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Corrigendum: Quantum steering ellipsoids, extremal physical states and monogamy (2014 New J. Phys. 16 083017)
View the table of contents for this issue, or go to the journal homepage for more 2015 New J. Phys. 17 019501
(http://iopscience.iop.org/1367-2630/17/1/019501)
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New J. Phys.17(2015) 019501 doi:10.1088/1367-2630/17/1/019501
CORRIGENDUM
Corrigendum: Quantum steering ellipsoids, extremal physical states and monogamy (2014 New J. Phys. 16 083017)
Antony Milne1, Sania Jevtic2, David Jennings1, Howard Wiseman3and Terry Rudolph1
1 Controlled Quantum Dynamics Theory, Department of Physics, Imperial College London, London SW7 2AZ, UK
2 Mathematical Sciences, Brunel University, Uxbridge UB8 3PH, UK
3 Centre for Quantum Computation and Communication Technology (Australian Research Council), Centre for Quantum Dynamics, Griffith University, Brisbane, Queensland 4111, Australia
E-mail:antony.milne@gmail.com
The proof of theorem 6(a) is incorrect, although the monogamy of steering result VA B∣ + VC B∣ ⩽ 4π 3
holds as stated. Here we give a corrected proof, which reveals a remarkable new result relating the volume of Aliceʼs steering ellipsoid to the centre of Charlieʼs:VA B∣ = πcC B
4 3
2 . We are very grateful to Michael Hall for identifying the mistake, independently verifying our numerical tests and assisting with the corrected proof.
Proof.The pure three-qubit state held by Alice, Bob and Charlie is∣ϕABC〉. The canonical transformation
ϕ ϕ ϕ
∣ 〉 → ∣∼ 〉 = ⊗ ⊗ ∣ 〉
( ρ )
ABC ABC ABC
1
2B leavesA B∣ andC B∣ invariant. To prove the monogamy of steering we therefore need consider only canonical states for which∼=
b 0. (WhenρBis singular and the canonical transformation cannot be performed, no steering by Bob is possible; we then haveVA B∣ =VC B∣ =0so that the bound holds trivially.)
We begin by showing thatVA B∣ = 4πcC B
3
2 . Denote the eigenvalues ofρ∼ = ∣ϕ∼ 〉〈ϕ∼ ∣
AB trC ABC ABC as{ }λi . For a canonical state Charlieʼs Bloch vector coincides withcC B∣ , and soρ∼ = ∣ϕ∼ 〉〈ϕ∼ ∣ = + σ
c ∣
tr ( · )
C AB ABC ABC C B
1
2 .
By Schmidt decomposition we therefore have{ }λi ={ (11 + cC B∣ ), (1−cC B∣ ), 0, 0}
2
1
2 .
From the expression forVA B∣ given in [1] we obtainVA B∣ =64π ∣detρ∼AB∣
3
TA sincedetρ∼AB= 0. Define the reduction map [2,3] asΛ( )X =trX− X. Following [4] we note thatdetρ∼ABTA = det ((σy ⊗)ρ∼ABTA(σy ⊗)) and that(σy ⊗)ρ∼ABT (σy ⊗)=(Λ⊗)(ρ∼AB)= 1⊗−ρ∼AB
2
A , where we have used the fact that Bobʼs local
state is maximally mixed. Since the eigenvalues of1⊗− ∼ρAB
2 are{1 −λi}
2 we obtaindetρ∼ABTA=∏i (1 − λi) 2
=(−1cC B∣ )( cC B∣ )( )( )
2 1 2
1 2
1
2 =−1cC B
16
2 , which givesVA B∣ =4πcC B
3 2 . From theorem 3 we haveVC B∣ ⩽ Vcmax
C B =4π(1−cC B∣ )
3
2. Hence VA B∣ + VC B∣ ⩽ 4πcC B∣
3 +
π − c ∣
(1 C B)
4
3 = 4π
3 .
□
References
[1] Jevtic S, Pusey M, Jennings D and Rudolph T 2014Phys. Rev. Lett.113020402 [2] Cerf N J and Adami C 1999Phys. Rev.A60893
[3] Horodecki M and Horodecki P 1999Phys. Rev.A594206
[4] Augusiak R, Demianowicz M and Horodecki P 2008Phys. Rev.A77030301 OPEN ACCESS
RECEIVED
22 December 2014
ACCEPTED FOR PUBLICATION
22 December 2014
PUBLISHED
27 January 2015
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