QFT is a nonlocal unitary transformation and so can generate entanglement in a system. It means a separable pure state can be converted into an entangled pure state. Now since the presence of entanglement can be witnessed via an increase in the entropy of the subsystems. Since all the subsystems witness a positive entropy change ,does the entropy of the complete system also increase (it seems to increase since entropy is additive) ? Now if it does increase , It seems to violate reversible nature of Quantum algorithms. I am very confused.
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Just because the entropy of the subsystems increases, that doesn't mean that the entropy of the whole system increases. This is possible here because of entanglement: an entangled pure state has zero overall entropy, but the subsystems have non zero entropy. A simple example is the state \begin{equation} \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle). \end{equation} Both qubits have subsystem states \begin{equation} \frac{1}{2}(|0\rangle\langle 0|+|1\rangle\langle 1|), \end{equation} and, if you compute the entropy of these states, via \begin{equation} S=-\textrm{tr}[\rho\log(\rho)], \end{equation} you get $\log(2)$ as the entropy for both qubits. But, if you compute the entropy of the full system, you get zero.
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$\begingroup$ what about additive nature of Entropy ? $\endgroup$ Apr 19, 2014 at 5:55
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$\begingroup$ @KishorBharti that's only true if the systems are independent. In this case, if the two qubits have the overall state $\rho_1\otimes\rho_2$. In the usual statistical mechanics derivations of the additivity of entropy, people are probably looking at that situation. $\endgroup$ Apr 19, 2014 at 10:03