3
$\begingroup$

Let us see the one-qubit case of different noise channel, the depolarizing channel is $\Lambda_1^{depol}(\rho_1)=(1-\frac{4}{3}\epsilon_1)\rho_1+\frac{1}{3}\epsilon_1\sum_{i=0}^{3}\sigma_i\rho_1\sigma_i$, the dephasing noise is $\Lambda_1^{depha}(\rho_1)=(1-2\epsilon_1)\rho_1+\epsilon_1\sum_{i\in\{0,3\}}\sigma_i\rho_1\sigma_i$, the bit flip channel is $\Lambda_1^{bit}(\rho_1)=(1-\epsilon_1)\rho_1+\epsilon_1\sigma_x\rho_1\sigma_x$, and the Pauli channel is $\Lambda^{Pau}(\rho)=\sum_jp_jP_j\rho P_j$, where $p_j$ is the Pauli error rate associated with the Pauli operator $P_j$. There are also other quantum noise models.

Can I say the common of those quantum noises is that they all have a probability distribution which generated by the error rates, for example, for the depolarizing channel, the rates $\{1-\frac{4}{3}\epsilon_1, \frac{1}{3}\epsilon_1,\frac{1}{3}\epsilon_1,\frac{1}{3}\epsilon_1,\frac{1}{3}\epsilon_1\}$ form a probability distribution; and for Pauli channel, $p_j$ forms a probability distribution thus $\sum _jp_j=1$. Am I right?

$\endgroup$
0

2 Answers 2

2
$\begingroup$

Operator sum representation

Quantum channels may be represented in an operator-sum form: For a channel $\Lambda$ acting on a $d$-dimensional state $\rho$, there exists some set of $d$-dimensional operators $\{E_1, \dots, E_k, \dots\}$ such that \begin{align} \Lambda(\rho) &= \sum_k E_k \rho E_k^\dagger \tag{1} \\ \sum_k E_k^\dagger E_k &= I_d, \tag{2} \end{align}

where $I_d$ is the identity operator (see Sec. 8.2.3 of Nielsen and Chuang or Corollary 2.27 of Watrous' TQI). In particular, if you can express the channel using only Pauli operators then we can choose to define $$ E_k = \sqrt{p_k}\sigma_k \tag{3}, $$ where we set $d=2$ for a single qubit and $p_k>0$ without loss of generality. In this case, tracing the LHS of Eq. (2) gives us \begin{align} \text{Tr}\left(\sum_k E_k^\dagger E_k\right) &= \sum_k p_k \text{Tr}\left(\sigma_k^\dagger \sigma_k \right) \tag{4a-c} \\&= \sum_k p_k \text{Tr}\left( I_2\right) \\ &= 2\sum_k p_k, \end{align} where we have used $\sigma_k^2 = I$ for all four Pauli operators. Meanwhile tracing the RHS of Eq. (2) gives $\text{Tr}(I_2) = 2$. Then, $$ \sum_{k=0}^3p_k = 1 \tag{5} $$ is sufficient for $\Lambda$ to be a valid channel. So under the identification of Eq. (3), if you are given a channel that permits an operator sum form that uses only Paulis (or you choose to define a channel in that way), then the coefficients will form a probability distribution. From Lines (4a-c) this holds for an operator sum form written in terms of an orthonormal basis for $2\times 2$ Hermitian matrices.

Not all channels obey have coefficients summing to one

If you are given an operator sum representation consisting of operators other than Paulis, the above does not necessarily hold. For example, the "replace with $|0\rangle$" channel has operator sum representation \begin{align} E_0 &= |0\rangle \langle 0|\\ E_1 &= |0\rangle \langle 1|. \end{align} The corresponding channel sends every input $\rho$ to $|0\rangle$. Clearly the coefficients for each $E_k$ do not form a probability distribution. Indeed, using $ |0\rangle \langle 0|= \frac{1}{2}(I + \sigma_3)$ and $|0\rangle \langle 1| = \frac{1}{2} (\sigma_1 - i\sigma_2)$ you can verify that there's no way to write this channel in the form of Eq (1) that uses only Pauli operators, and so the above analysis doesn't hold.

$\endgroup$
3
  • $\begingroup$ Thanks a lot. Several questions: 1) Is it sufficient to use only the Pauli operator (Eq. 3) if I want to represent quantum noise? Since we describe quantum noise in terms of quantum channels, I will now focus only on operator sum representations of quantum noise, and can all the types of quantum noise I described above (although it is not known which category they fall into) characterized by Pauli operators, right? 2) Thanks for your note, and I am curious about why this paper uses the non-standard depolarizing channel arxiv.org/abs/2201.00752 ? $\endgroup$
    – karry
    Oct 24, 2022 at 3:14
  • $\begingroup$ 1 - No, it is not sufficient to use only the Pauli operator, the "replace with $|0\rangle$" channel I describe does not have a Pauli-only representation. Off the top of my head I don't know how to check for whether a channel admits this form $\endgroup$
    – forky40
    Oct 24, 2022 at 12:32
  • $\begingroup$ 2 - Actually, I think the form of the depolarizing channel they use is fine, the use of $I$ twice in the sum just made it look weird to me. $\endgroup$
    – forky40
    Oct 24, 2022 at 12:39
1
$\begingroup$

In this paper, the authors use $$ \Lambda_1^{depol}(\rho_1)=(1-\frac{4}{3}\epsilon_1)\rho_1+\frac{1}{3}\epsilon_1\sum_{i=0}^{3}\sigma_i\rho_1\sigma_i$$ instead of $$ \Lambda_1^{depol}(\rho_1)=(1-\epsilon_1)\rho_1+\frac{1}{3}\epsilon_1\sum_{i=1}^{3}\sigma_i\rho_1\sigma_i$$ since they try to derive a general expression for an arbitrary number of qubits, such that the summation will contain all possible Pauli strings including $IIII$, $IXXX$, $XYZI$, otherwise the summation of these terms cannot be expressed by a direct product of each qubit.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.