# Wave functions and equations: a summary

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.

Original post:

Schrödinger’s wave equation for spin-zero, spin-1/2, and spin-one particles in free space differ from each other by a factor two:

1. For particles with zero spin, we write: ∂ψ/∂t = i·(ħ/m)·∇2ψ. We get this by multiplying the ħ/(2m) factor in Schrödinger’s original wave equation – which applies to spin-1/2 particles (e.g. electrons) only – by two. Hence, the correction that needs to be made is very straightforward.
2. For fermions (spin-1/2 particles), Schrödinger’s equation is what it is: ∂ψ/∂t = i·[ħ/(2m)]·∇2ψ.
3. For spin-1 particles (photons), we have ∂ψ/∂t = i·(2ħ/m)·∇2ψ, so here we multiply the ħ/m factor in Schrödinger’s wave equation for spin-zero particles by two, which amounts to multiplying Schrödinger’s original coefficient by four.

Look at the coefficients carefully. It’s a strange succession:

1. The ħ/m factor (which is just the reciprocal of the mass measured in units of ħ) works for spin-0 particles.
2. For spin-1/2 particles, we take only half that factor: ħ/(2m) = (1/2)·(ħ/m).
3. For spin-1 particles, we double that factor: 2ħ/m = 2·(ħ/m).

I describe the detail on my Deep Blue page, so please go there for more detail. What I did there, can be summarized as follows:

• The spin-one particle is the photon, and we derived the photon wavefunction from Maxwell’s equations in free space, and found that it solves the ∂ψ/∂t = i·(2ħ/m)·∇2ψ equation, not the ∂ψ/∂t = i·(ħ/m)·∇2ψ or ∂ψ/∂t = i·[ħ/(2m)]·∇2ψ equations.
• As for the spin-zero particles, we simplified the analysis by assuming our particle had zero rest mass, and we found that we were basically modeling an energy flow.
• The analysis for spin-1/2 particles is just the standard analysis you’ll find in textbooks.

We can speculate how things would look like for spin-3/2 particles, or for spin-2 particles, but let’s not do that here. In any case, we will come back to this. Let’s first focus on the more familiar terrain, i.e. the wave equation for spin-1/2 particles, such as protons or electrons. [A proton is not elementary – as it consists of quarks – but it is a spin-1/2 particle, i.e. a fermion.]

#### The phase and group velocity of the wavefunction for spin-1/2 particles (fermions)

We’ll start with the very beginning of it all, i.e. the two equations that the young Comte Louis de Broglie presented in his 1924 PhD thesis, which give us the temporal and spatial frequency of the wavefunction, i.e. the ω and k in the θ = ω·t − k·t argument  of the a·ei·θ wavefunction:

1. ω = E/ħ
2. k = p/ħ

This allows to calculate the phase velocity of the wavefunction:

vp = ω/k = (E/ħ)/(p/ħ) = E/p

This is an elementary wavefunction, several of which we would add with appropriate coefficients, with uncertainty in the energy and momentum ensuring our component waves have different frequencies, and, therefore, the concept of a group velocity does not apply. In effect, the a·ei·θ wavefunction does not describe a localized particle: the probability to find it somewhere is the same everywhere. We may want to think of our wavefunction being confined to some narrow band in space, with us having no prior information about the probability density function, and, therefore, we assume a uniform distribution. Assuming our box in space is defined by Δx = x2 − x1, and imposing the normalization condition (all probabilities have to add up to one), we find that the following logic should hold:

(Δx)·a2 = (x2−x1a= 1 ⇔ Δx = 1/a2

However, we are, of course, interested in the group velocity, as the group velocity should correspond to the classical velocity of the particle. The group velocity of a composite wave is given by the vg = ∂ω/∂k formula. Of course, that formula assumes an unambiguous relation between the temporal and spatial frequency of the component waves, which we may want to denote as ωn and kn, with n = 1, 2, 3,… However, we will not use the index as the context makes it quite clear what we are talking about.

The relation between ωn and kn is known as the dispersion relation, and one particularly nice way to calculate ω as a function of k is to distinguish the real and imaginary parts of the ∂ψ/∂t =i·[ħ/(2m)]·∇2ψ wave equation and, hence, re-write it as a pair of two equations:

1. Re(∂ψB/∂t) =   −[ħ/(2m)]·Im(∇2ψB) ⇔ ω·cos(kx − ωt) = k2·[ħ/(2m)]·cos(kx − ωt)
2. Im(∂ψB/∂t) = [ħ/(2m)]·Re(∇2ψB) ⇔ ω·sin(kx − ωt) = k2·[ħ/(2m)]·sin(kx − ωt)

Both equations imply the following dispersion relation:

ω = ħ·k2/(2m)

We can now calculate vg = ∂ω/∂k as:

vg = ∂ω/∂k = ∂[ħ·k2/(2m)]/∂k = 2ħk/(2m) = ħ·(p/ħ)/m = p/m = m·v/m = v

That’s nice, because it’s what we wanted to find. If the group velocity would not equal the classical velocity of our particle, then our model would not make sense.

We used the classical momentum formula in our calculation above: p = m·v. To calculate the phase velocity of our wavefunction, we need to calculate that E/p ratio and, hence, we need an energy formula. Here we have a lot of choice, as energy can be defined in many ways: is it rest energy, potential energy, or kinetic energy? At this point, I need to remind you of the basic concepts.

### The argument of the wavefunction as the proper time

It is obvious that the energy concept that is to be used in the ω = E/ħ is the total energy. Louis de Broglie himself noted that the energy of a particle consisted of three parts:

1. The particle’s rest energy m0c2, which de Broglie referred to as internal energy (Eint): it includes the rest mass of the ‘internal pieces’, as de Broglie put it (now we call those ‘internal pieces’ quarks), as well as their binding energy (i.e. the quarks’ interaction energy);
2. Any potential energy (V) it may have because of some field (so de Broglie was not assuming the particle was traveling in free space): the field(s) can be anything—gravitational, electromagnetic—you name it: whatever changes the energy because of the position of the particle;
3. The particle’s kinetic energy, which he wrote in terms of its momentum p: K.E. = m·v2/2 = m2·v2/(2m) = (m·v)2/(2m) = p2/(2m).

So the wavefunction, as de Broglie wrote it, can be written as follows:

ψ(θ) = ψ(x, t) = a·eiθ = a·e−i[(Eint + p2/(2m) + V)·t − p∙x]/ħ

This formula allows us to analyze interesting phenomena such as the tunneling effect and, hence, you may want to stop here and start playing with it. However, you should note that the kinetic energy formula that is used here is non-relativistic. The relativistically correct energy formula is E = mvc, and the relativistically correct formula for the kinetic energy is the difference between the total energy and the rest energy:

K.E. = E − E0 = mv·c2 − m0·c2 = m0·γ·c2 − m0·c2 = m0·c2·(γ − 1), with γ the Lorentz factor.

At this point, we should simplify our calculations by adopting natural units, so as to ensure the numerical value of = 1, and likewise for ħ. Hence, we assume all is described in Planck units, but please note that the physical dimensions of our variables do not change when adopting natural units: time is time, energy is energy, etcetera. But when using natural units, the E = mvc2 reduces to E = mv. As for our formula for the momentum, this formula remains p = mv·v, but is now some relative velocity, i.e. a fraction between 0 and 1. We can now re-write θ = (E/ħ)·t – (p/ħ)·x as:

θ = E·t – p·x = E·t − p·v·t = mv·t − mv·v·v·t = mv·(1 − v2)·t

We can also write this as:

ψ(x, t) = a·ei·(mv·t − p∙x) = a·ei·[(m0/√(1−v2))·t − (m0·v/√(1−v2)∙x) = a·ei·m0·(t − v∙x)/√(1−v2)

The (t − v∙x)/√(1−v2) factor in the argument is the proper time of the particle as given by  the formulas for the Lorentz transformation of spacetime:

However, both the θ = mv·(1 − v2)·t and θ = m0·t’ = m0·(t − v∙x)/√(1−v2) are relativistically correct. Note that the rest mass of the particle (m0) acts as a scaling factor as we multiply it with the proper time: a higher m0 gives the wavefunction a higher density, in time as well as in space.

Let’s go back to our vp = E/p formula. Using natural units, it becomes:

vp = E/p = mv/mv·v = 1/v

Interesting! The phase velocity is the reciprocal of the classical velocity! This implies it is always superluminal, ranging from vp = ∞ to vp= 1 for going from 0 to 1 = c, as illustrated in the simple graph below.

Let me note something here, as you may also want to use the dispersion relation, i.e. ω = ħ·k2/(2m), to calculate the phase velocity. You’d write:

vp = ω/k = [ħ·k2/(2m)]/k = ħ·k/(2m) = ħ·(p/ħ)/(2m) = m·v/(2m) = v/2

That’s a nonsensical result. Why do we get it? Because we are mixing two different mass concepts here: the mass that’s associated with the component wave, and the mass that’s associated with the composite wave. Think of it. That’s where Schrödinger’s equation is different from all of the other diffusion equations you’ve seen: the mass factor in the ∂ψ/∂t = i·[ħ/(2m)]·∇2ψ equation is the mass of the particle that’s being represented by the wavefunction that solves the equation. Hence, the diffusion constant ħ/(2m) is not a property of the medium. In that sense, it’s different from the κ/k factor in the ∂T/∂t = (κ/k)·∇2T heat diffusion, for example. We don’t have a medium here and, therefore, Schrödinger’s equation and the associated wavefunction are intimately connected.

