QuantumCore

Lesson 9 · 45 min · core

Postulate 1: State Space & Global Phase

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The first postulate of quantum mechanics says that the state of an isolated system is a unit vector in a complex Hilbert space. This lesson states the postulate precisely, proves that multiplying a state by a global phase eiγe^{i\gamma} changes no measurement statistic (so physical states are rays, not vectors), shows by explicit matrix computation that a relative phase between components is observable — H∣+⟩=∣0⟩H\lvert+\rangle=\lvert0\rangle but H∣−⟩=∣1⟩H\lvert-\rangle=\lvert1\rangle — and derives P(0)=cos⁡2(φ/2)P(0)=\cos^2(\varphi/2) for the state (∣0⟩+eiφ∣1⟩)/2(\lvert0\rangle+e^{i\varphi}\lvert1\rangle)/\sqrt2 after a Hadamard. It closes with the superposition principle and the normalisation condition ∑i∣ci∣2=1\sum_i\lvert c_i\rvert^2=1.

Prerequisites: Dirac Bra–Ket Notation & Outer Products

Objectives and key results
  • State Postulate 1: a pure state of an isolated system is a unit vector ∣ψ⟩\lvert\psi\rangle of a complex Hilbert space H\mathcal H, and ∣ψ⟩\lvert\psi\rangle and eiγ∣ψ⟩e^{i\gamma}\lvert\psi\rangle describe the same physical state.

  • Prove that a global phase is unobservable: ∥Peiγψ∥2=∥Pψ∥2\lVert Pe^{i\gamma}\psi\rVert^2=\lVert P\psi\rVert^2 for every projector PP and ⟨eiγψ∣A∣eiγψ⟩=⟨ψ∣A∣ψ⟩\langle e^{i\gamma}\psi\vert A\vert e^{i\gamma}\psi\rangle=\langle\psi\vert A\vert\psi\rangle for every operator AA.

  • Show that a relative phase is observable by computing H∣±⟩H\lvert\pm\rangle and P(0)=cos⁡2(φ/2)P(0)=\cos^2(\varphi/2) for H(∣0⟩+eiφ∣1⟩)/2H(\lvert0\rangle+e^{i\varphi}\lvert1\rangle)/\sqrt2.

  • Apply the superposition principle and normalise an arbitrary vector ∑ici∣i⟩\sum_ic_i\lvert i\rangle so that ∑i∣ci∣2=1\sum_i\lvert c_i\rvert^2=1.

definition · Postulate 1 (state space)
∣ψ⟩∈H,⟨ψ∣ψ⟩=1,∣ψ⟩∼eiγ∣ψ⟩\lvert\psi\rangle\in\mathcal H,\quad \langle\psi\vert\psi\rangle=1,\quad \lvert\psi\rangle\sim e^{i\gamma}\lvert\psi\rangle
theorem · Global phase is unobservable
∥P eiγψ∥2=∥Pψ∥2,⟨eiγψ∣A∣eiγψ⟩=⟨ψ∣A∣ψ⟩\lVert P\,e^{i\gamma}\psi\rVert^2=\lVert P\psi\rVert^2,\qquad \langle e^{i\gamma}\psi\vert A\vert e^{i\gamma}\psi\rangle=\langle\psi\vert A\vert\psi\rangle
identity · Hadamard on the X basis
H∣+⟩=∣0⟩,H∣−⟩=∣1⟩H\lvert+\rangle=\lvert0\rangle,\qquad H\lvert-\rangle=\lvert1\rangle
formula · Relative phase read out by H
P(0)=∣⟨0∣H12(∣0⟩+eiφ∣1⟩)∣2=cos⁡2φ2P(0)=\Big\lvert\langle0\vert H\tfrac{1}{\sqrt2}(\lvert0\rangle+e^{i\varphi}\lvert1\rangle)\Big\rvert^2=\cos^2\tfrac{\varphi}{2}
formula · Normalisation
∣ψ⟩=∑ici∣i⟩,∑i∣ci∣2=1\lvert\psi\rangle=\sum_ic_i\lvert i\rangle,\qquad \sum_i\lvert c_i\rvert^2=1

