# Maximum value of Von Neumann Entropy

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1. Apr 22, 2016

### Whitehole

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. Apr 23, 2016

### Whitehole

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?