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This is from exercise 4.11 in Nielsen and Chuang's Quantum Computation and Quantum Information (10th Anniversary Edition), which I think might be in error.

Background

A quantum unitary operator on the state of a single qubit (a 2-level system whose state space is represented by a 2D complex Hilbert space) is a member of $SU(2)$. $SU(2)$ is related to $SO(3)$ in such a way that we can think of members of $SU(2)$ as rotations for the present application. In fact, they have the effect of rotating the Bloch vector representation of the quantum state: http://www.vcpc.univie.ac.at/~ian/hotlist/qc/talks/bloch-sphere-rotations.pdf

It is known that an arbitrary such operation $U$ may be represented as follows:

\begin{eqnarray*}U &\equiv& R_{\hat{n}}(\theta) \equiv e^{i \alpha} e^{i \frac{\theta}{2} \hat{n} \cdot \vec{\sigma}} \\ &=& \cos\left(\theta/2\right) I + i \sin\left(\theta/2\right) \hat{n} \cdot \vec{\sigma} \\ &=& \cos\left(\theta/2\right) I + i \sin\left(\theta/2\right) (n_x \sigma_x + n_y \sigma_y + n_z \sigma_z) \, \, , \end{eqnarray*}

where the $\sigma_j$ are the Pauli matrices for $j \in \{x, y, z\}$, $\hat{n}$ is a unit vector in $\mathbb{R}^3$, and $\alpha$ is some phase angle (that incidentally is irrelevant to quantum mechanics, but its relevance might not be a consideration for this question). Note that this rotates the Bloch vector representation of a quantum state (let's call it $\vec{q}$) by an angle $\theta$ around $\hat{n}$, as is described in the link above and elsewhere.

A theorem states that an arbitrary rotation may be decomposed into $z$ and $y$ rotations as follows: $U = e^{i\alpha} R_{\hat{z}}(\beta)\, R_{\hat{y}}(\gamma)\, R_\hat{z}(\delta) \equiv e^{i\alpha} e^{i\beta \sigma_z} e^{i\gamma \sigma_y} e^{i\delta \sigma_z}$, where $\alpha$, $\beta$, $\gamma$, and $\delta$ are phase or rotation angles.

Claim

The claim (which the exercise asks the reader to prove using the theorem mentioned in the previous paragraph) is that given some non-equal fixed axes $\hat{n}$ and $\hat{m}$, an arbitrary operation $U \equiv R_{\hat{r}}(\theta) = e^{i \phi} e^{i \frac{\theta}{2} \hat{r} \cdot \vec{\sigma}}$ may be decomposed as follows:

\begin{equation*} U = e^{i\alpha} R_{\hat{m}}(\beta)\, R_{\hat{n}}(\gamma)\, R_\hat{m}(\delta) \, \, , \end{equation*}

for some angles $\alpha$, $\beta$, $\gamma$, and $\delta$.

Contention

Consider given $\hat{m}$ and $\hat{n}$ that are very close to each other (e.g. their dot product is very close to but not equal to 1, perhaps 0.99). Consider $\hat{r}$ that is very far away from either $\hat{m}$ or $\hat{n}$ (e.g. their dot product is close to 0, perhaps 0 or 0.01). Consider trying to rotate a vector $\vec{v}$ around $\hat{r}$, $\vec{v}$ being initially very close to $\hat{m}$ and $\hat{n}$ prior to the rotation (where $\vec{v}$ may be thought of as a Bloch vector in the quantum context we're discussing). Surely, we can think of rotations around $\hat{r}$ that cannot be achieved by rotations around $\hat{m}$ and $\hat{n}$ as claimed in the exercise.

Note

This set of errata mention this exercise having an error, but it refers to the printing from the year 2000 (see erratum referring to p. 176), and doesn't show what the original exercise had stated: http://www.michaelnielsen.org/qcqi/errata/errata/errata.html . I have the 2010 edition and so would expect that the error would have been corrected in my edition (I can't tell if the original erroneous exercise was the same as printed in the later 2010 edition).

Thanks!

