# Tag Info

4

The controlled dot doesn't do anything: it merely observes the bottom qubit in order to decide whether to apply the 𝑋 gate onto the top qubit. In the answer below, the qubit that appears first is the control qubit, it maps to the bottom qubit in your diagram, and the second one is the target qubit. The CNOT gate works on two qubits at once. You can't ...

3

You cannot directly send information using only entanglement like DaftWullie excellently explained. But here's what you can do Superdense coding: a) pre-share 2 entangled qubits between you and a friend b) then by sending a single qubit you can send 2 classical bits Teleportation: a) pre-share 2 entangled qubits between you and a friend b) when you want to ...

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An entangled quantum state, where one subsystem is held by each party, and there is no other communication, cannot be used to achieve communication. The two classes of operation that one party could perform on their quantum system are unitaries and measurements. A unitary performed on one quantum subsystem (Alice) does not change the other one (Bob), so ...

3

You can't associate some pure states $|x\rangle$ and $|y\rangle$ to the subsystems of the 2-qubit system in the entangled state ($a\vert 00\rangle+b\vert11\rangle, a\ne 0, b \ne 0$ in our case). This is the definition of entanglement. You can verify that there is no vectors $|x\rangle$, $|y\rangle$ such that $|x\rangle \otimes |y\rangle = a\vert 0\rangle \... 2 As everyone knows, all functions of quantum computing are inverses of each other. Hence, the 2 H gates cancel out. Quantum gates all have inverses, but the inverse of a gate is not necessarily the same gate, though Hadamard gates, which are the ones being most considered here, are their own inverse. I'm not sure if by "2 H gates" you mean the two gates ... 2 There could be a few things going on here. However, I think the fundamental issue is the tensor product that you write in the second expression. It implies an issue of identification. Your second expression is effectively saying: there are two different positions at which I can create a fermion. Call them P1 and P2. create a spin-up at P1 and a spin down at ... 2 I am not sure what the second question is about, but in the first one you are probably inquiring about the concept of "phase kickback": Why does the "Phase Kickback" mechanism work in the Quantum phase estimation algorithm? 1 On one hand, the oracle can use an arbitrary subset of the register qubits to decide whether it marks the state as the solution to the encoded problem. For example, if you're looking for any 4-qubit state with the first 2 qubits in state$|11\rangle\$, you can do a CCNOT with those qubits as controls and the marked qubit as target to implement this condition -...

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We have, \begin{aligned} S &= - \operatorname { Tr } \left( \varrho \log _ { 2 } \varrho \right) = \log _ { 2 } \left( \frac { \left| \gamma _ { B } \right| ^ { \left( 2 \left| \gamma _ { B } \right| ^ { 2 } \right) / \left( \left| \gamma _ { B } \right| ^ { 2 } - 1 \right) } } { 1 - \left| \gamma _ { B } \right| ^ { 2 } } \right) = \...

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