qubit.js

A very silly simple quantum computing circuit simulator.
(the implementation could be wrong though)
By Yiyi Wang

Example

const {QuantumComputing} = require('qubit.js')
 
const qc = new QuantumComputing()
 
qc
  .boot(3)   // simulate 3 qubits
  .x(0)      // apply Pauli X Gate (Not Gate) to qubit[0]
  .measure() // measure all
 
console.log(qc.toQASM())    // convert to QASM code  
console.log(qc.getResult()) // print probabilities  

3 Qubtis

8 possible quantum states

3 Qubtis

8 possible quantum states

qc
  .boot(3)
 
// =>
amplitudes = [1, 0, 0, 0, 0, 0, 0, 0]

Pauli X Gate (Not Gate)

It is a $unary\ quantum\ gate$
matrix: $$X = \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$$

qc
  .boot(2)
  .x(0)
 
amplitudes = [1, 0, 0, 0, 0, 0, 0, 0]
// =>
// what should the amplitudes be ?

We are applying the matrix to a single qubit q[0]...

So we found all possible pairs.

$\alpha|0\rangle + \beta|1\rangle$

• $|00\color{red}{0}\rangle$ and $|00\color{red}{1}\rangle$

• $|01\color{red}{0}\rangle$ and $|01\color{red}{1}\rangle$

• $|10\color{red}{0}\rangle$ and $|10\color{red}{1}\rangle$

• $|11\color{red}{0}\rangle$ and $|11\color{red}{1}\rangle$

For the pair:

• $|00\color{red}{0}\rangle$ and $|00\color{red}{1}\rangle$
  We have amplitudes $\alpha = 1$ and $\beta = 0$
  Therefore $$\begin{bmatrix}\alpha' \\ \beta' \end{bmatrix}= X\begin{bmatrix}\alpha \\\beta\end{bmatrix}= \begin{bmatrix}0 & 1 \\1 & 0\end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} =\begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

qc
  .boot(2)
  .x(0)
 
amplitudes = [1, 0, 0, 0, 0, 0, 0, 0]
// =>
amplitudes = [0, 1, 0, 0, 0, 0, 0, 0]

Measure

qc
  .boot(2)
  .x(0)
  .measure() // one qubit register to one classic register
 
amplitudes = [1, 0, 0, 0, 0, 0, 0, 0]
// =>
amplitudes = [0, 1, 0, 0, 0, 0, 0, 0]

$\because probability\ =\ amplitude ^ 2$
$\therefore p(|001\rangle) = 1^2 = 1$

Another example

qc
  .boot(3)   // simulate 3 qubits
  .x(0)      // apply Pauli X Gate (Not Gate) to qubit[0]
  .cnot(0, 1) // apply Controlled Not (CNot Gate)
              // q[0] is control, q[1] is target.
  .measure()

After applying Pauli X Gate to q[0]

qc
  .boot(2)
  .x(0)
  .cnot(0, 1) // q[0] is control, q[1] is target
 
amplitudes = [1, 0, 0, 0, 0, 0, 0, 0]
// =>
amplitudes = [0, 1, 0, 0, 0, 0, 0, 0]
// =>
// what should the amplitudes be ?

Recall the Controlled Gate

(CNot Gate)

it is a Binary Quantum Gate that operates on two qubits

$$CNOT = \begin{bmatrix}1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 0 & 1 \\0 & 0 & 1 & 0\end{bmatrix}$$
transforms the quantum state: ($|control, target \rangle$)
$a|00\rangle + b|01\rangle + c|10\rangle + d|11\rangle$
into:
$a|00\rangle + b|01\rangle + c|11\rangle + d|10\rangle$

We are applying the matrix to a two qubits (control) q[0] and (target) q[1]...

$|control, target \rangle$
$a|00\rangle + b|01\rangle + c|10\rangle + d|11\rangle$
So we found all possible pairs

• $|0\color{red}{00}\rangle$ and $|0\color{red}{01}\rangle$ and $|0\color{red}{10}\rangle$ and $|0\color{red}{11}\rangle$

• $|1\color{red}{00}\rangle$ and $|1\color{red}{01}\rangle$ and $|1\color{red}{10}\rangle$ and $|1\color{red}{11}\rangle$

For the pair:

• $|0\color{red}{00}\rangle$ and $|0\color{red}{01}\rangle$ and $|0\color{red}{10}\rangle$ and $|0\color{red}{11}\rangle$
  $\because |control, target \rangle$ and $q[0]$ is control, $q[1]$ is target.
  $\therefore a|0\color{red}{00}\rangle + b|0\color{red}{10}\rangle + c|0\color{red}{01}\rangle + d|0\color{red}{11}\rangle$

We have $a = 0$, $b = 0$, $c = 1$, $d = 0$ $$\therefore\begin{bmatrix}a' \\ b' \\ c' \\ d'\end{bmatrix} = CNOT \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} =\begin{bmatrix}1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 0 & 1 \\0 & 0 & 1 & 0\end{bmatrix}\begin{bmatrix}0\\0\\1\\0\end{bmatrix} =\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$$

qc
  .boot(2)
  .x(0)
  .cnot(0, 1) // q[0] is control, q[1] is target
 
amplitudes = [1, 0, 0, 0, 0, 0, 0, 0]
// =>
amplitudes = [0, 1, 0, 0, 0, 0, 0, 0]
// =>
amplitudes = [0, 0, 0, 1, 0, 0, 0, 0]

Measure

qc
  .boot(2)
  .x(0)
  .measure() // one qubit register to one classic register
 
amplitudes = [1, 0, 0, 0, 0, 0, 0, 0]
// =>
amplitudes = [0, 1, 0, 0, 0, 0, 0, 0]
// =>
amplitudes = [0, 0, 0, 1, 0, 0, 0, 0]

$\because probability\ =\ amplitude ^ 2$
$\therefore p(|011\rangle) = 1^2 = 1$

I implemented two helper functions for applying matrix.

• applySingleBitMatrix(offset, matrix)

  • apply unary quantum gate

• applyTwoBitsMatrix(control, target, matrix)

  • apply binary quantum gate

For example, Hadamard Gate:
$$H = \begin{bmatrix} 1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & -1/\sqrt{2} \end{bmatrix}$$

qc.h(0)
 
// is the same as
 
const t = 1 / Math.sqrt(2)
qc.applySingleBitMatrix(0, [[t, t], [t, -t]])
 

For more information about this project, check the qubit.js github repo:
https://github.com/shd101wyy/qubit.js

IBM Q

Thank you