We know that a single-qubit Unitary can be defined as a single rotation of angle $\theta$ around some axis $\hat{n}$, together with a global phase $\alpha$ (see Nielsen & Chuang Eq. 4.9):

$$ U = e^{i\alpha} R_{\hat{n}}(\theta) $$

My question is, does this generalize to higher-dimensional Unitaries on multiple qubits? Can such a Unitary be thought of as a rotation with one angle (or a set of angles?) around some higher-dimensional axis?

One idea, assuming that a Unitary can be thought of as a single rotation, is to look at the number $k$ such that:

$$ U^k = \mathbb{I} $$

Assuming that $U$ performs some rotation of angle $\theta$, then a rotation of $2\pi$ would effect the identity. The rotation angle could then be thought of as $\theta = 2\pi/k$.

  • 1
    $\begingroup$ related: quantumcomputing.stackexchange.com/a/23366/55 and links therein. The gist is that unitaries do always correspond to ${\bf SO}(n)$ operations in the Bloch representation, but those don't generally look like "rotations" in the sense of there being a periodic motion around some fixed axis. $\endgroup$
    – glS
    Commented Jul 13, 2022 at 8:15

1 Answer 1


A single-qubit rotation looks like $U(\theta,\mathbf{n})=\exp(-i \theta\mathbf{n}\cdot\pmb{\sigma})$ for the vector of Pauli matrices $\pmb{\sigma}=(\sigma_1,\sigma_2,\sigma_3)$. Here we assume $\mathbf{n}$ is normalized to unity; otherwise we can equivalently define some vector $\mathbf{r}=\theta\mathbf{n}$ and say $U(\mathbf{r})=\exp(-i \mathbf{r}\cdot\pmb{\sigma})$.

It indeed holds that every unitary can be written as the exponential of some Hermitian "generator" $$U(G)=\exp(-iG).$$ There is always a generalized basis in which this generator can be written for any number of qubits (see Wikipedia), with $G=\mathbf{r}\cdot\pmb{\Sigma}$, where I'm using $$\pmb{\Sigma}=(\sigma_0\otimes\sigma_0\otimes\cdots,\sigma_0\otimes\sigma_1\otimes\cdots,\cdots,\sigma_3\otimes\sigma_3\otimes\cdots)$$ to represent all possible combinations of Pauli matrices. If we normalize $\mathbf{r}$, we get to write $$U(\theta,\mathbf{n})=\exp(-i\theta \mathbf{n}\cdot\pmb{\Sigma})$$ in any dimension, so long as we use all of the basis generators in $\pmb{\Sigma}$ and we make $\mathbf{n}$ have sufficient dimensions ($4^N$ for $N$ qubits, minus one for the global phase).

Can we interpret $\theta$ as a rotation angle in the sense that the same unitary is enacted for $\theta+2\pi$? All unitaries have eigenvalues with modulus one, of the form $e^{i\lambda}$, so we know that adding phases will do something sensible. Start by finding an eigenbasis of the Hermitian operator $\mathbf{n}\cdot\boldsymbol{\Sigma}$: $$\mathbf{n}\cdot\boldsymbol{\Sigma}|\psi_k\rangle=\lambda_k|\psi_k\rangle,$$ where each eigenvalue $\lambda_k$ and eigenvector $|\psi_k\rangle$ may depend on $\mathbf{n}$ and $\boldsymbol{\Sigma}$. We can then write the spectral decomposition of $U$ in this eigenbasis as $$U(\theta,\mathbf{n})=\sum_{k}e^{-i\theta \lambda_k}|\psi_k\rangle\langle\psi_k|.$$

Let's stare at this for a few moments. If $\theta\to\theta +2\pi$, does $U$ stay the same? Only if $\exp(-2\pi i\lambda_k)=1$ for all $\lambda_k$! Maybe we'll permit $U$ to change by a global phase and still enact the same transformation, so we need the slightly more generous but still restrictive condition $\exp(-2\pi i\lambda_k)=e^{i\varphi}$ for all $\lambda_k$ and some real phase $\varphi$. Not so easy for a general distribution of eigenvalues.

If we keep staring, we realize that each $e^{-i\theta\lambda_k}$ is rotating around in the complex plane with period $2\pi/\lambda_k$, so we are seeking another rotation angle $\theta+\phi$ that matches up all of those rotations to each return to their starting point. If all of the ratios $\lambda_k/\lambda_l$ are rational numbers [or even more technically if this is true for any $(\lambda_k+E)/(\lambda_l+E)$ all for the same $E$ because the absolute "eigenenergies" of the generator just lead to global phases], there will always be some angle $\theta+\phi$ for which the phases all align and the unitary returns to the same value as $\theta$ (mathematically, this is still just looking for $\exp(-i\phi\lambda_k)=1$ for all $\lambda_k$). In general, the ratios may be irrational, so the unitary may never return to the same point, but in many special cases it does.

What kind of special cases are we talking about? Well if you've heard of angular momentum operators, you know the eigenenergies are spaced by half integers, so you know the rotations will eventually return to the same starting point. If all you tell me is that you have qubits, I can concoct an arbitrary unitary that either does or does not return to its starting point: $$U_{\mathrm{return}}=e^{-i\theta}\left(|0\rangle\langle 0|\right)^{\otimes N}+e^{-2i\theta}\left(|1\rangle\langle 1|\right)^{\otimes N}+e^{-3i\theta}\left(|0\rangle\langle 0|\right)^{\otimes N/2}\otimes \left(|1\rangle\langle 1|\right)^{\otimes N/2}$$ returns to its starting point when $\theta\to \theta+2\pi$ and $$U_{\mathrm{no\,return}}=e^{-i\theta}\left(|0\rangle\langle 0|\right)^{\otimes N}+e^{-\sqrt{2}i\theta}\left(|1\rangle\langle 1|\right)^{\otimes N}+e^{-\sqrt{3}i\theta}\left(|0\rangle\langle 0|\right)^{\otimes N/2}\otimes \left(|1\rangle\langle 1|\right)^{\otimes N/2}$$ never returns to its starting point for any $\theta\to\theta^\prime$.

  • $\begingroup$ Hi @quantum-mechanic, thank you for your answer! I think this makes sense to me. Looking at your equation for $\Sigma$, should the indices for the Pauli matrices not run from $0$ to $3$ instead of $0$ to $N$, with each element of the vector being a product of $N$ terms? $\endgroup$
    – David
    Commented Jul 13, 2022 at 18:40
  • $\begingroup$ Thanks @David, edited to fix that! $\endgroup$ Commented Jul 14, 2022 at 19:13
  • $\begingroup$ I like your answer, it shows a nice generalization of the angle-unit vector form of the 1 qubit case. I think what is missing is a discussion of the parameter theta. Is it periodic, and can it be thought of as a rotation angle? The comment by @glS seems to disagree with this notion. $\endgroup$
    – David
    Commented Jul 14, 2022 at 20:11
  • $\begingroup$ @David sometimes there is still periodicity, in general there is not because of multiple phases needed to align simultaneously, I updated the answer $\endgroup$ Commented Mar 15, 2023 at 14:04

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