Kraus and Pauli Operators

A few things in summary

  1. There is nothing wrong with Ike’s theorem
  2. I would like to retract the assertion regarding rectangular vs. square matrices being the key difference between Kraus and Pauli operators. Whether a matrix is full rank is is not enough to determine expansion in Pauli basis, though having rectangular matrices do make physically implementation a little more tricky
  3. The key difference between Kraus and Pauli comes from the fact that Kraus operator, when used to describe open quantum system/error induced by the environment, are not necessarily trace preserving, but Pauli operators are
  4. I will offer an example at the end

For a generic set of Kraus operators $\{K_i\}$ that gives rise to a quantum channel $\mathcal{K}$, it is required that (by Chuang Thm 8.1) that

$$ \sum K^\dagger_i K_i\leq I. $$

This is equivalent to saying $\mathcal{K}$ is not necessarily trace preserving. ( trace preserving would be $1=\text{tr}\big(\mathcal{E}(\rho)\big)=\text{tr}\big(\sum K_i^\dagger \rho K_i\big)=\text{tr}\big(\rho\sum K_i^\dagger K_i\big)\rightarrow\sum K_i^\dagger K_i=I$ )

And for a set of properly normalized Pauli operators $\{P_i\}$, $\sum P_i^\dagger P_i=I$, which makes it trace preserving.

Example

A concrete example is the amplitude damping channel used to model, for example, spontaneous emission, where the set of Kraus operator is

$$ K_0 = \begin{pmatrix} 1 & 0 \\ 0 & \sqrt{1 - \gamma} \end{pmatrix}, \quad K_1 = \begin{pmatrix} 0 & \sqrt{\gamma} \\ 0 & 0 \end{pmatrix}  $$

$K_1$ can easily be written as superposition of Pauli basis, $K_1=\sqrt{\gamma}(X+iY)$ but $K_0$ cannot. In this sense, we cannot invoke Ike’s Theorem on equivalence of Kraus representations, reproduced below

<aside> 💡

Suppose $\{E_1, . . . , E_m\}$ and $\{F_1, . . . , F_n\}$ are operation elements giving rise to quantum operations $\mathcal{E}$ and $\mathcal{F}$, respectively. By appending zero operators to the shorter list of operation elements we may ensure that $m=n$. Then $E = F$ if and only if there exist complex numbers $u_{ij}$ such that $E_i=\sum_ju_{ij}F_j$ , and $u_{ij}$ is an $m$ by $m$ unitary matrix.

</aside>

We realize that any linear combination of $I$ and $Z$ (the only ones with access to diagonal elements) will yield a matrix of trace 1, and produce something like $K_0$.

Two things to note:

  1. $K_0$ is full rank square matrix, but it does not have Pauli expansion.

  2. $K_1$ is not full rank but has Pauli expansion.

To shed more light, this is the same as having decaying fidelity through the loss of probability mass. For example, we prepare our single qubit in $|0\rangle$ and $|1\rangle$, but our error drives the qubit to $|2\rangle$ with some probability, so the probability for measuring it in $|0\rangle$ or $|1\rangle$ both decreases, and the trace of the 2x2 density matrix for this qubit is no longer 1. To model this kind of error, we need operators that are not trace preserving, and Pauli operators, being trace preserving, would not fulfill this request.