The state-space postulate#

Stage 1 so far has been pure mathematics: complex vector spaces, inner products, Dirac notation, operators and the spectral theorem. Quantum mechanics is the physical theory that tells us how to use that mathematics to describe a real system — a trapped ion, a superconducting circuit, a photon's polarisation. It is organised as a short list of postulates. They are not derived from anything more basic; they are the axioms whose consequences have survived a century of experiments. The first postulate answers the most basic question: what mathematical object is the state of a system?

The answer is surprisingly economical. A classical bit is one of two values; a classical probabilistic bit is a probability vector (p0,p1)(p_0,p_1) with p0+p1=1p_0+p_1=1. A quantum bit is a vector of complex amplitudes (α,β)(\alpha,\beta) with ∣α∣2+∣β∣2=1\lvert\alpha\rvert^2+\lvert\beta\rvert^2=1. The change from real, non-negative numbers summing to one, to complex numbers whose squared moduli sum to one, is the whole source of interference — and, as we shall see, it also introduces a redundancy: the overall complex phase of the vector carries no physical information.

Definition

Postulate 1 (state space)

Associated with any isolated physical system is a complex vector space with an inner product — a Hilbert space H\mathcal H — called the state space of the system. The system is completely described by its state vector ∣ψ⟩∈H\lvert\psi\rangle\in\mathcal H, a unit vector: ⟨ψ∣ψ⟩=1\langle\psi\vert\psi\rangle=1. Two unit vectors that differ only by a complex factor of modulus one, ∣ψ′⟩=eiγ∣ψ⟩\lvert\psi'\rangle=e^{i\gamma}\lvert\psi\rangle with γ∈R\gamma\in\mathbb R, describe the same physical state. For a qubit H=C2\mathcal H=\mathbb C^2 with computational basis ∣0⟩=(1,0)T\lvert0\rangle=(1,0)^T, ∣1⟩=(0,1)T\lvert1\rangle=(0,1)^T; for nn qubits H=(C2)⊗n≅C2n\mathcal H=(\mathbb C^2)^{\otimes n}\cong\mathbb C^{2^n}.

Definition

Ray (projective state)

The ray through a unit vector ∣ψ⟩\lvert\psi\rangle is the equivalence class [ψ]={eiγ∣ψ⟩:γ∈R}[\psi]=\{e^{i\gamma}\lvert\psi\rangle:\gamma\in\mathbb R\}. The relation ∣ψ⟩∼∣ϕ⟩  ⟺  ∣ϕ⟩=eiγ∣ψ⟩\lvert\psi\rangle\sim\lvert\phi\rangle\iff\lvert\phi\rangle=e^{i\gamma}\lvert\psi\rangle for some real γ\gamma is an equivalence relation (reflexive with γ=0\gamma=0, symmetric with −γ-\gamma, transitive because eiγ1eiγ2=ei(γ1+γ2)e^{i\gamma_1}e^{i\gamma_2}=e^{i(\gamma_1+\gamma_2)}). Physical pure states are rays; state vectors are convenient representatives of them. For one qubit the set of rays is exactly the Bloch sphere.

Quick check

Which of the following vectors of C2\mathbb C^2 is a valid state vector under Postulate 1 without rescaling?

Select one answer

Global phase is unobservable#

Why may we identify ∣ψ⟩\lvert\psi\rangle with eiγ∣ψ⟩e^{i\gamma}\lvert\psi\rangle? Because nothing we can ever measure distinguishes them. Every experimental statistic in quantum mechanics has one of two forms: a probability ∥Pψ∥2=⟨ψ∣P∣ψ⟩\lVert P\psi\rVert^2=\langle\psi\vert P\vert\psi\rangle of an outcome whose projector is PP (the Born rule, Postulate 3), or an expectation value ⟨ψ∣A∣ψ⟩\langle\psi\vert A\vert\psi\rangle of an observable AA (a weighted sum of such probabilities). The next theorem shows that both are invariant under ∣ψ⟩↦eiγ∣ψ⟩\lvert\psi\rangle\mapsto e^{i\gamma}\lvert\psi\rangle. The key algebraic fact is that the inner product is conjugate-linear in its first slot, so the phase appears once as e−iγe^{-i\gamma} and once as eiγe^{i\gamma}, and e−iγeiγ=∣eiγ∣2=1e^{-i\gamma}e^{i\gamma}=\lvert e^{i\gamma}\rvert^2=1.