It’s an interesting point, because if we’re going to be measuring the mass as multiples of ħ/2 (as suggested by the ħ/(2m) = 1/[m/[ħ/2)] factor itself), then its possible values (for ħ = 1) will be 1/2, 1, 3/2, 2, 5/2,… Now that should remind you of a few things—things like harmonics, or allowable spin values, or… Well… So many things. 🙂

Let’s do the exercise for bosons now.

### The phase and group velocity of the  wavefunction for spin-0 particles

My Deep Blue page explains why we need to drop the 1/2 factor in Schrödinger’s equation to make it fit the wavefunction for bosons. We distinguished two bosons: (1) the (theoretical) zero-mass particle (which has spin zero), and the (actual) photon (which has spin one). Let’s first do the analysis for the spin-zero particle.

• A zero-mass particle (i.e. a particle with zero rest mass) should be traveling at the speed of light: both its phase as well as its group velocity should be equal to = 1. In fact, we’re not talking composite wavefunctions here, so there’s no such thing as a group velocity. We’re not adding waves: there is only one wavefunction. [Note that we don’t need to add waves with different frequencies in order to localize our particle, because quantum mechanics and relativity theory come together here in what might well be the most logical and absurd conclusion ever: as an outside observer, we’re going to see all those zero-mass particles as point objects whizzing by because of the relativistic length contraction. So their wavefunction is only all over spacetime in their proper space and time, but not in ours!]
• Now, it’s easy to show that, if we choose our time and distance units such that c = 1, then the energy formula reduces to E = m∙c2 = m. Likewise, we find that p = m∙c = m. So we have this strange condition: E = p = m.
• Now, this is not consistent with the ω = ħ·k2/(2m) we get out of the ∂ψ/∂t = i·[ħ/(2m)]·∇2ψ equation, because E/ħ = ħ·(p/ħ)2/(2m) ⇔ E = m2/(2m) = m/2. That does not fit the E = p = m condition. The only way out is to drop the 1/2 factor, i.e. to multiply Schrödinger’s coefficient with 2.

Let’s quickly check if it does the trick. We assume E, p and m will be multiples of ħ/2 (E = p = m = n·(ħ/2), so the wavefunction is ei∙[t − x]n·/2, Schrödinger’s constant becomes 2/n, and the derivatives for ∂ψ/∂t = i·(ħ/m)·∇2ψ are:

• ∂ψ/∂t = −i·(n/2)·ei∙[t − x]·n/2
• 2ψ = ∂2[ei∙[t − x]·n/2]/∂x= i·(n/2)·∂[ei∙[t − x]·n/2]/∂x = −(n2/4)·ei∙[t − x]·n/2

So the Schrödinger equation becomes:

i·(n/2)·ei∙[t − x]n·/2) = −i·(2/n)·(n2/4)·ei∙[t − x]·n/2 ⇔  n/2 = n/2 ⇔ 1 = 1

As Feynman would say, it works like a charm, and note that n does not have to be some integer to make this work.

So what makes spin-1/2 particles different? The answer is: they have both linear as well as angular momentum, and the equipartition theorem tells us the energy will be shared equally among both , so they will pick up linear and angular momentum. Hence, the associated condition is not E = p = m, but E = p = 2m. We’ll come back to this.

Let’s now summarize how it works for spin-one particles

### The phase and group velocity of the  wavefunction for spin-1 particles (photons)

Because of the particularities that characterize an electromagnetic wave, the wavefunction packs two waves, capturing both the electric as well as the magnetic field vector (i.e. E and B). For the detail, I’ll refer you to the mentioned page, because the proof is rather lengthy (but easy to follow, so please do check it out). I will just briefly summarize the logic here.

1. For the spin-zero particle, we measured E, m and p in units of – or as multiples of – the ħ/2 factor. Hence, the elementary wavefunction (i.e. the wavefunction for E = p = m = 1) for the zero-mass particle is ei(x/2 − t/2).

2. For the spin-1 particle (the photon), one can show that we get two of these elementary wavefunctions (ψand ψB), and one can then prove that we can write the sum of the electric and magnetic field vector as:

E + BE + B = ψ+ ψ= E + i·E

= √2·ei(x/2 − t/2+ π/4) = √2·ei(π/4)·ei(x/2 − t/2) = √2·ei(π/4)·= √2·ei(π/4)·ei(x/2 − t/2)

Hence, the photon has a special wavefunction. Does it solve the Schrödinger equation? It does when we use the 2ħ/m diffusion constant, rather than the ħ/m or ħ/(2m) coefficient. Let us quickly check it. The derivatives are:

• ∂ψ/∂t = −√2·ei(π/4)·ei∙[t − x]/2·(i/2)
• 2ψ = ∂2[√2·ei(π/4)·ei∙[t − x]/2]/∂x= √2·ei(π/4)·∂[ei∙[t − x]/2·(i/2)]/∂x = −√2·ei(π/4)·ei∙[t − x]/2·(1/4)

Note, however, that we have two mass, energy and momentum concepts here: EE, pE, mE and EB, pB, and mB respectively. Hence, if E= p= mE = E= p= mB = 1/2, then E = E+ EB, p = p+ pB and m = m+ mare all equal to 1. Hence, because E = p = m = 1 and we measure in units of ħ, the 2ħ/m factor is equal to 2 and, therefore, the modified Schrödinger’s equation ∂ψ/∂t = i·(2ħ/m)·∇2ψ becomes:

i·√2·ei(π/4)·ei∙[t − x]/2·(1/2) = −i·√2·2·ei(π/4)·ei∙[t − x]/2·(1/4) ⇔ 1/2 = 2/4 = 1/2

It all works out. Let’s quickly check it for E, m and p being multiples of ħ, so we write: E = p = m = n·ħ = n, so the wavefunction is √2·ei(π/4)·ei∙[t − x]n·/2, Schrödinger’s 2ħ/m constant becomes 2ħ/m = 2/n, and the derivatives for ∂ψ/∂t = i·(ħ/m)·∇2ψ are:

• ∂ψ/∂t = −i·(n/2)·√2·ei(π/4)·ei∙[t − x]·n/2
• 2ψ = ∂2[ei∙[t − x]·n/2]/∂x= i·√2·ei(π/4)·(n/2)·∂[ei∙[t − x]·n/2]/∂x = −√2·(n2/4)·ei(π/4)·ei∙[t − x]·n/2

So the Schrödinger equation becomes:

i·√2·ei(π/4)·(n/2)·ei∙[t − x]·n/2) = −i·√2·ei(π/4)·(2/n)·(n2/4)·ei∙[t − x]·n/2 ⇔  n/2 = n/2 ⇔ 1 = 1

It works like a charm again. Note the subtlety of the difference between the ħ/(2m) and 2ħ/m factor: it depends on us measuring the mass (and, hence, the energy and momentum as well) in units of ħ/2 (for spin-0 particles) or, alternatively (for spin-1 particles), in units of ħ. This is very deep—but it does make sense in light of the En =n·ħ·ω = n·h·f formula that solves the black-body radiation problem, as illustrated below. [The formula next to the energy levels is the probability of an atomic oscillator occupying that energy level, which is given by Boltzmann’s Law. You can check things in my post on it.]

It is now time to look at something else.

### Schrödinger’s equation as an energy propagation mechanism

The Schrödinger equations above are not complete. The complete equation includes force fields, i.e. potential energy:

To write the equation like this, we need to move the on the right-hand side of our ∂ψ/∂t = i·(2ħ/m)·∇2ψ equation to the other side, and multiply both sides with −1. [Remember: 1/i = −i.] Now, it is very interesting to do a dimensional analysis of this equation. Let’s do the right-hand side first. The ħfactor in the ħ/(2m) is expressed in J2·s2. Now that doesn’t make much sense, but then that mass factor in the denominator makes everything come out alright. Indeed, we can use the mass-equivalence relation to express m in J/(m/s)2 units. So we get: (J2·s2)·[(m/s)2/J] = J·m2. But so we multiply that with some quantity (the Laplacian) that’s expressed per m2. So −(ħ2/2m)·∇2ψ is something expressed in joule, so it’s some amount of energy! Interesting, isn’t it? [Note that it works out fine with the addition Vψ term, which is also expressed in joule.] On the left-hand side, we have ħ, and its dimension is the action dimension: J·s, i.e. force times distance times time (N·m·s). So we multiply that with a time derivative and we get J once again, the unit of energy. So it works out: we have joule units both left and right. But what does it mean?

Well… The Laplacian on the right-hand side works just the same as for our heat diffusion equation: it gives us a flux density, i.e. something expressed per square meter (1/m2). Likewise, the time derivative on the left-hand side gives us a flow per second. But so what is it that is flowing here? Well… My interpretation is that it is energy, and it’s flowing between a real and an imaginary space—but don’t be fooled by the terms here: both spaces are equally real, as both have an actual physical dimension. Let me explain.

Things become somewhat more comprehensible when we remind ourselves that the Schrödinger equation is equivalent to the following pair of equations:

1. Re(∂ψ/∂t) =   −(ħ/2m)·Im(∇2ψ) ⇔ ω·cos(kx − ωt) = k2·(ħ/2m)·cos(kx − ωt)
2. Im(∂ψ/∂t) = (ħ/2m)·Re(∇2ψ) ⇔ ω·sin(kx − ωt) = k2·(ħ/2m)·sin(kx − ωt)

So what? Let me insert an illustration here. See what happens. The wavefunction acts as a link function between our physical spacetime and some other space whose dimensions – in my humble opinion – are also physical. We have those sines and cosines, which mirror the energy of the system at any point in time, as measured by the proper time of the system.

