Dynamics of the Wave Function: Heisenberg, Schrödinger, Collapse

My articles on complex numbers and linear algebra are prerequisites to this article.

This article follows my article on the essence of quantum mechanics, where I pointed out that everything is made of elementary waves. This is a big assumption of quantum mechanics, which then describes how these waves behave through time. This dynamics of wave functions is what will be discussed here. We will be using visualizations of my own creation based on the mathematics of quantum mechanics.

The Wave Function

Recall that these waves are fields which map each point of space with a number. This number is called the amplitude of the wave at that point. To simplify our understanding of wave functions, let’s consider a 1-dimension space. In this setting, waves are commonly represented as signals through space, as follows.

Wave Function

Now, the figure above represents a real-valued field. But the amplitude of the wave function is rather a complex number.

If you’re not familiar with complex numbers, most of what you need to know is that each complex number is a dot in a plane, called the complex plane.
Does this mean that the representation as a signal done above is wrong?

It’s not wrong. But it’s incomplete. Thus, let me introduce a new visualization of wave functions of my own creation. In this setting, each color corresponds to a position in space. Then, the amplitude of the wave function at a position in space is represented by its associated dot in the complex plane. This dot is colored by the color associated to the position in space. The wave can then be seen as a colored path in the complex plane.

Wave in Complex Plane

For instance, the amplitude of the wave function drawn above at the green position in space is -1, that is, the point of distance 1 from the origin on its left. Since this point is one of the farthest for the origin, the square of the norm of the amplitude is relatively great. This means that the green position is one of the most likely position where the wave may collapse to if observed.

I’m not sure I really understand the second representation… What would a collapsed wave function look like in this complex plane representation?

Good question. Let’s assume that the wave function has collapsed around the green position. This means that the wave function is very localized around this position, and nearly nil anywhere else. Then, we’d obtain the following figure:

Localized Wave in Complex Plane

Is the localized wave necessarily horizontal?

No! Wave functions are defined up to a phase. This means that a rotation of the complex plane change nothing to the physical nature of the wave. So a vertical localized wave would be essentially identical to the one above!

Momentum Space Wave Function

So far, we have described the position of the particle. In classical physics, we’d only need to know about its velocity to know everything about it. Equivalently, we can focus on its momentum. Classically, the momentum is the velocity multiplied by the mass of the particle. It’s a much more significative concept in physics because of the law of conservation of momentum, which states that the momentum of a closed system is always conserved. This has been essential to explain the photoelectric effect, which boils down to a transfer of momentum from photons to electrons.

But what is the momentum of a photon or of an electron?

What Planck and Einstein showed is that the momentum of light depends on the frequency of its wave and its direction of motion. What we usually mean by frequency of light is the time frequency of the wave, but since the speed of light is constant, we may equivalently consider the spatial frequency, also known as the wavenumber. A similar definition occurs in quantum physics for all particles.

So the momentum of a particle is related to the wavenumber of its wave function?

Basically, yes! Now, we have to be a little more careful before actually talking about frequencies. The basis of all frequencies comes from trigonometric waves. In the complex plane, the trigonometric waves are the simplest periodic paths one can think of: Circles. In this setting, the wavenumber corresponds to the number of cycles of the trigonometric wave in 1 meter. Equivalently, it is the inverse of the spatial period, as displayed below.

Wavenumber of Trigonometric Wave

Beware that I’m going to use the word spectrum to talk about the range in space described by the figure. This spectrum has absolutely nothing to do with the physical spectrum of frequencies of light.

Note that the trigonometric function in the figure turns anti-clockwise. This is important. It implies that the wavenumber of the trigonometric wave is positive, and that the associated wave function goes from left to right. If it went clockwise, it would be negative.

So, can actual wave functions be trigonometric waves?

No, because they don’t start nor end at 0. This is incompatible with waves representing probabilities.

OK… But if physical wave functions aren’t trigonometric, how can we find their wavenumbers?