Theorem

Global phase leaves every outcome probability and post-measurement ray unchanged

Let ∣ψ⟩∈H\lvert\psi\rangle\in\mathcal H, γ∈R\gamma\in\mathbb R, and let PP be any orthogonal projector (P=P†=P2P=P^\dagger=P^2). Then ∥P eiγψ∥2=∥Pψ∥2\lVert P\,e^{i\gamma}\psi\rVert^2=\lVert P\psi\rVert^2. Moreover, if P∣ψ⟩≠0P\lvert\psi\rangle\neq0, the normalised post-measurement vectors satisfy Peiγ∣ψ⟩∥Peiγψ∥=eiγP∣ψ⟩∥Pψ∥\dfrac{Pe^{i\gamma}\lvert\psi\rangle}{\lVert Pe^{i\gamma}\psi\rVert}=e^{i\gamma}\dfrac{P\lvert\psi\rangle}{\lVert P\psi\rVert}, i.e. they lie on the same ray.

Proof (6 steps)
  1. Step 1:
    P(eiγ∣ψ⟩)=eiγ P∣ψ⟩P\big(e^{i\gamma}\lvert\psi\rangle\big)=e^{i\gamma}\,P\lvert\psi\rangle
    PP is a linear operator, so a scalar factor passes through it.
  2. Step 2:
    ∥P eiγψ∥2=⟨eiγPψ ∣ eiγPψ⟩\lVert P\,e^{i\gamma}\psi\rVert^2=\big\langle e^{i\gamma}P\psi\,\big\vert\,e^{i\gamma}P\psi\big\rangle
    Definition of the norm induced by the inner product: ∥v∥2=⟨v∣v⟩\lVert v\rVert^2=\langle v\vert v\rangle.
  3. Step 3:
    =eiγ‾ eiγ ⟨Pψ∣Pψ⟩=e−iγeiγ ⟨Pψ∣Pψ⟩=\overline{e^{i\gamma}}\,e^{i\gamma}\,\langle P\psi\vert P\psi\rangle=e^{-i\gamma}e^{i\gamma}\,\langle P\psi\vert P\psi\rangle
    The inner product is conjugate-linear in the first argument and linear in the second; eiγ‾=e−iγ\overline{e^{i\gamma}}=e^{-i\gamma} because γ\gamma is real.
  4. Step 4:
    =e0 ∥Pψ∥2=∥Pψ∥2=e^{0}\,\lVert P\psi\rVert^2=\lVert P\psi\rVert^2
    Exponents add: e−iγeiγ=e−iγ+iγ=1e^{-i\gamma}e^{i\gamma}=e^{-i\gamma+i\gamma}=1, i.e. ∣eiγ∣2=cos⁡2γ+sin⁡2γ=1\lvert e^{i\gamma}\rvert^2=\cos^2\gamma+\sin^2\gamma=1.
  5. Step 5:
    ∥Pψ∥2=⟨ψ∣P†P∣ψ⟩=⟨ψ∣P∣ψ⟩=Pr⁡(outcome P∣ψ)\lVert P\psi\rVert^2=\langle\psi\vert P^\dagger P\vert\psi\rangle=\langle\psi\vert P\vert\psi\rangle=\Pr(\text{outcome }P\mid\psi)
    Using P†P=P⋅P=PP^\dagger P=P\cdot P=P: this is exactly the Born-rule probability, which is therefore the same for ∣ψ⟩\lvert\psi\rangle and eiγ∣ψ⟩e^{i\gamma}\lvert\psi\rangle.
  6. Step 6:
    P eiγ∣ψ⟩∥P eiγψ∥=eiγP∣ψ⟩∥Pψ∥=eiγ P∣ψ⟩∥Pψ∥\frac{P\,e^{i\gamma}\lvert\psi\rangle}{\lVert P\,e^{i\gamma}\psi\rVert}=\frac{e^{i\gamma}P\lvert\psi\rangle}{\lVert P\psi\rVert}=e^{i\gamma}\,\frac{P\lvert\psi\rangle}{\lVert P\psi\rVert}
    First step for the numerator, the norm identity just proved for the denominator: the collapsed states differ by the same global phase, so the equivalence survives the measurement.

