hindsight, I would recommend you to immediately read the more recent聽expos茅聽on the matter that is being presented here, which you can find by clicking on the provided link.

Original post:

This post is, essentially, a continuation of my previous post, in which I juxtaposed the following images:

Animation 5d_euler_f

Both are the same, and then they鈥檙e not. The illustration on the right-hand side is a regular quantum-mechanical wavefunction, i.e. an amplitude聽wavefunction. You’ve seen that one before. In this case, the x-axis represents time, so we’re looking at the wavefunction at some particular point in space. ]You know we can just switch the dimensions and it would all look the same.] The illustration on the left-hand side looks similar, but it’s not聽an amplitude wavefunction. The animation聽shows how the electric field vector (E) of an electromagnetic wave travels through space. Its shape is the same. So it’s the same function. Is it also the same reality?

Yes and no. And I would say: more no than yes鈥攊n this case, at least. Note that the animation does聽not聽show the accompanying magnetic field vector (B). That vector is equally essential in the electromagnetic propagation mechanism according to Maxwell鈥檚 equations, which鈥攍et me remind you鈥攁re equal to:

  1. B/鈭倀 = 鈥撯垏脳E
  2. E/鈭倀 = 鈭嚸B

In fact, I should write聽the second equation as 鈭E/鈭倀 = c2鈭嚸B, but then I assume we measure time and distance in equivalent units, so c = 1.

You know that E and B are two aspects of one and the same thing: if we have one, then we have the other. To be precise, B is always orthogonal to E聽in the direction that鈥檚 given by the right-hand rule for the following vector cross-product: B = exE,聽with ex聽the unit vector pointing in the x-direction (i.e. the direction of propagation). The reality behind is illustrated below for a聽linearly polarized electromagnetic wave.

E and b

The聽B = exE聽equation is equivalent to writing B= iE, which is equivalent to:

B聽=聽iE = ei(蟺/2)ei(kx 鈭 蠅t)聽= cos(kx 鈭 蠅t + 蟺/2) +聽i路sin(kx 鈭 蠅t + 蟺/2)

=聽鈭抯in((kx 鈭 蠅t) + i路cos(kx 鈭 蠅t)

Now,聽E and B have only two components: Ey聽and聽Ez, and By聽and Bz. That’s only because we’re looking at some ideal聽or聽elementary聽electromagnetic wave here but… Well… Let’s just go along with it. 馃檪 It is then easy to prove that the equation above amounts to writing:

  1. By聽= cos(kx 鈭 蠅t + 蟺/2) = 鈭抯in(kx 鈭 蠅t) = 鈭扙z
  2. Bz聽= sin(kx 鈭 蠅t + 蟺/2) = cos(kx 鈭 蠅t) = Ey

We should now think of Ey聽and聽Ez聽as the real and imaginary part of some wavefunction, which we’ll denote as 蠄E聽= ei(kx 鈭 蠅t). So we write:

E = (Ey,聽Ez) =聽Ey聽+ i路Ez聽= cos(kx 鈭 蠅t) + i鈭檚in(kx 鈭 蠅t) =聽Re(蠄E) +聽iIm(蠄E) = 蠄E聽= ei(kx 鈭 蠅t)

What about B? We just do the same, so we write:

B聽= (By, Bz) = By聽+ i路Bz聽= 蠄B聽= iE = i路蠄E聽= 鈭抯in(kx 鈭 蠅t) + i鈭檚in(kx 鈭 蠅t) = 鈭捖Im(蠄E) +聽iRe(蠄E)

Now we聽need to prove that 蠄E聽and聽蠄B聽are regular wavefunctions, which amounts to proving Schr枚dinger鈥檚 equation, i.e.聽鈭傁/鈭倀 =聽i路(魔/m)路鈭2蠄, for聽both E聽and聽蠄B. [Note I use the Schr枚dinger鈥檚 equation for a zero-mass spin-zero particle here, which uses the 魔/m factor rather than the 魔/(2m) factor.] To prove that 蠄E聽and聽蠄B聽are regular wavefunctions, we should prove that:

  1. Re(鈭傁E/鈭倀) = 聽鈭(魔/m)路Im(鈭2E) and聽Im(鈭傁E/鈭倀) = (魔/m)路Re(鈭2E), and
  2. Re(鈭傁B/鈭倀) = 聽鈭(魔/m)路Im(鈭2B) and聽Im(鈭傁B/鈭倀) = (魔/m)路Re(鈭2B).