We need to use the Fourier analysis. According to this, any wave is uniquely decomposable into an infinite superposition of trigonometric waves of different frequencies. This means that, for any wave and any frequency, the wave is composed of a certain amount of the trigonometric wave of that frequency. And the decomposition is given by the Fourier transform. Basically, the Fourier transform provides a description of any wave, like the one on the left, by an infinite combination of trigonometric waves, like those on the right:

Fourier Decomposition

In this figure, the greater the wavenumber, the more arrows I have put on the trigonometric functions. Note that the greater the wavenumber, the more energy is associated to it.

I’m not sure what the Fourier transform stands for in this picture?

The Fourier transform takes the wave function as an input and returns the weights of trigonometric waves on the right. Note that these weights can actually take complex numbers. In fact, by considering that the weight of trigonometric waves which don’t appear on the right is 0, we can consider this output as a matching of every wavenumber with an amplitude.

But how do wavenumbers relate to momentum?

The momentum is proportional to the wavenumber! This corresponds to the relation $p=hk/(2\pi)$, where $p$ is the momentum, $h$ is the Planck constant and $k$ is the wavenumber. From the Fourier transform that provides a mapping of every wavenumber with an amplitude, we can easily deduce a mapping of every momentum with an amplitude. We call this matching the momentum space wave function. By opposition, the wave function is sometimes called the position space wave function.

Does this mean that a wave function is associated with several momenta?

Precisely! A wave function has a superposition of momenta! And the fact that there are several momenta associated to a wave functions shouldn’t be the only thing shocking you, especially if you’re familiar with classical mechanics.


Classically, the position and the momentum of a particle are two distinct concepts which can be defined independently from each other. But it’s not the case in quantum mechanics! Now, the position of a wave function is distributed in space. Weirdly enough, the mere knowledge of the amplitudes of positions determine the amplitudes of momenta! The reciprocal phrase is true too: The amplitudes of momenta determine the amplitudes of positions. Mathematically, this corresponds to saying that the Fourier transform is more or less a bijection from the Hilbert space of wave functions and the Hilbert space of momentum space wave functions.

To be more accurate, the mapping of wave functions with momentum space wave function is even a linear isometry between two Hilbert spaces! I plan on writing an article on the amazing structure of Hilbert spaces later!

This mapping of wave functions with momentum space wave functions by the Fourier transform is displayed below. The following figure is definitely false since I haven’t made any actual computation to draw this. It rather aims at providing you an image of what the intimate relationship between position space and momentum space wave functions.

Fourier Transform

Heisenberg’s Principle

Is there a probabilistic interpretation of the momentum space wave function, just like for the wave function?

Yes! The square of the norm of the value of the momentum space wave function is the probability density function of finding the momenta in case of observation.

I’ve heard something about the impossibility of knowing both position and momentum… Is it true?

This is what’s called Heisenberg’s uncertainty principle. But I believe that the word uncertainty is misleading. Rather, you should think of Heisenberg’s principle as a spreading principle.

What does the principle say?

It says that the wave function and the momentum space wave function can’t be both localized. This is a direct consequence of their relationship through the Fourier transform. For instance, in the case of the figure above, where the momentum space wave function was localized around the pink-red momentum, while the wave function was spread all over the spectrum.

Humm… I don’t see why it’d be the case…

Imagine that the momentum space wave function were perfectly localized at one momentum. Recall that the wave function is obtained by summing all momentum space wave functions of all momenta. This implies that the wave function is nearly the trigonometric wave associated to the momentum at which the momentum space wave function is localized. This trigonometric wave circles over all space, and is therefore spread over all space. This is represented below:

Fourier Transform of Dirac

Conversely, were the wave function very localized, the momentum space wave function would be circling periodically, and would therefore be spread over all momentum space.

How is the spreading of a wave function measured?

