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Understanding the optimal approximate qubit cloning method

  1. Feb 16, 2016 #1


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    The paper Optimal Cloning of Pure States by R. F. Werner describes a method for approximately expanding an unknown state ##\rho## containing n copies of a qubit, so ##\rho = (\alpha \left| 0 \right\rangle + \beta \left| 1 \right\rangle)^{\otimes n}##, into a larger state with d more qubits approximating ##(\alpha \left| 0 \right\rangle + \beta \left| 1 \right\rangle)^{\otimes n+d}##.

    Specifically, the paper says that the optimal cloning map is to tensor ##d## randomized qubits onto the state then project onto the symmetric subspace and normalize. Symbolically: ##\widehat{T}(\rho) = \frac{n+1}{n+d+1} s_{n+d} (\rho \otimes I_2^{\otimes d}) s_{n+d}##.

    The thing I don't understand is what is meant operationally by "project onto the symmetric subspace".

    Does it mean that we're doing a projective measurement where the symmetric subspace has a unique eigenvalue, and we give up if the measurement result is some other eigenvalue? For example we could measure the observable ##S = \sum_k^{n+d} \widehat{\left| n+d \atop k \right\rangle} \widehat{\left\langle n+d \atop k \right|}##, which would return 1 if we're in the symmetric subspace and 0 otherwise. But then we'd only expect to succeed ##\frac{1}{2^d}## of the time, which is worse than what's reported in the paper.

    Does it mean we're tensor-factoring the space into a symmetric part and a non-symmetric part, and then discarding the parts of the state corresponding to the non-symmetric part (i.e. we trace over it)? I'm not sure how to compute that.

    Does it mean something else entirely?
  2. jcsd
  3. Feb 21, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
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