2
$\begingroup$

I am going through the paper Surface code with decoherence: An analysis of three superconducting architectures and I have a doubt about how the authors get what they refer to as the combined channel of amplitude and phase damping. In the section IIA, the authors discuss amplitude damping and dephasing channels with their Kraus operators in order to describe decoherence first, and afterwards they combine their effects. They state that the combined channel is represented by 3 Krauss operators depending on the amplitude damping and dephasing parameters.

I am wondering how such Kraus operators are obtained from the Kraus operators describing the individual channels, as the authors do not explain nor give any reference to such statement. My initial approach has been to consider a sequential combination of the channels, but my results are not successful (probably because the simultaneous action of the channels is not equivalent to their sequential actions on the quantum information).

$\endgroup$
  • $\begingroup$ Please add details. $\endgroup$ – Norbert Schuch Jul 12 at 8:25
4
$\begingroup$

You can obtain the Kraus operators of the combined channel by taking products of the Kraus operators of the individual channels (using the notation from the paper you linked):

Amplitude damping:

$E^{AD}_1 = \begin{bmatrix} 1 & 0 \\ 0 & \sqrt{1-p_{AD}} \end{bmatrix}$, $E^{AD}_2 = \begin{bmatrix} 0 & \sqrt{p_{AD}} \\ 0 & 0 \end{bmatrix}$

Phase damping:

$E^{PD}_1 = \begin{bmatrix} 1 & 0 \\ 0 & \sqrt{1-p_{PD}} \end{bmatrix}$, $E^{PD}_2 = \begin{bmatrix} 0 & 0 \\ 0 & \sqrt{p_{PD}} \end{bmatrix}$

Combined:

$E^{D}_1 = E^{PD}_1 E^{AD}_1 = \begin{bmatrix} 1 & 0 \\ 0 & \sqrt{1-p_{AD}}\sqrt{1-p_{PD}} \end{bmatrix} $

$E^{D}_2 = E^{PD}_1 E^{AD}_2 = \begin{bmatrix} 0 & \sqrt{p_{AD}} \\ 0 & 0 \end{bmatrix} $

$E^{D}_3 = E^{PD}_2 E^{AD}_1 = \begin{bmatrix} 0 & 0 \\ 0 & \sqrt{1-p_{AD}}\sqrt{p_{PD}} \end{bmatrix} $

This is the Kraus set given in the paper you linked. There is a fourth possible combination, which is

$E^{D}_4 = E^{PD}_2 E^{AD}_2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} $

which is the null channel. Since we first destroy any $|1\rangle$ states, the phase damping channel only has $|0\rangle$ states to act on, which are sent to 0.

The order in which you apply amplitude and phase damping does not actually matter, that is

$\mathcal{E}_{AD} \circ \mathcal{E}_{PD} (\rho) = \mathcal{E}_{PD} \circ \mathcal{E}_{AD} (\rho)$.

Thus, you could swap the products in the Kraus terms defined above, which would result in a different Kraus set (now with four non-null elements), which would also describe the channel (the Kraus representation is not unique).

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks a lot, I did not recall the non-uniqueness of the Krauss representation. $\endgroup$ – Josu Etxezarreta Martinez Jul 13 at 7:45
1
$\begingroup$

The other answer already uses this, but just to make the general fact more explicit: if $\mathcal E=\mathcal E_A\circ\mathcal E_B$, that is, $\mathcal E(\rho)=\mathcal E_A(\mathcal E_B(\rho))$, and the Kraus decompositions of the single channels read $$\mathcal E_A(\rho)=\sum_a A_a\rho A_a^\dagger, \qquad \mathcal E_B(\rho)=\sum_b B_b\rho B_b^\dagger,$$ then $\mathcal E(\rho)=\sum_{a,b} C_{ab}\rho C_{ab}^\dagger$, where $C_{ab}\equiv A_a B_b$ are the Kraus operators of the combined channel.

| improve this answer | |
$\endgroup$

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.