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-1

My understanding is that anything that qualifies as an "observation" or "measurement" will cause the "fuzziness" of the superposition to disappear to the local observer, and the results of said experiment will appear in the classical form that we are intuitively used to. The mathematics which controls quantum mechanics describes observed "fuzziness" by ...


8

$\renewcommand{\ket}[1]{\left \lvert #1 \right \rangle}$ $\renewcommand{\bra}[1]{\left \langle #1 \right \rvert}$ 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 ...


0

I had the same thought a few years before you did. As I understand it - and I'm still thinking about this one - your question is excellent - when this happens out and about, all information about what the system was is lost within the system itself when it 'collapses' due to a measurement. There is nothing in the system itself which retains information about ...


0

Let's consider a qubit that has two "classical states" $\left|0\right>$ and $\left|1\right>$, e.g. a current in a flux qubit that flows in one direction or in the other direction, while superpositions of these states are "non-classical" and will decohere into a mixed state of the classical states. What I'm going to demonstrate now is that a ...


-2

This description is based on the philosophy of the qm. In de broil interpretation bohmian mechanics is deterministic. But dehorence is normal in systems that can have lower preferred configuration. It is not forbidden so it can be achieved.


0

In any quantum experiment, as soon as the state of the system can in principal be known (be it because of emitted photons available to the experimenter, an interaction with the environment we may be able to read off the state, or any other means by which the observer can determine the state) it decays and ends up in one of the basis states. That's precisely ...


5

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 ...


9

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^...


2

In the context of solid-state physics, a closely related question has been an area of active research in the past few years. Most interacting systems do indeed thermalize (and thus delocalize) over long time scales. However, certain systems whose disorder is much stronger than their interactions experience "Many-Body Localization," in which the individual ...


3

To solve your problem exactly, you would have to solve the Schrödinger equation $$i \frac{\partial}{\partial t} \Psi (\vec r_1 \dots \vec r_N,t)= H \ \Psi(\vec r_1 \dots \vec r_N,t)$$ where $\Psi (\vec r_1 \dots \vec r_N,t)$ is the wave function of the $N$ particles and $$H=\sum_i^N \frac{p_i^2}{2 m} + \sum_{i<j}^N u_{ij}+V_{\text{ext}}$$ where $u_{ij}...


0

I suppose you mean that the gas is contained in a magic box. Otherwise the walls become part of the system, exchanging momentum/energy with the 'particles'. I have no answer for you; I don't know. What I do know is that none of the particle-particle collisions can be characterized other than by using a probability distribution. Common sense demands that ...


2

Quantum effects appear if the concentration of particles satisfies, $$\frac{N}{V} \ge n_q$$ where $N$ is the number of particles, $V$ is the volume, and $n_q$ is the quantum concentration, for which the interparticle distance is equal to the thermal de Broglie wavelength, so that the wavefunctions of the particles are barely overlapping. As the quantum ...


3

It really depends on the boundary conditions. For boundary conditions like a 3D box with reflecting walls, the initial quantum state $\Psi$ will stay a quantum state with the unique wave function depending on variables of each particle: $$\Psi({\bf{r}}_1,...,{\bf{r}}_n, t).$$ If the boundary conditions are such that allow exchange with the environment, then ...


0

The behaviour of the molecules in your gedanken experiment can be approached by using decoherence. But I do not believe you can get a definitive answer until somebody makes a full scale simulation (or until some expert's answer can make a formal proof of what really happens, but I am not skilled to do that). The decoherence effects can be argued ...



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