Let鈥檚 do the calculations for the second pair of equations. The聽time derivative on the left-hand side is equal to:

鈭傁B/鈭倀 = 鈭i蠅路iei(kx 鈭 蠅t)聽=聽蠅路[cos(kx 鈭 蠅t) + i路sin(kx 鈭 蠅t)] = 蠅路cos(kx 鈭 蠅t) +聽i蠅路sin(kx 鈭 蠅t)

The second-order derivative on the right-hand side is equal to:

2B聽=聽鈭2B/鈭倄2聽= i路k2ei(kx 鈭 蠅t)聽= k2路cos(kx 鈭 蠅t) + i路k2路sin(kx 鈭 蠅t)

So the two equations for聽蠄B聽are equivalent to writing:

  1. Re(鈭傁B/鈭倀) = 聽 鈭(魔/m)路Im(鈭2B)聽鈬斅犗壜穋os(kx 鈭 蠅t) =聽k2路(魔/m)路cos(kx 鈭 蠅t)
  2. Im(鈭傁B/鈭倀) = (魔/m)路Re(鈭2B)聽鈬 蠅路sin(kx 鈭 蠅t) = k2路(魔/m)路sin(kx 鈭 蠅t)

So we see that both conditions are fulfilled if, and only if, 蠅 = k2路(魔/m).

Now, we also demonstrated in that post of mine聽that Maxwell鈥檚 equations imply the following:

  1. 鈭侭y/鈭倀 =聽鈥(鈭嚸E)y聽= 鈭侲z/鈭倄 =聽鈭俒sin(kx 鈭 蠅t)]/鈭倄 = k路cos(kx 鈭 蠅t) = k路Ey
  2. 鈭侭z/鈭倀 =聽鈥(鈭嚸E)z聽=聽鈥 鈭侲y/鈭倄 =聽鈥 鈭俒cos(kx 鈭 蠅t)]/鈭倄 =聽k路sin(kx 鈭 蠅t) = k路Ez

Hence, using those By聽= 鈭扙z聽and聽Bz聽= Ey聽equations above, we can also calculate these derivatives as:

  1. 鈭侭y/鈭倀 = 鈭掆垈Ez/鈭倀 =聽鈭掆垈sin(kx 鈭 蠅t)/鈭倀 = 蠅路cos(kx 鈭 蠅t) = 蠅路Ey
  2. 鈭侭z/鈭倀 = 鈭侲y/鈭倀 = 鈭俢os(kx 鈭 蠅t)/鈭倀 = 鈭捪壜穂鈭抯in(kx 鈭 蠅t)] = 蠅路Ez

In other words, Maxwell鈥檚 equations imply that 蠅 = k, which is consistent with us measuring time and distance in equivalent units, so the phase velocity is 聽c聽= 1 =聽蠅/k.

So far, so good. We basically established that the propagation mechanism for an electromagnetic wave, as described by Maxwell鈥檚 equations, is fully coherent with the propagation mechanism鈥攊f we can call it like that鈥攁s described by Schr枚dinger鈥檚 equation. We also established the following equalities:

  1. 蠅 = k
  2. 蠅 = k2路(魔/m)

The second of the two聽de Broglie equations tells us that k = p/魔, so we can combine聽these two equations and re-write these two conditions as:

蠅/k = 1聽= k路(魔/m) = (p/魔)路(魔/m) = p/m聽鈬 p = m

What does this imply? The p here is the momentum: p = m路v, so this condition implies v聽must be equal to 1 too, so the wave velocity is equal to the speed of light. Makes sense, because we actually聽are聽talking light here. 馃檪 In addition, because it’s light, we also know E/p =聽c聽= 1, so we have – once again – the general E = p = m equation, which we’ll need!

