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3

It's true quite generally (for seperable Hilbert spaces at least). The original proof is in "Proof of the Strong Subadditivity of Quantum Mechanichal Entropy", by Lieb and Ruskai (J. Math. Phys. 14, 1938–1941 (1973)). The main idea is to prove the finite-dimensional case, and then extend it by taking a limit of the inequality on finite-dimensional subspaces ...


0

Yes, we can have entanglement between different degree of freedom of same particle or system. That is known as ''hybrid entanglement'' and that is experimentally demonstrated also. http://arxiv.org/pdf/1007.1322v1.pdf


7

First of all, the problem is technically difficult due to the fact that generally unbounded self-adjoint operators like those used in general QM have domain smaller that the whole Hilbert space. For this reason I will consider here only bounded self-adjoint operators whose domain, as is well known, is the full Hilbert space. Proposition. The elements of ...


0

In the context of your paper, the time-like Bell states are : $|\psi^\pm\rangle^{\tau, \tau'}_{ab} = \frac{1}{\sqrt{2}}(|h_a^\tau v_b^{\tau'}\rangle \pm |h_a^\tau v_b^{\tau'}\rangle)$ $|\phi^\pm\rangle^{\tau, \tau'}_{ab} = \frac{1}{\sqrt{2}}(|h_a^\tau h_b^{\tau'}\rangle \pm |v_a^\tau v_b^{\tau'}\rangle)$ So, equation $(3)$ just reexpress $|S\rangle = ...


0

Firstly the link you have provided for the paper is only accessible for those who have a subscription to the APS journal. So maybe instead you could just give a preview of the equation you are referring to. But I can give you a general overview on projections in quantum mechanics and Bell states. I will use Dirac's Bra-Ket notation for the rest of this ...


0

You have $2$ atomic ensembles. For each atomic ensemble, the ground state is $|\psi_0\rangle$, and the excited state is $|\psi_1\rangle$. The excited state is signaled by an idler photon. Without beam splitter, if you detect only one photon (for the whole set), you know which atomic ensemble is excited, so you have the state $|\psi_0\rangle |\psi_1\rangle$ ...


3

$X_i,Y_i,Z_i$ are three Pauli matrices acting on the $i$-th qubit where $i=1,2,3,4,5,6,7,8,9$ labels the qubit. In equation 4.1, the state is a superposition of tensor product of three states similar to $|000\rangle$. The latter is a state of three qubits, so if one takes the third power, it is a state of $3\times 3 = 9$ qubits. $X_1$ differs from $X_8$ by ...


0

I suspect that the reprint was withdrawn, rather than the original article; probably, for copyright reasons.


2

You've completely misunderstood the impact of adding a potential to an entangled state. Atoms a and b should roughly have the same distribution ... because each single measurement of position for each entangled pair in the ensemble should yeild $x_a=x_b$ with respect to their local axes ... because this is what it means to be entangled. (Need a check ...


-2

You describe a communication protocol exploiting EPR/Bell-type setups. If I understand your protocol correctly, you envisage a 'stream' of entangled ensembles, and preagreed measurements being performed on each block of the stream to yield a 'bit'. There are various ways we could do this: you use harmonic oscillator potential. Fine. We could do the ...


2

The relation between entropy and information is well established; indeed, Shannon entropy is the seminal measure of the information in a system. The other question, about determinacy and information, is more complex, and even more complex yet when extended to the entire universe. Let us set aside, for now, the fact that quantum mechanics would seem to ...


1

Your thought experiment does have a major flaw. According to quantum mechanics in any measurement of two spatially separated atoms a and b what happens to b has absolutely no effect at all on the probabilities of measurement outcomes on b. I'm not going to work out exactly what the flaw is in your proposed experiment, but just indicate why quantum theory ...


0

I just noticed that I forgot to show my solution to the problem. First, my measurement was wrong. In the paper they use $|0011><0011$ and $|1100><1100$. But that was not the main problem of the question. My first ansatz of applying an unitary operator on the qubits 1 and 2 at first and the same on the qubits 2 and 3 was wrong. One has to ...


0

I think I just figured this out. For cloning you want to do the following: (A|p>+B|q>)|0> goes to (A|p>+B|q>)(A|p>+B|q>) where |p> and |q> are orthogonal. For stimulated emission one gets: (A|p>+B|q>)|0> = A|p>|0>+B|q>|0> goes to A|p>|p>+B|q>|q> which is different from the cloning case. Hence, it does not violate the no-cloning theorem. It also ...


0

I don't think it is actually possible to have complete destructive interference everywhere in quantum mechanics (unless the state you started with has zero amplitude). The wavefunction of a particle contains all the information about that particle, including everything needed to calculate what it is going to do in the future. This means that a right ...


3

Why can't we just analyze the way they built, and see if it's quantum? because the environment can easily get in the way and prevent the quantum effects from creating the speedups they're meant to. Basically you can try to build a quantum computer and end up with a classical computer by mistake that does the same thing (but slower) if it wasn't built ...


2

With appropriate lab equipment, you can derive extremely narrow pulse shapes. A typical setup involves splitting the incoming beam of light and interfering it with itself. By shifting one of the path lengths, you can observe the change in the diffraction pattern and calculate the pulse width. This won't work, of course, for a single event. For that case, ...


0

As Trimok says, every photon - and for that matter, any physical two-level system - is immediately a qubit. In other words, qubits are abundant in nature and not very interesting by themselves. Also, a qubit alone is not entangled. The interesting part is, what you can do with your qubit: Can you create a qubit in always the same state? Can you entangle a ...


2

Any photon (pure) state may be described by a q-bit formalism: $$|photon\rangle = \alpha |0\rangle + \beta|1\rangle$$ where $|0\rangle$ and $|1\rangle$ represent the two possible polarizations of the photon. So, any photon "is" a q-bit. You don't have to "create" q-bits. Just prepare photons is some state. An entangled state of $2$ photons may be ...


2

Your understanding is correct. A destructive partial measurement will result in a deterministic bit. But it will never collapse to the state with zero probability as in your example. I have tried the example in their about page of the MeasureBit(b). However, after the partial measurement, the state always corresponds to the measurement result of 0, no matter ...



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