Let me more precise. The wavefunction, as a link function between two spaces here, associates every point in spacetime with some real as well as some imaginary energy here—but, as mentioned above, that imaginary energy is as real as the real energy. What it embodies really is the energy conservation law: at any point in time (as measured by the proper time) the sum of kinetic and potential energy must be equal to some constant, and so that’s what’s shown here. Indeed, you should note the phase shift between the sine and the cosine function: if one reaches the +1 or −1 value, then the other function reaches the zero point—and vice versa. It’s a beautiful structure.

Of course, the million-dollar question is: is it a physical structure, or a mathematical structure? The answer is: it’s a bit of both. It’s a mathematical structure but, at the same time, its dimension is physical: it’s an energy space. It’s that energy that explains why amplitudes interfere—which, as you know, is what they do. So these amplitudes are something real, and as the dimensional analysis of Schrödinger’s equation reveals their dimension is expressed in joule, then… Well… Then these physical equations say what they say, don’t they? And what they say, is something like the diagram below.

Note that the diagram above does not show the phase difference between the two springs. The animation below does a better job here, although you need to realize the hand of the clock will move faster or slower as our object travels through force fields and accelerates or decelerates accordingly.

We may relate that picture above to the principle of least action, which ensures that the difference between the kinetic energy (KE) and potential energy (PE) in the integrand of the action integral, i.e.

is minimized along the path of travel.

The spring metaphor should also make us think of the energy formula for a harmonic oscillator, which tells us that the total energy – kinetic (i.e. the energy related to its momentum) plus potential (i.e. the energy stored in the spring) – is equal to T + U = m·ω02/2. The ωhere is the angular velocity, and we have two springs here, so the total energy would be the sum of both, i.e. m·ω02, without the 1/2 factor. Does that make sense? It’s like an E = m·vequation, so that’s twice the (non-relativistic) kinetic energy. Does that formula make any sense?

In the context of what we’re discussing here, it does. Think about the limit situation by trying to imagine a zero-mass particle here (I am talking a zero-mass spin-1/2 particle this time). It would have no rest energy, so it’s only energy is kinetic, which is equal to:

K.E. = E − E0 = mv·c2 − m0·c2 = mc·c2

Why is mequal to mc? Zero-mass particles must travel at the speed of light, as the slightest force on them gives them an infinite acceleration. So there we are: the m·ω02 equation makes sense! But what if we have a non-zero rest mass? In that case, look at that pair of equations again: they give us a dispersion relation, i.e. a relation between ω and k. Indeed, using natural units again, so the numerical value of ħ = 1, we can write:

ω = k2/(2m) = p2/(2m) = (m·v)2/(2m) = m·v2/2

This equation seems to represent the kinetic energy but m is not the rest mass here: it’s the relativistic mass, so that makes it different from the classical kinetic energy formula (K.E. = m0·v2/2). [It may be useful here to remind you of how we get that classical formula. We basically integrate force over distance, from some start to some final point of a path in spacetime. So we write: ∫ F·ds = ∫ (m·a)·ds = ∫ (m·a)·ds = ∫ [m·(dv/dt)]·ds = ∫ [m·(ds/dt)]·d= ∫ m·v·ds. So we can solve that using the m·v2/2 primitive but only if m does not vary, i.e. if m = m0. If velocity are high, we need the relativistic mass concept.]

So we have a new energy concept here: m·v2, and it’s split over those two springs. Hmm… The interpretation of all of this is not so easy, so I will need to re-visit this. As for now, however, it looks like the Universe can be represented by a V-twin engine! 🙂

### Is it real?

You may still doubt whether that new ‘space’ has an actual energy dimension. It’s a figment of our mind, right? Well… Yes and no. Again, it’s a bit of a mixture between a mathematical and a physical space: it’s definitely not our physical space, as it’s not the spacetime we’re living in. But, having said that, I don’t think this energy space is just a figment of our mind. Let me give you some additional reasons, beside the dimensional analysis we did above.

For example, there is the fact that we need to to take the absolute square of the wavefunction to get the probability that our elementary particle is actually right there! Now that’s something real! Hence, let me say a few more things about that. The absolute square gets rid of the time factor. Just write it out to see what happens:

|reiθ|2 = |r|2|eiθ|2 = r2[√(cos2θ + sin2θ)]2 = r2(√1)2 = r2

Now, the gives us the maximum amplitude (sorry for the mix of terminology here: I am just talking the wave amplitude here – i.e. the classical concept of an amplitude – not the quantum-mechanical concept of a probability amplitude). Now, we know that the energy of a wave – anywave, really – is proportional to the amplitude of a wave. It would also be logical to expect that the probability of finding our particle at some point x is proportional to the energy densitythere, isn’t it? [I know what you’ll say now: you’re squaring the amplitude, so if the dimension of its square is energy, then its own dimension must be the square root, right? No. Wrong. That’s why this confusion between amplitude and probability amplitude is so bad. Look at the formula: we’re squaring the sine and cosine, to then take the square root again, so the dimension doesn’t change: it’s √J2 = J.]

The third reason why I think the probability amplitude represents some energy is that its real and imaginary part also interfere with each other, as is evident when you take the ordinary square (i.e. not the absolute square). Then the i2   = –1 rule comes into play and, therefore, the square of the imaginary part starts messing with the square of the real part. Just write it out:

(reiθ)2 = r2(cosθ + isinθ)2 = r2(cos2θ – sin2θ + 2icosθsinθ)2 = r2(1 – 2sin2θ + 2icosθsinθ)2

As mentioned above, if there’s interference, then something is happening, and so then we’re talking something real. Hence, the real and imaginary part of the wavefunction must have some dimension, and not just any dimension: it must be energy, as that’s the currency of the Universe, so to speak.

Let me add a philosophical note here—or an ontological note, I should say. When you think we should only have one physical space, you’re right. This new physical space, in which we relate energy to time, is not our physical space. It’s not reality—as we know, as we experience it. So, in that sense, you’re right. It’s not physical space. But then… Well… It’s a definitional matter. Any space whose dimensions are physical, is a physical space for me. But then I should probably be more careful. What we have here is some kind of projection of our physical space to a space that  lacks… Well… It lacks the spatial dimension. It’s just time – but a special kind of time: relativistic proper time – and energy—albeit energy in two dimensions, so to speak. So… What can I say? Just what I said a couple of times already: it’s some kind of mixture between a physical and mathematical space. But then… Well… Our own physical space – including the spatial dimension – is something like a mixture as well, isn’t it? We can try to disentangle them – which is what I am trying to do – but we’ll never fully succeed.

# The Imaginary Energy Space

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.

Original post:

Intriguing title, isn’t it? You’ll think this is going to be highly speculative and you’re right. In fact, I could also have written: the imaginary action space, or the imaginary momentum space. Whatever. It all works ! It’s an imaginary space – but a very real one, because it holds energy, or momentum, or a combination of both, i.e. action. 🙂

So the title is either going to deter you or, else, encourage you to read on. I hope it’s the latter. 🙂

In my post on Richard Feynman’s exposé on how Schrödinger got his famous wave equation, I noted an ambiguity in how he deals with the energy concept. I wrote that piece in February, and we are now May. In-between, I looked at Schrödinger’s equation from various perspectives, as evidenced from the many posts that followed that February post, which I summarized on my Deep Blue page, where I note the following:

1. The argument of the wavefunction (i.e. θ = ωt – kx = [E·t – p·x]/ħ) is just the proper time of the object that’s being represented by the wavefunction (which, in most cases, is an elementary particle—an electron, for example).
2. The 1/2 factor in Schrödinger’s equation (∂ψ/∂t = i·(ħ/2m)·∇2ψ) doesn’t make all that much sense, so we should just drop it. Writing ∂ψ/∂t = i·(m/ħ)∇2ψ (i.e. Schrödinger’s equation without the 1/2 factor) does away with the mentioned ambiguities and, more importantly, avoids obvious contradictions.

Both remarks are rather unusual—especially the second one. In fact, if you’re not shocked by what I wrote above (Schrödinger got something wrong!), then stop reading—because then you’re likely not to understand a thing of what follows. 🙂 In any case, I thought it would be good to follow up by devoting a separate post to this matter.

## The argument of the wavefunction as the proper time

Frankly, it took me quite a while to see that the argument of the wavefunction is nothing but the t’ = (t − v∙x)/√(1−v2)] formula that we know from the Lorentz transformation of spacetime. Let me quickly give you the formulas (just substitute the for v):

In fact, let me be precise: the argument of the wavefunction also has the particle’s rest mass m0 in it. That mass factor (m0) appears in it as a general scaling factor, so it determines the density of the wavefunction both in time as well as in space. Let me jot it down:

ψ(x, t) = a·ei·(mv·t − p∙x) = a·ei·[(m0/√(1−v2))·t − (m0·v/√(1−v2))∙x] = a·ei·m0·(t − v∙x)/√(1−v2)

Huh? Yes. Let me show you how we get from θ = ωt – kx = [E·t – p·x]/ħ to θ = mv·t − p∙x. It’s really easy. We first need to choose our units such that the speed of light and Planck’s constant are numerically equal to one, so we write: = 1 and ħ = 1. So now the 1/ħ factor no longer appears.

[Let me note something here: using natural units does not do away with the dimensions: the dimensions of whatever is there remain what they are. For example, energy remains what it is, and so that’s force over distance: 1 joule = 1 newton·meter (1 J = 1 N·m. Likewise, momentum remains what it is: force times time (or mass times velocity). Finally, the dimension of the quantum of action doesn’t disappear either: it remains the product of force, distance and time (N·m·s). So you should distinguish between the numerical value of our variables and their dimension. Always! That’s where physics is different from algebra: the equations actually mean something!]

