# Are states with same von Neumann entropy the same up to unitary equivalence?

Let $$H$$ be a finite-dimensional Hilbert space with identity operator $$I$$. It is well-known that the von Neumann entropy $$S$$ of a density operator $$\rho$$ on $$H$$ is maximized if and only if $$\rho = I$$. That is, $$S(\rho) = S(I) \Longleftrightarrow \rho = I.$$ Is this true in general for mixed states, up to unitary equivalence? In other words, if $$\rho$$ and $$\sigma$$ are two density matrices on $$H$$ representing mixed states such that $$S(\rho) = S(\sigma)$$, does it follow that $$\rho = U\sigma U^\dagger$$, for some unitary $$U$$?

Obviously, no, for a simple counting argument: The entropy is only a single number, while states for which $$\rho=U\sigma U^\dagger$$ must have identical spectrum, which is specified (up to trace) by $$d-1$$ numbers.
So if the Hilbert space has dimension $$d>2$$, there are counter-examples. (Just take any two diagonal states with different entries but same entropy.)
On the other hand, for $$d=2$$, your statement is correct, as the spectrum is specified by one number (since the trace is one).