# Implementing a depolarizing channel for 2 qubits on IBM Q

I am trying to use IBM Q to perform the following depolarizing channel on a state of 2 qubits $$\rho=|\psi \rangle \langle \psi |$$:

$$\rho \to (1-\lambda)\rho + \frac{\lambda}{4}I$$

This is within the limitations of the IBM Q. Namely 5 channels.

How can I come up with a combination of gates that will give this result?

A specific case I want to try would be

$$|\psi\rangle = \cos \theta |00\rangle + \sin \theta |11\rangle$$

• For example, I can do it with 5 cables if I use the cables $1$ and $2$ for the input state, cables $3$ and $4$ to prepare a completely mixed $I/4$ state (unsure how to do that too) and the last cable ($5$) for the $(1-\lambda)|0\rangle + \lambda|1\rangle$ state and then use two $\mathrm{CSWAP}$ gates to switch the $1 \leftrightarrow 3$ and $2 \leftrightarrow 4$ cables controlled by cable $5$. Apr 29 '19 at 11:02