I would like to rotate my $|\Psi\rangle$ state towards $|1\rangle$: $$ |\Psi\rangle= a|0\rangle + b|1\rangle \ \rightarrow \ |\Psi'\rangle= a'|0\rangle + b'|1\rangle$$ with $|a'| < |a|$, $|b'| > |b|$ and without prior knowledge of $a$, $b$.

I think a way to do it is to do a partial swap with an ancilla: $$ |\Psi\rangle|1\rangle_a \ \rightarrow \ |\Psi'\rangle|\phi\rangle_a $$

A criteria is that the strength of the rotation is parametrized $U(\theta)$. What happens to the ancilla $|\phi\rangle_a$ does not matter.

Is there something like a partial swap?

Thanks :)

  • $\begingroup$ Could you not just measure your qubit in the standard basis? If you get the answer 1, apply an $X$. Then, whatever state you started with, you end with $|0\rangle$. $\endgroup$
    – DaftWullie
    Commented Apr 11, 2022 at 6:51

1 Answer 1


$\newcommand{\1}{|1\rangle}$ $\newcommand{\0}{|0\rangle}$

seems to me like what you're looking for is a "reverse" amplitude damping channel.

Consider this circuit:

enter image description here

where the ancilla starts off at $\0$. If the initial state is $\1$, the first X gate will turn it into $|0\rangle$ and nothing will happen. If it's $\0$, the controlled Y will rotate the ancilla towards $\1$, which will then rotate the original qubit towards $|1\rangle$. The strength of the rotation will be determined by $t$.

If we calculate the action of this circuit on the data qubit $|\psi\rangle = a\0 + b\1$, the final state will be: $$ (a\cos\theta\0 +b\1)\0+(a\sin\theta\1)\1 $$ and so you can see that regardless of the state of the ancilla, the final amplitudes will be closer to $\1$.

  • $\begingroup$ Thanks! You call it a '(reverse) amplitude damping channel'. Is this circuit known? Can you point me to a source? :) $\endgroup$ Commented Apr 11, 2022 at 2:03
  • 1
    $\begingroup$ the amplitude damping channel is a well known representation of a T1 relaxation process as a quantum channel. See, for example, Nielsen and Chuang, anniversary edition, section 8.3.5. Since the amplitude damping channel will take any given state closer to zero, the idea here was to do an X gate on the qubit to bring it closer to 1. $\endgroup$
    – Lior
    Commented Apr 11, 2022 at 6:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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