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  • $\begingroup$ $SU(2)$ is not isomorphic to $SO(3)$! $SU(2)$ is isomorphic to the universal cover of $SO(3)$ and the kernel of the mapping $p: SU(2) \rightarrow SO(3)$ is $\{\pm 1\}$. $\endgroup$ – Mortified Through Math Jun 12 '17 at 15:32
  • $\begingroup$ @SZN perhaps this is the wrong relationship. I know that every rotation of the form I wrote down on qubits (2D complex Hilbert space) can represented by a rotation on a 3D real vector space). Do I have my terminology wrong? Am I confusing the group names? $\endgroup$ – Sherif F. Jun 12 '17 at 15:35
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    $\begingroup$ The double covering of $SU(2)$ over $SO(3)$ is close enough to an isomorphism for many purposes, so the objection may only amount to a minor quibble. Anyhow, you may wish to check out the Wikepedia article on Euler angles, which may answer your question, I suspect. $\endgroup$ – Harald Hanche-Olsen Jun 12 '17 at 15:37
  • $\begingroup$ An isomorphism is an equivalence functor in the category of groups. It means the two groups are algebraically identitcal. As @HaraldHanche-Olsen has said, $SU(2)$ is algebraically close to $SO(3)$, but that's because it's the universal covering group. I am not sure if there is a huge difference in your application, but isomorphism is a very special thing, so it's worth not confusing it. $\endgroup$ – Mortified Through Math Jun 12 '17 at 15:40
  • $\begingroup$ @SZN - see new edits that avoid reference to an isomorphism. $\endgroup$ – Sherif F. Jun 12 '17 at 15:44
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The Lie algebra of $SO(3)$ is spanned by $\{1,\sigma_i\}$ where the $\sigma_i$ are the Pauli matrices for $i = 1,2,3$. Thus your representation $$ U = e^{\alpha + \hat{n}\cdot \sigma} $$ is simply writng the rotation matrix as the exponential of an element in the Lie algebra. The center of $\mathfrak{so}(3)$ is $\mathbb{R}$, so we always have $$ e^{\alpha + \hat{n}\cdot \sigma} = e^{\alpha} e^{\hat{n}\cdot \sigma}. $$ This appears in your background section. The question is then whether or not $e^{\hat{n}\cdot \sigma}$ may be factored into rotations about single axes. The answer is generally no. For one, the fundamental group of $U(1) \times U(1) \times U(1)$ is not isomorphic to the fundamental group of $SO(3)$. However, there are charts on $SO(3)$ which allow many rotations to be represented by rotations about single axes. These are the so-called Euler angles.

You can also get closer approximations by applying Zassenhaus' expansion to $e^{\hat{n}\cdot \sigma}$ along each axis. Writing the pauli matrices as quaternions makes it clear that the adjoint endomorphism of $\mathfrak{so}(3)$ is just the cross product, so the computational burden of this approach is not so much.

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Let's denote by $N_{m,n}$ the smallest integer $p$ such that any $U\in SU(2)$ is the product of $p$ rotations about either $m$ or $n$. Then Lowenthal's [1] showed that

$$N_{m,n}=\left\lceil\frac{\pi}{\arccos|m\cdot n|}\right\rceil + 1.$$

So, following your intuition, when $m\cdot n$ goes to 1, $N_{m,n}$ goes to $+\infty$. That the decomposition you sought is impossible is then a trivial corollary.

Note the result holds for $SO(3)$ as well [2]. A recent and rather clear demonstration can be found in [3].

[1] Franklin Lowenthal, Uniform finite generation of SU(2) and SL(2,R), Can J. Math. 24 (1972), 713--727, https://cms.math.ca/openaccess/cjm/v24/cjm1972v24.0713-0727.pdf

[2] Franklin Lowenthal, Uniform finite generation of the rotation group, Rocky Mountain J. Math. 1 (1971), 575--586, http://projecteuclid.org/download/pdf_1/euclid.rmjm/1250131726

[3] Mitsuru Hamada, The minimum number of rotations about two axes for constructing an arbitrarily fixed rotation, Royal Society Open Science 1 (2014), http://rsos.royalsocietypublishing.org/content/1/3/140145

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