The State Postulate

Quantum mechanics is built on four postulates. The first answers the most basic question: what is the state of a system? The answer — a unit vector in a Hilbert space, identified up to a global phase — is deceptively simple, but it already encodes superposition, the irrelevance of global phase, and the physical reality of relative phase. Everything else in the theory acts on this object.

Learning Objectives

After this lesson you will be able to:

1. State Postulate 1 precisely and explain why states are unit vectors (rays), not arbitrary vectors.
2. Apply the normalization condition and normalize an arbitrary vector to a valid state.
3. Distinguish global phase (unobservable) from relative phase (observable) and justify the difference.
4. Use the superposition principle and interpret amplitudes via the (forthcoming) Born rule.
5. Represent and normalize states in NumPy and reason about phase equivalence.

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Intuition

A classical bit is in state $0$ or $1$. A quantum system is described instead by a vector of complex amplitudes — one amplitude per classically-distinguishable configuration. For a qubit there are two configurations, $\lvert 0\rangle$ and $\lvert 1\rangle$, so the state is

$$\lvert\psi\rangle = \alpha\lvert 0\rangle + \beta\lvert 1\rangle, \qquad \alpha,\beta\in\mathbb{C}.$$

The amplitudes are not probabilities — they're complex numbers whose squared magnitudes are probabilities, and whose relative phase governs interference. Two requirements pin down which vectors are physical: the probabilities must sum to $1$ (normalization), and multiplying the whole vector by a phase $e^{i\gamma}$ changes nothing measurable (global phase freedom). A physical state is therefore not quite a vector — it's a vector modulo these redundancies, a ray.

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Theory

Postulate 1 (State space)

Postulate 1. Associated to any isolated physical system is a complex Hilbert space $\mathcal{H}$, the state space. The state of the system is completely described by a unit vector $\lvert\psi\rangle\in\mathcal{H}$, i.e. $\langle\psi|\psi\rangle = 1$. Two vectors that differ only by a global phase, $\lvert\psi\rangle$ and $e^{i\gamma}\lvert\psi\rangle$ ($\gamma\in\mathbb{R}$), represent the same physical state.

For a single qubit, $\mathcal{H} = \mathbb{C}^2$ with computational basis $\{\lvert0\rangle,\lvert1\rangle\}$ (from 0.1.1). For $n$ qubits, $\mathcal{H} = (\mathbb{C}^2)^{\otimes n} = \mathbb{C}^{2^n}$ (Postulate 4, 1.1.4).

Normalization

Writing $\lvert\psi\rangle = \sum_k c_k\lvert k\rangle$ in an orthonormal basis, normalization is

$$\langle\psi|\psi\rangle = \sum_k |c_k|^2 = 1 .$$

By the (upcoming) Born rule, $|c_k|^2$ is the probability of obtaining outcome $k$ on measurement in that basis; normalization is exactly "the probabilities sum to one." Any nonzero vector $\lvert v\rangle$ becomes a state by normalizing: $\lvert\psi\rangle = \lvert v\rangle/\sqrt{\langle v|v\rangle}$ (Parseval, 0.1.2).

The superposition principle

Because $\mathcal{H}$ is a vector space, any linear combination of states is (after normalization) a state: if $\lvert\psi_1\rangle,\lvert\psi_2\rangle$ are valid states, so is $\alpha\lvert\psi_1\rangle + \beta\lvert\psi_2\rangle$for appropriate$\alpha,\beta$. This superposition principle is the mathematical heart of quantum computing — a single register can hold a superposition over all $2^n$ bit-strings — and is a direct consequence of states living in a linear space.

Global phase is unobservable; relative phase is not

These two facts look similar but are opposite in consequence.

Global phase. The states $\lvert\psi\rangle$ and $e^{i\gamma}\lvert\psi\rangle$ are physically identical. Why: every physical prediction is an expectation value $\langle\psi|A|\psi\rangle$ or a probability $|\langle\phi|\psi\rangle|^2$ (Postulate 2). Under $\lvert\psi\rangle\mapsto e^{i\gamma}\lvert\psi\rangle$,

$$\langle\psi|A|\psi\rangle \mapsto e^{-i\gamma}\langle\psi|A|\psi\rangle e^{i\gamma} = \langle\psi|A|\psi\rangle, \qquad |\langle\phi|\psi\rangle|^2 \mapsto |e^{i\gamma}|^2\,|\langle\phi|\psi\rangle|^2 = |\langle\phi|\psi\rangle|^2 .$$

The phase cancels in every observable quantity. Hence a physical state is really an equivalence class $\{e^{i\gamma}\lvert\psi\rangle\}$ — a ray in $\mathcal{H}$ (a point of projective Hilbert space).

