I'm very confused about what exactly causes the wavefunction collapse. I was under the impression that quantum objects propagate as waves and interact as particles. That is, upon interaction with another particle their wavefunction collapses. But I have heard that it is measurement which collapses the wavefunction or even that it may have to do with conscious observation. This seems very vague to me. What defines a "measurement"? Is it simply that the wavefunction collapses only when interacting with macroscopic objects, regardless of if they are measuring devices or connected to a conscious observer?

  • $\begingroup$ I will be careful with calling a measurement just an interaction. You can have many particles interacting like quantum many-body system without ever having a collapse of the wavefunction. I would mostly say that measurement is an interaction with a macroscopic object (but the limit of micro-macro is blurry defined). $\endgroup$
    – Mauricio
    Jul 5, 2022 at 15:45
  • $\begingroup$ @Mauricio is it known what causes this "blurriness"? Does a microscopic system interacting with a semi-macroscopic system only partially collapse the wavefunction? $\endgroup$
    – physBa
    Jul 5, 2022 at 15:47
  • 1
    $\begingroup$ Does this answer your question? Is there a natural principle that forbids real wavefunction collapse? $\endgroup$
    – Roger V.
    Jul 6, 2022 at 18:56

4 Answers 4


Every measurement is an interaction. That is an important statement from quantum theory. One way to think about collapse as follows:

The collapse of a wave function is triggered by the interaction with a bath in the measurement apparatus.

A bath is any system described by a temperature.

A measurement apparatus always has a bath inside, in order to be bale to show the measurement result.

  • $\begingroup$ Is this equivalent to the interaction with a macroscopic object? Can you have a microscopic bath? $\endgroup$
    – Mauricio
    Jul 6, 2022 at 12:42
  • $\begingroup$ Yes, all macroscopic bodies contain baths: they have a temperature. But also the surrounding radiation field has a temperature. And a body has to be really small to not have a temperature... So in practice, almost any body and alsmost any envrironment has a temperature. $\endgroup$
    – KlausK
    Jul 8, 2022 at 5:05

You are right to be confused. In those interpretations where wavefunction collapse occurs, there is no clear mechanism or explanation for how it works - just a variety of hand-waving speculations.

In the Everett Interpretation (sometimes called the Many Worlds Interpretation) there is no actual collapse of the wavefunction, but you get the appearance of wavefunction collapse when the system under study interacts with the observer's wavefunction, and the two wavefunctions become correlated. When a particle in a superposition of mutually orthogonal states interacts with an observer, the observer enters a superposition of mutually orthogonal states, each component being one observer observing the particle in one state.

Any interaction between quantum systems can constitute an observation, but it only looks like wavefunction collapse if you are one of the quantum systems, seeing it 'from the inside' so to speak.

I will try to explain using an example from classical physics - that of coupled oscillators and normal modes of vibration.

Suppose we have two independent simple harmonic oscillators, with equations $\ddot{x}=-\omega_1 x$ and $\ddot{y}=-\omega_2 y$. We can pack them both into a single matrix equation like so:

$\begin{pmatrix}\ddot{x} \\ \ddot{y}\end{pmatrix}=\begin{pmatrix}-\omega_1 & 0 \\ 0 & -\omega_2\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}$

These oscillators are independent, they do not affect one another, neither can see the other.

Now we introduce a weak interaction between them, such that the state of one applies a force on the other.

$\begin{pmatrix}\ddot{x} \\ \ddot{y}\end{pmatrix}=\begin{pmatrix}-\omega_1 & \epsilon \\ -\epsilon & -\omega_2\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}$

These oscillators are coupled. Each of them affects the other, they can sense one another. We can figure out what is going to happen if we apply the trick of diagonalisation to the matrix in the middle. We can represent a non-singular matrix $M$ as a product $M=U^{-1}DU$ where $D$ is a diagonal matrix of eigenvalues and $U$ is a unitary matrix of eigenvectors.

$\begin{pmatrix}\ddot{x} \\ \ddot{y}\end{pmatrix}=U^{-1}DU\begin{pmatrix}x \\ y\end{pmatrix}$


$U\begin{pmatrix}\ddot{x} \\ \ddot{y}\end{pmatrix}=DU\begin{pmatrix}x \\ y\end{pmatrix}$

and if we change variables to $\begin{pmatrix}u \\ v\end{pmatrix}=U\begin{pmatrix}x \\ y\end{pmatrix}$ which is a new pair of variables that are each some linear combination of $x$ and $y$, we get a separated pair of independent simple harmonic oscillators again. These are called the normal modes of vibration.