Now, because we’re working in natural units, the numerical value of both and cwill be equal to 1. It’s obvious, then, that Einstein’s mass-energy equivalence relation reduces from E = mvc2 to E = mv. You can work out the rest yourself – noting that p = mv·v and mv = m0/√(1−v2). Done! For a more intuitive explanation, I refer you to the above-mentioned page.

So that’s for the wavefunction. Let’s now look at Schrödinger’s wave equation, i.e. that differential equation of which our wavefunction is a solution. In my introduction, I bluntly said there was something wrong with it: that 1/2 factor shouldn’t be there. Why not?

## What’s wrong with Schrödinger’s equation?

When deriving his famous equation, Schrödinger uses the mass concept as it appears in the classical kinetic energy formula: K.E. = m·v2/2, and that’s why – after all the complicated turns – that 1/2 factor is there. There are many reasons why that factor doesn’t make sense. Let me sum up a few.

[I] The most important reason is that de Broglie made it quite clear that the energy concept in his equations for the temporal and spatial frequency for the wavefunction – i.e. the ω = E/ħ and k = p/ħ relations – is the total energy, including rest energy (m0), kinetic energy (m·v2/2) and any potential energy (V). In fact, if we just multiply the two de Broglie (aka as matter-wave equations) and use the old-fashioned v = λ relation (so we write E as E = ω·ħ = (2π·f)·(h/2π) = f·h, and p as p = k·ħ = (2π/λ)·(h/2π) = h/λ and, therefore, we have = E/h and p = h/p), we find that the energy concept that’s implicit in the two matter-wave equations is equal to E = m∙v2, as shown below:

1. f·λ = (E/h)·(h/p) = E/p
2. v = λ ⇒ f·λ = v = E/p ⇔ E = v·p = v·(m·v) ⇒ E = m·v2

Huh? E = m∙v2? Yes. Not E = m∙c2 or m·v2/2 or whatever else you might be thinking of. In fact, this E = m∙v2 formula makes a lot of sense in light of the two following points.

Skeptical note: You may – and actually should – wonder whether we can use that v = λ relation for a wave like this, i.e. a wave with both a real (cos(-θ)) as well as an imaginary component (i·sin(-θ). It’s a deep question, and I’ll come back to it later. But… Yes. It’s the right question to ask. 😦

[II] Newton told us that force is mass time acceleration. Newton’s law is still valid in Einstein’s world. The only difference between Newton’s and Einstein’s world is that, since Einstein, we should treat the mass factor as a variable as well. We write: F = mv·a = mv·= [m0/√(1−v2)]·a. This formula gives us the definition of the newton as a force unit: 1 N = 1 kg·(m/s)/s = 1 kg·m/s2. [Note that the 1/√(1−v2) factor – i.e. the Lorentz factor (γ) – has no dimension, because is measured as a relative velocity here, i.e. as a fraction between 0 and 1.]

Now, you’ll agree the definition of energy as a force over some distance is valid in Einstein’s world as well. Hence, if 1 joule is 1 N·m, then 1 J is also equal to 1 (kg·m/s2)·m = 1 kg·(m2/s2), so this also reflects the E = m∙v2 concept. [I can hear you mutter: that kg factor refers to the rest mass, no? No. It doesn’t. The kg is just a measure of inertia: as a unit, it applies to both mas well as mv. Full stop.]

Very skeptical note: You will say this doesn’t prove anything – because this argument just shows the dimensional analysis for both equations (i.e. E = m∙v2 and E = m∙c2) is OK. Hmm… Yes. You’re right. 🙂 But the next point will surely convince you! 🙂

[III] The third argument is the most intricate and the most beautiful at the same time—not because it’s simple (like the arguments above) but because it gives us an interpretation of what’s going on here. It’s fairly easy to verify that Schrödinger’s equation, ∂ψ/∂t = i·(ħ/2m)·∇2ψ equation (including the 1/2 factor to which I object), is equivalent to the following set of two equations:

1. Re(∂ψ/∂t) = −(ħ/2m)·Im(∇2ψ)
2. Im(∂ψ/∂t) = (ħ/2m)·Re(∇2ψ)

[In case you don’t see it immediately, note that two complex numbers a + i·b and c + i·d are equal if, and only if, their real and imaginary parts are the same. However, here we have something like this: a + i·b = i·(c + i·d) = i·c + i2·d = − d + i·c (remember i= −1).]

Now, before we proceed (i.e. before I show you what’s wrong here with that 1/2 factor), let us look at the dimensions first. For that, we’d better analyze the complete Schrödinger equation so as to make sure we’re not doing anything stupid here by looking at one aspect of the equation only. The complete equation, in its original form, is:

Notice that, to simplify the analysis above, I had moved the and the ħ on the left-hand side to the right-hand side (note that 1/= −i, so −(ħ2/2m)/(i·ħ) = ħ/2m). Now, the ħfactor on the right-hand side is expressed in J2·s2. Now that doesn’t make much sense, but then that mass factor in the denominator makes everything come out alright. Indeed, we can use the mass-equivalence relation to express m in J/(m/s)2 units. So our ħ2/2m coefficient is expressed in (J2·s2)/[J/(m/s)2] = J·m2. Now we multiply that by that Laplacian operating on some scalar, which yields some quantity per square meter. So the whole right-hand side becomes some amount expressed in joule, i.e. the unit of energy! Interesting, isn’t it?

On the left-hand side, we have i and ħ. We shouldn’t worry about the imaginary unit because we can treat that as just another number, albeit a very special number (because its square is minus 1). However, in this equation, it’s like a mathematical constant and you can think of it as something like π or e. [Think of the magical formula: eiπ = i2 = −1.] In contrast, ħ is a physical constant, and so that constant comes with some dimension and, therefore, we cannot just do what we want. [I’ll show, later, that even moving it to the other side of the equation comes with interpretation problems, so be careful with physical constants, as they really mean something!] In this case, its dimension is the action dimension: J·s = N·m·s, so that’s force times distance times time. So we multiply that with a time derivative and we get joule once again (N·m·s/s = N·m = J), so that’s the unit of energy. So it works out: we have joule units both left and right in Schrödinger’s equation. Nice! Yes. But what does it mean? 🙂

Well… You know that we can – and should – think of Schrödinger’s equation as a diffusion equation – just like a heat diffusion equation, for example – but then one describing the diffusion of a probability amplitude. [In case you are not familiar with this interpretation, please do check my post on it, or my Deep Blue page.] But then we didn’t describe the mechanism in very much detail, so let me try to do that now and, in the process, finally explain the problem with the 1/2 factor.

## The missing energy

There are various ways to explain the problem. One of them involves calculating group and phase velocities of the elementary wavefunction satisfying Schrödinger’s equation but that’s a more complicated approach and I’ve done that elsewhere, so just click the reference if you prefer the more complicated stuff. I find it easier to just use those two equations above:

1. Re(∂ψ/∂t) = −(ħ/2m)·Im(∇2ψ)
2. Im(∂ψ/∂t) = (ħ/2m)·Re(∇2ψ)

The argument is the following: if our elementary wavefunction is equal to ei(kx − ωt) = cos(kx−ωt) + i∙sin(kx−ωt), then it’s easy to proof that this pair of conditions is fulfilled if, and only if, ω = k2·(ħ/2m). [Note that I am omitting the normalization coefficient in front of the wavefunction: you can put it back in if you want. The argument here is valid, with or without normalization coefficients.] Easy? Yes. Check it out. The time derivative on the left-hand side is equal to:

∂ψ/∂t = −iω·iei(kx − ωt) = ω·[cos(kx − ωt) + i·sin(kx − ωt)] = ω·cos(kx − ωt) + iω·sin(kx − ωt)

And the second-order derivative on the right-hand side is equal to:

2ψ = ∂2ψ/∂x= i·k2·ei(kx − ωt) = k2·cos(kx − ωt) + i·k2·sin(kx − ωt)

So the two equations above are equivalent to writing:

1. Re(∂ψB/∂t) =   −(ħ/2m)·Im(∇2ψB) ⇔ ω·cos(kx − ωt) = k2·(ħ/2m)·cos(kx − ωt)
2. Im(∂ψB/∂t) = (ħ/2m)·Re(∇2ψB) ⇔ ω·sin(kx − ωt) = k2·(ħ/2m)·sin(kx − ωt)

So both conditions are fulfilled if, and only if, ω = k2·(ħ/2m). You’ll say: so what? Well… We have a contradiction here—something that doesn’t make sense. Indeed, the second of the two de Broglie equations (always look at them as a pair) tells us that k = p/ħ, so we can re-write the ω = k2·(ħ/2m) condition as:

ω/k = vp = k2·(ħ/2m)/k = k·ħ/(2m) = (p/ħ)·(ħ/2m) = p/2m ⇔ p = 2m

You’ll say: so what? Well… Stop reading, I’d say. That p = 2m doesn’t make sense—at all! Nope! In fact, if you thought that the E = m·v2  is weird—which, I hope, is no longer the case by now—then… Well… This p = 2m equation is much weirder. In fact, it’s plain nonsense: this condition makes no sense whatsoever. The only way out is to remove the 1/2 factor, and to re-write the Schrödinger equation as I wrote it, i.e. with an ħ/m coefficient only, rather than an (1/2)·(ħ/m) coefficient.

Huh? Yes.

