infparadox

3 ms·

1. 1.▲
   by infparadox 11y ago

   Well, 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
2. 2.▲
   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 t
3. 3.▲
   by infparadox 11y ago

   I am not expert in the field, I just took a course, but I know that a mixed state is not the same as an entangled state. In their case it is entangled not mixed and so there is no need to take the partial trace. For example, consider a two
4. 4.▲
   by infparadox 11y ago

   I don't think there is a problem in that, they are just considering a subspace of the system.