I was reading the paper No Universal Qubit Flipper. In this the paper they show inability to create a universal flipping machine. The method they follow is they take an entangled state between Alice and Bob. Perform a local operation on Bob's side. And then compare the greatest eigen values of Alice's reduced density operator before and after the operation on Bob's side. Initially the eigen values of Alice's reduced density operator are (Alice has a qubit entangled with Bob's sytem ) $$A+B \;\text{and}\; A-B$$ after operation on Bob's side eigen values of Alice's reduced density operator become $$A+C \;\text{and}\; A-C$$ As the greatest eigen values before and after are $A+B$ and $A+C$ it is said amount of entanglement has increased by local operation ( given $B>C$ ) which is not possible and thus universal qubit flipping is not possible by contradiction.
I know about the formula for entropy $S=-Tr\{\rho \log{\rho}\}$, but what is the formula comparing only maximum eigen values for amount of entanglement , that so only for Alice's reduced density operator, not the combined density operator for Alice and Bob's joint system ?

Qubit flipping is taking state from $\alpha|0\rangle+\beta|1\rangle$ to $\beta^*|0\rangle-\alpha^*|1\rangle$.

  • $\begingroup$ for an example and definition of reduced and conditional probabilities in terms for spin entanglement, see this: physics.stackexchange.com/a/170884/75518 $\endgroup$ – image Apr 3 '15 at 18:41
  • $\begingroup$ @MarcelKöpke Thanks for the link, I read the answer but it does not answer my question. $\endgroup$ – sashas Apr 4 '15 at 10:57
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    $\begingroup$ well, I hope it gave you an insight when to use reduced and conditional propabilities $\endgroup$ – image Apr 4 '15 at 10:59

A cursory reading tells me two things: a) we consider pure states as input states and b) both states on Alice's and Bob's side are qubit states.

Now, the reduced denisty matrix on either side is therefore a two-by-two matrix. If the state is pure, the entanglement is quantified by the von-Neumann entropy of the reduced density matrix, which in turn is completely determined by its eigenvalues. Since the reduced density matrix is normalized, this entropy is actually already determined by the largest eigenvalue, since there is only one other eigenvalue and the sum of the two must be one.

  • $\begingroup$ yeah i figured it out, thanks for the answer though :) $\endgroup$ – sashas Apr 13 '15 at 21:33

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