lecture3.tex

\section{Lecture 3: Quantum Bits and Quantum States} \label{sec:lecture3}

\dfn{Qubit}{A \textbf{qubit} \index{qubit} is the fundamental unit of quantum
  information. Unlike a classical bit, which is either $0$ or $1$, a qubit
  can exist in a \vocab{superposition} \index{superposition} of states:

  \[
    |\psi\rangle = \alpha|0\rangle + \beta|1\rangle, \quad \text{where }
    \alpha, \beta \in \mathbb{C} \text{ and } ||\alpha||^2 + ||\beta||^2 = 1
  \]
}

\vspace{0.3cm}

\noindent

Key features of qubits include: \index{qubit!properties@\textit{properties}}

\begin{itemize}
  \ii \textbf{Superposition:} A qubit can exist simultaneously in multiple
  basis states.

  \ii \textbf{Complex Amplitudes:} Coefficients $\alpha$ and $\beta$ are
  complex numbers carrying magnitude and phase information.

  \ii \textbf{Interference:} Quantum states can interfere constructively or
  destructively.

  \ii \textbf{Entanglement:} Qubits can be correlated in ways that
  classical bits cannot.

\end{itemize}

\subsection*{Classical Computing Paradigms} \index{classical computing}

Quantum computing introduces a fundamentally different computational model.
Here are some key paradigms in classical computing that quantum computing
challenges:

\begin{itemize}
  \ii \textbf{Deterministic Computing:} Uses discrete states (0 or 1) with
  predictable transitions.

  \ii \textbf{Analog Computing:} Uses continuous values susceptible to
  noise accumulation.

  \ii \textbf{Probabilistic Computing:} Represents probabilistic mixtures
  of states.

\end{itemize}

In contrast, for quantum computing:

\begin{itemize}
  \item \textbf{Quantum Computing:} Allows coherent superposition with
    complex amplitudes and quantum interference.
\end{itemize}

\dfn{Dirac Notation}{Quantum states are represented using \vocab{Dirac
  notation} (bra-ket notation): \index{vector!Dirac notation}
  \begin{itemize}
    \ii \textbf{Ket:} \( |0\rangle, |1\rangle \) represent computational
    basis states

    \ii Computational basis vectors:

    \[
      |0\rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad
      |1\rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}
    \]

    \ii General state: \( |\psi\rangle = \alpha|0\rangle + \beta|1\rangle \)

\end{itemize}}

\dfn{Basis States}{Common qubit bases include:
  \begin{itemize}
    \ii \textbf{Computational Basis:} \( |0\rangle, |1\rangle \)
    \index{universal bases!computational}

    \ii \textbf{Hadamard Basis:}
    \index{universal bases!Hadamard basis}

    \[
      |+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) = \begin{bmatrix}
      1/\sqrt{2} \\ 1/\sqrt{2} \end{bmatrix}
    \]

    \[
      |-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle) = \begin{bmatrix}
      1/\sqrt{2} \\ -1/\sqrt{2} \end{bmatrix}
    \]

    \ii \textbf{Phase/ Circular Polarization Basis:}
    \index{universal bases!phase basis}

    \[ |L\rangle = \ket{+i} = \frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle) \]

    \[|R\rangle = \ket{-i} = \frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle) \]

\end{itemize}}

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\subsection*{Bloch Sphere Representation}

\index{Bloch sphere}
\dfn{Bloch Sphere}{A geometric representation of a single qubit state:
  \[ |\psi\rangle = \lt[\cos\left(\frac{\theta}{2}\right)|0\rangle +
  e^{i\phi}\sin\left(\frac{\theta}{2}\right)|1\rangle\rt]e^{i\gamma} \]
  Where:
  \begin{itemize}
    \ii \( \theta \in [0,\pi] \) is the polar angle \index{Bloch sphere!polar angle}

    \ii \( \phi \in [0, 2\pi) \) is the azimuthal angle \index{Bloch sphere!azimuthal angle}

