# Tag Info

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 ...

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 ...

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, ...

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 ...

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

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 ...

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