8
$\begingroup$

I'm really new to quantum computation. Now, I'm going through a tutorial article Quantum Computation: a Tutorial (NB: PDF). I was confused by certain points over there.

So, on page 5, when the author was talking about the Bloch sphere, it mentioned the correspondences was pictured in the figure, where the index $s$ indicates spherical coordinates; index $c$ indicates 3-dimensional coordinate; and index $\mathcal{H}$ indicates coordinates in $\mathcal{C}^{2}$. My question is: how does the index $s$ and index $\mathcal{H}$ correspondence to each other?

Then, the author continues with talking about the three canonical "orthogonal" bases for a quantum bit, as the basis along $z$, the basis along $x$ and the basis along $y$ with the hadamard gate as an example. I do not get here either.

After digging a bit for hadamard gate, I found that, based on the book, this single quit gates correspond to rotations and reflections of the sphere. "The Hadamard operation is just a rotation of the sphere about the $\hat{y}$ axis by 90 degree, followed by a rotation about the $\hat{x}$ by 180 degree." So, I kindof understand the basis along $z$, basis along $x$ and basis along $y$. Still, any comments are greatly appreciated.

$\endgroup$
4
  • $\begingroup$ Just a side comment - what do you know about linear algebra? If you know it, great (sorry for asking). If you don't, I'd suggest you try to learn some - 3Blue1Brown's videos are a great place to start, and understanding some of the basics will help you a lot. $\endgroup$
    – auden
    Commented Feb 15, 2017 at 23:18
  • $\begingroup$ The correspondence between the s and H coordinates is given at the beginning of section 2.4 with the function $r(\theta, \phi)$, which takes your spherical coordinates as input and outputs an element of $C^2$. $\endgroup$ Commented Feb 16, 2017 at 13:51
  • 1
    $\begingroup$ Related : CR Drost's answer in Understanding the Bloch sphere $\endgroup$
    – Voulkos
    Commented Jul 12, 2018 at 16:22
  • $\begingroup$ Go here: quantumcomputing.stackexchange.com $\endgroup$
    – hft
    Commented Oct 21, 2023 at 4:08

1 Answer 1

1
$\begingroup$

The index $\mathcal H$ is just a notation that tells you the vector is in the Hilbert space $C^{2}$. The index $s$ Tells you the 2 numbers in the parentheses are 2 angles $\phi,\theta $ , which correspond to a certain point on the unit sphere in three dimensions. Every point on the unit sphere corresponds to a ray - an equivalence class of states in $\mathcal H$.

Every basis of $C^{2}$ is made out of 2 vectors, and the three bases are just standard choices, orthogonal in the sense of taking the scalar product of the 2 basis vectors $(a|0 \rangle +b|1 \rangle,c|0\rangle+d|1\rangle)=\bar ac+\bar bd$ and getting 0.

The figure in page 6 shows why those 3 bases are called "the bases along x/y/z".

$\endgroup$
4
  • $\begingroup$ Note that one really ought to use \rangle for the basis vectors, rather than >. $\endgroup$
    – Kyle Kanos
    Commented May 11, 2018 at 9:55
  • $\begingroup$ @KyleKanos note that one can edit it in pretty quick :-P $\endgroup$
    – DanielSank
    Commented Jun 10, 2018 at 21:24
  • $\begingroup$ @DanielSank true, I could have done that (a month ago!). But it's nice for OP too learn to do it too...I think, at least, that is what I was going for here... $\endgroup$
    – Kyle Kanos
    Commented Jun 10, 2018 at 21:45
  • 2
    $\begingroup$ @KyleKanos I think I learned to write better mathjax by other people editing my posts ;-) $\endgroup$
    – DanielSank
    Commented Jun 11, 2018 at 1:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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