# Tag Info

## New answers tagged measurement-problem

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One can indeed see the reason behind the Heisenberg uncertainty principle as a mathematical propeties. The link between the physics and the mathematics is provided by the foundational work of Planck and de Broglie. They established the link between energy/momentum and frequency/k-vector. The general textbook on quantum mechanics therefore always starts by ...

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The state is still $\psi (t=T)$ right after the rapid change of B field to y-direction because the system doesn't have enough time to response the change.

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It is not required to put the state in density matrix form prior to performing the unitary operation $U$. However, you want to learn properties about the reduced density matrix after tracing out some of the system post-unitary, so you need a density matrix eventually. So you could do the unitary, and then construct the density matrix: $$\newcommand{\ket}[1]{... 0 This is just a quick stab, and it might show my ignorance more than anything else. Since you are working with a two, level spin system i'm actually giessting m,n=\pm\frac{1}{2} . You can then explicitly write your density matrix as$$ \rho\left(t\right)=\begin{pmatrix}\rho_{\frac{1}{2},\frac{1}{2}} & \rho_{-\frac{1}{2},\frac{1}{2}}e^{-i\omega t}e^{-\...

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Among other things, you wrote: ...and according to the result from the first measurement (either $|+ \rangle_s$ or $|- \rangle_s$) the apparatus chooses either path $P_{ s^{+} \rightarrow f^{+} }$ or path $P_{ s^{-} \rightarrow f^{+} }$. Because the chosen path depends on the result of a measurement, your "transformation" isn't one transformation but ...

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The transition is due to the interaction of your small system that you want to experiment on with the environment (including the measurement apparatus). If the interaction is brief and the environment part of the system macroscopic, then what you see is the transition from the initial total wave function $\psi(t_0) = \psi_{\mathrm{in}} = \psi_{\mathrm{in,sys}... 0 Just wanted to add that the Simulation Hypothesis [1] [2] (SH) may suggest an answer. SH implies that there is a non-zero chance that we are in fact in a computer simulation. If this simulation is anything like the simulations we currently create, then it implies that the rules defined in the simulation code would appear to us as fundamental, irreducible ... 3 You're wrong in step 2. We don't go to$4\epsilon_0|\phi_2\rangle$. We go to$|\phi_2\rangle\$. Measuring an observable projects you into an eigenstate of that observable, but it does NOT mulitply your state by the eigenvalue of that observable. In most cases, as @WillO said, multiplying by the eigenvalue isn't even a measurable operation, since you're ...

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By "Copenhagen interpretation", I assume that you mean the interpretation with instantaneous "collapse" one usually encounters in an introductory quantum theory course. Such collapse is a useful rule to do calculation but it is only a fiction. What typically happens is that the quantum system is correlated with the macroscopic measurement device and other ...

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