{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

soln9_09

# soln9_09 - Ph219a/CS219a Solutions to Hw 9 June 5 2009...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Ph219a/CS219a Solutions to Hw 9 June 5, 2009 Problem 1 (a) Clearly for x = 1, ln x = x- 1 = 0. Since the function f ( x ) = ln x is strictly concave, ln x ≤ x- 1 for x 6 = 1, if x- 1 is a tangent at x = 1. It is easy to see that this is indeed the case, since their slopes match at x = 1: d (ln x ) dx | x =1 = 1 x | x =1 = 1 and d ( x- 1) dx = 1. (b) Using the above inequality, we have, ln p ( x ) q ( x ) ≥ 1- q ( x ) p ( x ) ⇒ H ( p k q ) = X x p ( x )log p ( x ) q ( x ) ≥ X x p ( x ) 1- q ( x ) p ( x ) = 0 (1) Equality holds iff q ( x ) p ( x ) = 1 for all x , ie. iff the distributions { p ( x ) } and { q ( x ) } are identical. (Note that in the second step, we have omitted a factor of ln2 in going from the natural loga- rithm to the log function.) (c) Expanding ρ and σ in their eigenbasis, as ρ = ∑ i p i | i ih i | and σ = ∑ a q a | a ih a | , we have, H ( ρ k σ ) = tr ρ (log ρ- log σ ) = X i p i log p i- X i,a p i log q a h i | a ih a | i i = X i p i log p i- X a D ia log q a ! (2) where we have defined the matrix with elements D ia = |h i | a i| 2 . It is easy to see that D is a doubly stochastic matrix: ∑ a D ia = ∑ a h i | a ih a | i i = h i | i i = 1 = h a | a i = ∑ i D ia . 1 (d) Since the log function is strictly concave, and the set ∑ a D ia = 1 for each i (implying D ia ≤ 1 ∀ i,a ), Jensen’s inequality directly gives, log X a D ia q a ! ≥ X a D ia log q a (3) with equality only if D ia = 1 for some a . (e) Putting Eqns.(2) and (3) together, we have, H ( ρ k σ ) = X i p i log p i- X a D ia log q a ! ≥ X i p i log p i- log X a D ia q a ! = X i p i (log p i- log r i ) = H ( p k r ) (4) with r i = ∑ a D ia q a , and equality iff D ia = 1 for some a , for each i . (f) Now that we have expressed the quantum relative entropy as a classical relative entropy between two classical distributions p = { p i } and r = { r i } , we can use the result of part(b), so that H ( ρ k σ ) ≥ H ( p k r ) ≥ (5) The second inequality saturates iff p i = r i = ∑ a D ia q a , and the first inequality saturates iff D ia = 1 for some a , for each i . Thus D has to be a permutation matrix, which implies that the set { p i } and { q a } are the same upto a permutation. Thus, equality holds iff ρ = σ . Problem 2 (a) Consider the relative entropy of ρ AB and the product state ρ A ⊗ ρ B . Using the positivity of relative entropy, we have, H ( ρ AB k ρ A ⊗ ρ B ) ≥ ⇒ Tr[ ρ AB log ρ AB ]- Tr[ ρ AB log( ρ A ⊗ ρ B )] ≥ (6) Using log( M + N ) = log...
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

soln9_09 - Ph219a/CS219a Solutions to Hw 9 June 5 2009...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online