5
$\begingroup$

Suppose I have a density matrix $\rho$ and an orthogonal projector $\Pi$. Is it true that, if $tr(\Pi \rho) = 1$ then it must hold that $$\Pi \rho \Pi = \rho$$?

If yes, how can I prove it?

$\endgroup$
6
$\begingroup$

Yes its true. Define another orthogonal projector $\Pi_\perp$ such that $\Pi + \Pi_\perp = I$ and write $\rho$ in terms of a spectral decomposition \begin{equation} \rho = \sum_k \lambda_k(\rho) |\lambda_k\rangle \langle \lambda_k| \tag{1} \end{equation}

where we're fine to just consider components such that $\lambda_k(\rho) >0$. Then we have $$ 1 = \text{Tr}(\Pi \rho) = \text{Tr}((I - \Pi_\perp) \rho) = 1 - \text{Tr}(\Pi_\perp \rho) \tag{2} $$ implying \begin{align} 0 &= \text{Tr}(\Pi_\perp \rho)\tag{3} \\&= \text{Tr}(\Pi_\perp \rho\Pi_\perp)\tag{4} \\&= \sum_k \lambda_k(\rho)\langle \lambda_k|\Pi_\perp\Pi_\perp |\lambda_k\rangle \tag{5}\\&= \sum_k \lambda_k(\rho) \lVert\Pi_\perp |\lambda_k\rangle \rVert^2\tag{6} \end{align} where line $(4)$ used the cyclic property of the trace and $\Pi_\perp \Pi_\perp = \Pi_\perp$. Now since $ \lambda_k(\rho) >0$ and $\lVert \cdot \lVert \geq 0$ the above only holds if $ \Pi_\perp |\lambda_k\rangle = 0$ for all $k$. And so \begin{align} \Pi \rho \Pi &= (I - \Pi_\perp) \rho (I - \Pi_\perp)\tag{7} \\&=\rho + \sum_k \lambda_k(\rho) \Bigl(-|\lambda_k\rangle \langle \lambda_k| \Pi_\perp -\Pi_\perp |\lambda_k\rangle \langle \lambda_k| + \Pi_\perp |\lambda_k\rangle \langle \lambda_k| \Pi_\perp\Bigr)\tag{8} \\&= \rho \tag{9} \end{align}

$\endgroup$
2
$\begingroup$

Write the eigendecomposition of the state as $\rho=\sum_k p_k u_k u_k^\dagger$, where $\{u_k\}_k$ is a family of orthonormal vectors in the underlying space.

Suppose there is some $u_k\notin \operatorname{supp}(\Pi)$, that is, some $u_k$ such that $\Pi u_k\neq u_k$. Then $$\operatorname{Tr}(\Pi u_k u_k^\dagger)=\| \Pi u_k\|^2<\|u_k\|^2=1,$$ and thus $$\operatorname{Tr}(\Pi\rho)=\sum_k p_k \operatorname{Tr}(\Pi u_k u_k^\dagger)< \sum_k p_k=1.$$ It follows that if $\operatorname{Tr}(\Pi\rho)=1$, each $u_k$ must be in the support of $\Pi$, and thus $\Pi\rho\Pi=\rho$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.