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?


1 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.

  • $\begingroup$ I tried doing it with the same ancillary qubit for both, but it seems that this is only possible in the simulator. I wanted to run it in the real processor if possible, but I managed to do it in the simulator the way you told me. Thank you so much! $\endgroup$
    – The Bosco
    Apr 30, 2019 at 3:53
  • $\begingroup$ Huh. Can you do it on the real processor using two ancillary qubits, one for each target qubit? I don't know much about the IBM-Q machine, but it seems like this at least should be possible. $\endgroup$
    – Will
    Apr 30, 2019 at 14:18

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