∎

Corollary

Global phase leaves every expectation value unchanged

For every linear operator AA on H\mathcal H (in particular every observable A=A†A=A^\dagger) and every real γ\gamma: ⟨eiγψ∣A∣eiγψ⟩=⟨ψ∣A∣ψ⟩\langle e^{i\gamma}\psi\vert A\vert e^{i\gamma}\psi\rangle=\langle\psi\vert A\vert\psi\rangle. Likewise, for every unitary UU, U(eiγ∣ψ⟩)=eiγU∣ψ⟩U(e^{i\gamma}\lvert\psi\rangle)=e^{i\gamma}U\lvert\psi\rangle, so time evolution maps rays to rays.

Proof (4 steps)
  1. Step 1:
    (eiγ∣ψ⟩)†=eiγ‾ ⟨ψ∣=e−iγ⟨ψ∣\big(e^{i\gamma}\lvert\psi\rangle\big)^\dagger=\overline{e^{i\gamma}}\,\langle\psi\vert=e^{-i\gamma}\langle\psi\vert
    The bra of a ket is its conjugate transpose; the scalar is complex-conjugated.
  2. Step 2:
    ⟨eiγψ∣A∣eiγψ⟩=e−iγ⟨ψ∣ A (eiγ∣ψ⟩)\langle e^{i\gamma}\psi\vert A\vert e^{i\gamma}\psi\rangle=e^{-i\gamma}\langle\psi\vert\,A\,\big(e^{i\gamma}\lvert\psi\rangle\big)
    Substitute the bra from the previous line.
  3. Step 3:
    =e−iγeiγ ⟨ψ∣A∣ψ⟩=⟨ψ∣A∣ψ⟩=e^{-i\gamma}e^{i\gamma}\,\langle\psi\vert A\vert\psi\rangle=\langle\psi\vert A\vert\psi\rangle
    Linearity of AA and of the bra moves eiγe^{i\gamma} out; the product of the two phases is 11.
  4. Step 4:
    U(eiγ∣ψ⟩)=eiγ U∣ψ⟩∈[Uψ]U\big(e^{i\gamma}\lvert\psi\rangle\big)=e^{i\gamma}\,U\lvert\psi\rangle\in[U\psi]
    Linearity of UU: the image of any representative of the ray [ψ][\psi] is a representative of the ray [Uψ][U\psi], so evolution is well defined on rays.

∎

Consequences. Since probabilities, expectation values, post-measurement states and time evolution all respect the equivalence ∣ψ⟩∼eiγ∣ψ⟩\lvert\psi\rangle\sim e^{i\gamma}\lvert\psi\rangle, a single-qubit state has only two real degrees of freedom, not four: writing α=∣α∣eia\alpha=\lvert\alpha\rvert e^{i a}, β=∣β∣eib\beta=\lvert\beta\rvert e^{ib}, normalisation removes one parameter and the global phase eiae^{ia} removes another, leaving ∣ψ⟩∼cos⁡θ2∣0⟩+eiϕsin⁡θ2∣1⟩\lvert\psi\rangle\sim\cos\tfrac\theta2\lvert0\rangle+e^{i\phi}\sin\tfrac\theta2\lvert1\rangle with θ∈[0,π]\theta\in[0,\pi], ϕ=b−a∈[0,2π)\phi=b-a\in[0,2\pi). These are exactly the polar and azimuthal angles of the Bloch sphere. The same argument shows that gates are only defined up to a global phase: ZZ and −Z=eiπZ-Z=e^{i\pi}Z are the same physical operation when applied to the whole register.

