33

It's a very good question, since indeed if the original Stern-Gerlach machine had a well-defined momentum, then you are right that there could be no coherence upon rejoining the beams! The rule of thumb for decoherence: a superposition is destroyed/decohered when information has leaked out. In this setting that would mean that if by measuring, say, the ...


26

If we do an interference experiment with a (charged) particle coupled to the electromagnetic field or a massive particle coupled to the gravitational field, we can see interference if no information gets stored in the environment about which path the particle followed (or at least, if the states of the environment corresponding to the two paths through the ...


25

Until we have an accepted solution of the Measurement Problem there is no definitive definition of quantum measurement, since we don't know exactly what happens at measurement. In the meanwhile, measurement is simply defined as part of the postulates and recipe associated with the notion of a quantum observable. Mostly an observable is thought of as an ...


24

Nobody is answering this question, so I'll take a stab at it. Consider the mirror. Suppose you started your experiment by (somehow) putting it in a nearly-exact momentum state, meaning there is a large uncertainty in its position. Now, when you send a photon at it, the photon either bounces off or passes through. If the photon bounces off the mirror, it ...


18

Unfortunately, physicists and philosophers disagree on what exactly the preferred basis problem is, and what would constitute a solution. Wojciech Zurek was my PhD advisor, and even he and I don't agree. I wish I could give you an objective answer, but the best I can do is state the problem as I see it. In my opinion, the most general version of the ...


17

Assuming wave-function collapse is correct (which can be a relatively hefty philosophical claim in some circles), then think of measurement this way: When you measure an observable on a system, you collapse the wave-function of the system into a Dirac delta function in the eigenbasis for that observable. If you measure position, you get a delta function in ...


17

I think most arguments in the literature can be boiled down to the point that decoherence does in no way touch the linearity of the Schrödinger equation, and thus cannot make an "or" from an "and". This is complicated in the literature by very technical discussions, which I would like to avoid. Let me explain the basic point in more details. A widely ...


17

What you describe is the process known as decoherence: any interaction of a quantum system with its environment (e.g. with photons or other particles passing by, and, yes, most likely interacting through gravity, although we don't have a theory to fully describe this yet) has the potential to destroy its genuinely quantum nature, turning quantum ...


14

$\renewcommand{\ket}[1]{\left \lvert #1 \right \rangle}$ $\renewcommand{\bra}[1]{\left \langle #1 \right \rvert}$ We can see how decoherence really works, why it messes up superposition states, and why it's particularly prone to messing up states of large objects all through a very simple example $^{[a]}$. Single two-level system Suppose we have a quantum ...


13

Correlation is first and foremost a term from statistics. Given a system that consists of two (or more parts), it quantifies how much I can predict about the second system if I have knowledge of the first in comparison to how much I can predict about the second system without that knowledge. For instance, if I have a bag of pieces of paper printed with ...


13

The many-worlds interpretation defines measurement as any physical procedure in which the observer gets entangled with a quantum system. Before the measurement, the universe containing the observer and the quantum system is in a direct product state, so the observer knows nothing about the quantum system. After the measurement, the two subsystems of the ...


11

Aren't the particles this quantum state consists of interacting with each other? Why doesn't that cause the state to collapse? We have a mathematical model for the observations we can make of any system in the micro world. This model is quantum mechanics and its predictions have been verified experimentally over and over again. Observables are quantities ...


11

As it turns out, the excited states of an atom are not really, strictly speaking, eigenstates. That is, they're eigenstates of the atomic hamiltonian, but they are not eigenstates of the atom-plus-EM-field hamiltonian. How do we know? Well, if you prepare such a state (excited atom, empty field) and then you leave it alone, then it will change: the atom will ...


10

Preliminaries: How do we define 'localized?' For a single particle, or for multiple non-entangled particles, it is easy to tell from the expressions for the wavefunctions whether they are localized or delocalized. For example, you might say that if the wavefunction is falling off exponentially or faster for large $x$, that is with a form like $\psi(x)\sim e^...


9

The collapse of the wavefunction is generally attributed to decoherence. This is time asymmetric in the same way the second law of thermodynamics is time asymmetric. I suppose it's theoretically possible for a wavefunction to uncollapse, but this is like saying it's theoretically possible for a broken egg to reassemble itself.


9

I agree in full with Marty Green except the explanations of chemistry in which I was unable to follow so well (that doesn't say that I disagree with them). But, let me put the things in short. The collapse is a phenomenon that is supposed to occur when a quantum object comes in contact with a quantum system. For instance, a quantum particle falls on a beam-...


