Purity, spectral weights and the criterion for a pure state

[(a) the weights are positive and sum to one] By [F1] the ${\lambda }_n$ are exactly the positive eigenvalues with multiplicity, so ${\lambda }_n>0$ for $n\in J$; the identity ${\sum }_{n\in J}{\lambda }_n=\mathrm{tr}\rho =1$ is [F1] together with [F3] applied to the positive trace-class operator $\rho$; since all terms are nonnegative and their sum is $1$, each single term satisfies ${\lambda }_n\le 1$.

[(b) the eigenvalues of ${\rho }^2$] By [F4] and [F5] the operator ${\rho }^2$ is self-adjoint, positive and trace class. For each distinct positive eigenvalue $\lambda$ of $\rho$, [F1] lists exactly its finite multiplicity of orthonormal eigenvectors among $(u_n)$; since an orthonormal family is linearly independent and $E_{\lambda }$ has that finite dimension by [F2], those vectors form an orthonormal basis of $E_{\lambda }$. Their union is therefore a complete orthonormal family of $M$ by [F2]. If ${\rho }^{2}x=\mu x$ with $\mu >0$, then $x={\mu }^{-1}{\rho }^{2}x\in \mathrm{ran}\rho \subseteq M$ by [F2], so [F6] gives the norm-convergent expansion $x={\sum }_{n\in J}{a}_n{u}_n$, where $a_n=⟨x,u_n⟩$. The expansion in [F1] gives $\rho u_n={\lambda }_n{u}_n$; applying the bounded operator $\rho$ twice to the finite partial sums of the expansion of $x$ yields ${\rho }^{2}x={\sum }_{n\in J}{\lambda }_n^2{a}_n{u}_n$ in norm. By uniqueness of coefficients in [F6], $({\lambda }_n^2-\mu )a_n=0$ for every $n$. Thus the $\mu$-eigenspace of ${\rho }^2$ is exactly the closed span of those $u_n$ with ${\lambda }_n^2=\mu$: the reverse inclusion holds because each such $u_n$ is an eigenvector of ${\rho }^2$, and the forward inclusion follows from the coefficient identity and completeness. Since ${\lambda }_n>0$, the positive eigenvalues of ${\rho }^2$ are exactly the numbers ${\lambda }_n^2$ with multiplicity. By [F3], $\mathrm{tr}({\rho }^2)={\sum }_{n\in J}{\lambda }_n^2$. The sum is positive because some ${\lambda }_n>0$ by $\sum {\lambda }_n=1$, and it is at most $1$ because ${\lambda }_n^2\le {\lambda }_n$ by step 1.1 and $\sum {\lambda }_n=1$.

[(c) equivalences] Statement (i) holds, that is ${\sum }_n{\lambda }_n^2=1={\sum }_n{\lambda }_n$, if and only if ${\sum }_n{\lambda }_n(1-{\lambda }_n)=0$; since $0\le {\lambda }_n\le 1$ by step 1.1, each term is nonnegative, so this happens exactly when ${\lambda }_n\in \{0,1\}$ for every $n$, and because ${\lambda }_n>0$ on the index set $J$ and ${\sum }_{n\in J}{\lambda }_n=1$, exactly when $J$ consists of a single index $n_0$ with ${\lambda }_{n_0}=1$; this is (ii). If (ii) holds, then the expansion of [F1] reads $\rho =T_{u_{n_0},u_{n_0}}$ with $u_{n_0}$ a unit vector of the orthonormal family, so $\rho$ is pure, which is (iii), by [F7]. If $\rho$ is pure, say $\rho =T_{u,u}$ with $\mathrm{\parallel }u\mathrm{\parallel }=1$, then ${\rho }^2=\rho$ by the product rule $T_{u,u}{T}_{u,u}=T_{T_{u,u}u,u}=T_{u,u}$ recalled in [F7], giving (iv). Finally, if ${\rho }^2=\rho$ and $\rho x=\lambda x$, then ${\lambda }^{2}x={\rho }^{2}x=\rho x=\lambda x$, so $\lambda \in \{0,1\}$ for the positive eigenvalues $n\in J$; since their sum is $1$ there is exactly one index $n_0$ with ${\lambda }_{n_0}=1$, which is (ii), and then $\mathrm{tr}({\rho }^2)=\mathrm{tr}\rho =1$ gives (i). The implications (i)$\Rightarrow$(ii)$\Rightarrow$(iii)$\Rightarrow$(iv)$\Rightarrow$(ii) show that all four statements are equivalent.

[(d) non-pure states have at least two positive eigenvalues] If $\rho$ is not pure, then $\mathrm{tr}({\rho }^2)\mathrm{\ne }1$ by step 3.1, and $\mathrm{tr}({\rho }^2)\le 1$ by step 2.1, so $\mathrm{tr}({\rho }^2)<1$. If $J$ had exactly one index $n_0$, then ${\sum }_{n\in J}{\lambda }_n={\lambda }_{n_0}=1$ by step 1.1, and condition (ii) of step 3.1 would hold, making $\rho$ pure by that step, a contradiction; hence $J$ has at least two indices. This proves (a)-(d).