[Preliminary note (added on 13 June 2019): When re-reading what I wrote below, I realize I would fundamentally re-write certain sections. I think I have found a comprehensive realist interpretation of quantum mechanics and, hence, I’d recommend you check my recent papers. The writings below are probably just good to illustrate how I got there. Lettura felice!]
In my post on the Schrödinger equation and the two de Broglie relations, I wrote the complete Schrödinger equation in a rather unconventional form:
I also showed how one could derive the quintessential wavefunction from it. All we had to do is to substitute ψ for ei(kx − ωt), and then calculate that time derivative and that Laplacian. To simplify the calculations, we assumed one-dimensional space, so ∇2ψ = ∂2ψ/∂x2. However, the result can easily be generalized. The development was as follows:
- The time derivative on the left-hand side is ∂ψ/∂t = −iω·ei(kx − ωt).
- The second-order derivative on the right-hand side is ∂2ψ/∂x2 = (ik)·(ik)·ei(kx − ωt) = −k2·ei(kx − ωt).
- The terms with V/ħ (i.e. the potential measured in units of ħ) and Eint = m0c2 (i.e. the rest energy) are just what they: (V/ħ)·ei(kx − ωt) and (Eint/ħ)·ei(kx − ωt).
- The ei(kx − ωt) factors on both sides cancel out and, hence, equating both sides gives us the following condition:
−iω = −(iħ/2m)·k2 − V/ħ − Eint/ħ ⇔ ω = (ħ/2m)·k2 + V/ħ + Eint/ħ
Using the ω = E/ħ equation (i.e. one of the two de Broglie relations), and substituting k for p/ħ (i.e. the second de Broglie relation), we find what’s obvious:
E = ħω = (ħ2/2m)·k2 + V + Eint = (p2/2m)·k2 + V + Eint
Substituting E for the above in the wavefunction gives us de Broglie‘s wavefunction, which is the complete one:
ψ(θ) = Ψ(x, t) = a·e−iθ = a·e−i[(Eint + p2/(2m) + V)·t − p∙x]/ħ
The only thing we added was some coefficient a, so as to remind us an actual wavefunction is a composite wave, made up of a potentially infinite number of aie−iθi functions, really. [And, yes, sorry for using i here as subscript, as well as for what it usually stands for, i.e. the imaginary unit. It’s just to confuse you. :-)]
This is all quite straightforward and so I won’t come back to that. What I do want to come back to is how that imaginary unit (i) in front of the right-hand side of our complete Schrödinger equation works. If we write dψ/dt as dψ/dt = a + ib, and the sum of terms within the brackets—i.e. (ħ/2m)·∇2ψ + (V/ħ)·ψ + (Eint/ħ)·ψ—as c + id, then the Schrödinger equation implies the following simple equality:
a + ib = i(c + id) = −d + ic
Now, two complex numbers are the same if, and only if, their real and imaginary parts are the same, so we find that the following must be true:
a = −d and b = c
You’ll say that’s obvious, and it is. But think of it. On the left-hand side, we’ve got a time derivative. It’s a complex-valued function, and its real part is minus the imaginary part of the expression between brackets, while its imaginary part is the real part of the same expression. In fact, when analyzing the dynamics of the situation, we may just treat the second and third term as some constant, so then we can write:
- Re(dψ/dt) = −(ħ/2m)·Im(∇2ψ)
- Im(dψ/dt) = (ħ/2m)·Re(∇2ψ)
This is very deep. In that post I mentioned, I mentioned the similarity with the heat diffusion equation, which we wrote as:
This equation modeled the flow, in space and in time, of heat, i.e. (thermal) energy. In fact, you’ll remember temperature is, in fact, a measure of the average kinetic energy of the constituent parts of the substance. So what is that ψ quantity then? We know it’s a complex number, and we also know its absolute square gives us probabilities. But that doesn’t tell us what ψ is, really.
In my humble opinion, we may look at it as a form of two-dimensional energy.
Yes. I know it’s hard to swallow, as we’re so used to thinking of energy as some kind of magnitude only, i.e. something without direction. So am I thinking of a vector quantity now?
Yes and no. I am not thinking of something directional, but I am thinking of something two-dimensional, like a simple array of two numbers, rather than just one. It’s hard to explain. For me, it’s sure there must be something there—if only because of the interference phenomenon—and it’s two-dimensional. Moreover, Schrödinger’s equation tells us its two dimensions are related to space and time respectively, but in a very special way: the imaginary unit (i) sort of mediates between its flow in space (which is captured by the Laplacian) and its flow in time (which is captured by the time derivative).
It’s a bit sad we’re stuck with the terms real and imaginary to describe those two dimensions, as I feel it contributes to us finding it difficult to understand what’s going on here in a more intuitive way. Some visualization may help, however. Compare the following two graps, for example. Just imagine we either look at how the wavefunction behaves at some point in space, with the time fixed at some point t = t0, or, alternatively, that we look at how the wavefunction behaves in time at some point in space x = x0. As you can see, increasing k = p/ħ or increasing ω = E/ħ gives the wavefunction a higher ‘density’ in space or, alternatively, in time.
That makes sense, intuitively. In fact, when thinking about how the energy, or the momentum, affects the shape of the wavefunction, I am reminded of an airplane propeller: as it spins, faster and faster, it gives the propeller some ‘density’, in space as well as in time, as its blades cover more space in less time. It’s an interesting analogy: it helps—me, at least—to think through what that wavefunction might actually represent.
So as to stimulate your imagination even more, you should also think of representing the real and complex part of that ψ = a·ei(k∙x−ω·t) = a·eiθ = a·(cosθ + i·sinθ) formula in a different way. In the graphs above, we just showed the sine and cosine in the same plane but, as you know, the real and the imaginary axis are orthogonal, so Euler’s formula a·eiθ = a·(cosθ + i·sinθ) = a·cosθ +i·a·sinθ = Re(ψ) + i·Im(ψ) = a·ei(kx − ωt) may also be graphed as follows:
I could write an awful lot about this because, the more you look at it, the more ideas come to mind. But I’ll let you be creative now yourself. 🙂