1
$\begingroup$

I know that a Hadamard states is a purely probabalistic one; e.g.

$$H\vert 0\rangle=a\vert 0\rangle+b\vert 1\rangle$$

where $a^2=0.5$ and $b^2=0.5$.

Are there any states in which the probabilities differ, and if there are how are they important?

$\endgroup$
5
$\begingroup$

Welcome to QCSE.

You already know that $a^2=b^2=0.5$. For a single qubit gate akin to the Hadamard gate you can achieve any two probabilities you want, as long as they add to $1$.

For example one trick that I learned was that you could choose ratios of Pythagorean triples, i.e. numbers $a$,$b$,$c$ such that $a^2+b^2=c^2$. Let's have a gate called $\mathrm{YOUSEF}$ defined as:

$$\mathrm{YOUSEF}\vert 0\rangle=\frac{3}{5}\vert 0\rangle+\frac{4}{5}\vert 1\rangle.$$

Such a gate may be useful in biasing your transition probabilities in a manner your algorithm dictates.

| improve this answer | |
$\endgroup$
  • $\begingroup$ for a qubit such as an electron, does that depend on the phase, for example does the probabilistic outcome of an electron spinning sideways e:g l+> differ from one that's halfway between the l1> and the l+> on a bloch sphere? if yes does l+> or l-> always have a 50 50 probability? $\endgroup$ – yousef elbrolosy Jun 1 at 2:03
  • $\begingroup$ @yousefelbrolosy First things first, dispense with the notion that the electron is actually physically spinning. It's not true, and will only confuse you later. Second, your example doesn't really describe a phase dependence. Usually, when we say "phase", we mean some $e^{i\delta}$ that is part of the coefficient on one or more of the basis states. $\endgroup$ – probably_someone Jun 1 at 13:33
  • $\begingroup$ @yousefelbrolosy Anyway, the state $\cos(3\pi/8)|0\rangle+\sin(3\pi/8)||1\rangle\approx 0.38|0\rangle+0.92|1\rangle$ is halfway between $|+\rangle$ and $|1\rangle$, and it clearly differs from $|+\rangle=\frac{1}{\sqrt{2}}|0\rangle+\frac{1}{\sqrt{2}}|1\rangle$. $\endgroup$ – probably_someone Jun 1 at 13:46
  • $\begingroup$ @probably_someone it is about the angular momentum of the electron but not physically spinning am i correct? $\endgroup$ – yousef elbrolosy Jun 1 at 14:15
  • 1
    $\begingroup$ @yousefelbrolosy Yes, that's basically it. $\endgroup$ – probably_someone Jun 1 at 14:24
6
$\begingroup$

You can use $Ry$ gate to prepare a qubit in superposition with arbitrary probabilities. When you apply the gate on qubit in state $|0\rangle$, you get a qubit in superposition $$ |\psi\rangle = \cos(\theta/2)|0\rangle + \sin(\theta/2)|1\rangle. $$

By chaning angle $\theta$ you can set any probability you want.

For setting $\theta = \pi/2$ you will get equally distributed superposition because $\cos(\pi/4) = \sin(\pi/4)=\frac{1}{\sqrt{2}}$, for setting $\theta = \pi$ you will get qubit $|\psi\rangle = |1\rangle$ because $\cos(\pi/2) =0$, etc.

Changing $\theta$ continously from $0$ to $\pi$, probability of measuring $|0\rangle$ is decreasing from $1$ to $0$ while probability of measuring $|1\rangle$ is increasing from $0$ to $1$.

| improve this answer | |
$\endgroup$

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.