4 ms·
I think I made a misunderstanding by the |q1>|q2> notation. Consider the entangled 2-qubits system a|00>+b|11>. The prob of the 2nd qubit is |0> is |a|^2 and th
by infparadox 11y ago
I think I made a misunderstanding by the |q1>|q2> notation. Consider the entangled 2-qubits system a|00>+b|11>. The prob of the 2nd qubit is |0> is |a|^2 and the prob of 2nd qubit is |1> is |b|^2. Now apply I(x)NOT, then the system is a|01>+b|10>. The prob of 2nd qubit to be |0> is now |b|^2 and prob of 2nd qubit is |1> is |a|^2. The same can be obtained by considering the 2nd qubit only as a|0>+b|1> and apply NOT only on the second qubit, without making any operation on the first qubit and without breaking the entanglement.
- irljf 11y agoYes, this is true for any unitary operation on a different subsystem, and is known as the no-signalling principle. But this doesn't let you remove the second system: you are still stuck with a mixed state on the first system, not a pure state (in your case the state is 1/2 |0><0| + 1/2 |1><1|).
- infparadox 11y agoWell, I still don't see any thing wrong with the math. Even the authors rewrite the equations as sum(|A_k>|C_k>) instead of sum(|C_k>) and apply I(x)M_x instead of applying only M_x on |C_k> alone, this will not change the following equations.