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  Principle behind fidelity balance in quantum cloning

+ 12 like - 0 dislike

If we do optimal state estimation on an unknown qubit, we can recreate a state with fidelity $F_c=2/3$ with respect to the original. Let us define the "quantum information content" $I_q=1-2/3=1/3$ as the "amount of fidelity lost" in this measurement procedure.

If we, instead of measuring, decide to clone the qubit using an optimal cloning machine, we can obtain two imperfect copies with fidelity $F_q=5/6$ each. The "quantum information content" of the two qubits is now $I_q=2 \times (5/6-2/3)=1/3$. Note that the value is the same as in the measurement procedure above.

This conservation of "quantum information content" holds more generally: it is true for symmetric, $N \to M$ system cloning, for systems of any dimensionality (see reference [1]). The question then is: is there a deeper principle or operational justification that can be invoked to justify this curious fidelity balance result? I originally raised this question in my PhD thesis (ref. [2] below, section 4.3.4).


[1] M. Keyl and R. F. Werner. Optimal cloning of pure states, testing single clones. J. Math. Phys., 40(7):3283–3299 (1999).

[2] E. F. Galvão, PhD thesis, http://arxiv.org/abs/quant-ph/0212124

This post has been migrated from (A51.SE)
asked Oct 8, 2011 in Theoretical Physics by Ernesto (135 points) [ no revision ]
retagged Mar 7, 2014 by dimension10

1 Answer

+ 8 like - 0 dislike

This may seem something of an obvious answer, and may well have already occured to you, but perhaps there is something interesting in it for you.

The input system has Holevo information of $\chi = \log_2 D$, where $D$ is the dimensionality of the system to be cloned. Applying the cloning procedure does not change this in any way, it is simply spreading this information out over a number of systems. You appear to be implicitly assuming that the input state is chosen uniformly at random from the pure states (since otherwise you can obtain $I_q = 0$ by taking the distribution only over orthogonal states). As the Holevo information is defined as $\chi = S(\sum_i p_i \rho_i) + \sum_i p_i S(\rho_i)$, and $p_i=p$ is fixed and $\rho_i$ are pure, you have $$\chi = S(p \sum_i \rho_i) = S(\bar{\rho})$$ where the bar denotes the average density matrix. Presumably any cloning scheme which is optimal must not decrease the Holevo information of the joint system, and so the entropy is necessarily preserved. When you make a projective measurement of all of final system you project onto a product state, and hence the entropy is zero, independent of the size of the system. Thus, the change in entropy is identical whether or not optimal cloning has been preformed. The reason that you can express this as a condition on the fidelity of the system, is then simply a consequence of the fact that fixing the joint entropy of the system imposes a constraint on the maximum fidelity of the clones.

This post has been migrated from (A51.SE)
answered Oct 10, 2011 by Joe Fitzsimons (3,575 points) [ no revision ]

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