$\begin{pmatrix}\ddot{u} \\ \ddot{v}\end{pmatrix}=\begin{pmatrix}-D_1 & 0 \\ 0 & -D_2\end{pmatrix}\begin{pmatrix}u \\ v\end{pmatrix}$

One of them will consist of $x$ and $y$ swinging in phase together, the other will have them swinging in anti-phase, in opposite directions. The result will be a superposition of the two - but because the $u$ and $v$ states are independent they cannot 'see' each other. Because the matrix $D$ is diagonal, they do not interact. It is as if they were in two separate worlds.

However, the $x$ and $y$ oscillators in either of the pure $u$ or $v$ states do see each other: they are correlated.

The wave equation works like there is something out there oscillating, too, and it is natural to expect that when two quantum oscillators interact, they will both enter a superposition of joint states in which they are correlated. In each component of the superposition, one state of $x$ 'observes' the corresponding state of $y$.

So to answer your questions, an observation is any interaction between you (the observer) and the system under study that causes you to enter a superposition of normal modes of vibration, and what causes wavefunction collapse is when each of your normal modes becomes unaware of the existence of all the others, and therefore assumes they all disappear.

And of course, for all we know, they might do. Since by definition we can't see them ($D$ being diagonal), there is no way to determine whether or not they exist.

It is essentially the same question as asking whether an electron passing through one slit can 'see' itself simultaneously passing through the other slit. (They're charged, so they ought to repel one another if they can interact.) For electrons, we have much less difficulty accepting that the answer is 'no'. Each possible history in Feynman's Sum-Over-Histories sees none of the others. We just have a lot more difficulty seeing ourselves as waves.


While there's still some (legitimate) debate over this, you can mostly think of it as interaction. When a quantum (microscopic) system interacts with a macroscopic system, "something happens" to its wavefunction that leads, one way or another, to collapse.

The macroscopic system can be an "active" measuring device built by humans for the purpose of measuring, but any macroscopic system capable of interactions will do.

What is known is that this interaction with the larger outside environment leads to decoherence, which is a process that selects a specific set of eigenstates. Most of the time, it'll be position eigenstates, because most interactions depend on position. For example, you can say that the Earth contributes to decoherence through its gravitational field.

What is not well understood is that, at some point, this interaction can also lead to the selection of one of those eigenstates. This process is hidden behind the "measurement" or "collapse of the wavefunction" moniker.

  • $\begingroup$ so would the interacting system have to be macroscopic in order to collapse the wavefunction? Is this because of decoherence effects require many interactions to become significant? I still don't understand how decoherence is related to the measurement problem $\endgroup$
    – physBa
    Jul 5, 2022 at 15:48
  • $\begingroup$ In the 50's and the 60's, yes people were mostly working with Bohr's decision to have microscopic quantum systems on one side, and macroscopic classical systems on the other. Now we know with mesoscopic systems that the phenomena becomes gradually more important and faster as size grows (see Nobel prize 2012). The "measurement problem" is just a name for the moment where a specific eigenstate is selected, after decoherence has done its work. $\endgroup$
    – Miyase
    Jul 5, 2022 at 23:04

This seems very vague to me. What defines a "measurement"?

This exact question is known as "the measurement problem". John Bell formulated the question in a way that I like a lot:

What exactly qualifies some physical systems to play the role of 'measurer'? Was the wavefunction of the world waiting to jump for thousands of millions of years until a single-celled living creature appeared? Or did it have to wait a little longer, for some better qualified system . . . with a PhD?

In the standard/textbook quantum formalism, sometimes known as the copenhagen interpretation of quantum mechanics, this distinction is not made clear. So there is no exact answer strictly in the standard formalism.

Having pursued it personally for a few years during my studies, the only framework which satisfactorily answered this question for me was Bohmian Mechanics. Since it has a particle position to begin with, there is no question of what the measurement outcome will show, and as for wave function collapse, there is no wave function collapse after any so-called measurement... so that making a distinction between measurement and non-measurement events is not necessary.

Physicists tend to have very differing (and often strong) views on interpretations/variants of quantum mechanics. There are more than a dozen of them, and surely there is some which someone else who sees this answer will like better.


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