As mentioned above, I could do those group and phase velocity calculations to show you what rubbish that 1/2 factor leads to – and I’ll do that eventually – but let me first find yet another way to present the same paradox. Let’s simplify our life by choosing our units such that = ħ = 1, so we’re using so-called natural units rather than our SI units. [Again, note that switching to natural units doesn’t do anything to the physical dimensions: a force remains a force, a distance remains a distance, and so on.] Our mass-energy equivalence then becomes: E = m·c= m·1= m. [Again, note that switching to natural units doesn’t do anything to the physical dimensions: a force remains a force, a distance remains a distance, and so on. So we’d still measure energy and mass in different but equivalent units. Hence, the equality sign should not make you think mass and energy are actually the same: energy is energy (i.e. force times distance), while mass is mass (i.e. a measure of inertia). I am saying this because it’s important, and because it took me a while to make these rather subtle distinctions.]

Let’s now go one step further and imagine a hypothetical particle with zero rest mass, so m0 = 0. Hence, all its energy is kinetic and so we write: K.E. = mv·v/2. Now, because this particle has zero rest mass, the slightest acceleration will make it travel at the speed of light. In fact, we would expect it to travel at the speed, so mv = mc and, according to the mass-energy equivalence relation, its total energy is, effectively, E = mv = mc. However, we just said its total energy is kinetic energy only. Hence, its total energy must be equal to E = K.E. = mc·c/2 = mc/2. So we’ve got only half the energy we need. Where’s the other half? Where’s the missing energy? Quid est veritas? Is its energy E = mc or E = mc/2?

It’s just a paradox, of course, but one we have to solve. Of course, we may just say we trust Einstein’s E = m·c2 formula more than the kinetic energy formula, but that answer is not very scientific. 🙂 We’ve got a problem here and, in order to solve it, I’ve come to the following conclusion: just because of its sheer existence, our zero-mass particle must have some hidden energy, and that hidden energy is also equal to E = m·c2/2. Hence, the kinetic and the hidden energy add up to E = m·c2 and all is alright.

Huh? Hidden energy? I must be joking, right?

Well… No. Let me explain. Oh. And just in case you wonder why I bother to try to imagine zero-mass particles. Let me tell you: it’s the first step towards finding a wavefunction for a photon and, secondly, you’ll see it just amounts to modeling the propagation mechanism of energy itself. 🙂

## The hidden energy as imaginary energy

I am tempted to refer to the missing energy as imaginary energy, because it’s linked to the imaginary part of the wavefunction. However, it’s anything but imaginary: it’s as real as the imaginary part of the wavefunction. [I know that sounds a bit nonsensical, but… Well… Think about it. And read on!]

Back to that factor 1/2. As mentioned above, it also pops up when calculating the group and the phase velocity of the wavefunction. In fact, let me show you that calculation now. [Sorry. Just hang in there.] It goes like this.

The de Broglie relations tell us that the k and the ω in the ei(kx − ωt) = cos(kx−ωt) + i∙sin(kx−ωt) wavefunction (i.e. the spatial and temporal frequency respectively) are equal to k = p/ħ, and ω = E/ħ. Let’s now think of that zero-mass particle once more, so we assume all of its energy is kinetic: no rest energy, no potential! So… If we now use the kinetic energy formula E = m·v2/2 – which we can also write as E = m·v·v/2 = p·v/2 = p·p/2m = p2/2m, with v = p/m the classical velocity of the elementary particle that Louis de Broglie was thinking of – then we can calculate the group velocity of our ei(kx − ωt) = cos(kx−ωt) + i∙sin(kx−ωt) wavefunction as:

vg = ∂ω/∂k = ∂[E/ħ]/∂[p/ħ] = ∂E/∂p = ∂[p2/2m]/∂p = 2p/2m = p/m = v

[Don’t tell me I can’t treat m as a constant when calculating ∂ω/∂k: I can. Think about it.]

Fine. Now the phase velocity. For the phase velocity of our ei(kx − ωt) wavefunction, we find:

vp = ω/k = (E/ħ)/(p/ħ) = E/p = (p2/2m)/p = p/2m = v/2

So that’s only half of v: it’s the 1/2 factor once more! Strange, isn’t it? Why would we get a different value for the phase velocity here? It’s not like we have two different frequencies here, do we? Well… No. You may also note that the phase velocity turns out to be smaller than the group velocity (as mentioned, it’s only half of the group velocity), which is quite exceptional as well! So… Well… What’s the matter here? We’ve got a problem!

What’s going on here? We have only one wave here—one frequency and, hence, only one k and ω. However, on the other hand, it’s also true that the ei(kx − ωt) wavefunction gives us two functions for the price of one—one real and one imaginary: ei(kx − ωt) = cos(kx−ωt) + i∙sin(kx−ωt). So the question here is: are we adding waves, or are we not? It’s a deep question. If we’re adding waves, we may get different group and phase velocities, but if we’re not, then… Well… Then the group and phase velocity of our wave should be the same, right? The answer is: we are and we aren’t. It all depends on what you mean by ‘adding’ waves. I know you don’t like that answer, but that’s the way it is, really. 🙂

Let me make a small digression here that will make you feel even more confused. You know – or you should know – that the sine and the cosine function are the same except for a phase difference of 90 degrees: sinθ = cos(θ + π/2). Now, at the same time, multiplying something with amounts to a rotation by 90 degrees, as shown below.

Hence, in order to sort of visualize what our ei(kx − ωt) function really looks like, we may want to super-impose the two graphs and think of something like this:

You’ll have to admit that, when you see this, our formulas for the group or phase velocity, or our v = λ relation, do no longer make much sense, do they? 🙂

Having said that, that 1/2 factor is and remains puzzling, and there must be some logical reason for it. For example, it also pops up in the Uncertainty Relations:

Δx·Δp ≥ ħ/2 and ΔE·Δt ≥ ħ/2

So we have ħ/2 in both, not ħ. Why do we need to divide the quantum of action here? How do we solve all these paradoxes? It’s easy to see how: the apparent contradiction (i.e. the different group and phase velocity) gets solved if we’d use the E = m∙v2 formula rather than the kinetic energy E = m∙v2/2. But then… What energy formula is the correct one: E = m∙v2 or m∙c2? Einstein’s formula is always right, isn’t it? It must be, so let me postpone the discussion a bit by looking at a limit situation. If v = c, then we don’t need to make a choice, obviously. 🙂 So let’s look at that limit situation first. So we’re discussing our zero-mass particle once again, assuming it travels at the speed of light. What do we get?

Well… Measuring time and distance in natural units, so c = 1, we have:

E = m∙c2 = m and p = m∙c = m, so we get: E = m = p

Waw ! E = m = p ! What a weird combination, isn’t it? Well… Yes. But it’s fully OK. [You tell me why it wouldn’t be OK. It’s true we’re glossing over the dimensions here, but natural units are natural units and, hence, the numerical value of c and c2 is 1. Just figure it out for yourself.] The point to note is that the E = m = p equality yields extremely simple but also very sensible results. For the group velocity of our ei(kx − ωt) wavefunction, we get:

vg = ∂ω/∂k = ∂[E/ħ]/∂[p/ħ] = ∂E/∂p = ∂p/∂p = 1

So that’s the velocity of our zero-mass particle (remember: the 1 stands for c here, i.e. the speed of light) expressed in natural units once more—just like what we found before. For the phase velocity, we get:

vp = ω/k = (E/ħ)/(p/ħ) = E/p = p/p = 1

Same result! No factor 1/2 here! Isn’t that great? My ‘hidden energy theory’ makes a lot of sense.

However, if there’s hidden energy, we still need to show where it’s hidden. 🙂 Now that question is linked to the propagation mechanism that’s described by those two equations, which now – leaving the 1/2 factor out, simplify to:

1. Re(∂ψ/∂t) = −(ħ/m)·Im(∇2ψ)
2. Im(∂ψ/∂t) = (ħ/m)·Re(∇2ψ)

Propagation mechanism? Yes. That’s what we’re talking about here: the propagation mechanism of energy. Huh? Yes. Let me explain in another separate section, so as to improve readability. Before I do, however, let me add another note—for the skeptics among you. 🙂

Indeed, the skeptics among you may wonder whether our zero-mass particle wavefunction makes any sense at all, and they should do so for the following reason: if x = 0 at t = 0, and it’s traveling at the speed of light, then x(t) = t. Always. So if E = m = p, the argument of our wavefunction becomes E·t – p·x = E·t – E·t = 0! So what’s that? The proper time of our zero-mass particle is zero—always and everywhere!?

Well… Yes. That’s why our zero-mass particle – as a point-like object – does not really exist. What we’re talking about is energy itself, and its propagation mechanism. 🙂

While I am sure that, by now, you’re very tired of my rambling, I beg you to read on. Frankly, if you got as far as you have, then you should really be able to work yourself through the rest of this post. 🙂 And I am sure that – if anything – you’ll find it stimulating! 🙂

## The imaginary energy space

Look at the propagation mechanism for the electromagnetic wave in free space, which (for = 1) is represented by the following two equations:

1. B/∂t = –∇×E
2. E/∂t = ∇×B

[In case you wonder, these are Maxwell’s equations for free space, so we have no stationary nor moving charges around.] See how similar this is to the two equations above? In fact, in my Deep Blue page, I use these two equations to derive the quantum-mechanical wavefunction for the photon (which is not the same as that hypothetical zero-mass particle I introduced above), but I won’t bother you with that here. Just note the so-called curl operator in the two equations above (∇×) can be related to the Laplacian we’ve used so far (∇2). It’s not the same thing, though: for starters, the curl operator operates on a vector quantity, while the Laplacian operates on a scalar (including complex scalars). But don’t get distracted now. Let’s look at the revised Schrödinger’s equation, i.e. the one without the 1/2 factor:

∂ψ/∂t = i·(ħ/m)·∇2ψ

On the left-hand side, we have a time derivative, so that’s a flow per second. On the right-hand side we have the Laplacian and the i·ħ/m factor. Now, written like this, Schrödinger’s equation really looks exactly the same as the general diffusion equation, which is written as: ∂φ/∂t = D·∇2φ, except for the imaginary unit, which makes it clear we’re getting two equations for the price of one here, rather than one only! 🙂 The point is: we may now look at that ħ/m factor as a diffusion constant, because it does exactly the same thing as the diffusion constant D in the diffusion equation ∂φ/∂t = D·∇2φ, i.e:

1. As a constant of proportionality, it quantifies the relationship between both derivatives.
2. As a physical constant, it ensures the dimensions on both sides of the equation are compatible.

