Quantum Computing

This repository documents my journey learning quantum computing from the ground up. I'm building practical understanding of quantum algorithms and circuit design through coding and mathematical exploration.

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Introduction

Dirac Notation

Quantum states are represented using the Dirac Notation, a shorthand for vector and matrix algebra.

The Ket ($\ket{\psi}$)

A Ket is a column vector representing the state of a qubit.

$$\displaystyle \ket{0} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad \ket{1} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

The Bra ($\bra{\psi}$)

A Bra is a row vector, specifically the conjugate transpose of a Ket.

$$\displaystyle \bra{0} = \begin{bmatrix} 1 & 0 \end{bmatrix}, \quad \bra{1} = \begin{bmatrix} 0 & 1 \end{bmatrix}$$

Dot Product ($\langle \phi | \psi \rangle$)

Combining a Bra and a Ket produces a scalar value representing the overlap between two states:

• $\displaystyle \langle 0 | 0 \rangle = 1$
• $\displaystyle \langle 0 | 1 \rangle = 0$

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Quantum Gates

The X-Gate (Pauli-X)

The X-Gate is the quantum equivalent of the classical NOT gate. It performs a $\pi$ rotation around the X-axis of the Bloch sphere, effectively flipping the state of a qubit.

Matrix:

$$ X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $$

Logic:

The X-Gate flips the state of the qubit:

• $X\ket{0} \rightarrow \ket{1}$
• $X\ket{1} \rightarrow \ket{0}$

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The Y-Gate (Pauli-Y)

The Y-Gate performs a $\pi$ rotation around the Y-axis of the Bloch sphere. Like the X-gate, it flips the state of the qubit, but it also introduces a complex phase shift.

Matrix:

$$ Y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} $$

Logic:

The Y-Gate flips the state and applies a phase:

• $Y\ket{0} \rightarrow i\ket{1}$
• $Y\ket{1} \rightarrow -i\ket{0}$

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The Z-Gate (Pauli-Z)

The Z-Gate performs a $\pi$ rotation around the Z-axis of the Bloch sphere. It is often called the Phase-Flip gate because it leaves the $\ket{0}$ state unchanged while flipping the phase of the $\ket{1}$ state.

Matrix:

$$ Z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} $$

Logic:

The Z-Gate preserves the state but flips the phase if it is $\ket{1}$:

• $Z\ket{0} \rightarrow \ket{0}$
• $Z\ket{1} \rightarrow -\ket{1}$

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The H-Gate (Hadamard) - Superposition Creator

The H-Gate is one of the most fundamental tools in quantum computing. It acts as a bridge between the computational basis ($\ket{0}$ and $\ket{1}$) and a state of superposition. On the Bloch sphere, it performs a $180^\circ$ rotation around a diagonal axis (the X+Z axis), effectively swapping the Z and X axes.

Matrix:

$$ H = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $$

Logic:

The H-Gate places a qubit into an equal superposition:

• $H\ket{0} \rightarrow \frac{1}{\sqrt{2}} (\ket{0} + \ket{1}) = \ket{+}$
• $H\ket{1} \rightarrow \frac{1}{\sqrt{2}} (\ket{0} - \ket{1}) = \ket{-}$

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The S-Gate (Phase Shift)

The S-Gate performs a $\pi/2$ ($90^\circ$) rotation around the Z-axis of the Bloch sphere. Like the Z-Gate, it only affects the phase of the $\ket{1}$ state, but it shifts it by a quarter-turn rather than a half-turn.

Matrix:

$$ S = \begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix} $$

Logic:

The S-Gate shifts the phase:

• $S\ket{0} \rightarrow \ket{0}$
• $S\ket{1} \rightarrow i\ket{1}$

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The T-Gate

The T-Gate performs a $\pi/4$ ($45^\circ$) rotation around the Z-axis of the Bloch sphere. It is the square root of the S-Gate ($T^2 = S$) and the fourth root of the Z-Gate ($T^4 = Z$).

