Set relative phase of qubit to zero without measurement?

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.

• 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. Jul 25 at 15:55
• @FreeAssange right, it sounds exactly like this question quantumcomputing.stackexchange.com/questions/27465/… :) Jul 25 at 16:45
• 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.
– cef
Jul 25 at 16:47
• @cef Thanks for the clarification, I've posted an answer. Jul 25 at 18:20
• @NikitaNemkov Haha, yeah, not sure why everyone is against unitary and linear QM today ;) Jul 25 at 18:20

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 $$U$$.