Survivor’s Guide

I am not sure if I survived my journey through the jungle of quantum mechanics in good enough shape. One day I feel like: I am getting this! The next day I feel like: do I understand anything at all? If you feel the same, then the following may help.

Math is a language – and language is always ambiguous

Physicists like to generalize. They talk about states and state vectors, or bosons and fermions, or wavefunctions, and forget to warn the reader they are actually talking about something very specific. The wavefunction for an electron orbital, or for a hypothetical particle traveling in free space, or for a photon, or for one of the states in a n-state system—and I can list many more instances of the wavefunction—are all very different beasts, and trying to cram them altogether in one generic lecture on quantum math will not necessarily further your understanding.

For example, I found it rather shocking to find out, after all of the lectures on the behavior of bosons (force carriers, as opposed to matter-particles), that the archetypal boson—the photon—always has spin up or down, no zero spin, although the theoretical spin-one boson has three possible spin states: −1, 0, and +1. In other words, the theoretical spin-one boson doesn’t exist, and so you wonder why physicists even talk about it.

Another thing that bothers me is that (real or possible) physical states are represented by so-called state vectors: the term suggests a resemblance with real vectors in three-dimensional space—think of vectors representing a direction, a force, a velocity, or whatever other reality that has a magnitude, a direction and something to grab onto—and… Well… No. State vectors are very different beasts too. In fact, some of the addition and multiplication rules—and the idea of base states—are encouragingly familiar concepts but you almost never feel like you get a satisfactory description of what a state actually is.

This Survivor’s Guide will, hopefully, give you some informal idea of what the physicists are talking about, which might be helpful in terms of… Well… Surviving a course on quantum mechanics… 🙂  


Dirac invented this notation for initial versus final states. The initial state might just be: I don’t have a clue. In that case, you’d probably write it like |ψ〉 or |s〉. The final state might be really be final, which means you’ve made an observation and so you’ve established the position of a particle, or you measured its linear or angular momentum, or its energy, or whatever else you wanted to know about it. Those states are final because the atomic or sub-atomic scale ensures your measurement will usually destroy whatever was there. Such state will usually be written as 〈x|, 〈|, 〈E|, 〈p|, 〈1| or whatever other else that looks a bit more specific than those Greek letters. To be specific, the 〈x|, 〈|, 〈E|, 〈p|, 〈1| might mean: we’ve intercepted our particle at x, its spin is up, its energy is E, its momentum is p, or it’s in state 1 (whatever state 1 might be). However, a final state can also be intermediary. In fact, an initial state might also be intermediary, so we can write something like this:


This probably means something like: our particle was at s, went to 1, and ended up at x. The combination of an initial and a final state, such as 〈x|1〉 or 〈1|s〉 here, is associated with an amplitude which, for all practical purposes, are just some complex number, and so we’re effectively multiplying two complex numbers here. To find what? Well… The amplitude for our particle to go from to via 1.

Now you’ll want to know the numbers: how do we calculate 〈x|1〉 and 〈1|s〉? Good question. We have a propagator function for that. In fact, there is a whole zoo of propagator functions out there, and it’s pretty hard to make sense of them. So you get stuck almost immediately. That makes quantum mechanics hard: even the simplest of ideas—the idea of a particle traveling from one place to another in a given time—is incredibly complicated in quantum mechanics. But let’s give it a try.

Feynman’s propagator

According to Feynman’s path integral formulation of quantum theory, a particle can travel from A to B following any path, as illustrated below.Three_paths_from_A_to_BIt’s even worse than that: for each path, the time that is needed to go from A to B can vary. So we have all of these random walks, so to speak, and Feynman tells us we should sum up all of the amplitudes for these various paths to get the amplitude for the particle to go from A to B. The Wikipedia article on this approach has a wonderful animation which shows how that works, and how that actually yields a pretty consistent picture! However, for a better understanding, it’s probably best to combine it with a thorough reading of Feynman’s Lecture on the Principle of Least Action. For each path, we have an action integral, which is going to give us the action for that path. The concept of physical action (Wirkung in German) is not very well known despite it being a rather logical concept: its physical dimension is force over a distance over some time. That’s equivalent to some energy (force times distance) being available for some time or, alternatively, some momentum (force times time) being available over some distance. Planck’s quantum of action is, effectively, the natural unit of action. Let’s quote Feynman here:

“Here is how it works: Suppose that for all paths, S is very large compared to ħ. One path contributes a certain amplitude. For a nearby path, the phase is quite different, because with an enormous even a small change in means a completely different phase—because ħ is so tiny. So nearby paths will normally cancel their effects out in taking the sum—except for one region, and that is when a path and a nearby path all give the same phase in the first approximation (more precisely, the same action within ħ). Only those paths will be the important ones.

In the limiting case in which Planck’s constant ħ goes to zero, the correct quantum-mechanical laws can be summarized by simply saying: ‘Forget about all these probability amplitudes. The particle does go on a special path, namely, that one for which does not vary in the first approximation. That’s the relation between the principle of least action and quantum mechanics.”

So how does it work exactly? As mentioned above, the action integral is going to give us S for each path in space and in time, and the amplitude for that path will be proportional to ei·S/ħ, so we’ll write that amplitude as a·ei·S/ħ. The coefficient is, obviously, a normalization constant. OK. That’s enough on this as for now! 🙂

Note: The story on how quantum mechanics becomes classical mechanics for the limiting case in which in which Planck’s constant ħ goes to zero, is related to another one, although I should do a better job of pointing out how exactly. Look at the following super-obvious equation in classical mechanics:

x·px − px·x = 0

In fact, this equation is so super-obvious that it’s almost meaningless. [In case you wonder, is just the position and px is the momentum in the x-direction.] Almost. It’s super-obvious because multiplication is commutative (for real as well for complex numbers). However, when going from classical to quantum mechanics, replacing the x and px by the position and momentum operator, we get an entirely different result. You can verify the following yourself:


As Feynman puts it: “If Planck’s constant were zero, the classical and quantum results would be the same, and there would be no quantum mechanics to learn!” Let me make two remarks here:

1. We should not put any dot (·) between our operators, because they do not amount to multiplying one with another. We just apply operators successively. Hence, commutativity is not what we should expect.

2. Also, note that, when doing the same calculations for the equivalent of the x·py − py·x expression, we do get zero, as shown below:

not strange

First Principles

Unfortunately, there are no First Principles in quantum mechanics. Those rules for adding or multiplying amplitudes are just mathematical rules: there are no real physics behind. So it’s not like classical mechanics, or electromagnetic theory, where you get a very limited number of equations (Newton’s Laws in classical mechanics, and Maxwell’s equations for electromagnetism) and then you write 200 pages on their consequences.

No. It’s not like that. In quantum mechanics, you constantly flip between rough heuristic arguments and a ton of incredibly complicated mathematical treatments of the same. The heuristic argument provides some rationale for whatever it is that you are trying to do, while the mathematical treatment tries to show you it might make some sense.

You’ll say: what could First Principles possibly look like? I’d say we should have some consistent idea of what a photon and an electron actually are, and then we can discuss why and how they interact the way they do. You’ll say: “Oh. You’re just another crazy wanting to present some physical interpretation of the wavefunction.” And… Well… Yes. I mean… Look at any equation involving some wavefunction, and they’ll also involve physical constants and whatever physical information we have about our object: its mass or its energy (these are equivalent, obviously), its (linear or angular) momentum, its position in space and time, and… Well… Whatever else you can think of. But then we say the (components of the) wavefunction itself have no physical dimension? They’re just some mathematical construct? Do a dimensional analysis of the equation below:

dispersion equation

All of the terms in this equation have a 1/m2 dimension if… Well… If the real and imaginary part of φ have no physical dimension. Come on ! That doesn’t make sense: (the components of) φ must have some physical dimension too.