9

Welcome to SE -- good question! Decoherence does not mean that there won't be a wavefunction anymore, it just means that if the electron becomes coupled to the surrounding environment, its state will be described by a probabilistic mixture of orbital wavefunctions rather than a (coherent) superposition thereof. The electron in an atom doesn't have some "non-...


8

I've never seen a single prediction based upon MWI. I've also never heard of the Cophenhagen interpretation called an approximation. If that were the case, then the Copenhagen interpretation must fail in at least one limit. Does Max provide such limits? Both of these statements seem to lean towards sensationalism than towards mathematical rigor.


8

It is because of quantum statistical irreversibility, which is closely related to entropy, as the OP suspected. Qualitatively it is quite easy to understand this. From the laws of quantum mechanics on the microscopic level emerges a classical behaviour for macroscopic (i.e. many particle objects). Of course this is not sufficient though and does not give a ...


8

When a particle passes through an ideal pair of slits, it's wave function can be written in the form $$\psi(x) = \alpha\, \psi_1(x) + \beta \,\psi_2(x)$$ where $\psi_{1,2}(x)$ are the wave-functions you would get by closing either slit 2 or slit 1, respectively. We say that the wavefunction is in a coherent superposition. What is coherent about this ...


7

Hopefully a sharper restatement of the question is: what's the difference between a mirror and a photocathode? Experimentally, in a Mach-Zender interferometer we can fold light paths with a mirror while maintaining coherent interference, but passing either beam into the photocathode of a photodetector destroys interference effects, even for photons that ...


7

Yes, everything is a detector, but you need to quantify which things your system interacts with (and how strongly). Gravity is in some sense a poor example, because the quantum details of gravity are still an unsettled question (and gravity is a weak force regardless), so let's bypass that red-herring by replacing gravity with the electromagnetic field: As ...


7

If you want to use nonrelativistic quantum mechanics you have to first start with the basics. Firstly it doesn't handle particle creation or destruction, so you need to fix how many particles you have of each type. Then you want a function from the configuration space $\mathbb R^{3n}$ into the joint spin state $\mathbb C^{k_1}\otimes\mathbb C^{k_2}\otimes .....


6

I'm probably straying into dangerour territory here, but let me venture an answer. Doing so is probably just asking to be shot down by John Preskill, or some other such expert, but let me stick my neck out. Despite Ron's comments, gravity and EM are different in this context, in the sense that you can't flip the sign of the gravitational interaction the way ...


6

A quantum system (for example a particle) is normally described by a wavefunction, which is a vector $|\psi\rangle$ from its Hilbert space $H$. But this description is not complete when the system is entangled with other system. In this case, it is more appropriate to use the density matrix (or density operator). For $|\psi\rangle$, the density matrix is $\...


6

You ask a lot, because these questions are very deep and require a very good understanding of the mathematical concepts and their physical meaning. That said, decoherence is an important aspect of the quantum to classical transition, because it shows how the environment can destroy the precisely defined phase relationship between different components of a ...


6

Narrowness is precisely the essence of the preferred basis problem. Consider: some states are narrow, some are not. Given that some are narrow and some are not, why should 'narrowness' come about as a meaningful concept at all? Why should this quality be an interesting one? Consider the position of a pointer. We don't interpret non-narrowly pointed states ...


6

I am going to answer this in a hurry because the question is on the edge of being closed. Quantum mechanics isn't just about "wavefunctions", it is also about "observables". An observable is something like: energy, position, momentum... i.e. it includes all the properties of classical physics. The wavefunction (or state vector or quantum state) is the ...


6

$\def\ii{{\rm i}} \def\dd{{\rm d}} \def\ee{{\rm e}} $ It turns out that the case of pure dephasing is exactly solvable, and one can obtain nice solutions under certain conditions. In particular, I will consider the case of Gaussian, stationary noise. Exact solution Let us define the noisy qubit Hamiltonian $(\hbar = 1)$ $$ \hat{H}(t) = \frac{1}{2}\left[\...


6

I am not aware of any experimental evidence, so this probably does not qualify as an answer. However I can offer a reference that addresses this question theoretically: Armen E. Allahverdyan, Roger Balian, Theo M. Nieuwenhuizen (2011) Understanding quantum measurement from the solution of dynamical models, https://arxiv.org/abs/1107.2138 and by the same ...


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