OK. Next. Let’s write the Schr枚dinger wave equation for both wavefunctions:

  1. 鈭傁E/鈭倀 =聽i路(魔/mE)路鈭2E, and
  2. 鈭傁B/鈭倀 =聽i路(魔/mB)路鈭2B.

Huh?What’s聽mE聽andmE? We should only associate one mass concept with our electromagnetic wave, shouldn’t we? Perhaps. I just want to be on the safe side now. Of course, if we distinguish聽mE聽and mB, we should probably also distinguish pE聽and pB, and EE聽and EB聽as well, right? Well… Yes. If we accept this line of reasoning, then the mass factor in Schr枚dinger’s equations is pretty much like the 1/c2聽= 渭00聽factor in Maxwell鈥檚 (1/c2)路鈭E/鈭倀 = 鈭嚸B聽equation: the mass factor appears as a property of the medium, i.e. the聽vacuum聽here! [Just check my post on physical constants in case you wonder what I am trying to say here, in which I explain why and how聽c聽defines聽the (properties of the) vacuum.]

To be consistent, we should also distinguish pE聽and pB, and EE聽and EB, and so we should write 蠄E聽and 蠄B聽as:

  1. E聽= ei(kEx 鈭 蠅Et), and
  2. B聽= ei(kBx 鈭 蠅Bt).

Huh?Yes.聽I know what you think: we鈥檙e talking one photon鈥攐r one electromagnetic wave鈥攕o there can be only one energy, one momentum and, hence, only one k, and one 蠅. Well鈥 Yes and no. Of course, the following identities should hold: kE聽= kB聽and, likewise, 蠅E聽=聽蠅B. So鈥 Yes. They鈥檙e the same: one k and one 蠅. But then鈥 Well鈥 Conceptually, the two k鈥檚 and 蠅鈥檚 are different. So we write:

  1. pE聽= EE聽= mE, and
  2. pB聽= EB聽= mB.

The obvious question is: can we just add them up to find the total聽energy and momentum of our photon? The answer is obviously positive: E = EE聽+ EB, p聽= pE聽+ pB聽and m聽= mE聽+ mB.

Let鈥檚 check a few things now. How does it work for the phase and group velocity of 蠄E聽and 蠄B? Simple:

  1. vg = 鈭傁E/鈭俴E聽= 鈭俒EE/魔]/鈭俒pE/魔] = 鈭侲E/鈭俻E = 鈭俻E/鈭俻E聽= 1
  2. vp = 蠅E/kE =聽(EE/魔)/(pE/魔) = EE/pE = pE/pE = 1

So we鈥檙e fine, and you can check the result for 蠄B聽by substituting the subscript E for B. To sum it all up, what we鈥檝e got here is the following:

  1. We can think of a photon having some energy that鈥檚 equal to E = p = m聽(assuming c = 1), but that energy would be split up in an electric and a magnetic wavefunction respectively:聽蠄E聽and 蠄B.
  2. Schr枚dinger鈥檚 equation applies to both聽wavefunctions, but the E, p and m in those two wavefunctions are the same and not the same: their numerical聽value is the same (pE聽=EE聽= mE聽= pB聽=EB聽= mB), but they鈥檙e conceptually聽different. They must be: if not, we鈥檇 get a phase and group velocity for the wave that doesn鈥檛 make sense.

Of course, the phase and group velocity for the聽sum聽of the 蠄E聽and 蠄B聽waves must also be equal to聽c. This is obviously the case, because we鈥檙e adding waves with the same phase and group velocity c, so there鈥檚 no issue with the dispersion relation.

