1
$\begingroup$

So the Pauli twirling approximation gives us a quantum channel $\Phi$ that transforms a density matrix $\rho$ to:

$\Phi(\rho)\mapsto\sum_{i=0}^3 \sigma^i \rho \sigma^i,$

where $\sigma^0 = \mathbb{I}, \sigma^1 = X, \sigma^2 = Y$ and $\sigma^3 = Z$. So the diagonal terms are omitted. From experience of people who use this approximation, they say that this is an excellent approximation in many cases. However, simply consider the Hadamard being applied to a qubit, so that

$\rho\mapsto H\rho H = \frac{1}{2}(X+Z)\rho(X+Z) = \frac{1}{2}\left(X\rho X+Z\rho Z + X\rho Z + Z \rho X\right)$.

Here we already see that omitting the diagonal terms loses half the information.

My question is: am I wrong in any of the assumptions of Pauli twirling, and if not, why is it so powerful still if many operations can not be approximated well by omitting diagonal terms.

$\endgroup$
4
  • 1
    $\begingroup$ What do you mean this is an approximation? An approximation to what? $\endgroup$
    – Rammus
    Nov 1, 2023 at 11:58
  • $\begingroup$ An approximation of the underlying loss mechanism that generally transforms like $\Phi(\rho)\mapsto \sum_{i,j} \sigma^i \rho \sigma^j$ $\endgroup$
    – JoJo P
    Nov 1, 2023 at 12:00
  • 1
    $\begingroup$ As far as I can tell, what you have written is not a quantum channel, can you please edit your question to make it more precise? $\endgroup$
    – Rammus
    Nov 1, 2023 at 12:53
  • 1
    $\begingroup$ Pauli twirling is usually done for channels. I think this is what you meant, but the twirled channel can be more complex than your example. Anyway, twirling is commonly enforced and I have never seen this as a type of approximation; this would be a bad one as you noted. $\endgroup$ Nov 1, 2023 at 17:31

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.