Physics, asked by dipak9964, 1 year ago

What can we do with just bell states and computational basis measurements?

Answers

Answered by RockyAk47
0
When we have just one qubit, there's nothing particularly special about the computational basis; it's just nice to have a canonical basis. In practice you could think that first you implement a gate ZZ with Z2=IZ2=I and Z≠IZ≠I, and then you say that the computational basis is the eigenbasis of this gate.

However, when we talk about multi-qubit systems, the computational basis ismeaningful. It comes from picking a basis for each qubit, and then taking the basis which is the tensor product of all these bases. Picking the same basis for each qubit is nice just to keep everything uniform, and calling them 00and 11 is a nice notational choice. What's really important is that our basis states are product states across our qubits: the computational basis states can be prepared by initializing our qubits separately and then bringing them together. This isn't true for arbitrary states! For example, the cat state 12√(|0n⟩+|1n⟩)12(|0n⟩+|1n⟩)requires a log-depth circuit in order to prepare it from a product state.

Answered by rockyak4745
0
Bell state measurement. The Bell measurement is an important concept in quantum information science: It is a joint quantum-mechanical measurement of two qubits that determines which of the four Bellstates the two qubits are in. ... If a Bell state is measured from this ambiguous class, the teleportation event fails.
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