So let’s聽insert those pE聽=EE聽= mE聽= pB聽=EB聽= mB聽values in the two wavefunctions. For 蠄E, we get:

E聽= ei[kEx 鈭 蠅Et)聽=聽ei[(pE/魔)路x 鈭 (EE/魔)路t]

You can do the calculation for 蠄B聽yourself. Let’s simplify our life a little bit and assume we’re using Planck units, so 魔 = 1, and so the wavefunction simplifies to 蠄E聽=聽ei路(pE路x 鈭 EE路t). We can now add the components of E and B聽using the summation formulas for sines and cosines:

1. By聽+ Ey聽= cos(pB路x聽鈭捖燛B路t + 蟺/2)聽+ cos(pE路x聽鈭捖燛E路t) = 2路cos[(p路x 鈭 E路t + 蟺/2)/2]路cos(蟺/4) = 鈭2路cos(p路x/2 鈭 E路t/2 + 蟺/4)

2. Bz聽+ Ez聽= sin(pB路x聽鈭捖燛B路t+蟺/2) + sin(pE路x聽鈭捖燛E路t) =聽2路sin[(p路x 鈭 E路t + 蟺/2)/2]路cos(蟺/4) =聽鈭2路sin(p路x/2 鈭 E路t/2 + 蟺/4)

Interesting!We find a聽composite聽wavefunction for our photon which we can write as:

E + B =聽蠄E聽+ 蠄B聽= E聽+聽iE聽=聽鈭2路ei(p路x/2 鈭 E路t/2 + 蟺/4)聽=聽鈭2路ei(蟺/4)ei(p路x/2 鈭 E路t/2)聽=聽鈭2路ei(蟺/4)E

What a great result! It鈥檚 easy to double-check, because we can see the E聽+聽iE聽= 鈭2路ei(蟺/4)E聽formula implies that 1 +聽i聽should equal聽鈭2路ei(蟺/4). Now that鈥檚 easy to prove, both geometrically (just do a drawing) or formally: 鈭2路ei(蟺/4)聽= 鈭2路cos(蟺/4) + i路sin(蟺/4ei(蟺/4)聽= (鈭2/鈭2) + i路(鈭2/鈭2) = 1 + i. We鈥檙e聽bang on!聽馃檪

We can double-check once more, because we should get the same from adding聽E and B聽= iE, right? Let鈥檚 try:

E聽+聽B聽=聽E聽+聽iE = cos(pE路x聽鈭捖燛E路t) + i路sin(pE路x聽鈭捖燛E路t) +聽i路cos(pE路x聽鈭捖燛E路t) 鈭捖爏in(pE路x聽鈭捖燛E路t)

= [cos(pE路x聽鈭捖燛E路t) 鈥 sin(pE路x聽鈭捖燛E路t)] +聽i路[sin(pE路x聽鈭捖燛E路t) 鈥 cos(pE路x聽鈭捖燛E路t)]

Indeed, we can see we鈥檙e going to obtain the same result, because the 鈭抯in胃 in the real part of our composite聽wavefunction is equal to聽cos(胃+蟺/2), and the 鈭抍os胃 in its imaginary part is equal to sin(胃+蟺/2). So the sum above is the same sum of cosines and sines that we did already.

So our electromagnetic wavefunction, i.e. the wavefunction for the聽photon, is equal to:

蠄聽=聽蠄E聽+ 蠄B聽= 鈭2路ei(p路x/2 鈭 E路t/2 + 蟺/4)聽= 鈭2路ei(蟺/4)ei(p路x/2 鈭 E路t/2)聽

What about the 鈭2 factor in front, and the聽蟺/4 term in the argument itself? No sure. It must have something to do with the way the magnetic force works, which is not聽like the electric force. Indeed, remember the Lorentz formula: the force on some unit charge (q = 1) will be equal to F = E + vB. So鈥 Well鈥 We鈥檝e got another cross-product here and so the geometry of the situation is quite complicated: it鈥檚聽not聽like adding two forces F1聽and聽F2聽to get some combined force F = F1聽and聽F2.

