0
$\begingroup$

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?

Improve this question
$\endgroup$
2
  • 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 $\endgroup$
    – Quantum Mechanic
    Jan 25, 2022 at 15:46

1 Answer 1

Reset to default
2
$\begingroup$

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, $$ \sigma_{10322}=X\otimes I\otimes Z\otimes Y\otimes Y $$

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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

Your Answer

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

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