I'm trying to understand the operation used for optimal cloning of pure qubits states from the paper Optimal Cloning of Pure States by R. F. Werner.

The paper describes the optimal cloning method $\widehat{T}$, which approximates the state $\sigma^{\otimes n+d}$ when given the state $\rho = \sigma^{\otimes n}$. $\sigma$ is promised to be a pure 2-level state. The optimal method is: tensor $d$ randomized qubits onto the state, project the system onto the symmetric subspace for $n+d$ qubits, and normalize. Symbolically:

$$\widehat{T}(\rho) = \frac{n+1}{n+d+1} s_{n+d} \cdot (\rho \otimes (I_2 / 2)^{\otimes d}) \cdot s_{n+d}$$

I don't understand what projecting onto the symmetric subspace means, exactly. If I was implementing this method on a quantum computer, or performing an experiment, how do I actually achieve the projection? And how do I compute its effects on paper?

What I've tried

The paper doesn't give a definition of $s_m$ more concrete than "a projection for the symmetric subspace". But I think this is the correct expression:

$$s_n = \sum_k^n \widehat{\left| n \atop k \right\rangle} \widehat{\left\langle n \atop k \right|}$$

Where $\widehat{\left| n \atop k \right\rangle}$ is a unit vector made up of a uniform superposition of all the classical states with exactly $k$ qubits on (out of $n$ total qubits). More specifically:

$$\widehat{\left| n \atop k \right\rangle} = {n \choose k}^{-1/2} \left| n \atop k \right\rangle$$

$$\left| n \atop k \right\rangle = \left\{ \begin{align} k = 0 & \rightarrow & \left| 0 \right\rangle^{\otimes n} \\ k = n & \rightarrow & \left| 1 \right\rangle^{\otimes n} \\ \text{else } & \rightarrow & \left| n-1 \atop k-1 \right\rangle \left| 1 \right\rangle + \left| n-1 \atop k \right\rangle \left| 0 \right\rangle \end{align}\right\}$$

I figured that the cloning method should work just as well on any symmetric state, so I picked a very simple one to analyze: $\left| 0 \right\rangle^{\otimes n}$.

However, when I simplify $s_{n+d} ((\left|0\right\rangle \left\langle 0 \right|)^{\otimes n} \otimes I_{2^d}/2^d) s_{n+d}$ I get the result $\frac{1}{2^d} \sum_k^d \widehat{\left| n+d \atop k \right\rangle} \widehat{\left\langle n+d \atop k \right|} {d \choose k}$. Which only has a $\frac{1}{2^d}$ chance of being in the correct state. Which is further from $\left| 0 \right\rangle^{\otimes n+d}$, in terms of trace distance or overlap, than reported by the paper. And I end up not needing the normalization factor.

Clearly I'm making a mistake somewhere. Thus why I'm asking:

What precisely is meant by "project onto the symmetric subspace"? How would it be realized in practice? How do I compute the result on paper?


As I understand from the discussion below, you are interested in the overlap between the cloned state and the perfect clone.

The cloned state is obtained as $$ \tau=\Pi(\sigma^{\otimes d}\otimes (I/2)^{\otimes n-d})\Pi\ , $$ where $\Pi$ is the projection onto the symmetric subspace, and you are interested in the overlap with $$ \gamma =\sigma^{\otimes n}\ . $$ This overlap is given by $$ \mathrm{tr}(\gamma\sigma) = \mathrm{tr}(\sigma^{\otimes n}\Pi(\sigma^{\otimes d}\otimes (I/2)^{\otimes n-d})\Pi)\ . $$ Now using cyclicity of the trace, you can apply the projection as well to $\gamma$, which already lives in the symmetric subspace. Thus, we have that $$ \mathrm{tr}(\gamma\sigma) = \mathrm{tr}(\sigma^{\otimes n}(\sigma^{\otimes d}\otimes (I/2)^{\otimes n-d}))\ . $$ For a pure state $\sigma$, this evaluates to $(1/2)^{n-d}$.

  • $\begingroup$ I don't think that's the right answer. The paper says that the overlap factor should be $\frac{N}{N+2} \frac{M+2}{M}$, where $N$ is the given number of copies and $M$ is the desired total number of copies at the end. Also, a naive cloning method I tried scales better than that (as you may recall). $\endgroup$ – Craig Gidney Feb 16 '16 at 8:48
  • $\begingroup$ I think the point is that the cloning is not simply adding maximally mixed states and projecting on the symmetric subspace. It is a CP map which has the same outcome as that procedure, but with optimal success probability. But I'll have to check more carefully. $\endgroup$ – Norbert Schuch Feb 16 '16 at 23:02

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