In any case, we need the energy, and we know that its proportional to the square of the amplitude, so鈥 Well鈥 We鈥檙e spot on: the square of the 鈭2 factor in the 鈭2路cos product and 鈭2路sin product is 2, so that鈥檚 twice鈥 Well鈥 What? Hold on a minute!聽We鈥檙e actually taking the聽absolute聽square of the E + B =聽蠄E聽+ 蠄B聽= E聽+聽iE聽=聽鈭2路ei(p路x/2 鈭 E路t/2 + 蟺/4)wavefunction here. Is that聽legal? I must assume it is鈥攁lthough鈥 Well鈥 Yes. You’re right. We should do some more explaining here.

We know聽that we usually measure the energy as some聽definite聽integral, from t = 0 to some other point in time, or over the cycle of the oscillation. So what鈥檚 the聽cycle聽here? Our combined wavefunction can be written as聽鈭2路ei(p路x/2 鈭 E路t/2 + 蟺/4)= 鈭2路ei(胃/2 + 蟺/4), so a full cycle would correspond to 胃 going from 0 to 4蟺 here, rather than from 0 to 2蟺. So that explains the 鈭2 factor in front of our wave equation.

Bingo! If you were looking for an interpretation of the Planck energy and momentum, here it is.:-) And, while everything that鈥檚 written above is not easy to understand, it鈥檚 close to the 鈥榠ntuitive鈥 understanding to quantum mechanics that we were looking for, isn鈥檛 it? The quantum-mechanical propagation model explains聽everything now. 馃檪聽I only need to show one more thing, and that鈥檚 the different behavior聽of bosons and fermions:

  1. The amplitudes of identitical聽bosonic particles interfere with a positive sign, so we have Bose-Einstein statistics here. As Feynman writes it: (amplitude direct) + (amplitude exchanged).
  2. The amplitudes of identical聽fermionic聽particles interfere with a negative sign, so we have Fermi-Dirac statistics here: (amplitude direct) 鈭 (amplitude exchanged).

I鈥檒l think about it. I am sure it鈥檚 got something to do with that B= iE聽formula or, to put it simply, with the fact that, when bosons are involved, we get two wavefunctions (蠄E聽and 蠄B) for the price of one. The reasoning should be something like this:

I. For a massless particle (i.e. a zero-mass聽fermion), our wavefunction is just 蠄 =聽ei(p路x 鈭 E路t). So we have no 鈭2 or 鈭2路ei(蟺/4)聽factor in front here. So we can just add any number of them 鈥撀犗1+ 蠄2+ 蠄3+ 鈥 鈥 and then take the absolute square of the amplitude to find a probability density, and we鈥檙e done.

II. For a photon (i.e. a zero-mass聽boson), our wavefunction is 鈭2路ei(蟺/4)ei(p路x 鈭 E路t)/2, which 鈥 let鈥檚 introduce a new symbol 鈥 we鈥檒l denote by 蠁, so 蠁 = 鈭2路ei(蟺/4)ei(p路x 鈭 E路t)/2. Now, if we add any number of these, we get a similar sum but with that 鈭2路ei(蟺/4)聽factor in front, so we write: 蠁1+ 蠁2+ 蠁3+ 鈥 = 鈭2路ei(蟺/4)路(蠄1+ 蠄2+ 蠄3+ 鈥). If we take the absolute square now, we鈥檒l see the probability density will be equal to twice聽the density for the 蠄1+ 蠄2+ 蠄3+ 鈥 sum, because

|鈭2路ei(蟺/4)路(蠄1+ 蠄2+ 蠄3+ 鈥)|2聽= |鈭2路ei(蟺/4)|2路|蠄1+ 蠄2+ 蠄3+ 鈥)|2=聽2路|蠄1+ 蠄2+ 蠄3+ 鈥)|2

So鈥 Well鈥 I still need to connect this to Feynman鈥檚 (amplitude direct) 卤 (amplitude exchanged) formula, but I am sure it can be done.

