Let $\rho\in M_n$ and $\sigma\in M_m$ be two quantum states. We denote the orthogonal projections onto $\text{range}(\rho)$ and $\text{range}(\sigma)$ by $P_\rho$ and $P_\sigma$, respectively. Now, if the two states are connected via the action of a quantum channel $\Psi:M_m\rightarrow M_n$, i.e., $\rho = \Psi(\sigma)$, is it true that

$$ \text{range}[\Psi(P_\sigma)] \subseteq \text{range}(P_\rho) = \text{range}(\rho) \,?$$

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    $\begingroup$ What exactly do you mean by the range of a quantum state? Also, note that any two states $\rho$ and $\sigma$ are connected by the channel $\Psi(\sigma) = \mathrm{tr}(\sigma)\rho$, which might be of interest to you. $\endgroup$
    – JSdJ
    Commented Mar 22, 2021 at 11:47
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    $\begingroup$ @JSdJ A quantum state is a positive semi-definite linear operator acting on a Hilbert space with unit trace. So its range is just the range of that linear operator. Thanks for your comment! $\endgroup$
    – mathwizard
    Commented Mar 22, 2021 at 11:57

1 Answer 1


Yes. In fact, equality can be shown to hold. Let us spectrally decompose $\sigma = \sum_{i}\lambda_i P_i$, where $\lambda_i > 0$ are the non-zero eigenvalues of $\sigma$ and $P_i$ are the orthogonal projectors onto the corresponding eigenspaces, so that $\rho=\Psi(\sigma)=\sum_i \lambda_i \Psi(P_i)$. Then it is clear that $P_\sigma = \sum_{i}P_i$ and $\Psi(P_\sigma)=\sum_{i}\Psi(P_i)$. Now, it is easy to see that

$$\text{range}[\Psi(P_\sigma)]=\text{range}[\sum_{i}\Psi(P_i)]=\text{range}[\sum_i \lambda_i \Psi(P_i)]=\text{range}(\rho), $$

since for arbitrary positive semi-definite matrices $A,B$ and arbitrary positive numbers $a,b>0$, we have $\text{range}(A+B) = \text{range}(aA+bB)$.


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