let's define 'a measurement device' as a system which is highly sensitive to the eigenstate of an observable. The sensitivity is quantified let's say by how irreversible and grand the small changes in the eigenstate result in the large scale, classical system's future. A wavefunction collapses when it interacts with such a device.

This seems like a simple and necessary definition, right?

But there's a problem with this interpretation. How do we calculate the 'sensitivity' of a given device, without first knowing when the wavefunction collapses? For example, in the double slit experiment, let's say that the wavefunction collapses into a sharp peak when it passes the slits. Then surely a small change in this eigenfunction will result in large changes in what happens at the screen? If the wavefunction collapses at the slits, then we can draw a midway line at the screen, and effectively use the screen as a which way device. However, since it does not collapse, we know that the screen is therefore not sensitive to the eigenstate of the wavefunction at the slits. Therefore the wavefunction should not collapse at the slits rather than at the screen.

The question is, this is clearly a case of circular logic. How do we know a priori what device will collapse the wavefunction when it passes.

  • $\begingroup$ Can you focus on one clear question? $\endgroup$ Commented Sep 2, 2020 at 18:43
  • $\begingroup$ which part are you saying is circular logic? In the double-slit example, you observe an output that is not compatible with the photons collapsible at the slits, thus you know/observed that it didn't collapse there but rather at the detector at the end. What's circular about this? $\endgroup$
    – glS
    Commented Sep 4, 2020 at 9:35
  • $\begingroup$ i don't even think this question makes sense. just being silly $\endgroup$
    – user86425
    Commented Sep 13, 2020 at 23:09
  • $\begingroup$ I traied to raise the issue of measurement and quantum properties not necessarily going hand in hand in a recent question (which unfortunately was closed): physics.stackexchange.com/questions/631584/… $\endgroup$
    – Roger V.
    Commented Apr 27, 2021 at 8:46
  • $\begingroup$ sorry, i would like this question closed! lol $\endgroup$
    – user86425
    Commented Apr 27, 2021 at 12:06

2 Answers 2


I think what you're missing is that there is also a wavefunction collapse when a detector fails to detect something. If the wavefunction is a superposition of position states A, B, and C, and there's a detector at A and none at B or C, then there is always a collapse. It leaves the wavefunction either in whatever state the detector leaves it or else in a superposition of B and C. If the detector isn't perfect and may miss the particle, then there is still always a collapse, leaving the wavefunction either in the detected state or in a superposition of A, B, and C, where the amplitude of A is lower than before (and B and C are higher, so it's still normalized).

If there is more than one detector (say A and B are pixels on the screen, and C misses the screen entirely), you can still always think of the detectors as attempting detection one by one, succeeding with a probability given by the Born rule, and leaving the wavefunction in a changed state whether they succeed or not. You'll get the same answer regardless of the order in which you consider the measurements, at least if they're spacelike separated.

This is called interaction-free measurement. That Wikipedia article gives the impression that it's a rare thing happening only in certain experiments, but it actually happens all the time.

  • $\begingroup$ @lucky-guess I suppose I'm answering "How do we know a priori what device will collapse the wavefunction when it passes" and the various questions in the next paragraph. I also said something about 'poor' measurement devices. $\endgroup$
    – benrg
    Commented Sep 2, 2020 at 18:26
  • $\begingroup$ @berng ' It leaves the wavefunction either in whatever state the detector leaves it or else in a superposition of B and C'. If the wavefunction collapses then why is it still in a superposition? $\endgroup$
    – user86425
    Commented Sep 2, 2020 at 18:40
  • $\begingroup$ @lucky-guess The state after collapse reflects what information about the system is in the environment. If the detector doesn't trigger then you know the particle isn't at A, but you don't know whether it's at B or C. $\endgroup$
    – benrg
    Commented Sep 2, 2020 at 19:30
  • $\begingroup$ @lucky-guess B+C isn't an eigenstate of the position operator, but it is an eigenstate of the is-it-at-A operator, which is what the detector at A actually measures. $\endgroup$
    – benrg
    Commented Sep 2, 2020 at 19:42

One need not to calibrate the sensitivity of the device in the experiment itself - it can be done beforehand using the appropriate eigenstates.

E.g., we can calibrate the device measuring position using particles in position eigenstates, the device measuring momentum using particles in momentum eigenstates, and then use the two devices to study something prepared in an energy eigenstate of a harmonic oscillator, where both position and momentum are uncertain.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.