3
$\begingroup$

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?

|improve this question
$\endgroup$
2
$\begingroup$

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.

|improve this answer
$\endgroup$
  • $\begingroup$ 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 '19 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 '19 at 14:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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