Post scriptum note added on 11 July 2016: This is one of the more speculative posts which led to my e-publication analyzing the wavefunction as an energy propagation. With the benefit of hindsight, I would recommend you to immediately the more recent exposé on the matter that is being presented here, which you can find by clicking on the provided link. In fact, I actually made some (small) mistakes when writing the post below.
In my previous post, I introduced the elementary wavefunction of a particle with zero rest mass in free space (i.e. the particle also has zero potential). I wrote that wavefunction as ei(kx − ωt) = ei(x/2 − t/2) = cos[(x−t)/2] + i∙sin[(x−t)/2], and we can represent that function as follows:
If the real and imaginary axis in the image above are the y- and z-axis respectively, then the x-axis here is time, so here we’d be looking at the shape of the wavefunction at some fixed point in space.
Now, we could also look at its shape at some fixed in point in time, so the x-axis would then represent the spatial dimension. Better still, we could animate the illustration to incorporate both the temporal as well as the spatial dimension. The following animation does the trick quite well:
Please do note that space is one-dimensional here: the y- and z-axis represent the real and imaginary part of the wavefunction, not the y- or z-dimension in space.
You’ve seen this animation before, of course: I took it from Wikipedia, and it actually represents the electric field vector (E) for a circularly polarized electromagnetic wave. To get a complete picture of the electromagnetic wave, we should add the magnetic field vector (B), which is not shown here. We’ll come back to that later. Let’s first look at our zero-mass particle denuded of all properties, so that’s not an electromagnetic wave—read: a photon. No. We don’t want to talk charges here.
OK. So far so good. A zero-mass particle in free space. So we got that ei(x/2 − t/2) = cos[(x−t)/2] + i∙sin[(x−t)/2] wavefunction. We got that function assuming the following:
The latter hypothesis (E = m = p = ħ/2) is somewhat strange at first but, as I showed in that post of mine, it avoids an apparent contradiction: if we’d use ħ, then we would find two different values for the phase and group velocity of our wavefunction. To be precise, we’d find v for the group velocity, but v/2 for the phase velocity. Using ħ/2 solves that problem. In addition, using ħ/2 is consistent with the Uncertainty Principle, which tells us that ΔxΔp = ΔEΔt = ħ/2.
OK. Take a deep breath. Here I need to say something about dimensions. If we’re saying that we’re measuring time and distance in equivalent units – say, in meter, or in seconds – then we are not saying that they’re the same. The dimension of time and space is fundamentally different, as evidenced by the fact that, for example, time flows in one direction only, as opposed to x. To be precise, we assumed that x and t become countable variables themselves at some point in time. However, if we’re at t = 0, then we’d count time as t = 1, 2, etcetera only. In contrast, at the point x = 0, we can go to x = +1, +2, etcetera but we may also go to x = −1, −2, etc.
I have to stress this point, because what follows will require some mental flexibility. In fact, we often talk about natural units, such as Planck units, which we get from equating fundamental constants, such as c, or ħ, to 1, but then we often struggle to interpret those units, because we fail to grasp what it means to write c = 1, or ħ = 1. For example, writing c = 1 implies we can measure distance in seconds, or time in meter, but it does not imply that distance becomes time, or vice versa. We still need to keep track of whether or not we’re talking a second in time, or a second in space, i.e. c meter, or, conversely, whether we’re talking a meter in space, or a meter in time, i.e. 1/c seconds. We can make the distinction in various ways. For example, we could mention the dimension of each equation between brackets, so we’d write: t = 1×10−15 s [t] ≈ 299.8×10−9 m [t]. Alternatively, we could put a little subscript (like t, or d), so as to make sure it’s clear our meter is a a ‘light-meter’, so we’d write: t = 1×10−15 s ≈ 299.8×10−9 mt. Likewise, we could add a little subscript when measuring distance in light-seconds, so we’d write x = 3×108 m ≈ 1 sd, rather than x = 3×108 m [x] ≈ 1 s [x].