So the question is… We’re modeling what per square meter (1/m2) here? Personally – but that’s just an uninformed opinion, of course – I think it’s an energy flow. Why? I don’t know. I need to work it out. But the m2c2/ħ2 factor on the right-hand side of the equation definitely re-scales stuff in terms of natural units. Why? Because it’s the (inverse) square of the Compton radius of an electron: ħ/mc.

Is the Compton radius a natural unit? In my flywheel model of an electron, it surely is. So… Well… What’s the model?

The crazy stuff

My physical interpretation of the wavefunction is inspired by the mathematical similarity between the E = m·a2·ω2 (the energy of two oscillators in a particular 90° degree combination) and the E = m·c2 formula. If this were to represent something real, then we need to give some meaning to the a·ω identity that comes out of it. Now, if we assume, just for fun, that E and m are the energy and mass of an electron, then the de Broglie relations suggest we should equate ω to E/ħ. As for a, the Compton scattering radius of the electron (ħ/(m·c) would be a more likely candidate than, say, the Bohr radius, or the Lorentz radius. Why? Because we’re not looking at an electron in orbit around a nucleus (Bohr radius), and we’re also not looking at the size of the charge itself (classical electron radius), because we assume the charge is pointlike. We get:

a·ω = [ħ/(m·c)]·[E/ħ] = E/(m·c) = m·c2/(m·c) = c

Wow! Did we just prove something? No. We don’t prove anything here. We only showed that our E = m·a2·ω2 = m·c2 equation might (note the emphasis: might) make sense. Let me show you something else. If this flywheel model of an electron makes sense, then we can, obviously, also calculate a tangential velocity for our charge. The tangential velocity is the product of the radius and the angular velocity: v = r·ω = a·ω = c. So that’s great visuals too. But where did we get our Compton scattering radius from?

Good question. Great question, actually. It comes out of the Great Depths of all of the quantum math, which I don’t want to explain to you because I don’t quite master it. However, I am happy to note that – in my search for First Principles –  the formula is consistent with my intuitive model of an electron, which is… Well… A pointlike electric charge rotating around some center, as illustrated below.


As for the mechanism, I’ve detailed that ad nauseam in a bunch of papers already, so I am not going to repeat that here. The basics are easy: our rotating charge should also have some angular momentum, right? In fact, the Stern-Gerlach experiment tells us the angular momentum should be equal to ± ħ/2. Is it? So we will need a formula for the angular mass, aka the moment of inertia. Let’s try the one for a rotating disk: I = (1/2)·m·a2. The formula for the angular momentum itself is just L = I·ω. Combining both gives us the following result:

L = ħ/2 = I·ω = (1/2)·m·a2·ω = (1/2)·(E/c2a2·(E/ħ) ⇔ = ħ/(m·c)

Did we just get the formula for the Compton scattering radius out of our model? Well… Yes. It is a rather Grand Result, I think. So… Well… It might make sense, but… Well… Does it, really?

There are several issues to be solved. For example, the idea of a force assumes some idea of inertia. If there is no inertia, a force will just give some object an infinite momentum in zero time. That doesn’t make sense ! So… Well… Perhaps the electron is just a charged version of the neutrino. 🙂

Certainty and Uncertainty

I find that a lot of the Uncertainty in quantum mechanics is actually suspiciously certain. For example, we know an electron will always have its spin up or down, in any direction along which we choose to measure it, and the value of the angular momentum will, accordingly, be measured as plus or minus ħ/2. That doesn’t sound uncertain to me. In fact, I feel it sounds remarkably certain: we know that the electron will be in either of those two states, and we also know that these two states are separated by ħ, Planck’s quantum of action, exactly.

Of course, the corollary of this is that the idea of the direction of the angular momentum is a rather fuzzy concept. As Feynman convincingly demonstrates, it is ‘never completely along any direction’. Why? Well… Perhaps it can be explained by the idea of precession?

In fact, the idea of precession might also explain the weird 720° degree symmetry of the wavefunction. Hmm… Now that is an idea to look into ! 🙂

[To be continued.]