Matrix:

$$ T = \begin{bmatrix} 1 & 0 \\ 0 & e^{\frac{i\pi}{4}} \end{bmatrix} $$

Logic:

The T-Gate shifts the phase:

• $T\ket{0} \rightarrow \ket{0}$
• $T\ket{1} \rightarrow e^{\frac{i\pi}{4}}\ket{1}$

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Calculating Probability

In quantum computing, a qubit's state is represented by amplitudes.

Given a state: $\displaystyle \ket{\psi} = \alpha\ket{0} + \beta\ket{1}$

The probability ($P$) is the square of the absolute value of the amplitude:

• Probability of $\ket{0}$: $\displaystyle P(0) = |\alpha|^2$
• Probability of $\ket{1}$: $\displaystyle P(1) = |\beta|^2$

Example:

After applying a Hadamard gate to $\ket{0}$, the qubit is in the state:

$\displaystyle \ket{+} = \frac{1}{\sqrt{2}}\ket{0} + \frac{1}{\sqrt{2}}\ket{1}$

The Math:

$\displaystyle P(0) = \left| \frac{1}{\sqrt{2}} \right|^2 = \frac{1}{2}$

$\displaystyle P(1) = \left| \frac{1}{\sqrt{2}} \right|^2 = \frac{1}{2}$

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The Tensor Product ($\otimes$)

To represent multiple qubits at once, we use the Tensor Product. This operation combines individual qubit vectors into a larger vector representing the entire system.

Mathematical Definition

For two qubits $\ket{a}$ and $\ket{b}$, the combined state is:

$$\displaystyle \ket{a} \otimes \ket{b} = \begin{bmatrix} a_0 \\ a_1 \end{bmatrix} \otimes \begin{bmatrix} b_0 \\ b_1 \end{bmatrix} = \begin{bmatrix} a_0 b_0 \\ a_0 b_1 \\ a_1 b_0 \\ a_1 b_1 \end{bmatrix}$$

The Computational Basis

Using the tensor product, we define the four possible states for a two-qubit system:

$$\displaystyle \ket{00} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \quad \ket{01} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \quad \ket{10} = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \quad \ket{11} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$$

Algorithms

Grover's Algorithm

Math Example

Goal: Search for $\ket{01}$

Initialization

$\ket{\psi_0}$ = $\ket{00}$

$\ket{\psi_1}$ = $H\ket{\psi_0}$ = $\frac{1}{2} (\ket{00} + \ket{01} + \ket{10} + \ket{11})$

Mark Target

$\ket{\psi_2}$ = $(X \otimes I) \ket{\psi_1}$ = $\frac{1}{2} (\ket{10} + \ket{11} + \ket{00} + \ket{01})$

$\ket{\psi_3}$ = $(CZ_{0,1}) \ket{\psi_2}$ = $\frac{1}{2} (\ket{10} - \ket{11} + \ket{00} + \ket{01})$

$\ket{\psi_4}$ = $(X \otimes I) \ket{\psi_3}$ = $\frac{1}{2} (\ket{00} - \ket{01} + \ket{10} + \ket{11})$

Diffusion Operator

$$ \begin{aligned} \ket{\psi_5} &= H\ket{\psi_4} = \frac{1}{2} \Bigg[ \frac{1}{2}(\ket{00} + \ket{01} + \ket{10} + \ket{11}) \\ &\quad - \frac{1}{2}(\ket{00} - \ket{01} + \ket{10} - \ket{11}) \\ &\quad + \frac{1}{2}(\ket{00} + \ket{01} - \ket{10} - \ket{11}) \\ &\quad + \frac{1}{2}(\ket{00} - \ket{01} - \ket{10} + \ket{11}) \Bigg] \end{aligned} $$

1. $\frac{1}{2}\Big(\overset{\color{green}{1}}{\ket{00}} +\overset{\color{green}{1}}{\ket{01}} + \overset{\color{green}{1}}{\ket{10}} + \overset{\color{green}{1}}{\ket{11}}\Big)$