If you wish, we could refer to the ‘light-meter’ as a ‘time-meter’ (or a meter of time), and to the light-second as a ‘distance-second’ (or a second of distance). It doesn’t matter what you call it, or how you denote it. In fact, you will never hear of a meter of time, nor will you ever see those subscripts or brackets. But that’s because physicists always keep track of the dimensions of an equation, and so they know. They know, for example, that the dimension of energy combines the dimensions of both force as well as distance, so we write: [energy] = [force]·[distance]. Read: energy amounts to applying a force over a distance. Likewise, momentum amounts to applying some force over some time, so we write: [momentum] = [force]·[time]. Using the usual symbols for energy, momentum, force, distance and time respectively, we can write this as [E] = [F]·[x] and [p] = [F]·[t]. Using the units you know, i.e. joule, newton, meter and seconds, we can also write this as: 1 J = 1 N·m and 1…
Hey! Wait a minute! What’s that N·s unit for momentum? Momentum is mass times velocity, isn’t it? It is. But it amounts to the same. Remember that mass is a measure for the inertia of an object, and so mass is measured with reference to some force (F) and some acceleration (a): F = m·a ⇔ m = F/a. Hence, [m] = kg = [F/a] = N/(m/s2) = N·s2/m. [Note that the m in the brackets is symbol for mass but the other m is a meter!] So the unit of momentum is (N·s2/m)·(m/s) = N·s = newton·second.
Now, the dimension of Planck’s constant is the dimension of action, which combines all dimensions: force, time and distance. We write: ħ ≈ 1.0545718×10−34 N·m·s (newton·meter·second). That’s great, and I’ll show why in a moment. But, at this point, you should just note that when we write that E = m = p = ħ/2, we’re just saying they are numerically the same. The dimensions of E, m and p are not the same. So what we’re really saying is the following:
What’s the quantum of mass? That’s where the equivalent units come in. We wrote: 1 kg = 1 N·s2/m. So we could substitute the distance unit in this equation (m) by sd/c = sd/(3×108). So we get: 1 kg = 3×108 N·s2/sd. Can we scrap both ‘seconds’ and say that the quantum of mass (ħ/2) is equal to the quantum of momentum? Think about it.
The answer is… Yes and no—but much more no than yes! The two sides of the equation are only numerically equal, but we’re talking a different dimension here. If we’d write that 1 kg = 0.527286×10−34 N·s2/sd = 0.527286×10−34 N·s, you’d be equating two dimensions that are fundamentally different: space versus time. To reinforce the point, think of it the other way: think of substituting the second (s) for 3×108 m. Again, you’d make a mistake. You’d have to write 0.527286×10−34 N·(mt)2/m, and you should not assume that a time-meter is equal to a distance-meter. They’re equivalent units, and so you can use them to get some number right, but they’re not equal: what they measure, is fundamentally different. A time-meter measures time, while a distance-meter measure distance. It’s as simple as that. So what is it then? Well… What we can do is remember Einstein’s energy-mass equivalence relation once more: E = m·c2 (and m is the mass here). Just check the dimensions once more: [m]·[c2] = (N·s2/m)·(m2/s2) = N·m. So we should think of the quantum of mass as the quantum of energy, as energy and mass are equivalent, really.
The beauty of the construct of the wavefunction resides in several mathematical properties of this construct. The first is its argument:
θ = kx − ωt, with k = p/ħ and ω = E/ħ
Its dimension is the dimension of an angle: we express in it in radians. What’s a radian? You might think that a radian is a distance unit because… Well… Look at how we measure an angle in radians below:
But you’re wrong. An angle’s measurement in radians is numerically equal to the length of the corresponding arc of the unit circle but… Well… Numerically only. 🙂 Just do a dimensional analysis of θ = kx − ωt = (p/ħ)·x − (E/ħ)·t. The dimension of p/ħ is (N·s)/(N·m·s) = 1/m = m−1, so we get some quantity expressed per meter, which we then multiply by x, so we get a pure number. No dimension whatsoever! Likewise, the dimension of E/ħ is (N·m)/(N·m·s) = 1/s = s−1, which we then multiply by t, so we get another pure number, which we then add to get our argument θ. Hence, Planck’s quantum of action (ħ) does two things for us:
In fact, I’d say the ħ in the (p/ħ)·x term in the argument is a different ħ than the ħ in the (E/ħ)·t term. Huh? What? Yes. Think of the distinction I made between s and sd, or between m and mt. Both were numerically the same: they captured a magnitude, but they measured different things. We’ve got the same thing here:
That’s why you need to watch out when writing θ = (p/ħ)·x − (E/ħ)·t as θ = (p·x − E·t)/ħ or – in the case of our elementary wavefunction for the zero-mass particle – as θ = (x/2 − t/2) = (x − t)/2. You can do it – in fact, you should do when trying to calculate something – but you need to be aware that you’re making abstraction of the dimensions. That’s quite OK, as you’re just calculating something—but don’t forget the physics behind!