Relative phase. Inside a superposition, the relative phase between components is physical. Compare

$$\lvert +\rangle = \tfrac1{\sqrt2}(\lvert0\rangle + \lvert1\rangle), \qquad \lvert -\rangle = \tfrac1{\sqrt2}(\lvert0\rangle - \lvert1\rangle).$$

These differ only by a relative sign ($e^{i\pi}$) on the $\lvert1\rangle$ component, yet they are orthogonal ($\langle +|-\rangle = 0$) — perfectly distinguishable states. A global phase rescales everything; a relative phase reweights components against each other, and that survives into measurable interference (e.g. measuring $\lvert\pm\rangle$ in the $X$ basis gives opposite certain outcomes). This distinction is the reason quantum mechanics needs $\mathbb{C}$, foreshadowed in 0.1.1.

Caution. "Global phase doesn't matter" is true for a state in isolation. A relative phase between two branches of a superposition — including one introduced by a controlled operation — absolutely matters. Phase-kickback (Term 2.4) weaponizes exactly this.

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Worked Examples

Example 1 — Normalizing a state

Let $\lvert v\rangle = 2\lvert0\rangle + (1-i)\lvert1\rangle$. Then $\langle v|v\rangle = |2|^2 + |1-i|^2 = 4 + 2 = 6$, so the normalized state is

$$\lvert\psi\rangle = \tfrac1{\sqrt6}\big(2\lvert0\rangle + (1-i)\lvert1\rangle\big).$$

Measurement-in-the-computational-basis probabilities (Born rule): $p_0 = 4/6 = 2/3$, $p_1 = 2/6 = 1/3$, summing to $1$. ✓

Example 2 — Global vs relative phase

Consider $\lvert\psi\rangle = \tfrac1{\sqrt2}(\lvert0\rangle + \lvert1\rangle) = \lvert+\rangle$.

• Apply a global phase $e^{i\pi/3}$: $e^{i\pi/3}\lvert+\rangle$. Overlap with any $\lvert\phi\rangle$ has the same modulus, so it is physically $\lvert+\rangle$ — same state.
• Apply a relative phase $e^{i\pi}=-1$ to the $\lvert1\rangle$ component: $\tfrac1{\sqrt2}(\lvert0\rangle - \lvert1\rangle) = \lvert-\rangle$. Now $\langle+|-\rangle = 0$ — a different, orthogonal state. Same-looking operation, opposite physical consequence.

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Hands-on (Python)

import numpy as np

ket0 = np.array([1, 0], dtype=complex)
ket1 = np.array([0, 1], dtype=complex)

def normalize(v):
    return v / np.linalg.norm(v)

# Example 1:
v = 2 * ket0 + (1 - 1j) * ket1
psi = normalize(v)
print(np.round(psi, 4))                       # [0.8165+0.j  0.4082-0.4082j]
print(np.abs(psi) ** 2)                        # [0.6667 0.3333] = Born probabilities, sum 1

def same_physical_state(a, b, tol=1e-9):
    """Equal up to GLOBAL phase iff |<a|b>| = 1 for normalized a, b."""
    a, b = normalize(a), normalize(b)
    return np.isclose(abs(np.vdot(a, b)), 1.0, atol=tol)

plus  = normalize(ket0 + ket1)
minus = normalize(ket0 - ket1)

# Global phase e^{iπ/3} on |+> -> same physical state:
print(same_physical_state(plus, np.exp(1j * np.pi / 3) * plus))   # True

# Relative phase (sign on |1>) turns |+> into |-> -> DIFFERENT, orthogonal state:
print(same_physical_state(plus, minus))                            # False
print(np.vdot(plus, minus))                                        # 0j  (orthogonal)

Looking ahead to Braket. A simulator stores the amplitude vector you just built. Because global phase is unobservable, two simulators may report state vectors differing by an overall phase yet be physically identical — always compare via $|\langle\phi|\psi\rangle|$, never raw components (Appendix B §8).

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Exercises

E1 (easy). Normalize $\lvert v\rangle = 3\lvert0\rangle + 4i\lvert1\rangle$ and give the computational-basis measurement probabilities.