So the diffusion constant for  Schrödinger’s equation is ħ/m. What is its dimension? That’s easy: (N·m·s)/(N·s2/m) = m2/s. [Remember: 1 N = 1 kg·m/s2.] But then we multiply it with the Laplacian, so that’s something expressed per square meter, so we get something per second on both sides.

Of course, you wonder: what per second? Not sure. That’s hard to say. Let’s continue with our analogy with the heat diffusion equation so as to try to get a better understanding of what’s being written here. Let me give you that heat diffusion equation here. Assuming the heat per unit volume (q) is proportional to the temperature (T) – which is the case when expressing T in degrees Kelvin (K), so we can write q as q = k·T  – we can write it as:

So that’s structurally similar to Schrödinger’s equation, and to the two equivalent equations we jotted down above. So we’ve got T (temperature) in the role of ψ here—or, to be precise, in the role of ψ ‘s real and imaginary part respectively. So what’s temperature? From the kinetic theory of gases, we know that temperature is not just a scalar: temperature measures the mean (kinetic) energy of the molecules in the gas. That’s why we can confidently state that the heat diffusion equation models an energy flow, both in space as well as in time.

Let me make the point by doing the dimensional analysis for that heat diffusion equation. The time derivative on the left-hand side (∂T/∂t) is expressed in K/s (Kelvin per second). Weird, isn’t it? What’s a Kelvin per second? Well… Think of a Kelvin as some very small amount of energy in some equally small amount of space—think of the space that one molecule needs, and its (mean) energy—and then it all makes sense, doesn’t it?

However, in case you find that a bit difficult, just work out the dimensions of all the other constants and variables. The constant in front (k) makes sense of it. That coefficient (k) is the (volume) heat capacity of the substance, which is expressed in J/(m3·K). So the dimension of the whole thing on the left-hand side (k·∂T/∂t) is J/(m3·s), so that’s energy (J) per cubic meter (m3) and per second (s). Nice, isn’t it? What about the right-hand side? On the right-hand side we have the Laplacian operator  – i.e. ∇= ·, with ∇ = (∂/∂x,  ∂/∂y,  ∂/∂z) – operating on T. The Laplacian operator, when operating on a scalar quantity, gives us a flux density, i.e. something expressed per square meter (1/m2). In this case, it’s operating on T, so the dimension of ∇2T is K/m2. Again, that doesn’t tell us very much (what’s the meaning of a Kelvin per square meter?) but we multiply it by the thermal conductivity (κ), whose dimension is W/(m·K) = J/(m·s·K). Hence, the dimension of the product is  the same as the left-hand side: J/(m3·s). So that’s OK again, as energy (J) per cubic meter (m3) and per second (s) is definitely something we can associate with an energy flow.

In fact, we can play with this. We can bring k from the left- to the right-hand side of the equation, for example. The dimension of κ/k is m2/s (check it!), and multiplying that by K/m(i.e. the dimension of ∇2T) gives us some quantity expressed in Kelvin per second, and so that’s the same dimension as that of ∂T/∂t. Done!

In fact, we’ve got two different ways of writing Schrödinger’s diffusion equation. We can write it as ∂ψ/∂t = i·(ħ/m)·∇2ψ or, else, we can write it as ħ·∂ψ/∂t = i·(ħ2/m)·∇2ψ. Does it matter? I don’t think it does. The dimensions come out OK in both cases. However, interestingly, if we do a dimensional analysis of the ħ·∂ψ/∂t = i·(ħ2/m)·∇2ψ equation, we get joule on both sides. Interesting, isn’t it? The key question, of course, is: what is it that is flowing here?

I don’t have a very convincing answer to that, but the answer I have is interesting—I think. 🙂 Think of the following: we can multiply Schrödinger’s equation with whatever we want, and then we get all kinds of flows. For example, if we multiply both sides with 1/(m2·s) or 1/(m3·s), we get a equation expressing the energy conservation law, indeed! [And you may want to think about the minus sign of the  right-hand side of Schrödinger’s equation now, because it makes much more sense now!]

We could also multiply both sides with s, so then we get J·s on both sides, i.e. the dimension of physical action (J·s = N·m·s). So then the equation expresses the conservation of actionHuh? Yes. Let me re-phrase that: then it expresses the conservation of angular momentum—as you’ll surely remember that the dimension of action and angular momentum are the same. 🙂

And then we can divide both sides by m, so then we get N·s on both sides, so that’s momentum. So then Schrödinger’s equation embodies the momentum conservation law.

Isn’t it just wonderfulSchrödinger’s equation packs all of the conservation laws! The only catch is that it flows back and forth from the real to the imaginary space, using that propagation mechanism as described in those two equations.

Now that is really interesting, because it does provide an explanation – as fuzzy as it may seem – for all those weird concepts one encounters when studying physics, such as the tunneling effect, which amounts to energy flowing from the imaginary space to the real space and, then, inevitably, flowing back. It also allows for borrowing time from the imaginary space. Hmm… Interesting! [I know I still need to make these points much more formally, but… Well… You kinda get what I mean, don’t you?]

To conclude, let me re-baptize my real and imaginary ‘space’ by referring to them to what they really are: a real and imaginary energy space respectively. Although… Now that I think of it: it could also be real and imaginary momentum space, or a real and imaginary action space. Hmm… The latter term may be the best. 🙂

Isn’t this all great? I mean… I could go on and on—but I’ll stop here, so you can freewheel around yourself. For  example, you may wonder how similar that energy propagation mechanism actually is as compared to the propagation mechanism of the electromagnetic wave? The answer is: very similar. You can check how similar in one of my posts on the photon wavefunction or, if you’d want a more general argument, check my Deep Blue page. Have fun exploring! 🙂

So… Well… That’s it, folks. I hope you enjoyed this post—if only because I really enjoyed writing it. 🙂

[…]

OK. You’re right. I still haven’t answered the fundamental question.

## So what about  the 1/2 factor?

What about that 1/2 factor? Did Schrödinger miss it? Well… Think about it for yourself. First, I’d encourage you to further explore that weird graph with the real and imaginary part of the wavefunction. I copied it below, but with an added 45º line—yes, the green diagonal. To make it somewhat more real, imagine you’re the zero-mass point-like particle moving along that line, and we observe you from our inertial frame of reference, using equivalent time and distance units.

So we’ve got that cosine (cosθ) varying as you travel, and we’ve also got the i·sinθ part of the wavefunction going while you’re zipping through spacetime. Now, THINK of it: the phase velocity of the cosine bit (i.e. the red graph) contributes as much to your lightning speed as the i·sinθ bit, doesn’t it? Should we apply Pythagoras’ basic r2 = x2 + yTheorem here? Yes: the velocity vector along the green diagonal is going to be the sum of the velocity vectors along the horizontal and vertical axes. So… That’s great.

Yes. It is. However, we still have a problem here: it’s the velocity vectors that add up—not their magnitudes. Indeed, if we denote the velocity vector along the green diagonal as u, then we can calculate its magnitude as:

u = √u2 = √[(v/2)2 + (v/2)2] = √[2·(v2/4) = √[v2/2] = v/√2 ≈ 0.7·v

So, as mentioned, we’re adding the vectors, but not their magnitudes. We’re somewhat better off than we were in terms of showing that the phase velocity of those sine and cosine velocities add up—somehow, that is—but… Well… We’re not quite there.

Fortunately, Einstein saves us once again. Remember we’re actually transforming our reference frame when working with the wavefunction? Well… Look at the diagram below (for which I  thank the author)

In fact, let me insert an animated illustration, which shows what happens when the velocity goes up and down from (close to) −c to +c and back again.  It’s beautiful, and I must credit the author here too. It sort of speaks for itself, but please do click the link as the accompanying text is quite illuminating. 🙂

The point is: for our zero-mass particle, the x’ and t’ axis will rotate into the diagonal itself which, as I mentioned a couple of times already, represents the speed of light and, therefore, our zero-mass particle traveling at c. It’s obvious that we’re now adding two vectors that point in the same direction and, hence, their magnitudes just add without any square root factor. So, instead of u = √[(v/2)2 + (v/2)2], we just have v/2 + v/2 = v! Done! We solved the phase velocity paradox! 🙂

So… I still haven’t answered that question. Should that 1/2 factor in Schrödinger’s equation be there or not? The answer is, obviously: yes. It should be there. And as for Schrödinger using the mass concept as it appears in the classical kinetic energy formula: K.E. = m·v2/2… Well… What other mass concept would he use? I probably got a bit confused with Feynman’s exposé – especially this notion of ‘choosing the zero point for the energy’ – but then I should probably just re-visit the thing and adjust the language here and there. But the formula is correct.

