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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?

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  • $\begingroup$ What is the spectral expansion of $\rho^2$? Compare with of $\rho$, studying the sign of $\lambda^2-\lambda$ for $\lambda\in [0,1]$... $\endgroup$ Commented Jun 27, 2019 at 5:53
  • $\begingroup$ Notice that the sum of all $\lambda$ must equal $1$. $\endgroup$ Commented Jun 27, 2019 at 6:03
  • $\begingroup$ 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 $tr(\rho)=1$? $\endgroup$
    – BGreen
    Commented Jun 27, 2019 at 12:05
  • $\begingroup$ I now posted a complete proof. $\endgroup$ Commented Jun 27, 2019 at 12:07

3 Answers 3

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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 $\rho = |\psi_j\rangle \langle \psi_j|$. $\Box$

With a similar route one easily proves that

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. $\Box$

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  • $\begingroup$ In other words, idempotency is not sufficient, but idempotency with unit trace is, and I had overlooked the assumption of unit trace before as it was hidden in the defining properties of the density operator. Retrospectively, this makes sense because it's essentially normalization of the state. Thank you! $\endgroup$
    – BGreen
    Commented Jun 27, 2019 at 12:16
  • $\begingroup$ Yes, I agree with you. $\endgroup$ Commented Jun 27, 2019 at 12:18
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if $\rho^2=\rho$, the only eigenvalues of $\rho$ can be $0$ or $1$, as a matter of fact if

$$ \rho|\psi\rangle=\lambda|\psi\rangle$$

then

$$ \rho^2|\psi\rangle=\lambda^2|\psi\rangle=\rho|\psi\rangle=\lambda|\psi\rangle$$

hence $\lambda^2=\lambda$, i.e. $\lambda\in\{0,1\}$. Since $\mathrm{Tr}(\rho)=1$, $\rho$ can have at most one eigenvalue $1$, hence $\rho=|\psi\rangle\langle \psi|$.

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Let $\rho\ge0$ be a positive semidefinite (finite-dimensional) operator, and consider the following three conditions:

  1. $\newcommand{\tr}{\operatorname{tr}}\tr(\rho)=1$
  2. $\tr(\rho^2)=1$
  3. $\rho^2=\rho$.

Any two of the above imply the remaining one.

(1 and 2 $\Longrightarrow$ 3) Positivity implies that we can eigedecompose $\rho$ as $\rho=\sum_k p_k P_k$ with $\tr(P_j P_k)=\delta_{jk}$, $P_j$ orthogonal projections, and $p_k\ge0$. Then $\tr(\rho)=1$ implies $\sum_k p_k=1$, and $\tr(\rho^2)=1$ implies $\sum_k p_k^2=1$. These two conditions are compatible only if $p_k=\delta_{k,1}$, that is, if $\rho=P_1$ for some trace-1 projection $P_1$.

(1 and 3 $\Longrightarrow$ 2) This is obvious.

(2 and 3 $\Longrightarrow$ 1) Also obvious.

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  • $\begingroup$ I was confused at first... I'd say "any two of the above imply the remaining one". $\endgroup$
    – pglpm
    Commented Dec 1, 2020 at 12:30
  • $\begingroup$ But note that condition 1. is automatically true, by definition, for a density matrix. $\endgroup$
    – pglpm
    Commented Dec 1, 2020 at 12:31
  • $\begingroup$ @pglpm sure, that's why I stated the conditions for a positive semidefinite operator, not for a state $\endgroup$
    – glS
    Commented Dec 1, 2020 at 13:46
  • $\begingroup$ My bad, I overlooked that. $\endgroup$
    – pglpm
    Commented Dec 1, 2020 at 15:33

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