Solution

$\langle v|v\rangle = 9 + 16 = 25$, so $\lvert\psi\rangle = \tfrac15(3\lvert0\rangle + 4i\lvert1\rangle)$. Probabilities $p_0 = 9/25 = 0.36$, $p_1 = 16/25 = 0.64$ (sum $1$).

E2 (easy). Are $\lvert\psi\rangle = \tfrac1{\sqrt2}(\lvert0\rangle + i\lvert1\rangle)$ and $\lvert\phi\rangle = \tfrac{i}{\sqrt2}(\lvert0\rangle + i\lvert1\rangle)$ the same physical state?

Solution

$\lvert\phi\rangle = i\lvert\psi\rangle$, a global phase $e^{i\pi/2}$. Same physical state. Check: $|\langle\psi|\phi\rangle| = |i|\,|\langle\psi|\psi\rangle| = 1$.

E3 (medium). Show that for normalized states, $|\langle\phi|\psi\rangle|^2$ is invariant if either state is multiplied by a global phase, and explain why this means rays (not vectors) are physical.

Solution

Under $\lvert\psi\rangle\mapsto e^{i\gamma}\lvert\psi\rangle$, $\langle\phi|\psi\rangle\mapsto e^{i\gamma}\langle\phi|\psi\rangle$, whose modulus squared is unchanged ($|e^{i\gamma}|=1$); same for a phase on $\lvert\phi\rangle$. Since all predictions are such overlaps/expectations, the global phase is operationally inaccessible — the physical object is the equivalence class $\{e^{i\gamma}\lvert\psi\rangle\}$, a ray.

E4 (medium). A student claims $\lvert0\rangle + \lvert1\rangle$ and $\lvert0\rangle - \lvert1\rangle$ "differ by a phase, so they're the same state." Where is the error?

Solution

The $-1$ is a relative phase on the $\lvert1\rangle$ component only, not a global phase on the whole vector. After normalization they are $\lvert+\rangle$ and $\lvert-\rangle$, which are orthogonal and perfectly distinguishable. A global phase would multiply both components equally; here only one flips.

E5 (hard). A qubit state up to global phase has how many real parameters? Derive the count and connect it to the Bloch sphere (Term 1.2).

Solution

A vector in $\mathbb{C}^2$ has $4$ real parameters (two complex amplitudes). Normalization $|\alpha|^2+|\beta|^2=1$ removes $1$; global-phase freedom removes another $1$. That leaves $4-2 = 2$ real parameters — exactly the two angles $(\theta,\varphi)$ of a point on the Bloch sphere $\lvert\psi\rangle = \cos\tfrac\theta2\lvert0\rangle + e^{i\varphi}\sin\tfrac\theta2\lvert1\rangle$ (developed in 1.2.2). This matches the $\dim_\mathbb{R}\mathbb{C}^2 = 4$ count from 0.1.1 E5.

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Checkpoint

1. State Postulate 1. Why a unit vector, and why "up to global phase"?
2. What does normalization mean physically?
3. Give an operation that is unobservable and one that is observable, both "just a phase."
4. State the superposition principle and why it follows from $\mathcal{H}$ being a vector space.
5. How many real parameters describe a qubit state, and why?

Answers

1. The state is a unit vector $\lvert\psi\rangle\in\mathcal H$ with $\langle\psi|\psi\rangle=1$, defined up to a global phase. Unit norm makes Born probabilities sum to $1$; global phase cancels in all predictions, so it carries no information.
2. The measurement-outcome probabilities $\sum_k|c_k|^2 = 1$ — total probability is one.
3. Unobservable: multiplying the whole state by $e^{i\gamma}$ (global phase). Observable: a relative phase between components, e.g. $\lvert0\rangle+\lvert1\rangle \to \lvert0\rangle-\lvert1\rangle$.
4. Any linear combination of states is (after normalization) a state, because $\mathcal H$ is closed under addition and scalar multiplication.
5. Two: $4$ real parameters of $\mathbb C^2$, minus one for normalization and one for global phase.

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Further Reading

• [NC] Nielsen & Chuang, §2.2.1 (Postulate 1) — the state-space postulate.
• [Sak] Sakurai & Napolitano, §1.1–1.3 — kets, states, and the physical setup.
• [Pre] Preskill, Ph219, Ch. 2 — axioms of quantum theory.

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