I am a layman and was wondering, the quantum observer effect. The regular notion to laymen seems to be literally "if you look at it", but as I am coming to understand the world I live in better I feel it means just coming in contact with something.

Is this an ongoing question? Whether a particle that is not interacting with anything undergoes changes in state? Though now that confuses me too, If it is in two states at once. What denote say spin to the left from spin to the right, what denotes the origins of the calculations we make?


"Observation" means interaction with anything, which might be a sentient being or might simply be matter.

At very small scales (sub-atomic) particles can exist in several states at once. When they interact with anything else - even as small as another electron - they can experience "wavefunction collapse".

A "wavefunction" in quantum mechanics is the maths used to describe how something behaves (and that means everything, though there is no point trying to work out a wavefunction for say a bus, as it would be a) impossibly complex, and b) quantum effects are so unlikely and so minuscule at that scale as to be irrelevant, so physicists only talk about wavefunctions for tiny things, such as an electron).

The point of these wavefunctions is to describe the world in terms of probability - for instance, one cannot talk about an electron being in a particular place, but rather only of the probability of it being there.

(this is entirely different from the uncertainty principle. That states that certain complementary metrics cannot be known with complete accuracy at small scales. For instance, the more one knows about the momentum of an electron, the less one can know about its location, and vice versa. The effect is similar - you can't know what you want to know - but the underlying phenomena is different)

Wavefunctions can allow the superposition of several states - for instance a photon's wavefunction can allow it to be in two places at once, and even to interfere with itself.

However, if you try to to observe the photon being in two places at once, then the wavefunction "collapses", and the photon will appear in only a single place.

This has very real implications. In two slit defraction, a photon can go through two separate openings at once and then in effect "bump" into itself on the other side.

However, any attempt to try to watch it going through both slits at once will require something to interact with that photon, whether that be a human observer or simply the matter that makes up a piece of apparatus. That interaction causes the photon's wavefunction to collapse, and the quantum effect of the two slit defraction to cease.

(the uncertainty principle by contrast is the same whether or not the photon is observed)

It is interesting to apply this explanation to Schrodinger's Cat. The radioactive decay that is to trigger the release of cyanide and kill the cat can exist in two states only as long as it interacts with nothing else. There is no need for a human observer, and so no paradox - the wavefunction of the radioactive decay collapses when it interacts with the detector and the cat in the closed box, meaning that the cat is already either alive or dead when the box is opened.

  • $\begingroup$ This is not an area of my expertise, but I think you should clarify that "sentient being" in the first sentence means more specifically something like "matter that is part of a sentient being". The current wording has a 'quantum woo' vibe and can easily be interpreted as quantum theory involving sentience and perception shaping objective reality. $\endgroup$ – Praxeolitic Jun 12 '18 at 1:30

This "observation" is actually just a more simple way to state the uncertainty principle. For a one dimensional case involving momentum and position it would be as follows. $$\Delta x \Delta p \ge h$$ This means that if we know the position exactly then we don't know the momentum.
For energy and time it would be $$\Delta E \Delta t \ge \frac{\hbar}{2}$$

Basically what these mean is that there is a certain amount of maximum precision that we can make. If we knew the energy exactly ($\Delta E = 0$) then we know that $\Delta t$ is AT LEAST $\frac{\hbar}{2}$. This is the problem with quantum mechanics (and is misinterpreted by many people). Observing is actually an act of measuring. There is a wave particle duality that goes on in qm. Think of a particle existing along a wave. It is not too difficult to find the momentum of a wave, but by measuring this you know that it is a wave, and lose information about where it is. The converse is true too, that if you measure it's position you have an inherent uncertainty in the momentum. I suggest looking up the double slit experiment if you would like a visual example of this. Many videos of this exist.

  • $\begingroup$ I like his question is about what counts as a 'measurement' rather than what the effect of a measurement is. $\endgroup$ – Kevin Driscoll Apr 11 '13 at 2:40
  • $\begingroup$ @Kevin, as far as I am aware any form of measurement does this. If you gain information about one aspect you are limited on your information on the other aspect. You can measure subatomic particles using magnetic detectors, bubble chambers, electron microscopes, and many other crafty things. Basically it comes down to things being acted upon. $\endgroup$ – Steven Walton Apr 11 '13 at 9:06
  • $\begingroup$ I was just wondering what counts as measurement. It arose because of the recent experiments of quantum information teleportation. I've read they wonder if sending the entangled photons through the atmosphere will cause them to decohere. I think i attributed "wave-function collapse" to decoherence between entangled particles(they are different; so my next idea is invalid). And realized that function collapse happens when one observes it. So I was thinking, if I have to look up and and look at a incident photon from the one sent, will I have ruined the experiment? $\endgroup$ – KDecker Apr 11 '13 at 12:03
  • $\begingroup$ And I have seen the double slit alot. The question I had also, 0 kelvin is when a particle is not moving, if we know this doesn't this mean it CAN be moving cause we know nothing of its momentum? So it could have momentum? $\endgroup$ – KDecker Apr 11 '13 at 12:04
  • $\begingroup$ 2 things. Actually, you may be right that decoherence and wavefunction collapse are related. Most of the 'modern' proposals to deal with the measurement problem propose that decoherence is basically what 'causes' wavefunction collapse. Or more accurately, I think, they posit that the reason we originally thought of wavefunction collapse was because decoherence existed, but we didn't understand it yet. $\endgroup$ – Kevin Driscoll Apr 11 '13 at 18:18

You have touched on something of an unsolved problem in physics or philosophy of physics (depending on who you ask). It is not at all clear what counts as a 'measurement' of a quantum state. When quantum mechanics was first formulated, the authors started with the mathematical definition of a 'measurement', namely projecting the wavefunction of the system on to a certain eigenvector. Since then, there have been attempts to understand what (if any) physical acts correspond to this mathematical definition. We have come up with some cases where what we are doing, physically, clearly does fit the mathematical definition, but still have not found a rule for classifying all actions as either 'measurements' or 'not measurements'.

