Quantum computing for the very curious——Part I: The state of a qubit

合集 - quantum computation(5)

1.Quantum computing for the very curious——Part I: The state of a qubit01-20

2.Part II: Introducing quantum logic gates01-293.Part III: Universal quantum computing02-044.mathematics of quantum mechanics02-075.Mathematics of Quantum Mechanics ii02-09

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Quantum computing for the very curious

Preface

qubit is short of the quantum bit, whereas the state of a bit is a number (0 or 1),the state of a qubit is a vector in a two-dimensional vector space

Maybe the state of the qubit is being stored somehow on an electron, or a photon, or an atom

Property

There are four properties we should take away:

1. qubits have a state;
2. much like a bit, that state is an abstract mathematical object;
3. whereas a bit's abstract state is a number, 0 or 1, the state of a qubit is a 2-dimensional vector
4. we call the 2-dimensional vector space where states live state space.

Connecting qubits to bits: the computational basis states

there are two special quantum states which correspond to the 0 and 1 states of a classical bit.

The quantum state corresponding to 0 is usually denoted $\vert 0 \rangle$ and corresponding to 1 is denoted $\vert 1 \rangle$

the fancy notations are the following vector respectively

$$\vert 0\rangle:=\begin{bmatrix}1\\0\end{bmatrix} , \vert 1 \rangle:=\begin{bmatrix}0\\1\end{bmatrix}$$

(this notation with$\vert$ and $\rangle$ is called the ket notation——a ket is just a vector)

the special state $\vert 0 \rangle$ and $\vert 1 \rangle$ ,called the computational basis state,have all the properties we expect of the state of a classical bit

General states of a qubit

with the computational basis states $\vert1\rangle$ and $\vert 0 \rangle$,many more states are possible.

In general,a quantum state is a two-dimensional vector.

• an example:
  

  then we can calculate:

  $$0.6\vert 0\rangle +0.8\vert 1\rangle =0.6 \begin{bmatrix}1\\0\end{bmatrix}+0.8\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}0.6\\0.8\end{bmatrix}$$

furthermore quantum states are also the complex vectors.

they can have complex numbers as entries.

• an instance:
  $$\frac{i+1}{2}\vert 0 \rangle + \frac{i}{\sqrt{2}}\vert 1\rangle=\begin{bmatrix} \frac{i+1}{2} \\ \frac{i}{\sqrt{2}} \end{bmatrix}$$

a quantum state is a two-dimensional vector in a complex vector space.

To understand it,we need to get familiar with some more nomenclature commonly used for quantum states

1. superposition
   people often say a state like $0.6\vert 0\rangle +0.8\vert 1\rangle$ is a superposition of $\vert 0 \rangle$ and $\vert 1 \rangle$(All they mean is that the state is a linear combination of $\vert 0 \rangle$ and $\vert 1 \rangle$ )

2. amplitude
   An amplitude is the coefficient for a particular state in superposition.
   for instance,in the state $0.6\vert 0\rangle +0.8\vert 1\rangle$ the amplitude for $\vert 0\rangle$ is 0.6 and the amplitude for $\vert 1 \rangle$ is 0.8

3. normalization
   a quantum state is a two-dimensional complex vector but there is a constraint :
   the sums of the squares of the amplitudes must be equal to 1.
   for a more general quantum state,the amplitudes can be complex numbers,we denote them by $\alpha$ and $\beta$.
   the state is $\alpha\vert 0 \rangle+\beta\vert 1 \rangle$.Obviously,$|\alpha|^2+|\beta|^2=1$
   this is called the normalization constraint
   for example:
   for the state 0.6$\vert 0 \rangle+0.8\vert 1 \rangle$ the sum of the squares of the amplitudes is $0.6^2+0.8^2$,which is equal to 1

   we consider $\vert1\rangle$ and $\vert 0 \rangle$ as orthonormal vectors.
   

SUMMARY:

the quantum state of a qubit is a vector of unit length in a two-dimensional complex vector space known as state space.

we’re taking is to start with the mathematics of quantum computing --we’ll keep getting familiar with qubits and the quantum state, and developing the consequences. Doing that is how we’ll build up intuition, and will give us the chops needed to come back and think harder about the meaning of the quantum state.