# How to write the Kraus representation for many-qubit states?

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

$$∑k(a_k^∗I+b_k^∗X+c_k^∗Y+d_k^∗Z)ρ(a_kI+b_kX+c_kY+d_kZ)$$

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?

• 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)?
– JSdJ
Jan 25, 2022 at 9:29
• 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 Jan 25, 2022 at 15:46

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, $$\sigma_{10322}=X\otimes I\otimes Z\otimes Y\otimes Y$$
• 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 Jan 26, 2022 at 2:34
• 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. Jan 26, 2022 at 7:53