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In your scheme, the measurement results will be equal if the plane normal to $v$ is between the measurement vectors, and unequal otherwise, so the chance that they'll be equal is linearly proportional to the angle between the measurement vectors. That's the red curve labeled "classical" in this diagram: The red curve makes Bell's inequality into ...


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There is no requirement that particles have to be newly created to be entangled. Essentially any time particles interact with each other they become partially entangled. Your quote shows that even this is not a requirement- two particles can become entangled without ever directly interacting. The reason you so often hear about particle creation when talking ...


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I found the answer. Actually I had made a mistake in the order of ket. (Thanks to the forum for pointing that out). So when Initial state ($t=0$) = $|\psi\rangle=|\alpha\rangle|0\rangle$ Final state=$|\chi\rangle= |\cos(t)\alpha\rangle|i\alpha\sin(t)\rangle$ I will get fidelity as $$F=e^{-|\alpha|^2}e^{-|\alpha\cos(t)|^2}|e^{|\alpha|^2cos(t)}|^2 e^{-|\alpha\...


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Well, if the formula you've written is correct, then $$|\alpha\sin(t)|^2 +|\alpha\cos(t)|^2 = |\alpha|^2,$$ quite trivially, as you should be able to see. Therefore: $$e^{-|\alpha|^2-(|\alpha\sin(t)|^2 +|\alpha\cos(t)|^2)} = e^{-2|\alpha|^2}.$$ The resulting function is certainly not a function of time, since the time-dependence cancels out, leaving only a ...


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Shortest answer: No! Short answer: No, in the many worlds interpretation measurement is described as an interaction between a "system to be measured" and a "measurement system". In quantum mechanics it is generally true that an interaction between 2 systems can lead to entanglement. Longer answer: No, the many worlds interpretation does ...


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Since only a very small amount of incoming photons are converted into two entangled photons we must use a coincidence circuit to distinguish them from the non-entangled majority. This rarity implies that it is not possible to detect the interference pattern instantly. No, you always have to use the coincidence circuit, even in an idealized version of the ...


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Is Copenhagen QM non-local? Consider an entangled system of two photons moving away from each other in the Schrödinger picture, e.g., given by the wavefunction: $$ \Psi(t,x) = \psi_1(t,x) + |\psi_2(t,x). $$ The photons have the same source, so at $t=0$ they are located at the same point. However, at large $t$ they will be far apart from each other and $\Psi(...


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The operator that'd check if the particle is on the left or the right can be written as $(1/2)\vert L\rangle\langle L\vert - (1/2) \vert R \rangle\langle R\vert$. Now, the same unitary transformations that takes $\sigma_z$ to $\sigma_{x,y}$ would take this operator to two other operators that don't commute with the original operator (as well as with each ...


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As soon as two atoms or electrons interact with each other, they become entangled. Say you shoot two atoms against each other in a collision experiment. At $t=0$ the atoms have a definite momentum $\hbar k$, $|\Psi \rangle = |k\rangle |-k \rangle$. After the collision, they will be in a superposition of different momenta, but the interaction is such that the ...


1

They are entangled. The electrons are in a Slater determinant of all the atomic orbitals, so if there are $n$ electrons in states $1,2,3,4,\cdots n$, where the basis state is written $|1,2,3,\cdots n\rangle$ is the first electron in the first state and so on, the atomic state is: $$|\psi\rangle = \frac 1 {\sqrt n}\sum_{\pi\in S_n}{\sigma(\pi)}|\pi(1,2,\cdots,...


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There's no way to affect the observable behavior of the "upper" photons by manipulating the "lower" photons after the initial generation of the entangled pairs in the Glan-Thompson prism (if there were, you could use it to send a signal faster than light). There's no interference pattern because the lower photons contain which-path ...


1

I am not an expert on this, but let me give this question a try. Here is an argument. Suppose that your system is composed of two particles in an entangled state: $$|\psi\rangle=\frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle +|\downarrow\uparrow\rangle$$ this is a pure state, so under von Neumman definition, it has entropy it has zero entropy. Now imagine ...


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Yes, entangled states certainly do exist for (products of) Hilbert spaces spanned by a continuous basis. In fact, the first example of entanglement that was presented by Einstein, Podolsky, and Rosen in their famous 1935 EPR paper was of this kind, which was only later popularized in the form the simpler example of the entanglement of spin-states by Bohm, ...


1

Let's expand: \begin{aligned} Z(t)|\psi\rangle=Z(t)\int_{-\infty}^\infty dp\, \psi(p)|p\rangle=\int_{-\infty}^\infty dp \,\psi(p)|p+t\rangle=\int_{-\infty}^\infty dp \,\psi(p-t)|p\rangle. \end{aligned} This could have also been written in the position basis as \begin{aligned} Z(t)|\psi\rangle=Z(t)\int_{-\infty}^\infty dq \,\tilde{\psi}(q)|q\rangle=\int_{-\...


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The Copenhagen Interpretation as formalised by John von Neumann is that when you make a measurement of an observable associated with a particle whose wave function is not an eigenfunction of that observable, the wave function of the particle changes as a consequence of the measurement to become an eigenfunction of the observable, namely an eigenfunction ...


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It is a postulate that subsequent measurement of the same physical quantity will give same result regardless of the measuring device. If this is not so, we have to identitfy both the measured quantity and the measuring device for a theoretical description. However, if two devices give different result after subsequent measurement, we can say (and we indeed ...


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Let me blunder in where angels fear to tread: What is a locus, hence locality. In my definition a locus is an (x,y,z) point in space at a time t, which theoretically has infinite precision and experimentally has a measurement error. Quantum mechanical solutions of the wave equation together with the postulates of quantum mechanics calculate with precision ...


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I think that it can be proved, and the proof is Bell's theorem (or a similar result). I suspect that Susskind had Bell's theorem in mind. In Bell's original paper, the outcomes of the two measurements were determined by arbitrary functions $A(\vec a, λ)$ and $B(\vec b, λ)$, where $λ$ was arbitrary hidden state information and $\vec a$ and $\vec b$ were the ...


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In quantum mechanics, possible states of a system are described by a Hilbert space. If you have two systems, with state spaces $\mathcal{H}_1, \mathcal{H}_2$ then the combined system is described by $\mathcal{H}_1 \otimes \mathcal{H}_2$, which consists of linear combinations of product states like $|\phi \rangle \otimes |\psi \rangle$. So e.g. for a pair of ...


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