2
$\begingroup$

Is there a quantum way to set the unknown relative phase of a qubit (assumed in a pure state) to zero, without measurement? The relative phase is not known, otherwise I would subtract it using a phase gate.

I tried dephasing but it decoheres the state, whereas I want the state to still be pure, but with a zero relative phase. In other words, the goal is to project the qubit state onto half xz-plane (x>0) of the Bloch sphere.

Thank you in advance.

$\endgroup$
6
  • 2
    $\begingroup$ Do you mean it would map both $\frac{1}{\sqrt{2}}(\vert 0\rangle\pm\vert 1\rangle)$ to $\frac{1}{\sqrt{2}}(\vert 0\rangle + \vert 1\rangle)$? If so, no, because it would violate the unitarity of quantum mechanics. PS: I'm leaving a comment because I'm not sure I understand the question. If the OP confirms that I understand the question correctly, I'd post an answer. $\endgroup$ Jul 25 at 15:55
  • 1
    $\begingroup$ @FreeAssange right, it sounds exactly like this question quantumcomputing.stackexchange.com/questions/27465/… :) $\endgroup$ Jul 25 at 16:45
  • $\begingroup$ Thank you for your comment @FreeAssange. Yes that's right. Or I can accept states like |-⟩ with phase equal to π if I know that their phase will be equal to π, this way I can subtract it with phase gate. This is why I originally said in my question to project the qubit state onto the xz-plane, I edited it now for more precision. $\endgroup$
    – cef
    Jul 25 at 16:47
  • $\begingroup$ @cef Thanks for the clarification, I've posted an answer. $\endgroup$ Jul 25 at 18:20
  • $\begingroup$ @NikitaNemkov Haha, yeah, not sure why everyone is against unitary and linear QM today ;) $\endgroup$ Jul 25 at 18:20

1 Answer 1

2
$\begingroup$

No, this cannot be done.

One naive way to go about constructing such a mapping would be to imagine that we just erase the phase and leave everything else untouched. This would map $\sqrt{p}\vert 0\rangle+e^{i\theta}\sqrt{1-p}\vert 1\rangle\to\sqrt{p}\vert 0\rangle+\sqrt{1-p}\vert1\rangle$. This is not an allowed mapping because it violates unitarity, e.g., as I pointed out in the comment, it would map both $\vert \pm\rangle\to\vert+\rangle$ -- making it so that the mapping is irreversible and thus, not unitary.

However, one might think that one can construct a more general mapping that maps $$\sqrt{p}\vert0\rangle+e^{i\theta}\sqrt{1-p}\vert 1\rangle\to\sqrt{q}\vert 0\rangle+\sqrt{1-q}\vert 1\rangle$$ that somehow encodes both $p,\theta$ into $q$ so as to avoid being non-unitary. I'm not sure if such a mapping exists or not -- however, it is easy to see that no such mapping exists that is both unitary and linear (which is what a quantum operation ought to be).

Let's say $U$ is such a mapping. Let's say $$U\vert 0\rangle =\sqrt{p}\vert 0\rangle + \sqrt{1-p}\vert 1\rangle$$ $$U\vert 1\rangle =\sqrt{q}\vert 0\rangle + \sqrt{1-q}\vert{1}\rangle$$

Now, by linearity,

\begin{align} U\bigg(\frac{\vert 0\rangle+i\vert 1\rangle}{\sqrt{2}}\bigg)&=\frac{\sqrt{p}+i\sqrt{q}}{\sqrt{2}}\vert 0\rangle+\frac{\sqrt{1-p}+i\sqrt{1-q}}{\sqrt{2}}\vert1\rangle\\ &=\sqrt{r}e^{i\theta_1}\vert 0\rangle + \sqrt{1-r}e^{i\theta_2}\vert 1\rangle \end{align} where \begin{align} &\ r=\frac{p+q}{2}\\ \sin\theta_1 = \frac{\sqrt{q}}{\sqrt{p+q}},&\ \cos\theta_1 = \frac{\sqrt{p}}{\sqrt{p+q}}\\ \sin\theta_2 = \frac{\sqrt{1-q}}{\sqrt{2-{p-q}}},&\ \cos\theta_2 = \frac{\sqrt{1-p}}{\sqrt{2-{p-q}}} \end{align} Now, the relative phase would be $\phi\equiv\theta_2-\theta_1$. It can be calculated that \begin{align} \sin\phi\equiv\sin(\theta_2-\theta_1)=\frac{\sqrt{1-q}\sqrt{p}-\sqrt{1-p}\sqrt{q}}{\sqrt{p+q}\sqrt{2-p-q}} \end{align} Thus, for $\phi=0$, we need \begin{align} &\sqrt{1-p}\sqrt{q}=\sqrt{p}\sqrt{1-q}\\ \implies &\sqrt{\frac{1}{p}-1}=\sqrt{\frac{1}{q}-1}\\ \implies &p=q \end{align} Thus, we have got $U\vert 0\rangle=U\vert 1\rangle$ meaning that $U$ is not unitary, i.e., a contradiction, i.e., such a mapping does not exist.

I've assumed throughout that $p+q\neq 0$ and $p+q\neq 2$. However, it is clear that these conditions can be violated only if $p=q=0$ or $p=q=1$ -- in which case, again, $U$ won't be unitary because $p=q$ implies non-unitary $U$.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you for your answer, that's helpful. $\endgroup$
    – cef
    Aug 2 at 17:13

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.