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First, introduce a UV cutoff $r_b\equiv (1/\epsilon)\to\infty$. Then perform the length integral on $[r_*,r_b]$ which you'll get a linear combination of natural logarithms in terms of $r_+, r_*$ and $r_b$ (let's denote the result by $\mathcal{L}\,$). Now by integrating the following equation $$\frac{dr}{dx}= r\,\sqrt{(r^{2}-r_{+}^{2})\left(\frac{r^{2}}{r^{2}... 1 When two particles interact, do they spontaneously start treating each other as particles in terms of potentials and so forth? I'm not sure what you mean here. There are no particles at this level, just localized quanta that happen to follow certain rules. If you're talking about the interactions of something like QFT, then yes, but this is more a ... 2 In the quantum eraser experiment, nothing is measured: If the which-way information would be measured, the interference would indeed be irreversibly gone. Rather, what is happening is that the which-way information is copied to another quantum system. If that system is measured (or just ignored - which means it could still be measured), the interference ... 1 So in the quantum eraser experiment you essentially have 3 qubits. When you measure one of them, it is no longer entangled with the other two, but rather the results of your measurement must color your understanding of the quantum state of those other two qubits, and the results of another measurement must color your understanding of the quantum state of the ... -4 Some processes are entangled but not suitable to show the uncertainty of entanglement. Using an electric circuit you may separat ions and electrons [QM objects) on the plates of a capacitors. Later you disconnect the plates and take them into different places. As long as you don’t know the direction of the potential difference of the circuit, your knowledge ... 4 Yes, one can break entanglement without getting entangled with something else. Consider e.g. two spins interacting via the Heisenberg interaction \vec S_1\cdot \vec S_2, which are initially in a state \lvert\uparrow,\downarrow\rangle. As time evolves, the system first gets entangled, and subsequently gets disentangled again, without any other system ... 0 When thinking of quantum states as resources for computation, one must consider the dimension of the system. Systems with more degrees of freedom have greater capacity for computation. In the case of quantum illumination (QI), the dimensionality of the composite system (i.e. d=dS dI where dS and dI are the dimensions of the signal and idler subsystems, ... 0 I'm thinking that light slows down in a medium because photons are being absorbed and then remitted by atoms or molecules in the beam. That is one possible process and it takes place for example with glass. It happens, what oleg told in his answer: Doing an experiment with polarized light, the glass doesn’t destroy the polarization. The light is coming out ... 0 Is it possible to set up a two photons entanglement so that their global state is$$|\psi\rangle = \frac{1}{\sqrt{3}}(|HH\rangle+|HV\rangle+|VV\rangle),$$where H indicates horizontal polarization and V indicates vertical polarization? Yes, this is perfectly possible. Does the first photon have a 50% or 33% chance to pass through a vertical ... 0 I'm no expert in weak measurement, so I might be missing something, but I'm pretty sure you can analyze both these examples by considering what the wavefunction should be after the glass optic, applying the projective measurement, and renormalizing. In which case, 1. If I add a mode label for the three spatial modes involved in the question, (i.e. |... -2 Question 1: No, this is not a state which can describe two photons. It is not symmetrized properly. Photons are bosons, so a two-photon wavefunction must be symmetric under exchange of the particle labels. Under exchange of your particle lables, your state becomes$$ \frac{1}{\sqrt{3}} (|HH\rangle + |VH\rangle + |VV\rangle)  which is not the same as the ...

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For two-qubit unitaries, these gates are known as "matchgates". They encompass all gates which have the block-diagonal form you list, and where the determinant of the even-parity $2\times 2$ matrix $(u_{ij})$ is the same as the one of the odd-parity $2\times 2$ submatrix $(v_{ij})$. For a detailed discussion, see, e.g., https://arxiv.org/abs/quant-ph/...

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That's a very interesting question, thank you. I'm not sure I have the answer but I would like to offer some points for consideration: Consider the easiest case, which is the Landau Hamiltonian. For an electron governed by the Landau Hamiltonian, there are no atoms at all. It's just kinetic energy in a magnetic field. Yet there is a topologically non-...

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First of, photons do not have identity. It is meaningless to say things like "photons which leave a lens or filter are not the ones which entered". More to the point of your question though: if a photon that is entangled with another system gets absorbed by an atom, that atom is then entangled with that other system. When an atom that is entangled with ...

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It may be misleading to say that entanglement refers to a particle reacting to the observation of another particle's state with which it is entangled. Instead, the entanglement refers to a quantum correlation that already exists between these particles even before the observation. The observation merely reveals the specific state of the other particle that ...

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It is called quantum mechanics with its postulates and its equations of state. Quantum mechanics is a complete mathematical model of the micro world of particles atoms molecules and nano states.And it has been discovered to be the theory for all experimental data up to now, i.e. it has not been falsified. A quantum mechanical solution of a system of ...

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Your question is expressing a muddle about the measurement problem in quantum mechanics. You ask "is there any fundamental demonstration that the state of the system cannot have been determined before the measurement, yet according to the probabilistic rules of quantum mechanics". This is a bit unclearly asked, I think, but I take it you want to know if ...

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Yes, it's the No communication theorem, the basic fact that local operations don't affect a distant system. Let $\sigma=\sum_i T_i^A\otimes S_i^B$ be a (possibly entangled) state shared between Alice and Bob. Suppose Alice does a quantum operation $\mathcal{E}$ on her system with the intent of sending information to Bob (for example, a measurement, or any ...

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Not an answer to the original question, but here is a general thought: Under these circumstances (and for more counter-intuitive ones such as quantum eraser) I tend to find solace in thinking of a wave function as a "static" function over a space-time manifold (instead of considering it as "dynamics" w.r.t. time as a special coordinate). It's a solution to a ...

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In the context of discussing quantum mechanics, realism means that the evolution of a system can be represented by a number or set of numbers, each of which represents the value of some measurable quantity. Note that this has nothing at all to do with the philosophical idea that reality is objective. It could easily be the case that reality is objective and ...

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The video doesn't give an explanation of the delayed choice quantum eraser - that is, an account of what is happening in reality to produce the results of the experiment. The video states the results of the experiment unclearly. The original paper https://arxiv.org/abs/quant-ph/9903047 or even the wikipedia entry on this experiment: https://en.wikipedia....

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