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5

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 ...


4

Let me rewrite $A=|H\rangle_1$, $B=|V\rangle_1$, $X=|H\rangle_2$, $Y=|V\rangle_2$ for typographical clarity. Then $A$ is (by assumption) orthogonal to $B$, so $A\otimes X$ is orthogonal to $B\otimes X$, and $A\otimes Y$ is orthogonal to $B\otimes Y$. Also $X$ is (by assumption) orthogonal to $Y$, so $A\otimes X$ is orthogonal to $A\otimes Y$, and $B\otimes ...


3

I think that 't Hooft's ideas about superdeterminism and Bell's theorem are relevant to this topic. If the universe is superdeterministic so that all experiments are determined by initial conditions, then the contra-factual arguments that lead to the Bell non-locality conclusions are ruled out. The universe only plays once - is how some have put this. In ...


3

An ensemble of interacting particles will, over time, develop entanglement between widely separated parts*, so this is similar to asking whether an interacting system can still be a BEC. The short answer is yes, but a subtlety is that various authors define BEC in slightly different ways. One way of defining BEC, as I mention in a recent answer, is the ...


3

Even if Alice and Bob are both first to measure their spin (according to their respective reference frames), two spins entangled into the singlet state will still give opposing results. That's what quantum mechanics predicts. Finding out that the entangled spins gave agreeing results would falsify a prediction of quantum mechanics. People would be very ...


2

This is one of the most misunderstood things about entanglement, which is that it doesn't matter who goes first. Neither measurement actually affects the other one, contrary to the intuitive implications of "wave function collapse". Entanglement is correlation, not causation.


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Your question has nothing to do with entanglement. You might as well ask this instead: Physics predicts that two positive charges will repel each other. Suppose I bring two positive charges into close proximity and find that they attract each other instead. How can this contradiction be resolved? Or you could posit any other experimental result that ...


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Can someone tell me the practical difference of "world splitting" in MWI, and the original "wavefunction collapse"? Even if there is no such thing as an "abrupt" split, I don't see why you couldn't also argue the same for the "collapse". There is no world splitting. A state that starts out as having a single value for some relevant observable gradually ...


2

Let $\lambda_i$ be the eigenvalues of $\rho_A$. Then $$ \log \text{tr} \rho_A^n = \log \left( \sum_i \lambda_i^n \right) $$ Now, let is differentiate w.r.t. $n$. We find $$ - \frac{\partial }{ \partial n} \log \text{tr} \rho_A^n \bigg|_{n=1} = - \frac{\sum_i \lambda_i^n \log \lambda_i }{\sum_i \lambda_i^n } \bigg|_{n=1} = - \frac{ \sum_i \lambda_i \log ...


2

This can be explained with a classical description of particles; quantum physics is overkill. The explanation stems from the physical constraints of the system, and therefore the specific details are completely system-dependent. Generally speaking, it involves some form of non-random selection of a subset of the random motion. To give you an example of how ...


1

Concerning the question: "Is it theoretically impossible to realize entanglement-like phenomena (e.g. non-local behavior or violation of some sort of Bell inequality) using a Couder-Fort experiment?", I have recently discussed this with John Bush from MIT, one of the experts of these experiments. I believe it is possible that a Bell inequality can indeed be ...


1

In all these discussions about entanglement, all the measured observables of Alice always commute with those of Bob. Their degrees of freedom describe two factors ${\mathcal H}_A$ and ${\mathcal H}_B$ of the overall Hilbert space of possibilities which is the tensor product $$ {\mathcal H} = {\mathcal H}_A\otimes {\mathcal H}_B $$ This factorization of the ...


1

Well, the trite answer is "no" because every bit of matter we see around us is in an entangled state. It's the normal classical world.



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