Thinking it all through, the ħ/2m constant in Schrödinger’s equation should be thought of as the reciprocal of m/(ħ/2). So what we’re doing basically is measuring the mass of our object in units of ħ/2, rather than units of ħ. That makes perfect sense, if only because it’s ħ/2, rather than ħthe factor that appears in the Uncertainty Relations Δx·Δp ≥ ħ/2 and ΔE·Δt ≥ ħ/2. In fact, in my post on the wavefunction of the zero-mass particle, I noted its elementary wavefunction should use the m = E = p = ħ/2 values, so it becomes ψ(x, t) = a·ei∙[(ħ/2)∙t − (ħ/2)∙x]/ħ = a·ei∙[t − x]/2.

Isn’t that just nice? 🙂 I need to stop here, however, because it looks like this post is becoming a book. Oh—and note that nothing what I wrote above discredits my ‘hidden energy’ theory. On the contrary, it confirms it. In fact, the nice thing about those illustrations above is that it associates the imaginary component of our wavefunction with travel in time, while the real component is associated with travel in space. That makes our theory quite complete: the ‘hidden’ energy is the energy that moves time forward. The only thing I need to do is to connect it to that idea of action expressing itself in time or in space, cf. what I wrote on my Deep Blue page: we can look at the dimension of Planck’s constant, or at the concept of action in general, in two very different ways—from two different perspectives, so to speak:

1. [Planck’s constant] = [action] = N∙m∙s = (N∙m)∙s = [energy]∙[time]
2. [Planck’s constant] = [action] = N∙m∙s = (N∙s)∙m = [momentum]∙[distance]

Hmm… I need to combine that with the idea of the quantum vacuum, i.e. the mathematical space that’s associated with time and distance becoming countable variables…. In any case. Next time. 🙂

Before I sign off, however, let’s quickly check if our a·ei∙[t − x]/2 wavefunction solves the Schrödinger equation:

• ∂ψ/∂t = −a·ei∙[t − x]/2·(i/2)
• 2ψ = ∂2[a·ei∙[t − x]/2]/∂x=  ∂[a·ei∙[t − x]/2·(i/2)]/∂x = −a·ei∙[t − x]/2·(1/4)

So the ∂ψ/∂t = i·(ħ/2m)·∇2ψ equation becomes:

a·ei∙[t − x]/2·(i/2) = −i·(ħ/[2·(ħ/2)])·a·ei∙[t − x]/2·(1/4)

⇔ 1/2 = 1/4 !?

The damn 1/2 factor. Schrödinger wants it in his wave equation, but not in the wavefunction—apparently! So what if we take the m = E = p = ħ solution? We get:

• ∂ψ/∂t = −a·i·ei∙[t − x]
• 2ψ = ∂2[a·ei∙[t − x]]/∂x=  ∂[a·i·ei∙[t − x]]/∂x = −a·ei∙[t − x]

So the ∂ψ/∂t = i·(ħ/2m)·∇2ψ equation now becomes:

a·i·ei∙[t − x] = −i·(ħ/[2·ħ])·a·ei∙[t − x]

⇔ 1 = 1/2 !?

We’re still in trouble! So… Was Schrödinger wrong after all? There’s no difficulty whatsoever with the ∂ψ/∂t = i·(ħ/m)·∇2ψ equation:

• a·ei∙[t − x]/2·(i/2) = −i·[ħ/(ħ/2)]·a·ei∙[t − x]/2·(1/4) ⇔ 1 = 1
• a·i·ei∙[t − x] = −i·(ħ/ħ)·a·ei∙[t − x] ⇔ 1 = 1

What these equations might tell us is that we should measure mass, energy and momentum in terms of ħ (and not in terms of ħ/2) but that the fundamental uncertainty is ± ħ/2. That solves it all. So the magnitude of the uncertainty is ħ but it separates not 0 and ± 1, but −ħ/2 and −ħ/2. Or, more generally, the following series:

…, −7ħ/2, −5ħ/2, −3ħ/2, −ħ/2, +ħ/2, +3ħ/2,+5ħ/2, +7ħ/2,…

Why are we not surprised? The series represent the energy values that a spin one-half particle can possibly have, and ordinary matter – i.e. all fermions – is composed of spin one-half particles.

To  conclude this post, let’s see if we can get any indication on the energy concepts that Schrödinger’s revised wave equation implies. We’ll do so by just calculating the derivatives in the ∂ψ/∂t = i·(ħ/m)·∇2ψ equation (i.e. the equation without the 1/2 factor). Let’s also not assume we’re measuring stuff in natural units, so our wavefunction is just what it is: a·ei·[E·t − p∙x]/ħ. The derivatives now become:

• ∂ψ/∂t = −a·i·(E/ħ)·ei∙[E·t − p∙x]/ħ
• 2ψ = ∂2[a·ei∙[E·t − p∙x]/ħ]/∂x=  ∂[a·i·(p/ħ)·ei∙[E·t − p∙x]/ħ]/∂x = −a·(p22ei∙[E·t − p∙x]/ħ

So the ∂ψ/∂t = i·(ħ/m)·∇2ψ = i·(1/m)·∇2ψ equation now becomes:

a·i·(E/ħ)·ei∙[E·t − p∙x]/ħ = −i·(ħ/m)·a·(p22ei∙[E·t − p∙x]/ħ  ⇔ E = p2/m = m·v2

It all works like a charm. Note that we do not assume stuff like E = m = p here. It’s all quite general. Also note that the E = p2/m closely resembles the kinetic energy formula one often sees: K.E. = m·v2/2 = m·m·v2/(2m) = p2/(2m). We just don’t have the 1/2 factor in our E = p2/m formula, which is great—because we don’t want it!  Of course, if you’d add the 1/2 factor in Schrödinger’s equation again, you’d get it back in your energy formula, which would just be that old kinetic energy formula which gave us all these contradictions and ambiguities. 😦

Finally, and just to make sure: let me add that, when we wrote that E = m = p – like we did above – we mean their numerical values are the same. Their dimensions remain what they are, of course. Just to make sure you get that subtle point, we’ll do a quick dimensional analysis of that E = p2/m formula:

[E] = [p2/m] ⇔ N·m = N2·s2/kg = N2·s2/[N·m/s2] = N·m = joule (J)

So… Well… It’s all perfect. 🙂

Post scriptum: I revised my Deep Blue page after writing this post, and I think that a number of the ideas that I express above are presented more consistently and coherently there. In any case, the missing energy theory makes sense. Think of it: any oscillator involves both kinetic as well as potential energy, and they both add up to twice the average kinetic (or potential) energy. So why not here? When everything is said and done, our elementary wavefunction does describe an oscillator. 🙂

# Schrödinger’s equation in action

This post is about something I promised to write about aeons ago: how do we get those electron orbitals out of Schrödinger’s equation? So let me write it now – for the simplest of atoms: hydrogen. I’ll largely follow Richard Feynman’s exposé on it: this text just intends to walk you through it and provide some comments here and there.

Let me first remind you of what that famous Schrödinger’s equation actually represents. In its simplest form – i.e. not including any potential, so then it’s an equation that’s valid for free space only—no force fields!—it reduces to:

i·ħ∙∂ψ/∂t = –(1/2)∙(ħ2/meff)∙∇2ψ

Note the enigmatic concept of the efficient mass in it (meff), as well as the rather awkward 1/2 factor, which we may get rid of by re-defining it. We then write: meffNEW = 2∙meffOLD, and Schrödinger’s equation then simplifies to:

• ∂ψ/∂t + i∙(V/ħ)·ψ = i(ħ/meff)·∇2ψ
• In free space (no potential): ∂ψ/∂t = i∙(ħ/meff)·∇2ψ

In case you wonder where the minus sign went, I just brought the imaginary unit to the other side. Remember 1/= −i. 🙂

Now, in my post on quantum-mechanical operators, I drew your attention to the fact that this equation is structurally similar to the heat diffusion equation – or to any diffusion equation, really. Indeed, assuming the heat per unit volume (q) is proportional to the temperature (T) – which is the case when expressing T in degrees Kelvin (K), so we can write q as q = k·T  – we can write the heat diffusion equation as:

Moreover, I noted the similarity is not only structural. There is more to it: both equations model energy flows. How exactly is something I wrote about in my e-publication on this, so let me refer you to that. Let’s jot down the complete equation once more:

∂ψ/∂t + i∙(V/ħ)·ψ = i(ħ/meff)·∇2ψ

In fact, it is rather surprising that Feynman drops the eff subscript almost immediately, so he just writes:

Let me first remind you that ψ is a function of position in space and time, so we write: ψ = ψ(x, y, z, t) = ψ(r, t), with (x, y, z) = r. And m, on the other side of the equation, is what it always was: the effective electron mass. Now, we talked about the subtleties involved before, so let’s not bother about the definition of the effective electron mass, or wonder where that factor 1/2 comes from here.

What about V? V is the potential energy of the electron: it depends on the distance (r) from the proton. We write: V = −e2/│r│ = −e2/r. Why the minus sign? Because we say the potential energy is zero at  large distances (see my post on potential energy). Back to Schrödinger’s equation.

On the left-hand side, we have ħ, and its dimension is J·s (or N·m·s, if you want). So we multiply that with a time derivative and we get J, the unit of energy. On the right-hand side, we have Planck’s constant squared, the mass factor in the denominator, and the Laplacian operator – i.e. ∇= ·, with ∇ = (∂/∂x,  ∂/∂y,  ∂/∂z) – operating on the wavefunction.

