Asymptotic Pure State Transformations PHYS 500 - Southern Illinois University April 18, 2017 PHYS 500 - Southern Illinois University Asymptotic Pure State Transformations April 18, 2017 1 / 15
Entanglement Distillation Definition Recall that a state s distillable entanglement is defined by: { } n E D (ρ = lim lim sup ɛ 0 m m ρ m LOCC ρ n, F (ρ n, Φ + n 1 ɛ. In today s lecture we will compute the distillable entanglement of an arbitrary bipartite pure state Ψ. Theorem: for any bipartite pure state Ψ. E D ( Ψ = E( Ψ = S(tr A Ψ Ψ PHYS 500 - Southern Illinois University Asymptotic Pure State Transformations April 18, 2017 2 / 15
To compute the distillable entanglement of a pure state, we rely heavily on its Schmidt decomposition: Ψ = r p α β. =1 Our goal is to understand when it becomes possible transformations of the form Ψ m Φ + n by LOCC. The entire problem can be easily solved once the structure of Ψ m is understood. PHYS 500 - Southern Illinois University Asymptotic Pure State Transformations April 18, 2017 3 / 15
In this lecture, we consider a two-qubit state: Ψ = p 00 + 1 p 11. Start taing copies: Ψ 2 = ( p 00 + 1 p 11 2 = ( p 00 AB + 1 p 11 AB ( p 00 AB + 1 p 11 AB = p 00 A 00 B + p(1 p 01 A 01 B + p(1 p 10 A 10 B + (1 p 11 A 11 B = p 00 A 00 B + (1 p 11 A 11 B + ( p(1 p 01 A 01 B + 10 A 10 B. PHYS 500 - Southern Illinois University Asymptotic Pure State Transformations April 18, 2017 4 / 15
In general we have Ψ m = =0 p 2 (1 p m 2 permutation π permutation π of permutation π of 0 s m- 1 s 0 s m- 1 s 00 0 11 1 A 00 0 11 1 B. How many terms are in the second sum? For each fixed value, the number of terms in the second sum is the number of ways to permute 0 s and m- 1 s. This number can be computed by counting the number of ways to place 0 s along the line 1 to m one at a time, and then dividing by! (since the order of placement does not matter and each particular distribution of 0 s will therefore be generated! times. PHYS 500 - Southern Illinois University Asymptotic Pure State Transformations April 18, 2017 5 / 15
The number of permutations is thus given by m m 1 m + 1! = m! (m!! = ( m. The number ( m is called a binomial coefficient. It is also described as m choose. These are the coefficients on the expansion of (x + y m : (x + y m = =0 ( m x y m. PHYS 500 - Southern Illinois University Asymptotic Pure State Transformations April 18, 2017 6 / 15
Ψ m = =0 p 2 (1 p m 2 permutation π permutation π of permutation π of 0 s m- 1 s 0 s m- 1 s 00 0 11 1 A 00 0 11 1 B. Each term in the sum is orthogonal (a Schmidt decomposition of Ψ m. For each {0, m}, (ρ A m = tr B Ψ Ψ m has an ( m -dimensional eigenspace associated with eigenvalue p (1 p m and spanned by permutation π of { 0 s m- 1 s } 00 0 11 1 : π is permutation of 0 s and m- 1 s. PHYS 500 - Southern Illinois University Asymptotic Pure State Transformations April 18, 2017 7 / 15
Let P be a projector onto this subspace; i.e. P = permutation π permutation π of 0 s m- 1 s 0 s 00 0 11 1 00 0 permutation π of m- 1 s 11 1. The P are projectors satisfying (i tr(p = ( m, (ii P P j = δ j P, and (iii P = I. It is easy to see that (ρ A m = (ρ B m = p (1 p m P. =0 PHYS 500 - Southern Illinois University Asymptotic Pure State Transformations April 18, 2017 8 / 15
We can write Ψ m = (m p 2 (1 p m 2 Ψ =0 where Ψ is an ( m -dimensional maximally entangled state, i.e. it is LU equivalent to the state 1 (m ( m ii, i=1 with P I Ψ j = I P Ψ j = P P Ψ j = δ j Ψ j. PHYS 500 - Southern Illinois University Asymptotic Pure State Transformations April 18, 2017 9 / 15
The LOCC distillation protocol consists of either Alice or Bob performing the projective measurement {P } m =0 and broadcasting the result. This will collapse the state Ψ m with probability q = ( m p (1 p m. LOCC Ψ The state Ψ has Entanglement entropy log ( m (since Ψ is maximally entangled, and so the average post-measurement entanglement is E = q E( Ψ = =0 =0 ( ( m m p (1 p m log. PHYS 500 - Southern Illinois University Asymptotic Pure State Transformations April 18, 2017 10 / 15
What is the value of E? Since the Entanglement Entropy is an entanglement monotone, we must have that E E ( Ψ m = mh(p. In deriving this formula, we are using a property of the von Neumann entropy nown as additivity: S(ρ σ = S(ρ + S(σ S(ρ m = ms(ρ. We therefore have the upper bound E = =0 ( ( m m p (1 p m log mh(p. PHYS 500 - Southern Illinois University Asymptotic Pure State Transformations April 18, 2017 11 / 15
A lower bound can be computed as ( m E = =0 ( m = =0 ( m =0 = H p (1 p m log p (1 p m log p (1 p m log ({( m p (1 p m } m =0 log(m + 1 + mh(p. ( m (( m p (1 p m (p (1 p m + mh(p PHYS 500 - Southern Illinois University Asymptotic Pure State Transformations April 18, 2017 12 / 15
The last term follows from the expectation value of detecting (or m of the same outcomes in n independent trials of a binomal distribution: =0 =0 ( m p (1 p m log p = log p and ( m p (1 p m = mp = mp = mp. =1 t=0 ( m p (1 p m, =0 ( m 1 p 1 (1 p (m 1 ( 1 1 n ( n p t (1 p n t t PHYS 500 - Southern Illinois University Asymptotic Pure State Transformations April 18, 2017 13 / 15
Combining both upper and lower bounds gives the average entanglement rate log(m + 1 h(p 1 E h(p. m m 1 Hence in the limit: m E h(p = E(Ψ. Let us summarize what we have shown thus far. There exists a multi-outcome LOCC transformation Ψ m LOCC {q, Ψ } such that each Ψ is a maximally entangled state of dimension ( m and the average post-measurement entanglement per copy of Ψ, 1 m m =0 q E(Ψ, is arbitrarily close to E(Ψ for sufficiently large m. PHYS 500 - Southern Illinois University Asymptotic Pure State Transformations April 18, 2017 14 / 15
Our final step will be to show that the ensemble of maximally entangled states {q, Ψ } can be converted into n copies of our standard resource state Φ + n with high fidelity and with m n E(Ψ. That is, in the next lecture we will see how to transform {q, Ψ } LOCC Φ + me(ψ. This will complete the desired transformation Ψ m LOCC Φ + n at the rate given by the entanglement entropy of Ψ. PHYS 500 - Southern Illinois University Asymptotic Pure State Transformations April 18, 2017 15 / 15