1  Quantum Mechanics Basics

Welcome to the quantum world. In this chapter, we’ll build the mathematical foundation you need to understand quantum computing—from first principles, with rigorous proofs and no hand-waving.

1.1 The Quantum State

In classical physics, a bit is either 0 or 1. In quantum mechanics, a qubit (quantum bit) can exist in a superposition of both states simultaneously.

1.1.1 Mathematical Representation

A quantum state is represented as a vector in a complex vector space called a Hilbert space. For a single qubit, this is a 2-dimensional space.

The computational basis states are:

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

Notation: We use Dirac notation (or “bra-ket” notation): - \(|0\rangle\) is pronounced “ket zero” - represents a column vector - \(\langle 0|\) is pronounced “bra zero” - represents a row vector (conjugate transpose)

1.1.2 General Qubit State

The most general state of a qubit is a superposition:

\[ |\psi\rangle = \alpha|0\rangle + \beta|1\rangle = \begin{pmatrix} \alpha \\ \beta \end{pmatrix} \]

where \(\alpha, \beta \in \mathbb{C}\) (complex numbers) and \(|\alpha|^2 + |\beta|^2 = 1\).

Key insight: The state exists in both \(|0\rangle\) and \(|1\rangle\) simultaneously until measured.

1.2 The Superposition Principle

One of the foundational principles of quantum mechanics:

Superposition Principle

If \(|\psi_1\rangle\) and \(|\psi_2\rangle\) are valid quantum states, then any linear combination:

\[ |\psi\rangle = \alpha|\psi_1\rangle + \beta|\psi_2\rangle \]

is also a valid quantum state (provided it’s normalized: \(|\alpha|^2 + |\beta|^2 = 1\)).

1.2.1 Example: The Hadamard State

One of the most important quantum states is:

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

This is an equal superposition of \(|0\rangle\) and \(|1\rangle\).

Similarly, we can define:

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

Together, \(\{|+\rangle, |-\rangle\}\) form an alternative basis for the qubit’s Hilbert space (the X-basis or Hadamard basis).

1.3 Measurement and the Born Rule

When we measure a quantum state, something dramatic happens: the superposition collapses to one of the basis states.

1.3.1 The Born Rule

Born Rule (Measurement Postulate)

Given a quantum state \(|\psi\rangle = \alpha|0\rangle + \beta|1\rangle\), when measured in the computational (Z) basis:

• Probability of measuring \(|0\rangle\): \(P(0) = |\alpha|^2\)
• Probability of measuring \(|1\rangle\): \(P(1) = |\beta|^2\)

After measurement, the state collapses to the measured outcome.

1.3.2 Example: Measuring \(|+\rangle\)

Consider the state \(|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\).

When measured in the Z-basis: - \(P(0) = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}\) - \(P(1) = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}\)

Interpretation: We get 0 or 1 with equal probability (50/50), and the state becomes either \(|0\rangle\) or \(|1\rangle\) after measurement.

1.4 Normalization Condition

The probabilities must sum to 1. This gives us the normalization condition:

\[ |\alpha|^2 + |\beta|^2 = 1 \]

Let’s prove this is required.

Proof: Normalization Ensures Valid Probabilities

Claim: For a quantum state to yield valid probabilities, it must satisfy \(|\alpha|^2 + |\beta|^2 = 1\).

Proof:

By the axioms of probability, the sum of all possible outcomes must equal 1:

\[ P(0) + P(1) = 1 \]

By the Born rule:

\[ P(0) = |\alpha|^2, \quad P(1) = |\beta|^2 \]

Therefore:

\[ |\alpha|^2 + |\beta|^2 = 1 \quad \square \]

This is called the normalization condition. Any valid quantum state must be normalized.

1.4.1 Inner Product and Normalization

We can express normalization using the inner product (or dot product) in Hilbert space:

\[ \langle\psi|\psi\rangle = 1 \]

where \(\langle\psi| = (\alpha^* \quad \beta^*)\) is the conjugate transpose of \(|\psi\rangle\).

Calculation:

\[ \langle\psi|\psi\rangle = (\alpha^* \quad \beta^*) \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \alpha^*\alpha + \beta^*\beta = |\alpha|^2 + |\beta|^2 = 1 \]

1.5 Measurement in Different Bases

The Born rule applies to any orthonormal basis, not just the computational basis.

1.5.1 Z-Basis vs X-Basis

Z-basis (computational basis): \[ \{|0\rangle, |1\rangle\} \]

X-basis (Hadamard basis): \[ \{|+\rangle, |-\rangle\} = \left\{\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle), \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)\right\} \]

1.5.2 Example: Measuring \(|0\rangle\) in X-Basis

To measure \(|0\rangle\) in the X-basis, we first express it in that basis:

\[ |0\rangle = \frac{1}{\sqrt{2}}|+\rangle + \frac{1}{\sqrt{2}}|-\rangle \]

Verification: \[ \frac{1}{\sqrt{2}}|+\rangle + \frac{1}{\sqrt{2}}|-\rangle = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle) \]

\[ = \frac{1}{2}(|0\rangle + |1\rangle) + \frac{1}{2}(|0\rangle - |1\rangle) = \frac{1}{2}(2|0\rangle) = |0\rangle \quad \checkmark \]

Now applying Born rule: - \(P(+) = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}\) - \(P(-) = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}\)

Key insight: The state \(|0\rangle\) is a superposition when viewed in the X-basis! This is called basis-dependent measurement or contextuality.

