Say there are two particles in the x axis. Initially, with high probability one staying at $x=-1$ and going to the right, and the other $x=1$ going to the left. Will they collapse when both arriving at $x=0$?

If one does not use the Born rule, one has only a combined $\psi(x_1,x_2,p_1,p_2,t):\mathbb{R}^4\times\mathbb{R} \rightarrow \mathbb{C}$ wave function, where $|\psi(x_1,x_2,p_1,p_2,t)|^2$ is the probability of particles being respectively at position $x_1$ and $x_2$ at time $t$ with momentum $p_1, p_2$. In this case, will the particles eventually collapse? If not, what does it mean to "measure"? Is "measuring" a primitive concept like "particle", not defined in terms of other concepts? So that it is implicitly understood what does it mean when physicists say "measuring".

Born rule says if a measurement is performed on the $x$ axis, then $\psi$ will become $\phi$, with $|\phi(x_1,x_2,p_1,p_2)|^2=1$ for some $x_1,x_2,p_1,p_2$. And this happens with probability $|\psi(x_1,x_2,p_1,p_2)|^2$.

Is it right to say that "if I am not there" no collapse occur?

So, if one considers the whole universe, who is making the universe collapse?

Is this question solved by defining another concept: a particle is either classical either quantum? And saying that "a collapse occurs when a classical system interacts with a quantum system". In this case, how "interaction" is defined? In the example, above, say the particle on the right is classical: its probability of being is some $x(t)$ with momentum $p(t)$ is 1 at all times. Will the two particles interact as soon the probability of the two of being near each other is non zero? $\exists \epsilon,x: \; |\psi(x-\epsilon,x+\epsilon)|^2 > 0$. And in this case the Born rule will roll the dice and take a definite position for the left particle?


3 Answers 3


"Measurement" is only fuzzily defined.

There are situations in which it's certainly safe to apply the Born rule and collapse the wave function. Those are situations in which the observable you care about has become so thoroughly entangled with the environment that there's no hope of undoing the entanglement precisely enough to do an interference experiment that could determine whether the wave function has collapsed or not. This is the case if, for instance, a Geiger counter has emitted a click, which is a sound wave that alters the positions of a vast number of atmospheric molecules. When people talk about a "classical system" in this context, they essentially mean a system so complicated that you'll never successfully disentangle your observable from it.

There are situations in which you definitely can't collapse the wave function. Those are situations that are similar to experiments we've done that found quantum interference that wouldn't have been there if the wave function had collapsed. Your example of two quantum particles interacting is one of those. If a mere interaction of two particles caused a collapse then we would never have formulated quantum mechanics because we would have no evidence for it.

In between those situations, there's a grey area. There have been attempts to narrow it by, for example, doing double-slit interference experiments with viruses, but I don't know the current status of those efforts. It's difficult because if you don't find interference, it's hard to know whether that's because a physical collapse happened or because you failed to erase all of the which-path information.

Incidentally, when you say that $ψ$ is a function of $x_1,x_2,p_1,p_2$, that's incorrect. It can be written as a function of $x_1$ and $x_2$ or as a function of $p_1$ and $p_2$ but not of all four.

  • $\begingroup$ Example: particle interaction=measurement. High energy particle A is fired towards a wall of particles $B_1..B_n$ (e.g Rutherford). A smashes into $B_k$ and produces a cascade of child particles. The interaction $A \rightarrow B_k$ represents a measurement even if N = 1. $\endgroup$ Jan 8, 2021 at 5:57
  • $\begingroup$ "When people talk about a "classical system" in this context" they usually refer to Bohr's belief that measurement is "essentially" classical as it measures classically defined quantities like position, momentum, ... $\endgroup$ Jan 8, 2021 at 5:59
  • $\begingroup$ I don't know what you mean by "grey area". The problem is that the waveform is itself undetectable, if it is real at all (doubtful). If double-slit diffraction of viruses failed, it would challenge QM itself and would not be simply dismissed as a simple waveform collapse issue. $\endgroup$ Jan 8, 2021 at 6:04
  • 1
    $\begingroup$ @shaunokane001 Safe to collapse = no chance an experiment will ever see other branches/worlds so you may as well prune them. Can't collapse = experiments show collapse doesn't happen. Grey area = experiments might be possible but haven't been done yet, so it's open. Yes, the experiment might "challenge QM itself", that's the point of doing it or any experiment. I expect a boring result (either it's impossible to do or they do find interference), as do you apparently, but occasionally people find parity violation or something. $\endgroup$
    – benrg
    Jan 8, 2021 at 7:45

I will concentrate on the first question in the title, "what is measurement". If after this there are other question please submit these.

A measurement involves entanglement of say an atom of which a radiative transition is measure and a detector. The detector part of the wave function has a very large phase space, because it is a many particle system and its states, say 0 and 1, are coupled to for example electrons and phonons. Let's say that the atom is initially in an excited state. After some time there is a substantial possibility to find the system in an arbitrary state. Since the detector has many more states than the atom, the probability that the atom ever returns to its initial state is negligibly small.

  • $\begingroup$ Very interesting concept. Do you have links to online resources expanding on this? Thank you. $\endgroup$ Dec 20, 2020 at 12:19
  • $\begingroup$ So, overall there is less probability of the atom in an excited state to remain as such compared to the probability of the atom emitting radiation and transferring its energy to others. $\endgroup$ Dec 20, 2020 at 12:31
  • $\begingroup$ Yes that is correct. I have not really a reference but I also hesitate to claim it as my own idea :-) $\endgroup$
    – my2cts
    Dec 20, 2020 at 14:26

"What is Measurement?" - You have asked one of the fundamental questions at the heart of the interpretation of quantum mechanics. Currently any answer is going to be subject to debate by adherents to different interpretations.

Personally as a (sort of) adherent to Copenhagen, I would say to treat measurement as a primitive concept (i.e. not necessary to explain in terms of anything else). Arguably Bohr effectively did this by claiming the existence of a quantum-classical boundary; we're happy about measurement on the classical side of the boundary. Please don't down-vote because you adhere to another interpretation.

I would say that the interaction of the (non-virtual) particle records information about the other particle and so is potentially a measurement. (Informational interpretation of QM). Copenhagen would require one particle to be considered as part of the measuring device for it to be a measurement!

Some other points:

  • Particles do not collapse, the waveform that describes the probability of finding them in a particular state collapses.
  • "If I am not there" does collapse occur? Again, a matter of interpretation. If you accept that the waveform is only a mathematical construct (Copenhagen) and not real then it is meaningless question. The answer can only be given by reference to a particular interpretation.
  • Copenhagen would not allow the Universe to be described by a single waveform as it requires an quantum-classical boundary and observer, although variants could put the observer in the future. Again the answer can only be given by reference to a particular interpretation.
  • ψ(x1,x2,p1,p2) is not the standard QM representation - a joint probability distribution like this would contain hidden constraints (uncertainty relations). Much better to represent the system as ψ(x1,x2) or ψ(p1,p2) or similar.
  • $\begingroup$ "Copenhagen would not allow the Universe to be described by a single waveform as it requires an quantum-classical boundary and observer". Nice point. $\endgroup$ Jan 15, 2021 at 22:25

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