I would say that the basic problem is that quantum mechanics is, at its core, a mathematical formalism. Trying to understand that formalism through intuition has always been a challenge and this includes understanding it through the concepts of language (like 'measurement' and 'observer'). This remains an unsettled problem because from the physicists view it simply isn't important. Not knowing what counts as a measurement and what doesn't turns out to not matter much. We can do all the calculations we care to without making the distinction.

You can read more about this if you like; Wikipedia is a good place to start: the measurement problem. As a preview, there are some interpretations of quantum mechanics where the distinction is entirely meaningless. There is no 'quantum observer effect' (or wavefunction collapse, as it is known technically). The Stanford Encyclopedia of Philosophy also has a number of good articles on the measurement problem and interpretations of quantum mechanics.

I don't understand you last few questions. Maybe if you rephrase a bit I could give an answer.

  • $\begingroup$ The last questions are. If there is a quantum "spin", to which way is left spin? Any particle spinning toward alpha centauri? (I think spin might literally not be spinning though) Does that make sense? What is the origin (not beginnings of;but a place or point in space) of some of the quantum variable we try to measure? $\endgroup$ – KDecker Apr 11 '13 at 11:53
  • $\begingroup$ Ah, the origin is arbitrary! What you call spin left, I could call spin right. It is just a label in this case. Although, you're right that spin isn't really a particle spinning, it just has some properties that make it act like a spinning particle, so we call it spin. What matters is that we establish an origin and coordinate system. If we use two different origins and coordinate systems, then we can always move between the two, as long as we tell each other how we measured. Exactly which one we use isn't important. $\endgroup$ – Kevin Driscoll Apr 11 '13 at 18:23

I am going to try to ask this question from the point of view of the Copenhagen interpretation of quantum mechanics (pioneered by Niels Bohr). However, regardless of the overall interpretation, the mathematical description of measurements in quantum mechanics is a well-established theory.

The interpretation of the quantum state in the Copenhagen view, is that it represents the amount of information available about a physical system. What is peculiar about quantum mechanics is, that this information is limited in a specific sense. This implies, e.g., that it is impossible to assign values to what is called complementary observables (such as momentum and position as mentioned above).

In the Copenhagen interpretation, it is only allowed to ask question regarding the probability for observing outcomes of a specific choices of measurement. Since the quantum state reflects the amount of information available it naturally changes when a measurement is performed, since we now know more... This is similar to Bayes theorem from regular probability theory, where a gain in information is reflected in an updated, changed, probability distribution for our unknown information. Thus, in this view, the quantum state is not so much a property of the system, but a reflection of the observers knowledge about the system. If you have heard of the Bayesian approach to classical probability theory, this is quite similar.

In order to try to answer you original question "what is an observation", the question can usually be resolved by realizing that any measurement action always entails the interaction of your measurement apparatus (another quantum system) with the system you are measuring. The subsequent measurement on the apparatus then modifies the original quantum system (since we extract information indirectly). What is perhaps less obvious is, that if someone were to measure our system in the way, and not tell us the result, the quantum system state of our original system is still modified. Simply the possibility of someone having measured our system (in the sense that information is extracted from it to another system) requires us to take the measurement into account. This has something to do with the complementarity mentioned above, and is a distinguishing feature of quantum mechanics opposed to classical probability theory. Thus any kind of interaction with an external system can be viewed as a kind of observation of our own system - the modification of the quantum system state depends on whether or not this other system is measured.

So I might have answered you question by moving the measurement away from the original system to another system (the measurement apparatus). The point is, that to some extent, it is not too important whether or not the additional system (the apparatus) is measured - in either case a kind of observation has taken place. Exactly how much knowledge is gained by the observer can be different though. This is of course not particularly satisfactory, but beyond this explanation, I think (I might be wrong), we would be moving away from the standard quantum theory. What I have presented above is an operational way of viewing measurements and the effect it has on quantum systems as it is generally used today.

Hope this clarified things a bit.

  • $\begingroup$ Yeah, I know somewhat most of what you said but you said it more eloquently and gave me more knowledge. $\endgroup$ – KDecker Apr 11 '13 at 12:10

The "observer effect" does not really have much to do with observation in the every day sense. The effect in question has to do with the fact that some kinds of interaction between systems cause them to stop exhibiting quantum interference: this process is called decoherence. The relevant kind of interaction copies information so that it is present in two or more systems where it was initially present only in one system. If an interfering system undergoes an interaction that copies an unsharp observable in this sense, the interference is prevented. For more see http://arxiv.org/abs/1212.3245


protected by Qmechanic May 1 '14 at 14:47

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