A warning. "Global" means multiplying the entire state vector. A phase that multiplies only some components is a relative phase and is physical. Likewise eiγUe^{i\gamma}U is indistinguishable from UU, but a controlled version of it is not: controlled-(−I)(-I) equals Z⊗IZ\otimes I, which flips the sign of ∣10⟩\lvert10\rangle and ∣11⟩\lvert11\rangle only — the phase kickback that powers phase estimation.

Quick check

Which pair of vectors represents the same physical state?

Select one answer

Relative phase is observable#

Consider ∣+⟩=12(∣0⟩+∣1⟩)\lvert+\rangle=\tfrac1{\sqrt2}(\lvert0\rangle+\lvert1\rangle) and ∣−⟩=12(∣0⟩−∣1⟩)\lvert-\rangle=\tfrac1{\sqrt2}(\lvert0\rangle-\lvert1\rangle). Measured in the computational basis both give P(0)=P(1)=12P(0)=P(1)=\tfrac12, so a ZZ-basis measurement alone cannot tell them apart. They are nonetheless different states — they are orthogonal. To reveal the difference we first rotate the basis with the Hadamard gate and then measure. The Hadamard matrix is unitary and Hermitian, H=H†=H−1H=H^\dagger=H^{-1}, as the unitarity check below confirms.

H=12(111−1),H†H=12(111−1)(111−1)=12(2002)=IH=\frac{1}{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix},\qquad H^\dagger H=\frac12\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}1&1\\1&-1\end{pmatrix}=\frac12\begin{pmatrix}2&0\\0&2\end{pmatrix}=I
Hadamard gate H2×2

Unitary verified: U†U=IU^\dagger U = I (max deviation 2.2e-16)

Derivation

Hadamard maps the X basis to the computational basis

  1. Step 1:
    H∣+⟩=12(111−1)12(11)=12(1⋅1+1⋅11⋅1+(−1)⋅1)=12(20)=(10)=∣0⟩H\lvert+\rangle=\frac{1}{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}\frac{1}{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix}=\frac12\begin{pmatrix}1\cdot1+1\cdot1\\1\cdot1+(-1)\cdot1\end{pmatrix}=\frac12\begin{pmatrix}2\\0\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}=\lvert0\rangle
    Row-times-column matrix multiplication; the two scalar prefactors combine to 12\tfrac12.
  2. Step 2:
    H∣−⟩=12(111−1)12(1−1)=12(1⋅1+1⋅(−1)1⋅1+(−1)(−1))=12(02)=(01)=∣1⟩H\lvert-\rangle=\frac{1}{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}\frac{1}{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix}=\frac12\begin{pmatrix}1\cdot1+1\cdot(-1)\\1\cdot1+(-1)(-1)\end{pmatrix}=\frac12\begin{pmatrix}0\\2\end{pmatrix}=\begin{pmatrix}0\\1\end{pmatrix}=\lvert1\rangle
    Same computation with the sign of the second amplitude flipped: now the first row cancels and the second row adds.
  3. Step 3:
    P+(0)=∣⟨0∣H∣+⟩∣2=∣⟨0∣0⟩∣2=1,P−(0)=∣⟨0∣H∣−⟩∣2=∣⟨0∣1⟩∣2=0P_{+}(0)=\lvert\langle0\vert H\vert+\rangle\rvert^2=\lvert\langle0\vert0\rangle\rvert^2=1,\qquad P_{-}(0)=\lvert\langle0\vert H\vert-\rangle\rvert^2=\lvert\langle0\vert1\rangle\rvert^2=0
    Born rule on the rotated states: the relative sign decides the outcome with certainty, whereas a global sign never could.
  4. Step 4:
    H(−∣−⟩)=−∣1⟩,∣⟨0∣−1⟩∣2=0, ∣⟨1∣−1⟩∣2=1H\big(-\lvert-\rangle\big)=-\lvert1\rangle,\qquad \lvert\langle0\vert{-1}\rangle\rvert^2=0,\ \lvert\langle1\vert{-1}\rangle\rvert^2=1
    Control experiment: a global sign on ∣−⟩\lvert-\rangle survives the gate as a global sign on ∣1⟩\lvert1\rangle and changes no probability, as the theorem predicts.

