# Why is the von Neumann entropy maximised when $p_n=1/N$?

For example, $$\hat \rho=\sum_{\substack{n}} p_n |E_n \rangle \langle E_n|$$ is a stationary mixed state of a given quantum system, where $$|E_n \rangle$$ are eigenstates of the Hamiltonian $$\hat H$$ with the eigenvalues $$E_n$$.

In every book that I read on quantum information, it says that $$p_n=1/N$$ for the maximal value of the von Neumann entropy for the given mean energy.

I don't see how we could end up with this value. And I would like to understand how to prove that.

Any references to books/websites where I can read about the derivation is also much appreciated.

Any help is much appreciated.

• Check Schur-concavity. Commented Feb 11, 2020 at 18:47
• $p_n=1/N$ is the maximal value of the entropy without constraining the mean energy. If you constrain the mean energy, $p_n=1/N$ might not even be a possible probability distribution
– glS
Commented Feb 15, 2020 at 18:38

I'm assuming you meant the say that $$p_n=1/N$$ is the maximal value of the von Neumann entropy without constraining the mean energy (otherwise the statement is not true).
One way to see it is using Lagrange multipliers. You want to maximise $$S(\mathbf p)\equiv-\sum_i p_i \log p_i$$ in the hyperplane $$\sum_i p_i=1$$. For this to be the case, $$\nabla_{\mathbf p} S$$ needs to be proportional to $$\nabla_{\mathbf p}(\sum_i p_i-1)=\sum_i \hat{\mathbf e}_i$$. This means that, for some $$\lambda$$, you have $$- (\log p_i + 1) = \lambda,$$ and thus $$p_i=e^{-1-\lambda}$$. Imposing $$\sum_i p_i=1$$ then gives $$N=e^{1+\lambda}$$. We conclude that $$p_i = 1/N$$ for all $$i$$.
The case constraining the average energy can also be worked out, but now you are maximising $$S(\mathbf p)$$ in the intersection of the hyperplanes $$\sum_i p_i=1$$ and $$\sum_i p_i E_i=E$$. A variation of Lagrange multipliers can handle this more general case (and you get a distribution of the type $$p_n\simeq e^{-\beta E_n}$$).