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I know there are various quantum noise channels, which include the depolarizing channel, the dephasing channel and the bit-flip channel; We can apply them in simulators easily.

However, is there any standard to choose the model we apply?

I mean, why we should (for instance) choose a flip channel over a depolarizing channel and when should we choose a specific channel? Why is the depolarizing channel commonly used?

glS
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Henry_Fordham
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  • I'm not aware of any 'standard' for choosing a channel over another. May depend on your specific scenario. Depolarizing channel is usually chosen as it is sort of the 'worst' that can happen to a travelling qubit. – QuestionEverything Jul 30 '20 at 02:39
  • what do you mean that: 'worst' that can happen to a travelling qubit – Henry_Fordham Jul 30 '20 at 02:41
  • Consider what happens in a depolarizing channel. There is a probability of bit-flip, phase-flip and both bit and phase flilp in a depolarizing channel. So it's sort of the worst case scenario. You can find more here: http://www.theory.caltech.edu/people/preskill/ph219/chap3_15.pdf – QuestionEverything Jul 30 '20 at 02:54
  • http://epubs.surrey.ac.uk/849158/1/__homes.surrey.ac.uk_home_.System_Desktop_08423050_Duality%20of%20Quantum%20and%20Classical%20Error%20Correction%20Codes-%20Design%20Principles%20and%20Examples.pdf in this page also mentioned the worst scenario, but is that neccessary to apply 3/p for X,Y,Z Error? – Henry_Fordham Jul 30 '20 at 03:02
  • I mean is that mean: only 3/p for X,Y,Z error called the depolarizing channel(because it told me that in the link i send), if so , why the circumstance when 3/p for X,Y,Z error is the worst case, but not other possibility for X,y,Z error? – Henry_Fordham Jul 30 '20 at 03:04
  • Assuming you've meant 'p/3' instead of '3/p', this just means, with probability (1-p), no error is occurring to the qubit, and then remaining probability, p, is being equally distributed to the X, Y, Z errors. – QuestionEverything Jul 30 '20 at 03:26
  • so, I just want to ask why "equally distributed to the X, Y, Z errors" is the worst case, but not "not-equal" cases? – Henry_Fordham Jul 30 '20 at 04:03
  • The 'worst-case' scenario is actually a little different. If you have Kraus operators $\sqrt{\frac{p}{3}}(X, Y, Z)$ and $\sqrt{1-p}I$, you get the maximally mixed state for $p = \frac{3}{4}$. – JSdJ Jul 30 '20 at 08:25

1 Answers1

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Some thoughts:

A theoretical perspective

From a theoretical perspective, the depolarizing channel is the 'standard' (if there is such a thing) or by some means the most applicable.

Because the Paulis (together with the identity operator) form a basis for $SU(2)$, if a code can correct the $X, Y$ and $Z$ flips on a certain qubit (and it it is able to correctly identify no error (i.e. '$I$-flip') having happened), it can correct all errors on that qubit. A theoretical analysis of a code cares a little less about the relative probabilities of these three flips happening, so we just as well can set them all to $\frac{p}{3}$, thereby obtaining the depolarizing channel.

The depolarizing channel $\Lambda_{\mathrm{depo}}$ can be written as:

$$ \Lambda_{\mathrm{deph}}\left(\begin{bmatrix}a & b \\ b^{*} & d\end{bmatrix}\right) = \begin{bmatrix} (1-\frac{2p}{3})a + \frac{2p}{3}d & (1-\frac{4p}{3}) b \\ (1-\frac{4p}{3})b^{*} & (1-\frac{2p}{3})d + \frac{2p}{3}a\end{bmatrix}. $$ Since $d = 1-a$, we can rewrite this to:

$$ \Lambda_{\mathrm{deph}}\left(\rho_{\mathrm{in}}\right) = (1-\frac{4p}{3})\rho_{\mathrm{in}} + \frac{4p}{3} \frac{I}{2}, $$ which is a convex combination of the input $\rho_{\mathrm{in}}$ and the maximally mixed state $\frac{I}{2}$. Furthermore, if you would equate $p$ to the elapsed time, $p$ would asymptotically go to $\frac{3}{4}$, thereby obtaining the maximally mixed state; therefore the depolarizing channel is in some way the 'worst' noise channel: it destroys both all quantum- (i.e. coherent superpositions) and classical information (there's literally only noise left) in the qubit.

A physical perspective

If you want your simulation to be more true to the physical world, the depolarizing channel is not a very good model, as much as theorists might like it to be. A good first model for noise in qubits is the combination of two channels, the dephasing channel $\Lambda_{\mathrm{deph}}$ and the amplitude damping channel $\Lambda_{\mathrm{amp}}$.

Loosely speaking, the dephasing channel map destroys the coherent phase between the $|0\rangle$ and $|1\rangle$ state:

$$ \Lambda_{\mathrm{deph}}\left(\begin{bmatrix}a & b \\ b^{*} & 1-a\end{bmatrix}\right) = \begin{bmatrix}a & e^{-\frac{t}{T_{2}}} b \\ e^{-\frac{t}{T_{2}}}b^{*} & 1-a\end{bmatrix}, $$

where $T_{2}$ is known as the characteristic qubit dephasing time.

The Kraus operators of $\Lambda_{\mathrm{deph}}$ are $A_{1} = \sqrt{1-p}I$ and $A_{2} = \sqrt{p}Z$, so it is still a Pauli channel, which can help in the analysis or simulation.

$\Lambda_{\mathrm{amp}}$ is a little bit trickier: it simulates the relaxation of the excited (by convention $|1\rangle$) state, and maps it to the $|0\rangle$ state:

$$ \Lambda_{\mathrm{amp}}\left(\begin{bmatrix}a & b \\ b^{*} & 1-a\end{bmatrix}\right) = \begin{bmatrix}a & e^{-\frac{t}{2T_{1}}} b \\ e^{-\frac{t}{2T_{1}}} b^{*} & e^{-\frac{t}{T_{1}}}(1-a)\end{bmatrix}, $$ where $T_{1}$ is known as the qubit relaxation time.

The Kraus operators are also a little trickier: $B_{1} = \begin{bmatrix}1 & 0 \\ 0 & \sqrt{1-p}\end{bmatrix}$ and $B_{2} = \begin{bmatrix}0 & \sqrt{p} \\ 0 & 0\end{bmatrix}$. This means that amplitude damping channel is not a Pauli channel (allthough its Kraus operators can of course be written as linear combinations of the Paulis).

A simulation of a system undergoing both an amplitude damping and dephasing is a good start. There is one big caveat, however: this does not take leakage errors into account whatsoever. Depending on the physical system you are trying to simulate, this might range from either not a very large issue (e.g. for quantum dots) to a gross oversimplification (e.g. for transmon qubits).

JSdJ
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