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) = [*e*^{−i·(θ/ħ)}]^{ħ }= Φ(φ) = [*e*^{−i·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/s^{2})·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·*c*^{2 }and p = m·*v *formulas still apply? They should: both *c *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·*c*^{2}∙t − m∙*v*·x = m·*c*^{2}[t – (*v*/*c*)∙(x/*c*)] = m·*c*^{2}[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·*c*^{2 }and p = m·*v *formulas.

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

- measure x and t in equivalent units (so
*c*= 1); - denote
*v*/*c*by β; and - 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.

- 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). - 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–*v*t, with*v*the velocity of the wave. Of course, we should note two major differences between the t–β∙x and x–*v*t expressions:

- 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*c*≈ 299,792,458 m/s). - The t–β∙x expression switches the time and distance variables as compared to the x–
*v*t 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–*v*t 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–*v*t expression. Remember its origin: if we have some wave function F(x–*v*t), and we *add *some time Δt to its argument – so we’re looking at F[x−*v*(t+Δt)] now, instead of F(x−*v*t) – 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+*v*Δt–*v*t−*v*Δt) = F(x–*v*t). 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–*v*t expressions is surely meaningful. The F(x–*v*t) 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) = [*e*^{−i·(θ/ħ)}]^{ħ }= [*e*^{−i·φ}]^{ħ} = Φ(φ) = Χ(χ) = [*e*^{−i·φ/m}]^{ħ·m} = [*e*^{−i·χ}]^{ħ·m }= [*e*^{−i·θ/(ħ·m)}]^{ħ·m}

Does that make sense? Maybe. Think of it: the spatial dimension of the wave pulse F(x–*v*t) – 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**:

- You may wonder what that ħ·m factor in that Χ(χ) = [
*e*^{−i·χ}]^{ħ·m }= [*e*^{−i·(t–β∙x)/(ħ·m)}]^{ħ·m }function actually stands for. Well… If we measure time and distance in equivalent units (so*c*= 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. - 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? 🙂