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Quantum state tomography suffers from the measurement effort increasing exponentially with the number of qubits.
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For six qubits, we have P s ( 6 ) ≥ 2 225 ( J x 2 + J y 2 + J z 2 ) − 1 90 ( J x 4 + J y 4 + J z 4 ) + 1 450 ( J x 6 + J y 6 + J z 6 ) P_{\rm s}^{(6)}\geq\frac{2}{225}(J_{x}^{2}+J_{y}^{2}+J_{z}^{2})-\frac{1}{90}(J_{x}^{4}+J_{y}^{4}+J_{z}^{4})+\frac{1}{450}(J_{x}^{6}+J_{y}^{6}+J_{z}^{6}) with J i = 1 2 ∑ k σ i ( k ) J_{i}=\frac{1}{2}\sum_{k}\sigma_{i}^{(k)} and σ i ( k ) \sigma_{i}^{(k)} the application of σ i \sigma_{i} on the k th k^{\textrm{th}} qubit
Cited in the paper.
See Supplemental Material for additional derivations, which includes Ref. TW09 ; LEONARDO_BERGMANN
Cited in the paper.
The overall runtime of the experiment was almost two weeks from which 50 h of useful data could be extracted
Cited in the paper.
We use the Uhlmann fidelity F ( ϱ 1 , ϱ 2 ) = Tr ( ϱ 1 ϱ 2 ϱ 1 ) 2 F(\varrho_{1},\varrho_{2})=\mathrm{Tr}{(\sqrt{\sqrt{\varrho_{1}}\varrho_{2}\sqrt{\varrho_{1}}})}^{2} between the states ϱ 1 \varrho_{1} and ϱ 2 \varrho_{2} which simplifies to F ( ϱ 1 , ϱ 2 ) = Tr ( ϱ 1 ϱ 2 ) F(\varrho_{1},\varrho_{2})=\mathrm{Tr}{(\varrho_{1}\varrho_{2})} if one of the two states is pure
Cited in the paper.
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For efficient entanglement detection methods for PI density matrices, see L. Novo, T. Moroder and O. Gühne, Phys. Rev. A 88
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In order to apply CS and PI tomography together, it suffices that the ρ j \rho_{j} matrices in Eq. ( 1
2013
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