    \ii \( \gamma \) is a global phase, often omitted since it cannot be
    represented on the Bloch sphere directly \index{Bloch sphere!global phase}
\end{itemize}}

\index{Bloch sphere!Cartesian coordinates conversion}
\aside{Bloch Sphere Conversion to Cartesian Coordinates:
  \[ x = \sin\theta \cos\phi, \quad
    y = \sin\theta \sin\phi, \quad
    z = \cos\theta
  \]
}

Rearranging the Bloch sphere formula, we obtain that $\theta$ and
$\phi$ can be expressed as:

\[
  \theta = 2\arccos(\alpha_1), \quad \phi = -i
  \ln\left(\frac{\alpha_2}{\sin\left(\frac{\theta}{2}\right)}\right)
\]

\ex{Example Bloch Sphere Representation}{
  \text{For the state } \(\theta = \frac{\pi}{2}, \phi = 0\):
  \bloch{90}{0}
}

\ex{Factoring Out the Global Phase}{
  Let's say that we have the following quantum state vector $\ket{\psi}$:

  \[
    \begin{aligned}
      \ket{\psi} & = \frac{1}{\sqrt{2}}\lt(i \zero + \one\rt) \\
      & = \frac{i}{\sqrt{2}}\zero + \frac{1}{\sqrt{2}}\one \\
      & = \underbrace{i}_{\text{global phase}}
      \lt(\frac{1}{\sqrt{2}}\zero + \frac{1}{\sqrt{2}}\one\rt) \\
    \end{aligned}
  \]
}


\subsection*{Quantum Measurement} \index{quantum measurement}

When a qubit is measured:

\begin{itemize}
  \ii The quantum state \textit{collapses} to an eigenstate

  \ii Measurement probability depends on squared amplitude

  \ii Computational basis measurement probabilities:
  \[ P(0) = |\alpha|^2, \quad P(1) = |\beta|^2 \]

  \ii Post-measurement state:
  \[
    \boxed{|\psi_{\text{new}}\rangle = \frac{|b\rangle \langle b | \psi
      \rangle}{\sqrt{P(b)}}
    }
  \]

\end{itemize}

\ex{Measurement Example}{For the state \( |\psi\rangle =
  \frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac{2}{3}}|1\rangle \):
  \begin{itemize}
    \ii Probability of measuring \( |0\rangle \): \( P(0) = \frac{1}{3} \)
    \ii Probability of measuring \( |1\rangle \): \( P(1) = \frac{2}{3} \)
\end{itemize}}

\qs{Orthonormality Check}{Verify the inner products of basis states:
  \begin{align*}
    \langle 0 | 1 \rangle &= 0 \\
    \langle 0 | 0 \rangle &= 1 \\
    \langle + | + \rangle &= 1 \\
    \langle + | - \rangle &= 0
\end{align*}}

\sol{These relations hold due to the orthonormal nature of quantum basis
states.}

\nt{Quantum Bases and Their $\theta$ and $\phi$ Values:

  \begin{itemize}
    \item \textbf{Computational Basis:} \( \zero \rightarrow \theta = 0, \phi
      = 0, \quad \one \rightarrow \theta = \pi, \phi = 0 \)
      \index{universal bases!computational!angles@\textit{angles}}

    \item \textbf{Hadamard Basis:} \( \ket{+} \rightarrow \theta =
      \frac{\pi}{2}, \phi = 0, \quad \ket{-} \rightarrow \theta =
      \frac{\pi}{2}, \phi = \pi \)
      \index{universal bases!Hadamard basis!angles@\textit{angles}}

  \item \textbf{Phase Basis:} \( \left|L\right\rangle = \ket{+i} \rightarrow
  \theta = \frac{\pi}{2}, \phi = \frac{\pi}{2}, \quad \left|R\right\rangle
  = \ket{-i} \rightarrow \theta = \frac{\pi}{2}, \phi = -\frac{\pi}{2} \)
  \index{universal bases!phase basis!angles@\textit{angles}}

\end{itemize}
}

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