Technically, the spreading of a wave function is measured by the standard deviation of the probabilistic distribution of position induced by the wave function. Now, I have been using the word probabilistic, but you don’t need to mention probability to describe the spreading of the wave function. Indeed, the mathematical concept matches the intuitive idea of how far in space the wave is spread.

So how does Heisenberg’s spreading principle translate into mathematics?

It states that the product of the spreading of the wave function by the spreading of the momentum space wave function must be greater than $h/(4\pi)$.

So why do people talk about the uncertainty principle rather than the spreading principle?

This is because people often focus on the measurement part of quantum mechanics. In this setting, what’s surprising is that the measurement of the position of a wave localizes the wave around a position. As a result, the wave function becomes very localized, which implies that the momentum space wave function is greatly spread. As quantum mechanics was approached with concepts of classical mechanics, this frustrating phenomenon was commented as the impossibility of knowing both position and momentum simultaneously. But quantum mechanics rather says that when the wave is very localized, then the wave has a superposition of a large range of momenta. This principle doesn’t involve any probability.

Can Heisenberg’s principle be tested?

Yes! This has been done by Derek Muller on Veritasium in a very simple experiment! Check out this awesome video:

As pointed out by Derek, Heisenberg’s principle only applies to spreadings along one dimension. In particular, there is no spreading relation between the position in the horizontal direction and the momentum in the vertical direction. Also, if you want to go further, there are other variables which are deduced from position and momenta which cannot be known simultaneously like angular momenta. But to understand this, I’d need to introduce operators, which would take too much time. If you can, please write about this!

Schrödinger Equation

Wait a minute… Momenta are defined as the Fourier transform of the wave functions… But do they still have the meaning they had in classical mechanics?

Yes! The momentum is still somehow the velocity of a particle multiplied by its mass. And the reason why such a reasoning holds is because of the most fundamental equation of quantum physics which describes the dynamics of wave functions. This equation is the Schrödinger equation.

What does Schrödinger equation say?

Hummm… That’s a tough one. I have a lot of trouble making sense out of it. Here’s the best explanation I’ve come up with. At any position of space, a wave function has a particular energy which is deduced from its global structure. This local energy, also known as the Hamiltonian, is a composition of some potential energy which is due to external forces like electromagnetism, and some kinetic energy which depends on the superposition of momenta which makes up the wave function. What’s particularly weird about this kinetic energy is that it can have a complex value. Now, what Schrödinger equation says is that, at every position, the wave function rotates around the origin of the complex plane at a speed proportional to the Hamiltonian. This corresponds to the following formula:

Schrödinger Equation Explained

This formulation is the most general form of Schrödinger equation. It includes Dirac’s equation for relativistic wave functions.
Humm…. I don’t see what this has to do with momenta…

From Schrödinger equation can be derived the fact that the average position varies according to the average momentum. This coincides with the classical setting of classical mechanics! This should sound surprising to you. At least, it does to me. Even though I can prove it mathematically, I have no understanding of the fundamental reason why Schrödinger equation links average position and average momentum.

In particular, I can’t seem to find a way to relate Schrödinger equation with the idea of superposition of momenta. This prevents me from describing the spreading of position through time. If you find a way to combine my representations of wave functions with Schrödinger equation and the ideas of superposition, I would be very interested in hearing about it!
This is disappointing…

Sorry for that. But anyways, the greatest applications of Schrödinger equation rather concern the calculations of stationary states. Indeed, at the scale we are looking at, distances are so short that stationary states are reached in such a small fraction of time that we can only observe these equilibria.

What’s a stationary state?

