**Pre-script** (dated 26 June 2020): This post got mutilated by the removal of some material by the dark force. You should be able to follow the main story line, however. If anything, the lack of illustrations might actually help you to think things through for yourself. In any case, we now have different views on these concepts as part of our realist interpretation of quantum mechanics, so we recommend you read our recent papers instead of these old blog posts.

**Original post**:

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*·*e*^{−iθ }= ψ = *a*·*e*^{−i·θ } = *a*·*e*^{−i}^{(}^{ω·t−k·x}^{)} = *a*·*e*^{−}^{(}^{i}^{/ħ)·}^{(E·}^{t−p·x}^{)} .We know this elementary wavefunction cannot* *represent a real-life particle. Indeed, the *a*·*e*^{−i·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}·1^{2}= *a*^{2} *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*) =

*e*

^{−i·(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

*e*

^{−i·θ}=

*e*

^{−i·(E/ħ)·t}=

*cos*θ +

*i*·

*sin*θ 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.Hence, if we don’t change the direction of the *y*– and *z*-axes – so we keep defining the *z*-axis as the axis pointing upwards, and the y-axis as the axis pointing *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.

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*·*e*^{−i}^{(}^{ω·t−k·x}^{)} rather than *a*·*e*^{i}^{(}^{ω·t+k·x}^{)} or *a*·*e*^{i}^{(}^{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·*c*^{2} 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/s^{2}. So… Because energy is force times distance, the unit of *energy *may be expressed in units of kg·m/s^{2}·m, or kg·m^{2}/s^{2}, 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·*c*^{2} 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.

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·ω^{2}. 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·*c*^{2} 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. 🙂The 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? 🙂

The 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. 🙂