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I understand that CNOT and Toffoli (CCNOT) gates are not universal for quantum computation since they (alone) cannot create superposition. Moreover, it is totally possible to perform a CNOT operation where the control qubit is in superposition, $\alpha\vert 0 \rangle + \beta\vert 1 \rangle$, and target qubit to be $\vert 0 \rangle$.

My question is can we have a control qubit to be in superposition while using CCNOT, something like the following?

If yes, what will be the values for $p, q, $ and $r$? If not, why not?

The related material I found are universality of the toffoli gate and CNOT in superposition.

Thank you in advance!

So if the 2nd qubit is $\vert 1 \rangle$, then $\vert q \rangle = \vert 1 \rangle$, $ \vert p \rangle = \alpha\vert 0 \rangle + \beta\vert 1 \rangle$ and $ \vert r \rangle= \vert pr \rangle = \alpha \vert 00 \rangle + \beta \vert 11 \rangle$? – M. Al Jumaily – 2019-12-18T19:01:32.103

No. $\vert pr \rangle$ is entangled 2-qubit state which cannot be factored into separate $\vert p \rangle$ and $\vert r \rangle$ states. In other words, there is $\vert pr \rangle$ but no $\vert p \rangle$ and $\vert r \rangle$ – kludg – 2019-12-18T19:05:55.303

So I can say that the input is three separate qubits and the output is a one separate qubit and one entangled 2-qubit state? – M. Al Jumaily – 2019-12-18T19:13:33.340

Yes, this is OK; or simply "one separate qubit and two entangled qubits". – kludg – 2019-12-18T19:18:38.437

One last thing, I had $\vert q \rangle = \vert 0 \rangle$ and you said it would be more interesting if $\vert q \rangle = \vert 1 \rangle$. Can $\vert q \rangle = \alpha_2 \vert 0 \rangle + \beta_2 \vert 1 \rangle$? How about all three qubits to be in superposition? is that allowed? – M. Al Jumaily – 2019-12-18T19:24:22.333

I guess you mean 2nd input; yes, if both control qubits are in superposition then all 3 qubits should be entangled; target qubit can be in superposition too; superposition is always allowed. – kludg – 2019-12-18T19:32:07.480