Inspired by the comments in this question How to approximate $Rx$, $Ry$ and $Rz$ gates?, there is the errata for question 4.11 pg 176 in N&C. The original form states that for any non parallel $m$ and $n$, then for an arbitrary $U$:

$U = e^{i\alpha}R_n(\beta)R_m(\gamma)R_n(\delta)$, for appropriate $\alpha,\beta, \gamma, \delta$

The errata, http://www.michaelnielsen.org/qcqi/errata/errata/errata.html, corrects this s.t.

$U = e^{i\alpha}R_n(\beta_1)R_m(\gamma_1)R_n(\beta_2)R_m(\gamma_2)\dots$.

However I found that other textbooks such as Kayes and Mosca (pg. 66, thm 4.2.2), and various online material still quotes the original form of the theorem. As such I am wondering is the errata correct, and is just that all the other material has 'incorrectly' quoted the result from N&C?


1 Answer 1


The errata is correct. I had a project student who erroneously took one of these mis-quotes and she spent ages working with it, realising it didn't make sense, and subsequently proving that the stated formula was incorrect, only later to find the N&C erratum. As you say, it has propagated far and wide!

If you want some insight about the problem, imagine $n$ and $m$ are two axes that are almost parallel (visualise this on the Bloch sphere). Now imagine I start with a state that is aligned with the $n$ axis. With a sequence $n-m-n$, so the claim goes, I should be able to produce any state on the surface of the Bloch sphere. But the first $n$ does nothing because of what our initial state is. Then the $m$ only creates a rotation preserving the angle of the initial state with the axis, and so it's always close to where it stated. The same again with the final rotation about the $n$ axis.

  • $\begingroup$ Thanks! That is what I thought! I just wanted to make sure I wasn't crazy seeing it in all these other reputable places! I was trying to expand out the corrected form to prove it using quaterions to prove the statement, but things got messy quickly. $\endgroup$
    – Sam Palmer
    Commented May 19, 2020 at 15:11

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