Now, we haven’t tested the complete聽鈭2路ei(蟺/4)ei(p路x 鈭 E路t)/2wavefunction. Does it respect Schr枚dinger’s 鈭傁/鈭倀 = i路(1/m)路鈭2蠄 or, including the 1/2 factor, the聽鈭傁/鈭倀 = i路[1/2m)]路鈭2蠄 equation? [Note we assume, once again, that 魔 = 1, so we use Planck units once more.] Let’s see. We can calculate the derivatives as:

  • 鈭傁/鈭倀 =聽鈭掆垰2路ei(蟺/4)ei鈭[p路x 鈭 E路t]/2路(i路E/2)
  • 2蠄 =聽鈭2[鈭2路ei(蟺/4)路ei鈭[p路x 鈭 E路t]/2]/鈭倄2聽= 鈭2路ei(蟺/4)鈭俒鈭2路ei(蟺/4)路ei鈭[p路x 鈭 E路t]/2路(i路p/2)]/鈭倄 = 鈭掆垰2路ei(蟺/4)路ei鈭[p路x 鈭 E路t]/2路(p2/4)

So Schr枚dinger’s equation becomes:

i路鈭2路ei(蟺/4)ei鈭[p路x 鈭 E路t]/2路(i路E/2) = 鈭i路(1/m)路鈭2路ei(蟺/4)路ei鈭[p路x 鈭 E路t]/2路(p2/4) 鈬 1/2 = 1/4!?

That’s funny ! It doesn’t work ! The E and m and p2聽are OK because we’ve got that E = m = p equation, but we’ve got problems with yet another factor 2. It only works when we use the 2/m coefficient in Schr枚dinger’s equation.

So… Well… There’s no choice. That’s what we’re going to do. The聽Schr枚dinger equation for the photon is 鈭傁/鈭倀 = i路(2/m)路鈭2蠄 !

It’s a very subtle point. This is all great, and very聽fundamental stuff! Let’s now move on to Schr枚dinger’s actual聽equation, i.e. the聽鈭傁/鈭倀 = i路(魔/2m)路鈭2蠄 equation.

Post scriptum on the Planck units:

If we measure time and distance in equivalent units, say seconds, we can re-write the quantum of action as:

1.0545718脳10鈭34聽N路m路s = (1.21脳1044聽N)路(1.6162脳10鈭35聽m)路(5.391脳10鈭44s)

鈬 (1.0545718脳10鈭34/2.998脳108) N路s2聽= (1.21脳1044聽N)路(1.6162脳10鈭35/2.998脳108聽s)(5.391脳10鈭44s)

鈬 (1.21脳1044聽N) = [(1.0545718脳10鈭34/2.998脳108)]/[(1.6162脳10鈭35/2.998脳108聽s)(5.391脳10鈭44s)]聽N路s2/s2

You’ll say: what’s this? Well… Look at it. We’ve got a much easier formula for the Planck force鈥攎uch easier than the standard formulas you’ll find on Wikipedia, for example. If we re-interpret the symbols聽魔 and聽c聽so they denote the numerical聽value of the quantum of action and the speed of light in standard SI units (i.e. newton, meter and second)鈥攕o 魔 and聽c become dimensionless, or聽mathematical聽constants only, rather than聽physical聽constants鈥攖hen the formula above can be written as:

FPnewton聽= (魔/c)/[(lP/c)路tP] newton聽鈬 FP聽= 魔/(lP路tP)

Just double-check it: 1.0545718脳10鈭34/(1.6162脳10鈭35路5.391脳10鈭44) = 1.21脳1044. Bingo!

You’ll say: what’s the point? The point is: our model is complete. We don’t need the other physical constants 鈥 i.e. the Coulomb, Boltzmann and gravitational constant 鈥撀爐o calculate the Planck units we need, i.e. the Planck force, distance and time units. It all comes out of our elementary wavefunction! All we need to explain the Universe聽鈥 or, let’s be more modest, quantum mechanics聽鈥 is two numerical constants (c and 魔) and Euler’s formula (which uses 蟺 and聽e, of course). That’s it.

If you don’t think that’s a great result, then… Well… Then you’re not reading this. 馃檪