Let’s start with the latter. The Laplacian works just the same as for our heat diffusion equation: it gives us a flux density, i.e. something expressed per square meter (1/m2). The ħfactor gives us J2·s2. The mass factor makes everything come out alright, if we use the mass-equivalence relation, which says it’s OK to express the mass in J/(m/s)2. [The mass of an electron is usually expressed as being equal to 0.5109989461(31) MeV/c2. That unit uses the E = m·cmass-equivalence formula. As for the eV, you know we can convert that into joule, which is a rather large unit—which is why we use the electronvolt as a measure of energy.] To make a long story short, we’re OK: (J2·s2)·[(m/s)2/J]·(1/m2) = J! Perfect. [As for the Vψ term, that’s obviously expressed in joule too.]

In short, Schrödinger’s equation expresses the energy conservation law too, and we may express it per square meter or per second or per cubic meter as well, if we’d wish: we can just multiply both sides by 1/m2 or 1/s or 1/mor by whatever dimension you want. Again, if you want more detail on the Schrödinger equation as an energy propagation mechanism, read the mentioned e-publication. So let’s get back to our equation, which, taking into account our formula for V, now looks like this:

Feynman then injects one of these enigmatic phrases—enigmatic for novices like us, at least!

“We want to look for definite energy states, so we try to find solutions which have the form: ψ (r, t) =  e−(i/ħ)·E·t·ψ(r).”

At first, you may think he’s just trying to get rid of the relativistic correction in the argument of the wavefunction. Indeed, as I explain in that little booklet of mine, the –(p/ħ)·x term in the argument of the elementary wavefunction ei·θ =  ei·[(E/ħ)·t – (p/ħ)·x] is there because the young Comte Louis de Broglie, back in 1924, when he wrote his groundbreaking PhD thesis, suggested the θ = ω∙t – kx = (E∙t – px)/ħ formula for the argument of the wavefunction, as he knew that relativity theory had already established the invariance of the four-vector (dot) product pμxμ = E∙t – px = pμ‘xμ‘ = E’∙t’ – p’x’. [Note that Planck’s constant, as a physical constant, should obviously not depend on the reference frame either. Hence, if the E∙t – px product is invariant, so is (E∙t – px)/ħ.] So the θ = E∙t – px and the θ = E0∙t’ = E’·t’ are fully equivalent. Using lingo, we can say that the argument of the wavefunction is a Lorentz scalar and, therefore, invariant under a Lorentz boost. Sounds much better, doesn’t it? 🙂

But… Well. That’s not why Feynman says what he says. He just makes abstraction of uncertainty here, as he looks for states with a definite energy state, indeed. Nothing more, nothing less. Indeed, you should just note that we can re-write the elementary a·ei[(E/ħ)·t – (p/ħ)·x] function as e−(i/ħ)·E·t·ei·(p/ħ)·x]. So that’s what Feynman does here: he just eases the search for functional forms that satisfy Schrödinger’s equation. You should note the following:

1. Writing the coefficient in front of the complex exponential as ψ(r) = ei·(p/ħ)·x] does the trick we want it to do: we do not want that coefficient to depend on time: it should only depend on the size of our ‘box’ in space, as I explained in one of my posts.
2. Having said that, you should also note that the ψ in the ψ(r, t) function and the ψ in the ψ(r) denote two different beasts: one is a function of two variables (r and t), while the other makes abstraction of the time factor and, hence, becomes a function of one variable only (r). I would have used another symbol for the ψ(r) function, but then the Master probably just wants to test your understanding. 🙂

In any case, the differential equation we need to solve now becomes:

Huh? How does that work? Well… Just take the time derivative of e−(i/ħ)·E·t·ψ(r), multiply with the i·ħ in front of that term in Schrödinger’s original equation  and re-arrange the terms. [Just do it: ∂[e−(i/ħ)·E·t·ψ(r)]/∂t = −(i/ħ)·E·e−(i/ħ)·E·t·ψ(r). Now multiply that with i·ħ: the ħ factor cancels and the minus disappears because i= −1.]

So now we need to solve that differential equation, i.e. we need to find functional forms for ψ – and please do note we’re talking ψ(r) here – not ψ(r, t)! – that satisfy the above equation. Interesting question: is our equation still Schrödinger’s equation? Well… It is and it isn’t. Any linear combination of the definite energy solutions we find will also solve Schrödinger’s equation, but so we limited the solution set here to those definite energy solutions only. Hence, it’s not quite the same equation. We removed the time dependency here – and in a rather interesting way, I’d say.

The next thing to do is to switch from Cartesian to polar coordinates. Why? Well… When you have a central-force problem – like this one (because of the potential) – it’s easier to solve them using polar coordinates. In fact, because we’ve got three dimensions here, we’re actually talking a spherical coordinate system. The illustration and formulas below show how spherical and Cartesian coordinates are related:

x = r·sinθ·cosφ; y = r·sinθ·sinφ; zr·cosθ

As you know, θ (theta) is referred to as the polar angle, while φ (phi) is the azimuthal angle, and the coordinate transformation formulas can be easily derived. The rather simple differential equation above now becomes the following monster:

Huh? Yes, I am very sorry. That’s how it is. Feynman does this to help us. If you think you can get to the solutions by directly solving the equation in Cartesian coordinates, please do let me know. 🙂 To tame the beast, we might imagine to first look for solutions that are spherically symmetric, i.e. solutions that do not depend on θ and φ. That means we could rotate the reference frame and none of the amplitudes would change. That means the ∂ψ/∂θ and ∂ψ/∂φ (partial) derivatives in our formula are equal to zero. These spherically symmetric states, or s-states as they are referred to, are states with zero (orbital) angular momentum, but you may want to think about that statement before accepting it. 🙂 [It’s not  that there’s no angular momentum (on the contrary: there’s lots of it), but the total angular momentum should obviously be zero, and so that’s what meant when these states are denoted as = 0 states.] So now we have to solve:

Now that looks somewhat less monstrous, but Feynman still fills two rather dense pages to show how this differential equation can be solved. It’s not only tedious but also complicated, so please check it yourself by clicking on the link. One of the steps is a switch in variables, or a re-scaling, I should say. Both E and r are now measured as follows:

The complicated-looking factors are just the Bohr radius (r= ħ2/(m·e2) ≈ 0.528 Å) and the Rydberg energy (E= m·e4/2·ħ2 ≈ 13.6 eV). We calculated those long time ago using a rather heuristic model to describe an atom. In case you’d want to check the dimensions, note eis a rather special animal. It’s got nothing to do with Euler’s number. Instead, eis equal to ke·qe2, and the ke here is Coulomb’s constant: ke = 1/(4πε0). This allows to re-write the force between two electrons as a function of the distance: F = e2/r2This, in turn, explains the rather weird dimension of e2: [e2] = N·e= J·m. But I am digressing too much. The bottom line is: the various energy levels that fit the equation, i.e. the allowable energies, are fractions of the Rydberg energy, i.e. E=m·e4/2·ħ2. To be precise, the formula for the nth energy level is:

E= − ER/n2.

The interesting thing is that the spherically symmetric solutions yield real-valued ψ(r) functions. The solutions for n = 1, 2, and 3 respectively, and their graph is given below.

As Feynman writes, all of the wave functions approach zero rapidly for large r (also, confusingly, denoted as ρ) after oscillating a few times, with the number of ‘bumps’ equal to n. Of course, you should note that you should put the time factor back in in order to correctly interpret these functions. Indeed, remember how we separated them when we wrote:

ψ(r, t) =  ei·(E/ħ)·t·ψ(r)

We might say the ψ(r) function is sort of an envelope function for the whole wavefunction, but it’s not quite as straightforward as that. However, I am sure you’ll figure it out.

### States with an angular dependence

So far, so good. But what if those partial derivatives are not zero? Now the calculations become really complicated. Among other things, we need these transformation matrices for rotations, which we introduced a very long time ago. As mentioned above, I don’t have the intention to copy Feynman here, who needs another two or three dense pages to work out the logic. Let me just state the grand result:

• We’ve got a whole range of definite energy states, which correspond to orbitals that form an orthonormal basis for the actual wavefunction of the electron.
• The orbitals are characterized by three quantum numbers, denoted as ln and m respectively:
• The is the quantum number of (total) angular momentum, and it’s equal to 0, 1, 2, 3, etcetera. [Of course, as usual, we’re measuring in units of ħ.] The l = 0 states are referred to as s-states, the = 1 states are referred to as p-states, and the = 2 states are d-states. They are followed by f, g, h, etcetera—for no particular good reason. [As Feynman notes: “The letters don’t mean anything now. They did once—they meant “sharp” lines, “principal” lines, “diffuse” lines and “fundamental” lines of the optical spectra of atoms. But those were in the days when people did not know where the lines came from. After f there were no special names, so we now just continue with g, h, and so on.]
• The is referred to as the ‘magnetic’ quantum number, and it ranges from −l to +l.
• The n is the ‘principle’ quantum number, and it goes from + 1 to infinity (∞).

How do these things actually look like? Let me insert two illustrations here: one from Feynman, and the other from Wikipedia.

The number in front just tracks the number of s-, p-, d-, etc. orbital. The shaded region shows where the amplitudes are large, and the plus and minus signs show the relative sign of the amplitude. [See my remark above on the fact that the ψ factor is real-valued, even if the wavefunction as a whole is complex-valued.] The Wikipedia image shows the same density plots but, as it was made some 50 years later, with some more color. 🙂

This is it, guys. Feynman takes it further by also developing the electron configurations for the next 35 elements in the periodic table but… Well… I am sure you’ll want to read the original here, rather than my summaries. 🙂

Congrats ! We now know all what we need to know. All that remains is lots of practical exercises, so you can be sure you master the material for your exam. 🙂