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Does quantum mechanics really predict that a particle prepared in a state of superposition of spin will result, after being measured by an appropriate instrument (Stern-Gerlach device), in a macroscopic superposition of the instrument? Is there some theorem that proves that when an object in an unknown quantum state (the quantum state of the instrument is not in fact calculated) interacts with a particle in a superposition state this object will be found in a superposition state itself? And not any superposition, but a superposition of its classical states, like pointer going to the right - for spin-up or to the left - for spin-down?

Such an amplification of superpositions from the quantum level to macroscopic level seems to be at odds with observation. The actual state of a macroscopic object is in principle available everywhere due to the gravitational field associated with that object. There is no way to block the gravitational field of an object. So, either QM makes a wrong prediction (very unlikely) or there is some unjustified assumption behind the preparation of macroscopic superpositions.

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    $\begingroup$ Related question here. $\endgroup$ – knzhou Jul 16 at 16:48
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The problem you are describing is the complicated problem of the emergence of classical reality from quantum theory. The most accepted line of reasoning for this porblem today is based on decoherence developed by W. Zurek (a beginner friendly review of this work can be found here)

The short version is that the quantum system and the measurement apparatus get entangled as a result of the measurement and are found in a state

$$\left| \psi \right>_{SA}=\sum_{x} \left|x \right>_S\left|p_x \right>_A$$

where $x$ labels the states of the system and $p_x$ labels the corresponding states of the pointer of the measurement apparatus.

Since the apparatus is a macroscopic system it interacts very strongly and uncontrollably with its environment and gets entangled with it, resulting in the state

$$\left| \psi \right>_{SAE}=\sum_{x} \left|x \right>_S\left|p_x \right>_A\left|e_x \right>_E$$

Since the environment is not accessible (it is the EM fields and the thermal baths all around), we trace it out from the state resulting in the mixed state

$$\rho_{SA}=\sum_{x} \left|x \right> \left|p_x \right> \left<x\right|\left<p_x\right|$$

The superpositions of the apparatus become a statistical distribution.

How do we end up seeing only one possible outcome from this? It's an open problem that is commonly known as the quantum measurement problem. Nevertheless it still shows that superpositions of macroscopic object are unobservable because they "leak" into the environment very rapidly and become unobservable.

There are many many nuances here that I did not touch on but there is only so much you can say in one post.

Adding the gravitational effects to this opens a whole new can of worms that I saw a few papers on but it is still highly debatable and poorly understood.

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  • $\begingroup$ oleg, thanks for your detailed answer! I understand to some extent the measurement problem and the attempted solution be decoherence. But my first issue starts with the assumptions that 1. px (pointer states) are indeed proper quantum states spanning a corresponding Hilbert space and 2. we get an entanglement between the particle and the pointer. To my limited knowledge, entanglement is a consequence of conservation principles + uncertainty about the subsystems. -to continue $\endgroup$ – Andrei Jul 17 at 6:12
  • $\begingroup$ cont: If we split a diatomic molecule with 0 angular momentum we get entangled atoms. We know that on any axis the sum of their spins must be 0. By this logic we can also say that when a particle interacts with an instrument the total angular momentum (particle + instrument) must stay the same, so the particle is entangled with the instrument. But clearly, the angular momentum of the instrument and the position of the pointer are different things. So how can you arrive at the conclusion that you get entanglement between particle states and pointer states? $\endgroup$ – Andrei Jul 17 at 6:19
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    $\begingroup$ Entanglement is a consequence of interaction. In general we describe dynamics with Hamiltonians which specify both how the individual systems evolve (self part of the Hamiltonian) and how their mutual states evolve together (interaction part). A measurement device is described by a Hamiltonian with tunable interaction part (usually some kind of EM field) that the experimenter turns on briefly (pulse) to generate time evolution that correlates (entangles) some degree of freedom of the apparatus (like polarization of atoms) with the measured system. You should look up concrete examples. $\endgroup$ – oleg Jul 17 at 16:14
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    $\begingroup$ You are correct to suspect pointer states as some arbitrarily selected degrees of freedom that happen to do the job. The resolution of that has to do with the nature of interaction of the macroscopic system with its environment. The pointer basis are not chosen by us, they are selected by which degrees of freedom interact with the environment and how. This information is again encoded in the Hamiltonian. Since we cannot control this Hamiltonian completely we cannot measure whatever we want. It is not easy to see how it works without carefully following the math (look up einselection). $\endgroup$ – oleg Jul 17 at 16:24
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You seem to be trying to play decoherence theory against the measurement problem, but the two are actually orthogonal. QM does not make wrong predictions, but we need to be precise about what a "prediction of QM" actually is:

The standard formalism of quantum mechanics gives a prescription for how to calculate the probabilities of the results of measurements of a quantum state (the Born rule), and it gives a prescription for how to calculate the evolution of quantum systems undisturbed by measurement (the Schrödinger equation). It does not, as such, explain what a measurement is or how it works. If we want to be precise, quantum mechanics without an interpretation is not a complete model of "how the world works" (because the Born rule and the Schrödinger equation contradict each other - the evolution induced by the first is not unitary, while the evolution induced by the latter is), but merely a tool that correctly predicts the results of measurements in a laboratory (or anywhere else) by chaining these to different evolutions together in the appropriate manner for the measurement being performed.

You can model the interaction between the measurement apparatus and the measured system as a quantum interaction itself, resulting in a superposition of states of the form $\sum_\lambda c_\lambda \lvert \lambda_s\rangle \otimes \lvert \lambda_a \rangle$ for the eigenvalues $\lambda$ of the quantity being measured, where the $s$ subscript denotes an eigenstate of the system for that quantity and the $a$ subscript denotes the corresponding "pointer state" of the measurement apparatus. This line of thinking began with things like the von Neumann measurement scheme and finally resulted in decoherence theory, explaining how measurement apparatus and measured system end up in this rather peculiar form of superposition. Decoherence does not explain away the measurement problem: It has no explanation for why we see only a single pointer state of the macroscopic measurement apparatus, nor does it claim to. It "merely" explains how the apparatus' state becomes entangled with the measured system in such a way that the pointer state always matches the actual value of the measured quantity for the corresponding state of the measured system.

But whether you model the interaction between the apparatus and the system thusly is irrelevant for the prediction of the probabilities of the results of the measurement (unless your measurement apparatus is imperfect and the $c_\lambda$ are not equal to the coefficients $c'_\lambda$ in the original state $\sum_\lambda c'_\lambda\lvert \lambda_s\rangle$ of the measured system alone) - the Born rule still holds. Adding gravity into the mix is likewise irrelevant - whether we "measure" the extended system of measured system + apparatus by looking at the apparatus it or by somehow deducing the position of the apparatus' pointer from its gravitational field doesn't change anything.

There is no universally accepted answer to the measurement problem. The lack of such an answer is one of the main reasons for the multitude of quantum interpretations, with each interpretation usually offering a different explanation of why we only see a single pointer state.

It is important to note that "the quantum state" of an object is inaccessible to us. All we can observe are the individual pointer states of quasi-classical measurement apparati, we cannot observe an actual quantum state. So whether a system is "in a superposition or not" is not, strictly speaking, a prediction of quantum mechanics. The predictions of quantum mechanics are the probabilities for results of measurements, and nothing else.

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  • $\begingroup$ Thank you for your answer! As your point is very similar to what oleg said, please see my questions to him! My very first problem is not what happens with the entangled state between the particle and the instrument (where decoherence is involved) but with the assumption that such an entanglement is in fact implied by QM. $\endgroup$ – Andrei Jul 17 at 6:24
  • $\begingroup$ @Andrei It's not an assumption. If you treat the apparatus as a quantum system, then decoherence theory shows how this entangled state comes to be. $\endgroup$ – ACuriousMind Jul 17 at 6:30
  • $\begingroup$ You might be right, but I fail to connect this assumption with my experience of QM. I am a chemist, so QM was used to calculate things like molecular orbitals, atomic spectra and so on. In order to do that you start with the Hamiltonian of the atom or molecule, solve Schrodinger's equation and determine the eigenstates. This is the procedure I understand. Sure, an instrument is a quantum system, but why should I accept that some more or less arbitrary classical states (pointer positions) should be treated as quantum states? to cont- $\endgroup$ – Andrei Jul 17 at 7:02
  • $\begingroup$ cont: Are you sure that by solving Schrodinger's equation for the instrument (in principle possible, if not so in practice) you will get those pointer states as solutions with the right probabilities? $\endgroup$ – Andrei Jul 17 at 7:05

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