1.6 Global Phase and Physical Equivalence

Consider two states:

\[ |\psi\rangle = \alpha|0\rangle + \beta|1\rangle \]

\[ |\psi'\rangle = e^{i\theta}(\alpha|0\rangle + \beta|1\rangle) = e^{i\theta}|\psi\rangle \]

where \(\theta \in \mathbb{R}\) and \(e^{i\theta}\) is a global phase factor.

Claim: These states are physically indistinguishable.

Why? Measurement probabilities are unchanged:

\[ P(0) = |e^{i\theta}\alpha|^2 = |e^{i\theta}|^2|\alpha|^2 = |\alpha|^2 \]

since \(|e^{i\theta}| = 1\) for any real \(\theta\).

Global Phase Irrelevance

States that differ only by a global phase factor are physically equivalent:

\[ |\psi\rangle \equiv e^{i\theta}|\psi\rangle \]

However, relative phases (phases between different components) are observable and physically meaningful.

1.6.1 Example: Relative Phase Matters

These two states are different:

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

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

The relative phase between \(|0\rangle\) and \(|1\rangle\) components (\(+1\) vs \(-1\)) is observable when measured in the X-basis!

1.7 Quantum State Space: The Bloch Sphere

For a single qubit, we can visualize the state space geometrically.

Any single-qubit state can be written as:

\[ |\psi\rangle = \cos\frac{\theta}{2}|0\rangle + e^{i\phi}\sin\frac{\theta}{2}|1\rangle \]

where \(\theta \in [0, \pi]\) and \(\phi \in [0, 2\pi)\).

This parameterization maps quantum states to points on a unit sphere in 3D space called the Bloch sphere.

Key points on the Bloch sphere: - North pole (\(\theta=0\)): \(|0\rangle\) - South pole (\(\theta=\pi\)): \(|1\rangle\) - Equator (\(\theta=\pi/2\)): - \(\phi=0\): \(|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\) - \(\phi=\pi\): \(|-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)\) - \(\phi=\pi/2\): \(|+i\rangle = \frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle)\) - \(\phi=3\pi/2\): \(|-i\rangle = \frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle)\)

Why the Bloch Sphere Matters

The Bloch sphere gives us geometric intuition about quantum states. Antipodal points (opposite ends of a diameter) represent orthogonal states - states that are maximally distinguishable by measurement.

1.8 Pure States vs Mixed States

So far we’ve discussed pure states - states with definite (though possibly superposed) quantum information.

In reality, quantum systems often exist in mixed states due to: - Incomplete knowledge about the system - Interaction with the environment (decoherence) - Entanglement with other systems

Mixed states are described by density matrices (we’ll cover this in later chapters).

1.9 Summary

In this chapter, you learned:

✅ Quantum states are vectors in complex Hilbert space ✅ Superposition: \(|\psi\rangle = \alpha|0\rangle + \beta|1\rangle\) ✅ Born rule: Measurement probabilities \(P(i) = |\alpha_i|^2\) ✅ Normalization: \(|\alpha|^2 + |\beta|^2 = 1\) ensures valid probabilities ✅ Basis-dependent measurement: States look different in different bases ✅ Global phase is irrelevant, relative phase is observable ✅ Bloch sphere: Geometric representation of qubit states

1.10 Exercises

1.10.1 Exercise 1.1: Normalization

Given the (unnormalized) state:

\[ |\psi\rangle = 2|0\rangle + 2i|1\rangle \]

(a) Calculate \(\langle\psi|\psi\rangle\).

(b) Normalize the state to obtain \(|\psi_{\text{norm}}\rangle\).

(c) Verify that \(\langle\psi_{\text{norm}}|\psi_{\text{norm}}\rangle = 1\).

1.10.2 Exercise 1.2: Measurement Probabilities

Consider the state:

\[ |\psi\rangle = \frac{1}{2}|0\rangle + \frac{\sqrt{3}}{2}|1\rangle \]

(a) Calculate \(P(0)\) and \(P(1)\) when measured in the Z-basis.

(b) Verify that \(P(0) + P(1) = 1\).

(c) After measuring and obtaining outcome 1, what is the post-measurement state?

1.10.3 Exercise 1.3: Basis Transformation

Express the state \(|1\rangle\) in the X-basis (i.e., as a linear combination of \(|+\rangle\) and \(|-\rangle\)).

Hint: Use the inverse relations: \[ |0\rangle = \frac{1}{\sqrt{2}}(|+\rangle + |-\rangle), \quad |1\rangle = \frac{1}{\sqrt{2}}(|+\rangle - |-\rangle) \]

1.10.4 Exercise 1.4: Cross-Basis Measurement

Consider the state \(|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\).

(a) What are \(P(0)\) and \(P(1)\) when measured in the Z-basis?

(b) What are \(P(+)\) and \(P(-)\) when measured in the X-basis?

(c) Explain why the probabilities differ between the two bases.

1.10.5 Exercise 1.5: Global Phase

Show that the states \(|\psi\rangle = |0\rangle\) and \(|\psi'\rangle = -|0\rangle = e^{i\pi}|0\rangle\) are physically equivalent by proving they yield identical measurement probabilities in both Z-basis and X-basis.

1.10.6 Exercise 1.6: Bloch Sphere Coordinates

A qubit is in the state:

\[ |\psi\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{e^{i\pi/4}}{\sqrt{2}}|1\rangle \]

Determine the Bloch sphere angles \(\theta\) and \(\phi\) for this state.

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Next: Chapter 2: Entanglement - Where quantum mechanics gets truly strange