It’s a wave function which will appear to be non-moving. Now, recall that the wave function is defined up to a rotation in the complex plane. Thus, stationary states involve only such rotations. The following figure shows a wave function rotating. Arrows drawn on the wave represent the velocity at which each point of the wave moves:

Stationary Wave

Now, for a wave to be stationary, the velocity at any given point must be perpendicular to the vector from the origin to that point. Also, the farther to the origin a point is, the greater the speed must be. These two conditions correspond to the Hamiltonian at position being proportional to the wave function at this position with a real-valued proportionality coefficient independent from this position. In mathematical terms, this means that the wave function should be an eigenvector of the Hamiltonian. The eigenvalue, which is the real proportionality coefficient, is then the energy of the wave. This corresponds to what’s known as the time-independent Schrödinger equation:

Time-Independent Schrödinger Equation

Because of this equation, eigenvectors and eigenvalues plays an essential role in quantum mechanics. These objects are studied by linear algebra, and, more precisely, the theory of eigenvalues. If you can, please write about these!
So the energy of a wave is the speed at which it rotates?

Exactly! The faster the wave turns, the greater the energy. More precisely the energy of the wave is the number of turns it makes in a second multiplied by Planck’s constant. In other words, Planck’s constant is the energy of a wave which turns at the speed of 1 turn per second. Since rotational speed is often rather measured in radians per second, physicists often prefer to use reduced Planck’s constant $\hbar$, which is therefore $\hbar = h/(2\pi)$.

Does this have something to do with quantum levels?

Yes! As it turns out, in any defined system, not all energies can be reached by a stationary wave. The set of all possible energies, which is called the spectrum of the Hamiltonian, is made of discrete values more often than not. In particular, this is what happens if we consider wave functions confined in a box.

OK… That’s not really realistic, is it?

No indeed. More interestingly, in an atom, the positive charges of a nucleus affect the potential energies of wave functions of electrons, which in turns modify the Hamiltonian. The eigenvalues and eigenvectors of the obtained Hamiltonian, especially the lowest eigenvalues, now correspond to possible orbits of electrons around the nucleus. The lower energies refer to stationary wave functions localized closer to the nucleus. This phenomenon accounts for Niels Bohr’s weird model of the hydrogen atom, as explained in this video from the Science Channel:

Does that mean that, as energy decreases, electrons all fall down to the lowest energy level?

Fortunately, no. Because of Pauli’s exclusion principle, two electrons cannot be in the same quantum state. A quantum state is defined by the wave function and the spin of the electron, which can be up or down. Since there are only two spins, this means that there are at most two electrons with the same wave function. In particular, the energy levels cannot contain too many electrons, which implies that there are sorts of blanks around nuclei that can each contain one electron only.

What do you mean by “fortunately”?

Without Pauli’s exclusion principle, electrons would not be able to combine atoms together, which means that there would be no chemistry. No chemistry, no life. No life, none of us…


If you look carefully at what we’ve discussed until now, no probability is involved. Indeed, Schrödinger equation is perfectly deterministic and time-revertible.

I thought that quantum mechanics was a probabilistic description of the world…

Half of quantum mechanics has absolutely nothing to do with probabilities. And this half accounts for plenty of counter-intuitive observations. But there are other observations which can’t be explained without the other half of quantum mechanics. This is the case, for instance, of the troubling double slit experiment and its variants you can read about in my article introducing quantum mechanics. This other half is what people often refer to as quantum weirdness.

So what’s quantum weirdness?

What people often refer to as quantum weirdness is the wave function being transformed by its mere observation. We say that the wave function collapses. This collapse is a probabilistic dynamics of the wave function. Now, even though the result of the collapse is not determined deterministically, it follows some probabilistic distribution which is defined by what the wave function was before collapsing.

In mathematical terms, this corresponds to saying that the collapsing of the wave function corresponds to a Markov chain. At least, that’s how I like to think about this, since, apparently, at least on the web, this is controversial. Quite often though, this idea is rejected because a restricted concept of Markov chain is discussed.
So what’s the probabilistic distribution of a wave function?

What’s extremely weird is that this probabilistic distribution depends on what’s being measured. Assume the position is measured. Then, the wave function will collapse into a localized wave function. The position of this localized wave function follows a probabilistic distribution whose density is the square of the norm of the wave function before it collapses. The following figure displays the possible collapses of a wave function, where bolder arrows correspond to more likely collapses.