2. $\frac{1}{2}\Big(-\overset{\color{red}{-1}}{\ket{00}} +\overset{\color{green}{1}}{\ket{01}} -\overset{\color{red}{-1}}{\ket{10}} +\overset{\color{green}{1}}{\ket{11}}\Big)$

3. $\frac{1}{2}\Big(\overset{\color{green}{1}}{\ket{00}} +\overset{\color{green}{1}}{\ket{01}} -\overset{\color{red}{-1}}{\ket{10}} -\overset{\color{red}{-1}}{\ket{11}}\Big)$

4. $\frac{1}{2}\Big(\overset{\color{green}{1}}{\ket{00}} -\overset{\color{red}{-1}}{\ket{01}} -\overset{\color{red}{-1}}{\ket{10}} +\overset{\color{green}{1}}{\ket{11}}\Big)$

• $\ket{00} = \frac{1}{2} - \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 1$

• $\ket{01} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} - \frac{1}{2} = 1$

• $\ket{10} = \frac{1}{2} - \frac{1}{2} - \frac{1}{2} - \frac{1}{2} = -1$

• $\ket{11} = \frac{1}{2} + \frac{1}{2} - \frac{1}{2} + \frac{1}{2} = 1$

$= \frac{1}{2}(\ket{00} + \ket{01} - \ket{10} + \ket{11})$

$\ket{\psi_6} = X \ket{\psi_5} = \frac{1}{2}(\ket{11} + \ket{10} - \ket{01} + \ket{00})$

$\ket{\psi_7} = (CZ_{0,1}) \ket{\psi_6} = \frac{1}{2}(-\ket{11} + \ket{10} - \ket{01} + \ket{00})$

$\ket{\psi_8} = X \ket{\psi_7} = \frac{1}{2}(-\ket{00} + \ket{01} - \ket{10} + \ket{11})$

$$ \begin{aligned} \ket{\psi_9} &= H\ket{\psi_8} = \frac{1}{2} \Bigg[ -\frac{1}{2}(\ket{00} + \ket{01} + \ket{10} + \ket{11}) \\ &\quad + \frac{1}{2}(\ket{00} - \ket{01} + \ket{10} - \ket{11}) \\ &\quad - \frac{1}{2}(\ket{00} + \ket{01} - \ket{10} - \ket{11}) \\ &\quad + \frac{1}{2}(\ket{00} - \ket{01} - \ket{10} + \ket{11}) \Bigg] \end{aligned} $$

1. $\frac{1}{2}\Big(-\overset{\color{red}{-1}}{\ket{00}} -\overset{\color{red}{-1}}{\ket{01}} -\overset{\color{red}{-1}}{\ket{10}} -\overset{\color{red}{-1}}{\ket{11}}\Big)$

2. $\frac{1}{2}\Big(\overset{\color{green}{1}}{\ket{00}} -\overset{\color{red}{-1}}{\ket{01}} +\overset{\color{green}{1}}{\ket{10}} -\overset{\color{red}{-1}}{\ket{11}}\Big)$

3. $\frac{1}{2}\Big(-,\overset{\color{red}{-1}}{\ket{00}} -\overset{\color{red}{-1}}{\ket{01}} +\overset{\color{green}{1}}{\ket{10}} +\overset{\color{green}{1}}{\ket{11}}\Big)$

4. $\frac{1}{2}\Big(\overset{\color{green}{1}}{\ket{00}} -\overset{\color{red}{-1}}{\ket{01}} -\overset{\color{red}{-1}}{\ket{10}} +\overset{\color{green}{1}}{\ket{11}}\Big)$

• $\ket{00} = - \frac{1}{2} + \frac{1}{2} - \frac{1}{2} + \frac{1}{2} = 0$

• $\ket{01} = -\frac{1}{2} - \frac{1}{2} - \frac{1}{2} - \frac{1}{2} = -2$

• $\ket{10} = -\frac{1}{2} - \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 0$

• $\ket{11} = \frac{1}{2} - \frac{1}{2} - \frac{1}{2} + \frac{1}{2} = 0$

$= \frac{1}{2}(-2\ket{01})$

$= -\ket{01}$