Depolarization of density operator with zeros in diagonal

Question

I suppose a quantum state with density matrix like the following is not valid. $$ \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}. $$

Now, let's say I have a valid density operator representing the state $|\psi \rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1 \rangle)$. $$ |\psi \rangle \langle\psi | = \frac{1}{2}(|0\rangle \langle 0| + |0\rangle \langle 1| + |1\rangle \langle 0| + |1\rangle \langle 1|) = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}. $$

Now I send this state to depolarizing channel $\mathcal{E}$. Because $\mathcal{E}$ is linear: $$ \mathcal{E}(|\psi \rangle \langle\psi |) = \frac{1}{2}(\mathcal{E}(|0\rangle \langle 0|) + \mathcal{E}(|0\rangle \langle 1|) + \mathcal{E}(|1\rangle \langle 0|) + \mathcal{E}(|1\rangle \langle 1|)). $$

I'm wondering what the depolarization of $\mathcal{E}(|0\rangle \langle 1|)$ would mean. By definition of depolarizing channel, for noise parameter $p$,

$$ \mathcal{E}(\rho) = (1 - p)\rho + \frac{pI}{2}. $$

But then, what is the meaning of $\mathcal{E}(|0\rangle \langle 1|)$?

Answer 1

Quantum channels are foremost, linear operators. So given a basis for the Hilbert-Schmidt operator space (for example the states $\{|0\rangle\langle 0|,|0\rangle\langle 1|,|1\rangle\langle 0|,|1\rangle\langle 1|\}$ that you've chosen above), where density matrices reside, it acts linearly on the basis elements. Perhaps, the easiest way to see it is to write it in the Kraus form, $$\mathcal{E}(X) = \sum\limits_{j}^{} K_{j} X K_{j}^{\dagger}$$ with $\sum\limits_{j}^{} K_{j}^{\dagger} K_{j} = \mathbb{I}$, where its linear action is more "transparent" (I'm assuming the convex-combination way of writing is what's confusing you here). With this, the action of $\mathcal{E}$ on $| 0 \rangle \langle 1 | $ is $\mathcal{E}(| 0 \rangle \langle 1 | ) = \sum\limits_{j}^{} K_{j} | 0 \rangle \langle 1 | K_{j}^{\dagger}$, whatever that's equal to.

There is no "meaning" of $\mathcal{E}(|0\rangle\langle 1|)$ because $|0\rangle\langle 1|$ is not a state; so don't worry about interpreting it.