Result

H∣+⟩=∣0⟩,H∣−⟩=∣1⟩\boxed{H\lvert+\rangle=\lvert0\rangle,\quad H\lvert-\rangle=\lvert1\rangle}
Derivation

General relative phase: P(0) = cos²(φ/2) after a Hadamard

  1. Step 1:
    ∣ψφ⟩=12(∣0⟩+eiφ∣1⟩)=12(1eiφ),φ∈[0,2π)\lvert\psi_\varphi\rangle=\frac{1}{\sqrt2}\big(\lvert0\rangle+e^{i\varphi}\lvert1\rangle\big)=\frac{1}{\sqrt2}\begin{pmatrix}1\\e^{i\varphi}\end{pmatrix},\qquad \varphi\in[0,2\pi)
    The general equatorial Bloch state; φ=0\varphi=0 gives ∣+⟩\lvert+\rangle, φ=π\varphi=\pi gives ∣−⟩\lvert-\rangle, φ=π/2\varphi=\pi/2 gives ∣+i⟩\lvert{+i}\rangle.
  2. Step 2:
    H∣ψφ⟩=12(111−1)12(1eiφ)=12(1+eiφ1−eiφ)H\lvert\psi_\varphi\rangle=\frac{1}{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}\frac{1}{\sqrt2}\begin{pmatrix}1\\e^{i\varphi}\end{pmatrix}=\frac12\begin{pmatrix}1+e^{i\varphi}\\1-e^{i\varphi}\end{pmatrix}
    Matrix–vector product, row by row.
  3. Step 3:
    1+eiφ=eiφ/2(e−iφ/2+eiφ/2)=2eiφ/2cos⁡φ2,1−eiφ=eiφ/2(e−iφ/2−eiφ/2)=−2i eiφ/2sin⁡φ21+e^{i\varphi}=e^{i\varphi/2}\big(e^{-i\varphi/2}+e^{i\varphi/2}\big)=2e^{i\varphi/2}\cos\tfrac{\varphi}{2},\qquad 1-e^{i\varphi}=e^{i\varphi/2}\big(e^{-i\varphi/2}-e^{i\varphi/2}\big)=-2i\,e^{i\varphi/2}\sin\tfrac{\varphi}{2}
    Factor out the half-angle phase and use Euler's formulas eix+e−ix=2cos⁡xe^{ix}+e^{-ix}=2\cos x and eix−e−ix=2isin⁡xe^{ix}-e^{-ix}=2i\sin x.
  4. Step 4:
    H∣ψφ⟩=eiφ/2(cos⁡φ2 ∣0⟩−isin⁡φ2 ∣1⟩)H\lvert\psi_\varphi\rangle=e^{i\varphi/2}\Big(\cos\tfrac{\varphi}{2}\,\lvert0\rangle-i\sin\tfrac{\varphi}{2}\,\lvert1\rangle\Big)
    Substitute both factorisations; the common factor eiφ/2e^{i\varphi/2} is a global phase of the output state.
  5. Step 5:
    P(0)=∣eiφ/2cos⁡φ2∣2=cos⁡2φ2,P(1)=∣−ieiφ/2sin⁡φ2∣2=sin⁡2φ2P(0)=\Big\lvert e^{i\varphi/2}\cos\tfrac{\varphi}{2}\Big\rvert^2=\cos^2\tfrac{\varphi}{2},\qquad P(1)=\Big\lvert -ie^{i\varphi/2}\sin\tfrac{\varphi}{2}\Big\rvert^2=\sin^2\tfrac{\varphi}{2}
    Born rule; ∣eiφ/2∣=∣−i∣=1\lvert e^{i\varphi/2}\rvert=\lvert-i\rvert=1.
  6. Step 6:
    P(0)=∣1+eiφ∣24=(1+cos⁡φ)2+sin⁡2φ4=2+2cos⁡φ4=1+cos⁡φ2=cos⁡2φ2P(0)=\frac{\lvert1+e^{i\varphi}\rvert^2}{4}=\frac{(1+\cos\varphi)^2+\sin^2\varphi}{4}=\frac{2+2\cos\varphi}{4}=\frac{1+\cos\varphi}{2}=\cos^2\tfrac{\varphi}{2}
    Independent check without factoring: expand the modulus squared, use cos⁡2φ+sin⁡2φ=1\cos^2\varphi+\sin^2\varphi=1 and the half-angle identity cos⁡2φ2=1+cos⁡φ2\cos^2\tfrac\varphi2=\tfrac{1+\cos\varphi}{2}.
  7. Step 7:
    P(1)=1−P(0)=sin⁡2φ2=1−cos⁡φ2P(1)=1-P(0)=\sin^2\tfrac{\varphi}{2}=\frac{1-\cos\varphi}{2}
    Normalisation of the output, consistent with the direct computation.

