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Reading the Nielsen and Chuang, I saw that every unitary operator $U$ can be written as $e^{i\alpha} R_n(\theta)$ for some well chosen $n \in \mathbb{R}^3$ and $0 \leq \theta < 2\pi$.

I would like to know if there is a simple criterion to ensure that $\alpha=0$, without knowing the value of $n$ or $\theta$.

As a consequence of it, using the fact that $e^{i\alpha}$ "disappear" when $U$ is seen as a rotation of the Bloch Sphere, and the fact that the composition of two rotations on $\mathbb{R}^3$ is also a rotation, I will have the proof that the composition of two morphisms $R_{n_1}(\theta_1) R_{n_2}(\theta_2)$ is on the form $R_{n_3}(\theta_3)$.

This problem is thus linked to the fact that the Bloch Sphere identify the unitary transformations that are equal up to a global phase (if I am right : $U_1$ and $U_2$, seen as morphisms on the Bloch sphere, are equal iff there are equal up to a global phase).

Edit: I precise I am interested by the result even if the global phase can be removed in one-qubit gates. One of the reasons is because of the multi-qubits gates, such as for exemple the two-bits gate that is a controled one-qubit gate.

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2 Answers 2

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As to your first question, I believe global phases can always be ignored, and we generally only concern ourselves with relative phases. So we don't really require that $\alpha=0$, we just ignore it.

As to your second point, when I read

are equal iff there are equal up to a global phase

I understand it as saying that they can have different global phases (since they don't matter). So you can have $U_1 = e^{i\alpha}R_n(\theta)$ and $U_2 = e^{i\beta}R_n(\theta)$, with $\alpha \neq \beta$, and we would say $U_1=U_2$ up to a global phase.

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  • $\begingroup$ Actually, I am interested for the strict equality. One of the reasons is because of the multi-qubit gates, as I precised it in my edit. But, more generally, I would also be happy to know if it is true, even if it isn't relevant for the quantum computation applications. $\endgroup$ Feb 6, 2023 at 21:59
  • $\begingroup$ Not sure I understand your edit; we can still ignore global phases for n-qubit operations. Check out my answer here quantumcomputing.stackexchange.com/a/30077/19585, where we have an explicit example of ignoring global phases in a two-qubit gate. $\endgroup$
    – Bebotron
    Feb 6, 2023 at 22:40
  • $\begingroup$ Thanks for the link! I think there is indeed some point where I have a difficulty. My interest to keep the global phase for a one-qubit gate $U$ is to deal with controlled bit gate because I think that the global phase of $U$ becomes a relative phase if I study the controlled U gate. More generally, I think there is probably examples of multi-qubit gates V that are constructed using U, in such a way that the global phase of U only becomes relative phases in V and thus, has an effect. $\endgroup$ Feb 7, 2023 at 8:29
  • $\begingroup$ I'd be interested to see specific examples of gates you're looking at where this may be the case. It should not affect a controlled-$U$ operation. Consider the controlled operation $C(U)|a\rangle|b\rangle \rightarrow |a\rangle U|b\rangle$. You can always take out an arbitrary factor from $U$ to make it $e^{-i\alpha}U$, which again just gives you the same state up to a global phase factor $ |a\rangle e^{-i\alpha}U|b\rangle$, which can be ignored. Chapter 2 of Nielsen and Chuang should help you out. $\endgroup$
    – Bebotron
    Feb 7, 2023 at 9:23
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    $\begingroup$ For example, if I consider $U$ acting on the second qubit, controlled by the first qubit, and if I apply this gate to a state $\alpha |0\rangle| \phi \rangle + \beta | 1\rangle | \psi\rangle$, I think the result will be $| u\rangle= \alpha |0\rangle |\phi\rangle + \beta |1\rangle U(|\psi\rangle)$. If I replace $U$ by $e^{i\theta}U$, the result will instead be $|v\rangle = \alpha | 0\rangle |\phi\rangle + e^{i\theta}\beta |1\rangle U(|\psi\rangle)$, which is a change of relative phase, and thus $|u\rangle$ can be different of $|v\rangle$. $\endgroup$ Feb 8, 2023 at 20:01
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$$ \text{det}(U)=1 $$ The rotations $R_n(\theta)$ are in the special unitary group, meaning that they have determinant 1. Hence, if you want $\alpha=0$, that would be achieved by having $\text{det}(U)=1$.

Sufficiency: let's assume that $\text{det}(U)=1$. This means that its eigenvalues must be of the form $e^{\pm i\theta}$. Let $P$ be the projector onto one of these eigenvalues, so $$ U=e^{i\theta}P+e^{-i\theta}(I-P). $$ Since $P$ is a projector (Hermitian, rank 1), and the Pauli matrices are a basis, we can always write $$ P=\frac{1}{2}(I+\vec{n}\cdot\vec{\sigma}) $$ where $\vec{n}$ is real-valued and has length 1. Thus, $$ U=I\cos\theta+i\sin\theta \vec{n}\cdot\vec{\sigma}=e^{i\theta\vec{n}\cdot\vec{\sigma}}. $$

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  • $\begingroup$ By homogenity of the determinant, $\mathrm{det}(e^{i\alpha} U) = e^{i\alpha d}$ if $U\in SU(d)$. Hence, we can choose $\alpha = 2\pi k /d$ ($k\in\mathbb{Z}$) to get $\mathrm{det}(e^{i\alpha} U)=1$, i.e. $\alpha = 0, \pi$ for $d=2$. This is because the center of $SU(2)$ is $\{\pm I\}$. $\endgroup$ Mar 9, 2023 at 8:03
  • $\begingroup$ But if $\alpha=\pi$, there is an equivalent way of writing it such that $\alpha=0$, right? $\endgroup$
    – DaftWullie
    Mar 9, 2023 at 8:08
  • $\begingroup$ Yes indeed, we have $ - R_n(\theta) = R_n(\theta-\pi)$. I was missing that. Anyway, I doubt that a similar statement is true for for $d > 2$. $\endgroup$ Mar 9, 2023 at 13:35
  • $\begingroup$ @MarkusHeinrich agreed. $\endgroup$
    – DaftWullie
    Mar 9, 2023 at 13:57

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