Suppose there is an electron in a state $|\psi \rangle$, and there is a measurement apparatus whose atoms have a joint wavefunction $|m\rangle$. In experiments, we always know the initial value of $|\psi\rangle$, but we never know the initial value of $|m\rangle$, because it's too complicated. Is the unknown nature of $|m\rangle$ the the source of randomness in quantum mechanics?

More specifically, if we take the initial state $|\psi \rangle \otimes |m\rangle$ and evolve it with the Schrodinger equation, could it evolve into $|e\rangle \otimes |m(t)\rangle$, where $|e\rangle$ is the eigenstate that we measure, and $|m(t)\rangle$ is the final state of the measurement apparatus? If we randomise the initial value of $|m\rangle$, we may be able to obtain random final $|e\rangle$ states, which is what's observed in experiments.

Have physicists ruled this out? If not, how realistic is this as the explanation of collapse?

P.S. There is one very weird thing about this hypothesis. The final state $|e\rangle$ is always an eigenstate of the observable that the apparatus is designed to measure. So, in a way, the external potential of the measurement apparatus on the electron "favors" the eigenbasis of the variable being measured. So we need to incorporate this into the potential term in the Schrodinger equation. Different measurement apparatuses should apply very different potentials. This is weird because all measurement apparatuses are composed of similar particles. And the interaction potential between particles does not care about what measurement apparatus the particle is a part of.

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    $\begingroup$ To rule it out requires an experimental result, but you've framed the problem in terms that make an experiment impossible. $\endgroup$
    – John Doty
    Oct 11, 2022 at 16:16
  • $\begingroup$ @JohnDoty I think this can be proved or disproved on paper by proving or disproving if unitary evolution can evolve a superposition into an eigenstate, with probability given by the Born rule, provided that the randomness in the initial state of the measurement apparatus is account for. I was asking if this has been disproved, or how seriously this is taken as the solution to the measurement problem $\endgroup$ Oct 11, 2022 at 16:25
  • $\begingroup$ You can only prove it on paper from mathematical axioms. But the physical world doesn't give us axioms: it gives us phenomena. There is no guarantee that any axioms you use match physical reality, and history suggests that all axiom systems that purport to capture the phenomena eventually need revision. $\endgroup$
    – John Doty
    Oct 11, 2022 at 16:51

3 Answers 3


"Is the unknown nature of $\left|m\right>$ the the source of randomness in quantum mechanics?"

No. If the randomness came from the measurement apparatus, it would be virtually impossible for the EPR experiment to give perfectly correlated random outcomes when two different measurement apparatus measure each particle of a widely-separated entangled pair.

"More specifically, if we take the initial state $|\psi\rangle \otimes |m\rangle$ and evolve it with the Schrodinger equation, could it evolve into $|e\rangle \otimes |m(t)\rangle$, where $|e\rangle$ is the eigenstate that we measure, and $|m(t)\rangle$ is the final state of the measurement apparatus?"

Maybe, something a bit like that. The following argument is very vague and hand-wavey - as is inevitable when discussing measurement apparatus with on the order of $10^{23}$ particles. It is intended only to illustrate the idea. We can combine the Schrodinger equations for two independent systems prior to measurement using a matrix equation as follows:

$$i\hbar \frac{\partial}{\partial t}\pmatrix{\psi_1 \\ \psi_2}=\pmatrix{H_1 & 0 \\ 0 & H_2}\pmatrix{\psi_1 \\ \psi_2}$$

$\psi_1$ and $\psi_2$ are the observer and observed system states respectively, and may have many dimensions; even infinitely many. The matrix $H$ should be thought of as a block diagonal matrix. The zero blocks top-right and bottom-left in $H$ express the fact that the two systems do not interact. Each can vary completely independently of the other.

Now we modify the equation to introduce a weak interaction between the systems. The off-diagonal terms are set to non-zero values, expressing how each system affects the other. This represents the measurement process, and the details of the constant $\epsilon$ values define what sort of measurement it is.

$$i\hbar \frac{\partial}{\partial t}\pmatrix{\psi_1 \\ \psi_2}=\pmatrix{H_1 & \epsilon_1 \\ \epsilon_2 & H_2}\pmatrix{\psi_1 \\ \psi_2}$$

Now the equations are no longer independent, they are coupled. There is a classical technique we can use to disentangle coupled linear differential equations, which is to diagonalise the Hamiltonian. We write $H$ as a product $U^TDU$ where $U$ is a unitary matrix of orthogonal eigenvectors, and $D$ is a diagonal matrix of eigenvalues.

$$i\hbar \frac{\partial}{\partial t}\pmatrix{\psi_1 \\ \psi_2}=U^T\pmatrix{D_1 & 0 \\ 0 & D_2}U\pmatrix{\psi_1 \\ \psi_2}$$

We can shift the $U^T$ over to the left hand side, using the fact that $U^TU=I$.

$$i\hbar \frac{\partial}{\partial t}U\pmatrix{\psi_1 \\ \psi_2}=\pmatrix{D_1 & 0 \\ 0 & D_2}U\pmatrix{\psi_1 \\ \psi_2}$$

