Question

For a mixed state given by the density matrix p, show that Tr(e) < 1, with equality when p is a pure state. (Hint: You can solve this either by considering p as a statistical mixture of some ensemble of states {Pi, [Wi)} so p= EpilWi) (vil, or simply using the properties of the density matrix, i.e. that it is hermitian, positive semi-definite and that Tr(e) = 1)

Answer 1

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.