# Necessary and sufficient conditions for a pure state

I've seen some claims that idempotency ($$\rho^2=\rho$$) is necessary and sufficient to guarantee the existence of some state $$\psi$$ such that $$\rho=|\psi\rangle\langle\psi|$$, as well as claims on the trace such as here. However, I have so far been unable to prove a necessary and sufficient condition for a density matrix to represent a pure state.

It is easy to see that if $$\rho=|\psi\rangle\langle\psi|$$ then $$\rho$$ is idempotent: $$\rho^2=|\psi\rangle\langle\psi|\psi\rangle\langle\psi|=|\psi\rangle\langle\psi|=\rho$$ by normalization. However, the converse has proven more difficult I have been able to show, by expanding in a basis and noting that the eigenvalues of an idempotent matrix are either 0 or 1, that if $$\rho$$ is idempotent then $$\rho=\sum_j \lambda_j |\psi_j\rangle\langle\psi_j|$$ where each $$\lambda_j$$ is either 0 or 1. How can I complete the proof? Alternatively, is idempotency not a sufficient condition for a pure state?

• What is the spectral expansion of $\rho^2$? Compare with of $\rho$, studying the sign of $\lambda^2-\lambda$ for $\lambda\in [0,1]$... – Valter Moretti Jun 27 at 5:53
• Notice that the sum of all $\lambda$ must equal $1$. – Valter Moretti Jun 27 at 6:03
• That is not true - for example, $\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0.5 & 0.5 & 0 \\ 0 & 0.5 & 0.5 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$ is idempotent and has the sum of all eigenvalues equal to 2. Are you saying that another condition needs to be imposed, that the trace must be 1 so that there can only be 1 nonzero eigenvalue? This gives me a proof that idempotency together with unit trace is a sufficient condition, at least, but is it necessary to have $tr(\rho)=1$? – BGreen Jun 27 at 12:05
• I now posted a complete proof. – Valter Moretti Jun 27 at 12:07

THEOREM 1 If $$\rho$$ is a density matrix (i.e., a positive, unit-trace, trace-class operator, also in an infinite dimensional Hilbert space), then $$\rho$$ is a pure state iff $$\rho^2=\rho$$.
Proof. If $$\rho$$ is pure, then $$\rho^2=\rho$$. Let us prove the converse implication. Suppose that $$\rho^2 = \rho$$ ($$\rho^2$$ is trace class if $$\rho$$ is because the set of trace class operators is a $$^*$$ ideal of the $$C^*$$-algebra of bounded operators) then $$0=\rho^2-\rho = \sum_{j} (\lambda_j^2 -\lambda_j) |\psi_j\rangle \langle \psi_j|\tag{1}$$ where, from the definition of density matrix (positive unit-trace trace-class operator) $$\lambda_j \in [0,1]\tag{2}$$ and $$\sum_j \lambda_j =1\tag{3}\:.$$ Let us assume that there are at least two $$j\neq j'$$ with $$\lambda_j,\lambda_{j'}>0$$. We conclude from (2) that both $$\lambda_j^2-\lambda_j<0$$ and $$\lambda_{j'}^2-\lambda_{j'}<0$$. Since $$\langle \psi_k|\psi_h \rangle = \delta_{hk}$$, (1) leads to $$0 = \langle \psi_{j'}|0 \psi_{j'}\rangle = \lambda_{j'}^2-\lambda_{j'} <0$$ that is impossible. We conclude that the assumption that there are at least two $$j\neq j'$$ with $$\lambda_j,\lambda_{j'}>0$$ is untenable so that $$\rho = |\psi_j\rangle \langle \psi_j|$$. $$\Box$$
THEOREM 2 If $$\rho$$ is a density matrix (also in an infinite dimensional Hilbert space), then $$\rho$$ is a pure state iff $$tr(\rho^2)=tr(\rho)$$ ($$=1$$).
Proof. If $$\rho$$ is pure the thesis it trivial. Let us pass to the converse implication. Since $$\sum_j \lambda_j =1$$ and $$\lambda_j\in [0,1]$$, if more than one $$\lambda_j$$ does not vanish, every $$\lambda_j$$ is strictly less than $$1$$, so that we have in particular $$\lambda^2_j < \lambda_j$$ for all $$j$$, which implies $$\sum_j \lambda_j^2 < \sum_j \lambda_j$$. This meas that if $$tr(\rho^2) = tr(\rho)$$, then only one $$\lambda_j$$ does not vanish so that $$\rho$$ is pure. $$\Box$$