The trace of a density matrix squared, Tr(p^2), is always less than or equal to 1, with equality when p is a pure state.
Proof:
A density matrix p is positive semi-definite and hermitian. This means that it can be diagonalized with positive eigenvalues. Let's denote these eigenvalues as {λi}, with i ranging from 1 to N.
Then, we have:
Tr(p^2) = Σ λi^2
Since all λi are positive and sum up to 1 (because Tr(p) = 1), the maximum value of Tr(p^2) is achieved when one of the λi equals 1 and all others equal 0. This corresponds to a pure state.
In all other cases, when p is a mixed state, we have more than one non-zero λi and hence:
Tr(p^2) < 1
This shows that Tr(p^2) is always less than or equal to 1, with equality when p is a pure state.