# Depolarization of density operator with zeros in diagonal

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 $$\mathcal{E}(|0\rangle \langle 1|)$$?

• I'm not exactly sure what your question is. Anyway, if you want to apply your channel on arbitrary matrices, you have to extend the map linearly to the full matrix space. In this case, this would be $\tilde{\mathcal{E}}(M) = (1-p)M + p \mathrm{tr}(M) I/2$. Now, you can use linearity and evaluate the terms individually. With your definition of $\mathcal{E}$, the second equation does NOT hold. Nov 9 '20 at 8:58
• The depolarising channel is a convex mixture between the identity channel and the fully depolarising channel which resets any input to the maximally mixed state. The "linear" form of the latter channel involves the trace of the input matrix, see also quantumcomputing.stackexchange.com/questions/14456/… Nov 9 '20 at 9:00

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.