# Freewheeling once more…

You remember the elementary wavefunction Ψ(x, t) = Ψ(θ), with θ = ω·t−k∙x = (E/ħ)·t − (p/ħ)∙x = (E·t−p∙x)/ħ. Now, we can re-scale θ and define a new argument, which we’ll write as:

φ = θ/ħ = E·t−p∙x

The Ψ(θ) function can now be written as:

Ψ(x, t) = Ψ(θ) = [ei·(θ/ħ)]ħ = Φ(φ) = [ei·φ]ħ with φ = E·t−p∙x

This doesn’t change the fundamentals: we’re just re-scaling E and p here, by measuring them in units of ħ.

You’ll wonder: can we do that? We’re talking physics here, so our variables represent something real. Not all we can do in math, should be done in physics, right? So what does it mean? We need to look at the dimensions of our variables. Does it affect our time and distance units, i.e. the second and the meter? Well… I’d say it’s OK.

Energy is expressed in joule: 1 J = 1 N·m. [In SI base units, we write: J = N·m = (kg·m/s2)·m = kg·(m/s)2.] So if we divide it by ħ, whose dimension is joule-second (J·s), we get some value expressed per second, i.e. a (temporal) frequency. That’s what we want, as we’re multiplying it with t in the argument of our wavefunction!

Momentum is expressed in newton-second (N·s). Now, 1 J = 1 N·m, so 1 N = 1 J/m. Hence, if we divide the momentum value by ħ, we get some value expressed per meter: N·s/J·s = N/J = N/N·m = 1/m. So we get a spatial frequency here. That’s what we want, as we’re multiplying it with x!

So the answer is yes: we can re-scale energy and momentum and we get a temporal and spatial frequency respectively, which we can multiply with t and x respectively: we do not need to change our time and distance units when re-scaling E and p by dividing by ħ!

The next question is: if we express energy and momentum as temporal and spatial frequencies, do our E = m·cand p = m·formulas still apply? They should: both and v are expressed in meter per second (m/s) and, as mentioned above, the re-scaling does not affect our time and distance units. Hence, the energy-mass equivalence relation, and the definition of p (p = m·v), imply that we can re-write the argument (φ) of our ‘new’ wavefunction – i.e. Φ(φ) – as:

φ = E·t−p∙x = m·c2∙t − m∙v·x = m·c2[t – (v/c)∙(x/c)] = m·c2[t – (v/c)∙(x/c)]

In effect, when re-scaling our energy and momentum values, we’ve also re-scaled our unit of inertia, i.e. the unit in which we measure the mass m, which is directly related to both energy as well as momentum. To be precise, from a math point of view, m is nothing but a proportionality constant in both the E = m·cand p = m·formulas.

The next step is to fiddle with the time and distance units. If we

1. measure x and t in equivalent units (so c = 1);
2. denote v/c by β; and
3. re-use the x symbol to denote x/c (that’s just to simplify by saving symbols);

we get:

φ = m·(t–β∙x)

This argument is the product of two factors: (1) m and (2) t–β∙x.

1. The first factor – i.e. the mass m – is an inherent property of the particle that we’re looking at: it measures its inertia, i.e. the key variable in any dynamical model (i.e. any model – classical or quantum-mechanical – representing the motion of the particle).
2. The second factor – i.e. t–v∙x – reminds one of the argument of the wavefunction that’s used in classical mechanics, i.e. x–vt, with v the velocity of the wave. Of course, we should note two major differences between the t–β∙x and x–vt expressions:
1. β is a relative velocity (i.e. a ratio between 0 and 1), while v is an absolute velocity (i.e. a number between 0 and ≈ 299,792,458 m/s).
2. The t–β∙x expression switches the time and distance variables as compared to the x–vt expression, and vice versa.

Both differences are important, but let’s focus on the second one. From a math point of view, the t–β∙x and x–vt expressions are equivalent. However, time is time, and distance is distance—in physics, that is. So what can we conclude here? To answer that question, let’s re-analyze the x–vt expression. Remember its origin: if we have some wave function F(x–vt), and we add some time Δt to its argument – so we’re looking at F[x−v(t+Δt)] now, instead of F(x−vt) – then we can restore it to its former value by also adding some distance Δx = v∙Δt to the argument: indeed, if we do so, we get F[x+Δx−v(t+Δt)] = F(x+vΔt–vt−vΔt) = F(x–vt). Of course, we can do the same analysis the other way around, so we add some Δx and then… Well… You get the idea.

Can we do that for for the F(t–β∙x) expression too? Sure. If we add some Δt to its argument, then we can restore it to its former value by also adding some distance Δx = Δt/β. Just check it: F[(t+Δt)–β(x+Δx)] = F(t+Δt–βx−βΔx) = F(t+Δt–βx−βΔt/β) = F(t–β∙x).

So the mathematical equivalence between the t–β∙x and x–vt expressions is surely meaningful. The F(x–vt) function uniquely determines the waveform and, as part of that determination (or definition, if you want), it also defines its velocity v. Likewise, we can say that the Φ(φ) = Φ[m·(t–β∙x)] function defines the (relative) velocity (β) of the particle that we’re looking at—quantum-mechanically, that is.

You’ll say: we’ve got two variables here: m and β. Well… Yes and no. We can look at m as an independent variable here. In fact, if you want, we could define yet another variable –χ = φ/m = t–β∙x – and, hence, yet another wavefunction here:

Ψ(θ) = [ei·(θ/ħ)]ħ = [ei·φ]ħ = Φ(φ) = Χ(χ) = [ei·φ/m]ħ·m = [ei·χ]ħ·m = [ei·θ/(ħ·m)]ħ·m

Does that make sense? Maybe. Think of it: the spatial dimension of the wave pulse F(x–vt) – if you don’t know what I am talking about: just think of its ‘position’ – is defined by its velocity v = x/t, which – from a math point of view – is equivalent to stating: x – v∙t = 0. Likewise, if we look at our wavefunction as some pulse in space, then its spatial dimension would also be defined by its (relative) velocity, which corresponds to the classical (relative) velocity of the particle we’re looking at. So… Well… As I said, I’ll let you think of all this.

Post Scriptum:

1. You may wonder what that ħ·m factor in that Χ(χ) = [ei·χ]ħ·m = [ei·(t–β∙x)/(ħ·m)]ħ·m function actually stands for. Well… If we measure time and distance in equivalent units (so = 1 and, therefore, E = m), and if we measure energy in units of ħ, then ħ·m corresponds to our old energy unit, i.e. E measured in joule, rather than in terms of ħ. So… Well… I don’t think we can say much more about it.
2. Another thing you may want to think about is the relativistic transformation of the wavefunction. You know that we should correct Newton’s Law of Motion for velocities approaching c. We do so by integrating the Lorentz factor. In light of the fact that we’re using the relative velocity (β) in our wave function, do you think we still need to apply such corrections for the wavefunction? What’s your guess? 🙂