Result

P(0)=cos⁡2φ2:φ=0⇒1,φ=π2⇒12,φ=π⇒0\boxed{P(0)=\cos^2\tfrac{\varphi}{2}:\quad \varphi=0\Rightarrow1,\quad \varphi=\tfrac{\pi}{2}\Rightarrow\tfrac12,\quad \varphi=\pi\Rightarrow0}

Circuit · 1 qubit

Quantum circuit on 1 qubitX on q0; H on q0q0XH

XX then HH on one qubit: X∣0⟩=∣1⟩X\lvert0\rangle=\lvert1\rangle and H∣1⟩=∣−⟩H\lvert1\rangle=\lvert-\rangle, so the simulator shows exactly the amplitudes (12, −12)\big(\tfrac1{\sqrt2},\,-\tfrac1{\sqrt2}\big) — not merely a vector on the same ray. The alternative HH then ZZ gives the same vector, while HH, ZZ, XX gives XZ∣+⟩=−∣−⟩XZ\lvert+\rangle=-\lvert-\rangle: identical physics, different global phase.

∣ψ⟩=12∣0⟩−12∣1⟩\lvert\psi\rangle = \tfrac{1}{\sqrt{2}}\lvert 0\rangle - \tfrac{1}{\sqrt{2}}\lvert 1\rangle

In the PhaseExplorer, move the γ\gamma slider: every amplitude rotates in the complex plane by the same angle, and both the ZZ-basis and the XX-basis histograms stay frozen. Now move φ\varphi: the ZZ-basis histogram still shows 12,12\tfrac12,\tfrac12, but the XX-basis histogram (equivalently, "apply HH then measure") follows cos⁡2φ2\cos^2\tfrac\varphi2 and sin⁡2φ2\sin^2\tfrac\varphi2. This is the operational meaning of Postulate 1: information stored in relative phases is invisible in one basis and fully visible in another.

Quick check

The state 12(∣0⟩+eiπ/2∣1⟩)=∣+i⟩\tfrac1{\sqrt2}(\lvert0\rangle+e^{i\pi/2}\lvert1\rangle)=\lvert{+i}\rangle is sent through HH and measured. What is P(0)P(0)?

