0
$\begingroup$

In section 4.5.1 of Nielsen and Chuang, two-level unitary matrices are defined as unitary matrices which act non-trivially only on two or fewer vector components. I'm not sure that I understand this definition. For instance, is $$\begin{pmatrix} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ 0 & 1 & 0\end{pmatrix}\quad$$ a two level unitary matrix? It only acts non-trivially on the first and third vector components, but it doesn't leave this linear space invariant, unlike the examples given in Nielsen and Chuang.

$\endgroup$
  • $\begingroup$ "It only acts non-trivially on the first and third vector components..." That's not correct. Enumerating the basis vectors as $\{e_1, e_2, e_3\}$, that matrix transforms $T e_2 = e_3$, which is not "trivial". $\endgroup$ – DanielSank Aug 4 '18 at 17:36
  • $\begingroup$ Yes, I guess that acting trivially on component $j$ means that $Te_j = e_j$ I thought that it might mean that $U_{ij} = 0$ or $1$. I have continued reading and in the end you prove that unitaries such that $Te_j = e_j$ except for at most two j are universal, which is the aim of the section, so I guess that the definition you suggest is the right one. Thanks! $\endgroup$ – Alvaro Ortega Aug 4 '18 at 18:27
1
$\begingroup$

In addition to the error pointed out by DanielSank in his comment, note that your matrix is not even unitary.

Indeed, in particular for a unitary matrix any two distinct columns (resp. rows) form orthogonal vectors, and you can quickly check that it is not the case with the first and third columns (resp. first and second rows) of your matrix.

But, the following matrix works, i.e. is a two-level unitary matrix: $$\begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0\\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1\end{pmatrix}\quad$$

$\endgroup$
  • $\begingroup$ Oops sorry. Well that can be fixed by changing any of the $\frac{1}{\sqrt{2}}$ into $-\frac{1}{\sqrt{2}}$ $\endgroup$ – Alvaro Ortega Aug 4 '18 at 18:15
  • $\begingroup$ @AlvaroOrtega: exactly, I have added an example of a two-level unitary matrix in my answer. $\endgroup$ – Anthony Bordg Aug 4 '18 at 19:13
0
$\begingroup$

Unitary matrices must satisfy $U^\dagger = U^{-1}$. Your example is not unitary because it isn't invertible so $U^{-1}$ doesnt exist. In that entire section, he is only talking about unitary (and therefore invertible) matrices.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.