You’ll immediately ask: what are the physics behind here? Well… I don’t know. Perhaps nobody knows. As Feynman once famously said: “I think I can safely say that nobody understands quantum mechanics.” But then he never wrote that, and I am sure he didn’t really mean that. And then he said that back in 1964, which is 50 years ago now. 🙂 So let’s try to understand it at least. 🙂
Planck’s quantum of action – 1.0545718×10−34 N·m·s – comes to us as a mysterious quantity. A quantity is more than a a number. A number is something like π or e, for example. It might be a complex number, like eiθ, but that’s still a number. In contrast, a quantity has some dimension, or some combination of dimensions. A quantity may be a scalar quantity (like distance), or a vector quantity (like a field vector). In this particular case (Planck’s ħ or h), we’ve got a physical constant combining three dimensions: force, time and distance—or space, if you want. It’s a quantum, so it comes as a blob—or a lump, if you prefer that word. However, as I see it, we can sort of project it in space as well as in time. In fact, if this blob is going to move in spacetime, then it will move in space as well as in time: t will go from 0 to 1, and x goes from 0 to ± 1, depending on what direction we’re going. So when I write that E = p = ħ/2—which, let me remind you, are two numerical equations, really—I sort of split Planck’s quantum over E = m and p respectively.
You’ll say: what kind of projection or split is that? When projecting some vector, we’ll usually have some sine and cosine, or a 1/√2 factor—or whatever, but not a clean 1/2 factor. Well… I have no answer to that, except that this split fits our mathematical construct. Or… Well… I should say: my mathematical construct. Because what I want to find is this clean Schrödinger equation:
∂ψ/∂t = i·(ħ/2m)·∇2ψ = i·∇2ψ for m = ħ/2
Now I can only get this equation if (1) E = m = p and (2) if m = ħ/2 (which amounts to writing that E = p = m = ħ/2). There’s also the Uncertainty Principle. If we are going to consider the quantum vacuum, i.e. if we’re going to look at space (or distance) and time as count variables, then Δx and Δt in the ΔxΔp = ΔEΔt = ħ/2 equations are ± 1 and, therefore, Δp and ΔE must be ± ħ/2. In any case, I am not going to try to justify my particular projection here. Let’s see what comes out of it.
Schrödinger’s equation for my zero-mass particle (with energy E = m = p = ħ/2) amounts to writing:
Now that reminds of the propagation mechanism for the electromagnetic wave, which we wrote as ∂B/∂t = –∇×E and ∂E/∂t = ∇×B, also assuming we measure time and distance in equivalent units. However, we’ll come back to that later. Let’s first study the equation we have, i.e.
ei(kx − ωt) = ei(ħ·x/2 − ħ·t/2)/ħ = ei(x/2 − t/2) = cos[(x−t)/2] + i∙sin[(x−t)/2]
Let’s think some more. What is that ei(x/2 − t/2) function? It’s subject to conceiving time and distance as countable variables, right? I am tempted to say: as discrete variables, but I won’t go that far—not now—because the countability may be related to a particular interpretation of quantum physics. So I need to think about that. In any case… The point is that x can only take on values like 0, 1, 2, etcetera. And the same goes for t. To make things easy, we’ll not consider negative values for x right now (and, obviously, not for t either). But you can easily check it doesn’t make a difference: if you think of the propagation mechanism – which is what we’re trying to model here – then x is always positive, because we’re moving away from some source that caused the wave. In any case, we’ve got a infinite set of points like:
In my previous post, I calculated the real and imaginary part of this wavefunction for x going from 0 to 14 (as mentioned, in steps of 1) and for t doing the same (also in steps of 1), and what we got looked pretty good:
I also said that, if you wonder how the quantum vacuum could possibly look like, you should probably think of these discrete spacetime points, and some complex-valued wave that travels as illustrated above. In case you wonder what’s being illustrated here: the right-hand graph is the cosine value for all possible x = 0, 1, 2,… and t = 0, 1, 2,… combinations, and the left-hand graph depicts the sine values, so that’s the imaginary part of our wavefunction. Taking the absolute square of both gives 1 for all combinations. So it’s obvious we’d need to normalize and, more importantly, we’d have to localize the particle by adding several of these waves with the appropriate contributions. But so that’s not our worry right now. I want to check whether those discrete time and distance units actually make sense. What’s their size? Is it anything like the Planck length (for distance) and/or the Planck time?
