John Schanck
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 Jun 29 awarded Critic Jun 17 awarded Enthusiast Jun 15 awarded Scholar Jun 15 accepted Constructing a maximally entangled qutrit state from $n$ Bell states Jun 15 awarded Student Jun 15 asked Constructing a maximally entangled qutrit state from $n$ Bell states Jun 12 comment Is this algorithm for simulating a quantum computer accurate? You're certainly correct if Dan intends to iterate through his steps. But is it not the case, as I mentioned in my answer, that the simulation is fine (albeit inefficient) as long as the unitary in step 3 represents the entire computation? Jun 12 comment Is this algorithm for simulating a quantum computer accurate? Sure, consider what happens if you apply 2 sequential Hadamard gates to the zero state and measure the result. A quantum computer will always give the answer zero since $\mathbf{H}\mathbf{H}\vert 0 \rangle = \mathbf{I}\vert 0 \rangle = \vert 0 \rangle$. Using your simulator in a gate-by-gate manner, a measurement would be made after the first Hadamard, the input to the second Hadamard would then be $\vert 1 \rangle$ with probability 1/2 and the final measurement will be of the state $\mathbf{H}\vert 1 \rangle$. This measurement will yield 1 with probability 1/2. Jun 12 answered Is this algorithm for simulating a quantum computer accurate? Jun 10 comment Relation between classical and quantum information One minor point: $N$ bits (or $N\log{2}$ nats if you prefer) is an upper bound on the entropy of $N$ qubits, but it is only the case that the von Neumann entropy of an $N$-level system is $N$ when that system is in a completely mixed state. When describing the number of bits which may be encoded (and subsequently extracted) from a system, the measure you want to use is the log of the dimension of the system, not the entropy. A Bell state has 4 orthogonal vectors on which 2 bits may be encoded, but the entropy of a particular Bell state is zero. Jun 6 comment Quantum circuit decomposition Can you provide a little more detail in your question? The answer is going to depend on a number of factors: the form of your 2-qubit unitary, the set of gates you are allowing in the final circuit, and whether you want an exact or approximate solution. There are "recipes" for a number of decompositions, but for a three qubit operation these might give you a circuit with hundreds of gates. I do not believe there is a recipe, at this time, for an optimal decomposition. Jun 3 comment Is a weak measurement the same as an unsharp measurement or POVM? I don't have a firm enough grasp of these concepts to give you a proper answer, but the weak measurements here are those of Aharonov, Albert and Vaidman. They are not POVMs. Jun 2 comment Can two entanglement particles satisfy at same time two different wave functions? The $\cos^2({\theta})$ result can be derived from the fidelity of pure states (en.wikipedia.org/wiki/Fidelity_of_quantum_states#Pure_states) and from the Chaucy-Schwarz inequality (see unapologetic.wordpress.com/2009/04/17/inner-products-and-angles to learn about the geometric interpretation of Chaucy-Schwarz) applied to corresponding elements of each parties' basis. The fidelity of pure states is the absolute value of their inner product - the result is a probability amplitude, so it must be squared to obtain a true probability. May 27 awarded Editor May 27 revised Can two entanglement particles satisfy at same time two different wave functions? added 1665 characters in body May 24 awarded Supporter May 23 awarded Teacher May 23 answered Can two entanglement particles satisfy at same time two different wave functions?