The most general formula of Kraus operator on density matrix is:

$$\rho\to \sum_k A_k^\dagger\rho A_k.$$ If I want to write this equation for one qubit, the most general way will be:

$\rho_f = (a^*I+b^*X+c^*Y+d^*Z)\rho(aI+bX+cY+dZ)$ Thanks to @Quantum Mechanic, the formula should be


Now, I want to write it for n qubit. (For 2 and 3 qubit and then generalized it)

How to write these equation for 2, 3 and multi qubits?

  • 3
    $\begingroup$ In general those $a,b,c,d$ are complex numbers and should be (complex) conjugated on the right: $\rho_{f} = \big(\alpha I + \beta X + \gamma Y + \delta Z\big)\rho\big(\alpha^{*} I + \beta^{*} X + \gamma^{*} Y + \delta^{*} Z\big)$. Or is this what you mean already (but you have them on the left)? $\endgroup$
    – JSdJ
    Jan 25, 2022 at 9:29
  • 3
    $\begingroup$ This equation for one qubit is misleading; it is better to write something like $\sum_k (a_k^*I+b_k^*X+c_k^*Y+d_k^*Z)\rho(a_k I+b_k X+c_k Y+d_k Z)$ because the sum over $k$ is essential in general $\endgroup$ Jan 25, 2022 at 15:46

1 Answer 1


The Pauli matrices and their tensor products form a basis for however many qubits you want. The way that I would choose to write this is $$ A_k=\sum_{x\in\{0,1,2,3\}^n}a_x\sigma_x $$ for $n$ being the number of qubits you want. I'm using the notation as, for example, $$ \sigma_{10322}=X\otimes I\otimes Z\otimes Y\otimes Y $$

  • $\begingroup$ Thanks for the answer. well it is my bad but I did not understand how to open this formula for 2 qubits. It is obvious for $\sigma$ how to write from $x∈{0,1,2,3}^n $ but I did not understand how to use n = number of qubit in this formula and a $\endgroup$
    – quest
    Jan 26, 2022 at 2:34
  • 1
    $\begingroup$ So if you want two qubits, you set $n=2$, which means that you have to consider all possibilities for $x$ from 00,01,02,03,10,11,...,33 in the summation. The $a_x$ are just numbers (possibly complex), the equivalent of your a,b,c,d in the question. $\endgroup$
    – DaftWullie
    Jan 26, 2022 at 7:53

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