3
$\begingroup$

In the paper "Quantum Expectation-Maximization for Gaussian Mixture Models" I encountered the following proposition :

Consider two vectors x,y and $\theta$ the angle between x and y, $\theta < \frac{\pi}{2}$. Then $||x - y|| \leq \epsilon$ implies $|| |x\rangle - |y\rangle || \leq \frac{\sqrt 2 \epsilon}{||x||}$.

I'm not sure how to interpret the link between $x$ and $| x \rangle $ ? In the case x is an integer, for example $x = 4$ then I'd say that $| x \rangle = | 100 \rangle $. But what about $x = 3.5$ or $x = (2.5, 2.5)$ ? In those cases what $| x \rangle$ would be ?

$\endgroup$
5
$\begingroup$

Generally it's context-specific what the label means. For example, if $x$ and $y$ are integers, then if $x\neq y$ then $|x\rangle$ and $|y\rangle$ are orthogonal (since they are different binary strings).

I haven't read the paper you mentioned but in the "Preliminaries" section of 3.1, they describe what they mean by $| x\rangle$. Specifically: They assume $x\in \mathbb{R}^d$, and they define $$|x\rangle = \frac{1}{\Vert x\Vert}\sum_{j=1}^d x_j|j\rangle$$

So the $|j\rangle$ will be computational basis vectors, i.e. for $j=1$, $|j\rangle = |0\dots 01\rangle$, etc. The components of $x$ give the amplitudes of each of those basis vectors in some superposition.

With this definition of $|x\rangle$, the "Claim 3.4" should follow.

$\endgroup$
1
  • $\begingroup$ thank you, I missed this definition $\endgroup$ Mar 16 '20 at 13:01

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.