I've been reading about different ideas of getting the observer out of quantum theory. One Sean Carroll seems to like is Everett's many worlds concept.

I had a thought that I need someone here to shoot down. It could be something as basic as the double slit experiment that proves me to be wrong. Here's my thought:

As I see it, we never really "measure" the position of something. We sense a photon that is emitted by what we are "looking" at. Why couldn't the wave function of a particle just break down at the moment it emits the photon at a specific point in space-time regardless of whether there is an observer or not. After the emission, the particle location would go back to being a probability function.

What am I missing?

  • 1
    $\begingroup$ FWIW, quantum mechanics itself doesn't posit that a quantum measurement requires a conscious observer, but some interpretations of QM do. $\endgroup$
    – PM 2Ring
    Jun 24, 2018 at 5:04

2 Answers 2


Quantum mechanics exists as a theory because there have been experimental observations that could not be modeled mathematically using Newtonian physics and Maxwell's equations.

It has a number of postulates intrinsic in every quantum mechanical theory,(theory= a general mathematical model which is descriptive and predictive) and in particle physics the symmetries and elementary particles in the table are also a priori assumptions equivalent to axioms. They are necessary to fit existing data and very successful in predicting future measurements.

In classical physics a measurement is an observation. It does not need an observer, most often is an instrument and data registered, now digitally in computers. Then the physicist thinking about those data is a live observer, but he/she has nothing to do with the origins of the data, except in setting up the original experiment.

This disposes of the "observer" as a conscious entity.

Quantum mechanics is a probabilistic theory. It can only predict probabilities of interactions . One gets scattering events accumulated so as to check a theoretically predicted distribution so as to validate or falsify the theory:

Here is one pi mu e event in the bubble chamber.


The beam , probably a K- 10GeV/c beam , hits a proton in the hydrogen bubble chamber , and several particles come out. One of them is identified as a pi mu e decay (by the ionization track the mass of the backwards curling track can be identified as a pi , etc).

This is one event, and probably thousands of people like us have observed it. In the film tapes where it is recorded there are probably tens of thousand like this.

From this one instance we may start building up a probability distribution for the decays of charged pions and muons. One instance. We would have to accumulate versus energy and momentum a statistically significant number of these decays to be able to check against energy and decay lifetimes as calculated/predicted quantum mechanically.

One is forced to the conclusion that , quantum mechanically, observer= recorded-interaction. Once an interaction is recorded it has been observed. The mathematics leading to the theoretical fit should not be anthropomorphized. One can visualize with Feynman diagrams the interactions and how their use will build up the probability distribution, but in the end the Lydian stone are the measurements, one by one building the probability distributions which fit the quantum mechanical predictions.

  • $\begingroup$ thank you for your eloquent answer once again. what I am wondering is whether the "break down of the wave function" for a single measurement is at the point of "observation" or at the point of emission of the photon by the particle (or does it make no sense to ask this question)? $\endgroup$ Jun 26, 2018 at 0:04
  • $\begingroup$ I think my last bold should clear that, at the the point of interaction, which is what you call 'emission", possible to become an observation $\endgroup$
    – anna v
    Jun 26, 2018 at 3:31

I think it is impossible to find out why an observer is needed. Because it is nearly impossible to have a strict definition of "observer".

Lets say that you are in the box with Schrodinger's cat.Although you may see the cat as being alive,the person outside still cannot be sure whether you and the cat are alive or dead.ie."looking" constitutes the presence of an observer.However,this rule seems to only apply to humans,for a dog cannot understand the outcome of the Double slit experiment.It should be tried though.

In the end though,the inability for a concrete definition of observer serves as the biggest problem to Quantum Physics and in truth,your question cannot be answered yet.


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