And now we change basis by renaming $U\psi=\phi$.

$$i\hbar \frac{\partial}{\partial t}\pmatrix{\phi_1 \\ \phi_2}=\pmatrix{D_1 & 0 \\ 0 & D_2}\pmatrix{\phi_1 \\ \phi_2}$$

Now this is again a system of completely independent equations, each acting as an independent system that does not interact with any of the others. But each component of $\phi$ is a mixture of $\psi_1$ and $\psi_2$ states in linear combination. The observer/observed system states in each component are correlated. Each of these corresponds to one possible outcome of the quantum measurement. Each consists of the observed system in a particular state, and the observer in a corresponding state of having observed it. But because the off-diagonal terms in the interaction Hamiltonian are all zero in this basis, none of the correlated observer/observed states can interact with any of the others. They are orthogonal to one another. To each, it is as if none of the others existed.

When we use this technique to decouple linear equations in classical physics, the orthogonal components are called normal modes of vibration. 'Normal' because they are orthogonal to one another. We can solve each independent sub-equation separately, and then the general solution is a linear combination of the normal modes.

Because an observer in such a 'normal mode' state cannot interact with, and hence perceive, any of the other normal modes, she has no way to tell whether they are really there. Maybe they all disappear, and get zeroed out? The assumption that all but one of them disappears for some unknown and unspecified reason is what we call wavefunction collapse. The assumption that they don't, but are simply unobservable is called the Everett Interpretation (or Many Worlds Interpretation). Each normal mode acts like a separate 'world'.

Because the observer system is usually very complicated, the split is generally into lots and lots of component states - the diagonal matrix $D$ might have $>10^{23}$ dimensions. They are splitting not just based on the state of the observed system, but all the quantum decisions related to processes going on inside the observer. So each possible observation outcome probably doesn't have just one world where it happens, but trillions upon trillions of worlds. This aspect of the process is called decoherence and is where we think things revert to 'Born Rule' classical probabilities - although this bit of the theory is still somewhat ill-defined and controversial. However, everything is still quantum, and the superposition between alternative outcome states still exists.

So in this interpretation of the observation process, the randomness comes from each of the observers seeing only one outcome. The entire process is deterministic - you always get observers seeing 'up' and observers seeing 'down', every time. But from the point of view of each observer, only one outcome happens, and which one appears to be random. The randomness can still be correlated over long distances, though, because the measurement process throws away no alternatives, but instead just introduces correlations between systems. One observer becomes correlated to one particle, which is correlated to the other particle, which is correlated to the other observer. Observers can only perceive other observers who make compatible observations. And the correlations can be created one at a time, so no signals have to travel faster than light.

All we have to do is assume the standard rules for unitary quantum evolution applies at the macroscopic level without exception, and apply the 'normal modes' technique from classical mechanics, and the appearance of a classical world drops out of the equations automatically. It's not perfect, it's not fully understood, but none of the alternatives are either.

  • $\begingroup$ Great answer. So, the Schrodinger equation cannot evolve a superposition into an eigenstate, even after you account for the measurement apparatus. But it can evolve into a superposition with non-interacting components... Would you say that collapse, or the perception of collapse, happens only when a consciousness becomes entangled? This is because only consciousness is forced to be in one of the worlds. $\endgroup$ Oct 12, 2022 at 2:58

I think this can be ruled out by unitarity. Suppose that $|\psi\rangle$ describes a free particle, which lives in an infinite-dimensional vector space. You can encode an arbitrarily large amount of information in this state. But the measuring device could easily be one that is describable by a finite-dimensional vector space. Therefore the measurement process would be projecting an infinite-dimensional vector space down to a finite-dimensional subspace. A projection is not a unitary operator. But the Schrodinger equation is unitary, so that's not possible if the evolution of the whole system is governed by the Schrodinger equation.

  • $\begingroup$ You're assuming that the measuring device has no state beyond what it reports. But real measuring devices are extremely complicated systems that report very simple things. $\endgroup$
    – John Doty
    Oct 12, 2022 at 0:44

Have physicists yet ruled out if wavefunction collapse happens due to the potential from the measurement apparatus?

A wavefunction for a specific experimental setup is the solution of the quantum mechanical equation with the boundary conditions of the problem.

The double slit experiment is a simple example of this: "electron of given four vector scattering off two slits given width and distance apart"

The detector apparatus, lets consider it an electron sensitive screen for the discussion, is not within the initial wave function for the electron probability on its way after the slits.

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The original wave function describes electron tracks that would go to infinity, if not disrupted with a new scatter, the interaction of the electron with the screen , which deposits enough energy to show a dot.

This would need a completely different wavefunction solution because the boundary conditions of "electron + atom in screen" is a different scattering problem. It cannot be solved in one go ,"electron of given four vector scattering off two slits given width and distance apart + screen scatter"

So the potentials and location from the measuring apparatus is what "collapses" the original wavefunction, it does not define it and is irrelevant to the measurement, except for this example, the distances between peaks of the interference pattern being larger the further away the detector is set up.


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