# Ranges of quantum states that are related via a quantum channel

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

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

• 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.
– JSdJ
Commented Mar 22, 2021 at 11:47
• @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! Commented Mar 22, 2021 at 11:57

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 $$\text{range}(A+B) = \text{range}(aA+bB)$$.