Let’s see. What are the implications of our model? The question here is: if ħ/2 is the quantum of energy, and the quantum of momentum, what’s the quantum of force, and the quantum of time and/or distance?
Huh? Yep. We treated distance and time as countable variables above, but now we’d like to express the difference between x = 0 and x = 1 and between t = 0 and t = 1 in the units we know, this is in meter and in seconds. So how do we go about that? Do we have enough equations here? Not sure. Let’s see…
We obviously need to keep track of the various dimensions here, so let’s refer to that discrete distance and time unit as tP and lP respectively. The subscript (P) refers to Planck, and the l refers to a length, but we’re likely to find something else than Planck units. I just need placeholder symbols here. To be clear: tP and lP are expressed in meter and seconds respectively, just like the actual Planck time and distance, which are equal to 5.391×10−44 s (more or less) and 1.6162×10−35 m (more or less) respectively. As I mentioned above, we get these Planck units by equating fundamental physical constants to 1. Just check it: (1.6162×10−35 m)/(5.391×10−44 s) = c ≈ 3×108 m/s. So the following relation must be true: lP = c·tP, or lP/tP = c.
Now, as mentioned above, there must be some quantum of force as well, which we’ll write as FP, and which is – obviously – expressed in newton (N). So we have:
Let’s try to divide both formulas: E/p = (FP·lP N·m)/(FP·tP N·s) = lP/tP m/s = lP/tP m/s = c m/s. That’s consistent with the E/p = c equation. Hmm… We found what we knew already. My model is not fully determined, it seems. 😦
What about the following simplistic approach? E is numerically equal to 0.527286×10−34, and its dimension is [E] = [F]·[x], so we write: E = 0.527286×10−34·[E] = 0.527286×10−34·[F]·[x]. Hence, [x] = [E]/[F] = (N·m)/N = m. That just confirms what we already know: the quantum of distance (i.e. our fundamental unit of distance) can be expressed in meter. But our model does not give that fundamental unit. It only gives us its dimension (meter), which is stuff we knew from the start. 😦
Let’s try something else. Let’s just accept that Planck length and time, so we write:
Now, if the quantum of action is equal to ħ N·m·s = FP·lP·tP N·m·s = 1.0545718×10−34 N·m·s, and if the two definitions of lP and tP above hold, then 1.0545718×10−34 N·m·s = (FP N)×(1.6162×10−35 m)×(5.391×10−44 s) ≈ FP 8.713×10−79 N·m·s ⇔ FP ≈ 1.21×1044 N.
Does that make sense? It does according to Wikipedia, but how do we relate this to our E = p = m = ħ/2 equations? Let’s try this:
Is EP = pP? Let’s substitute: 1.956×109 N·m = 1.956×109 N·(s/c) = 1.956×109/2.998×109 N·s = 0.6525 N·s. So, yes, it comes out alright. In fact, I omitted the 1/2 factor in the calculations, but it doesn’t matter: it does come out alright. So I did not prove that the difference between my x = 0 and x = 1 points (or my t = 0 and t = 1 points) is equal to the Planck length (or the Planck time unit), but I did show my theory is, at the very least, compatible with those units. That’s more than enough for now. And I’ll come surely come back to it in my next post. 🙂
Post Scriptum: One must solve the following equations to get the fundamental Planck units:
We have five fundamental equations for five fundamental quantities respectively: tP, lP, FP, mP, and EP respectively, so that’s OK: it’s a fully determined system alright! But where do the expressions with G, kB (the Boltzmann constant) and ε0 come from? What does it mean to equate those constants to 1? Well… I need to think about that, and I’ll get back to you on it. 🙂