# Why is quantum measurement not happening all the time? [duplicate]

I have a question which may be very naive yet I have no answer. I studied undergraduate quantum mechanics 4 years ago now and even if I studied more advanced stuff like QFT I feel like I don't understand the basics yet, so feel free to answer me with "take a QM book and study it" and close the question.

My problem is about quantum measurements and superposition of states. I really don't understand why measurement is not happening all the time. I feel like I may be confusing measurement with interaction, but, for example if we have a magnetic field and a an electron passing through it, depending on how it is deflected it means the interaction with the magnetic field fixed a component of the spin: the electron now has spin up, it's not anymore in a superposition of spin up and down.

Now, these kind of interactions happen all the time, every particle interacts with its environment constantly so I feel like the successions of all these interactions constantly fixes the quantum states of all the particles in the universe.

Why is this wrong? What am I missing and what are the flaws of this view?

I'm not sure this is the final version of my question and edits may arrive as soon as some comments/answers clarifies for me what I am actually not understanding.

• Congratulations, you just rediscovered quantum decoherence – user1379857 Jun 8 '20 at 21:13
• @user1379857 I've often heard the name and thought it would be one of the interesting to study when I'll finally study QM seriously. So you think reading some chapters of some books about quantum decoherence may help clarify my misunderstanding? – AnOrAn Jun 8 '20 at 21:15
• I recommend reading the first few chapters of the book "Decoherence and the Quantum-To-Classical Transition" by Maximilian Schlosshauer – user1379857 Jun 8 '20 at 21:16
• @user1379857 quantum decoherence may be happening with no measurements going on, decoherence only destroy interference, leaving you with a "classical" probability distribution, but it doesn't make states acquire/collapse into definitive eigenstates – lurscher Jun 9 '20 at 10:23

Your intuition is correct: from the point of view of physics, there is no distinction between interaction and measurement. There is some argument about this, but as long as all interactions are accounted for, all the way to an observer's awareness of a measurement outcome, so that the observer is included as part of the system, then there is no distinction.

But that does not mean that all states are fixed. Let's say there are two particles, one of which is in a superposition of UP/DOWN states, and the other of which is in a fixed UP state. They interact in such a way that the second particle stays UP when the first particle is UP, and flips DOWN if the first particle is DOWN. The result is that the second particle, after the interaction, is in a superposition of states. In a sense, the superposed state of the first particle is transferred to the second particle.

Now let's complicate the scenario. Let the first particle instead interact with an instrument that measures its state. If the first particle is UP, the instrument panel displays "UP"; and if the first particle is DOWN, the instrument panel displays "DOWN". Now, after the measurement (remember, measurement = interaction), the instrument is in a superposition of states because the particle was in a superposition of states. BUT: the instrument has two "perspectives" corresponding to its two states: in the "UP" state, it "knows" that the particle was UP, and in the "DOWN" state it "knows" the particle was DOWN. The instrument never sees the particle as being in both states.

However, the instrument itself is in a mixed state. Now along comes an observer (who is just a really complicated instrument from the point of view of physics) who looks at the instrument and sees the instrument displaying either "UP" or "DOWN". The observer is put into a superposed state by looking at (interacting with) the instrument. The observer state that sees "UP" can only see "UP"; and the observer state that sees "DOWN" can only see "DOWN". The observer can never see the instrument displaying a mix of both.

This would all seem very abstract and unnecessary, except for the fact that it is supported by experiment. Young's double slit experiment, and other related experiments, demonstrate very solidly that a particle actually exists in a superposition of states until it is detected.

It's very difficult to design an experiment to prove that anything much larger than a small molecule can exist in a superposition of states, but it has been done. Proving that Schroedinger's cat - or a human observer - is in a mixed state may well be beyond our reach; but there is plenty of theoretical basis to assume that each quantum measurement puts the lab tech in a superposition of states.

• But doesn't the text of my question contradicts the existence of superposition of states? From what I've written I think I can conclude that no particle has ever been, or will ever be in a superposition of states. – AnOrAn Jun 8 '20 at 21:26
• I see what confuses you. I'll edit my answer to address it. – S. McGrew Jun 8 '20 at 21:30
• @AnOrAn Measuring a state doesn't mean that it no longer remains in a superposition. If you measure the spin to be up in $z$ direction, it is a superposition of eigenstates of the spin operator in $x$ direction. Now, if you measure it in $x$ direction, it becomes a superposition of eigenstates of the spin operator in $z$ direction. You will have to reduce the dimensionality of the Hilbert space to one to make sure your state is not in a non-trivial superposition of states. ;) – Dvij D.C. Jun 8 '20 at 21:48
• -1: a. This answer confuses the concept of superposition of states with the concept of a mixed state, something that is super important to not confuse when talking about decoherence/entanglement. b. "from the point of view of physics, there is no distinction between interaction and measurement.", this is not true. Not all interactions destroy coherence. You require a macroscopic environment for decoherence, moreover, an interaction with even a macroscopic environment also involves subtleties around its relationship with measurement. See physics.stackexchange.com/q/282410/20427. – Dvij D.C. Jun 9 '20 at 0:07
• c. This answer completely ignores the issue of preferred basis for decoherence. In particular, what is not a superposition in one basis is very much a superposition in another basis. And any state can always be viewed as a single eigenstate (i.e. not a superposition) of some observable. This is another reason why simply saying "interactions = measurements" is misleading. Decoherence picks out a basis in which to decohere based on as to in what basis the interactions are local. – Dvij D.C. Jun 9 '20 at 0:12

Measurement is happening all the time only in the macroscopic world. The fundamental difference between a quantum particle and a macroscopic body is that a quantum particle is not interacting with its environment, or if it is interacting then it is not interacting in such a way as to generate the measured property you are interested in.

The conceptual difficulty in quantum mechanics is that in the absence of measured properties, particles do not behave in a way which we find intuitive. Indeed, one must go into the mathematical foundations, not just study the results, if one wants to understand it. I have sought to clarify this in a published paper The Hilbert space of conditional clauses.

I thought of an interpretation about this question and would like to know if it is meaningful:

Saying that there is no superposition of states is only true for a given orientation of coordinates. And the notion of superposition has only meaning for a given frame.

If a particle $$A$$ has spin up in a given direction it has a superposition of up and down for an axis not at that direction. It is not possible to choose a frame where all particles have no superposition of states.

It is simmilar to an elastic field inside a body in equilibrium. It is always possible for each point, to find an axis orientation such that there is no shear stresses in the stress tensor.

But if I choose that orientation, it is perfectly possible to have shear stresses at points in the neighborhood.