[Physics] How to prove that $\mathrm{tr}(\rho^2)=1$ if and only if the state is pure

density-operatorhilbert-spacehomework-and-exercisesoperatorsquantum mechanics

How can I prove that $\mathrm{tr}(\rho^2) $ = 1 if and only if the state is pure?

My idea:
I know how to show that $\mathrm{tr}(\rho^2) \leq 1$ and from there I am trying to show by contradiction that $\mathrm{tr}(\rho^2) = 1$ can only but true for pure state, but I am kind of stuck.

I just need a hint on how to prove this.

Best Answer

Since an arbitrary $\rho$ is self-adjoint, it has the spectral decomposition $\rho = \sum_n \rho_n |\psi_n><\psi_n|$, in terms of an orthonormal basis $\{ |\psi_n>\}$, which here we pick discrete for simplicity.

Hermiticity implies $\rho_n = \rho^*_n$. $\mathrm{Tr} \rho = 1$ implies $\sum_n \rho_n =1$. Semi-positivity implies $0 \leq \rho_n$. Together they imply $0 \leq \rho_n \leq 1$, which implies $\rho^2_n \leq \rho_n$. Hence, $\mathrm{Tr} \rho^2 = \sum_n \rho^2_n \leq \sum_n \rho_n = 1$ and so $\mathrm{Tr} \rho^2 \leq 1$ for a generic state, as you mention.

Now let's start assuming that $\mathrm{Tr} \rho^2 =1$. Following the inequalities we just wrote, this implies that $\rho^2_n = \rho_n$ for all $n$ i.e. $\rho^2 = \rho$. In particular this implies that $\rho_n = 1$ or $\rho_n = 0$. More precisely, due to the trace condition $\mathrm{Tr} \rho = 1$, only one $\rho_n$ is equal to one while the others vanish. This is a pure state.

The reverse implication is direct.