Making a maximally mixed 2-qubit state in the IBM Q

Question

I am trying to make a 2-qubit maximally mixed state $\mathbb{I}/4$ where $\mathbb{I}$ is the identity $4\times 4$ matrix.

I know that, for a maximally mixed 1-qubit state I can use a Hadamard gate, and a CNOT gate with an ancilla, and then trace out the ancilla as follows:

$$(H\otimes \mathrm{id})(|0\rangle \otimes |0\rangle)=\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$$

Therefore, the density matrix would be

$$\rho=\frac{1}{2}(|0\rangle\langle0| + |1\rangle\langle1|)=\frac{1}{2}\begin{pmatrix}1&0\\0&1\end{pmatrix}$$

How can I do the same for 2 qubits?

Answer 1

Prepare two qubits each in the single-qubit maximally mixed state. Their state together will be the two-qubit maximally mixed state ;) You could prepare them sequentially and use the same ancillary qubit for both.