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Maximum value of Von Neumann Entropy

  1. Apr 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Prove that the maximum value of the Von Neumann entropy for a completely random ensemble is ##ln(N)## for some population ##N##

    2. Relevant equations
    ##S = -Tr(ρ~lnρ)##
    ##<A> = Tr(ρA)##

    3. The attempt at a solution
    Using Lagrange multipliers and extremizing S

    Let ##~S = -Tr(ρ~lnρ) + γ (<1> - 1) = -Tr(ρ~lnρ) + γ (Tr(ρ) - 1)##

    ##δS = Tr( -δρ~ lnρ - ρ \frac{δρ}{ρ} + γ δρ) = 0##

    ## ∑_k [-lnρ_{kk} - 1 + γ] δρ = 0 ##

    ⇒ ##ρ_{kk} = e^{γ-1}##

    Let ##γ = \frac{1}{N}##

    ##ρ_{kk} = e^{\frac{1}{N}-1} → \frac{1}{N}## for N very large.

    Thus, ##S = -Tr(ρ~lnρ) = -∑_k \frac{1}{N} ln(\frac{1}{N}) = lnN##

    Is my solution valid?
     
  2. jcsd
  3. Apr 23, 2016 #2
    Actually, I think after the line ##ρ_{kk} = e^{γ-1}## everything is wrong. I should use the result ##Tr(ρ)=1## for a completely random ensemble.

    So we take the sum from 1 to N, ##∑_k ρ_{kk} = ∑_k e^{γ-1} =Ne^{γ-1} = 1~## ⇒ ##~e^{γ-1} = \frac{1}{N}~##

    Thus, ##S = -Tr(ρ~lnρ) = -∑_k \frac{1}{N} ln(\frac{1}{N}) = lnN##

    I think this is kinda correct. Any comment?
     
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