Quantum Computing 1 -- The qubit

The qubit is the smallest storage unit in the quantum computer and represents the simplest nontrivial quantum system. The qubit can assume one of two distinguishable states. These eigenstates of the qubit are denoted by 0| 0 \rangle and 1| 1 \rangle and correspond to the two setting possibilities of a classical bit. They can also be represented as two-dimensional unit vectors 0=(10),1=(01)|0\rangle = \left( \begin{array}{c} 1 \\ 0 \end{array} \right), \quad |1\rangle = \left( \begin{array}{c} 0 \\ 1 \end{array} \right) In contrast to the classical bit, the qubit can also assume a superposition of the states 0| 0 \rangle and 1| 1 \rangle . The general state ψ| \psi \rangle of the qubit is therefore a linear combination of the two eigenstates ψ=α0+β1=(αβ),α,βC |\psi\rangle = \alpha |0\rangle + \beta |1 \rangle = \left( \begin{array}{c} \alpha \\ \beta \end{array} \right), \quad \alpha, \beta \in \mathbb{C} The vectors 0| 0 \rangle and 1| 1 \rangle thus form a basis of the vector space. The state ψ| \psi \rangle is characterized by the two complex numbers α\alpha and β\beta .

If the qubit is in such a superposition and you take a measurement, you get a random result. The probability of finding the qubit in the state 0| 0 \rangle or 1| 1 \rangle is proportional to the absolute square α2| \alpha | ^ 2 or β2| \beta | ^ 2 of the respective pre-factor. After the measurement, the qubit is in one of the eigenstates 0| 0 \rangle or 1| 1 \rangle , so that the information about the original state is lost (wave function collapse)

before measurement:ψ=α0+β1after measurement:ψ={0with probability α2α2+β21with probability β2α2+β2 \begin{aligned} \text{before measurement:} & & |\psi \rangle &= \alpha |0 \rangle + \beta |1\rangle \\ \text{after measurement:} & & |\psi'\rangle &= \begin{cases} |0 \rangle & \text{with probability } \frac{|\alpha|^2}{|\alpha|^2 + |\beta|^2} \\ |1 \rangle & \text{with probability } \frac{|\beta|^2}{|\alpha|^2 + |\beta|^2} \end{cases} \end{aligned} Usually, state vectors will be normalized (i.e. α2+β2=1| \alpha | ^ 2 + | \beta | ^ 2 = 1 ), so the probabilities for the states 0| 0 \rangle and 1| 1 \rangle are given directly by absolute squares α2| \alpha | ^ 2 and β2| \beta | ^ 2 . In addition, a phase factor can be chosen freely, so that the qubit can be described by two real numbers θ\theta and φ\varphi: ψ=cosθ20+eiφsinθ21|\psi\rangle = \cos \frac{\theta}{2} |0\rangle + e^{i \varphi} \sin \frac{\theta}{2}|1\rangle If one interprets these numbers θ\theta and φ\varphi as angles, the qubit can be represented as a point on a sphere surface (Bloch sphere).

A qubit in the normalized state ψ=320+121|\psi\rangle = \frac{\sqrt{3}}{2} |0\rangle + \frac{1}{2} |1\rangle is measured. What are the probabilities of the two different outcomes 0| 0 \rangle and 1| 1 \rangle ?

0 with probability 322=34=75%,1 with probability 122=14=25% \begin{aligned} |0\rangle & \quad \text{ with probability } \left|\frac{\sqrt{3}}{2}\right|^2 = \frac{3}{4} = 75\,\%,\\ |1\rangle & \quad \text{ with probability } \left|\frac{1}{2}\right|^2 = \frac{1}{4} = 25\,\% \end{aligned}

Conclusion: The state of a qubit is described by a two-dimensional vector, so that a qubit contains much more information than a classical bit. However, if the qubit is read out, most information is deleted, so that the result is only one of the two states 0| 0 \rangle or 1| 1 \rangle . The readable part of the quantum information therefore only corresponds to one classical bit. However, the full quantum information be can used for quantum operations, as we will see later.

Note by Markus Michelmann
3 years, 1 month ago

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I hope to see more additions to the series :)

Agnishom Chattopadhyay - 3 years, 1 month ago

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After almost two weeks break there is non a sequel to this series with a new chapter. I hope you like it :D

Markus Michelmann - 3 years ago

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