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?


2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$ Jun 1, 2020 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. $\endgroup$ Jun 1, 2020 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 $|+\rangle=\frac{1}{\sqrt{2}}|0\rangle+\frac{1}{\sqrt{2}}|1\rangle$. $\endgroup$ Jun 1, 2020 at 13:46
  • $\begingroup$ @probably_someone it is about the angular momentum of the electron but not physically spinning am i correct? $\endgroup$ Jun 1, 2020 at 14:15
  • 1
    $\begingroup$ @yousefelbrolosy Yes, that's basically it. $\endgroup$ Jun 1, 2020 at 14:24

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 $1$.


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