tolerance ±0.000001

Superposition and normalisation#

Proposition

Superposition principle

If ∣ψ1⟩,…,∣ψm⟩\lvert\psi_1\rangle,\dots,\lvert\psi_m\rangle are state vectors and a1,…,am∈Ca_1,\dots,a_m\in\mathbb C satisfy ∑jaj∣ψj⟩≠0\sum_ja_j\lvert\psi_j\rangle\neq0, then ∣χ⟩=∑jaj∣ψj⟩∥∑jajψj∥\lvert\chi\rangle=\dfrac{\sum_ja_j\lvert\psi_j\rangle}{\lVert\sum_ja_j\psi_j\rVert} is again a valid state. In an orthonormal basis {∣i⟩}i<d\{\lvert i\rangle\}_{i<d} every state is ∣ψ⟩=∑ici∣i⟩\lvert\psi\rangle=\sum_ic_i\lvert i\rangle with ci=⟨i∣ψ⟩c_i=\langle i\vert\psi\rangle and ∑i∣ci∣2=1\sum_i\lvert c_i\rvert^2=1.

Proof (4 steps)
  1. Step 1:
    v=∑jaj∣ψj⟩∈Hv=\sum_ja_j\lvert\psi_j\rangle\in\mathcal H
    H\mathcal H is a complex vector space, so it is closed under linear combinations.
  2. Step 2:
    ∥v∥v∥∥=∥v∥∥v∥=1(v≠0)\Big\lVert\frac{v}{\lVert v\rVert}\Big\rVert=\frac{\lVert v\rVert}{\lVert v\rVert}=1\quad(v\neq0)
    Homogeneity of the norm, ∥λv∥=∣λ∣∥v∥\lVert\lambda v\rVert=\lvert\lambda\rvert\lVert v\rVert with λ=1/∥v∥>0\lambda=1/\lVert v\rVert>0: the rescaled vector is a unit vector, hence a state by Postulate 1.
  3. Step 3:
    ∣ψ⟩=∑i∣i⟩⟨i∣ψ⟩=∑ici∣i⟩,ci=⟨i∣ψ⟩\lvert\psi\rangle=\sum_i\lvert i\rangle\langle i\vert\psi\rangle=\sum_ic_i\lvert i\rangle,\qquad c_i=\langle i\vert\psi\rangle
    Insert the completeness relation ∑i∣i⟩⟨i∣=I\sum_i\lvert i\rangle\langle i\rvert=I of the orthonormal basis.
  4. Step 4:
    1=⟨ψ∣ψ⟩=∑i,jci‾ cj ⟨i∣j⟩=∑i,jci‾ cj δij=∑i∣ci∣21=\langle\psi\vert\psi\rangle=\sum_{i,j}\overline{c_i}\,c_j\,\langle i\vert j\rangle=\sum_{i,j}\overline{c_i}\,c_j\,\delta_{ij}=\sum_i\lvert c_i\rvert^2
    Expand the inner product bilinearly (conjugating the left coefficients) and use orthonormality ⟨i∣j⟩=δij\langle i\vert j\rangle=\delta_{ij}.

∎

Derivation

Worked example: normalising (1+i)|0⟩ + 2|1⟩

  1. Step 1:
    v=(1+i)∣0⟩+2∣1⟩=(1+i2)v=(1+i)\lvert0\rangle+2\lvert1\rangle=\begin{pmatrix}1+i\\2\end{pmatrix}
    An unnormalised vector, e.g. the output of a superposition of two preparations.
  2. Step 2:
    ∥v∥2=∣1+i∣2+∣2∣2=(12+12)+4=6\lVert v\rVert^2=\lvert1+i\rvert^2+\lvert2\rvert^2=(1^2+1^2)+4=6
    ∣a+ib∣2=a2+b2\lvert a+ib\rvert^2=a^2+b^2 for each component, then sum.
  3. Step 3:
    ∣ψ⟩=v∥v∥=1+i6∣0⟩+26∣1⟩\lvert\psi\rangle=\frac{v}{\lVert v\rVert}=\frac{1+i}{\sqrt6}\lvert0\rangle+\frac{2}{\sqrt6}\lvert1\rangle
    Divide by the norm 6\sqrt6.
  4. Step 4:
    P(0)=∣1+i∣26=26=13,P(1)=46=23,P(0)+P(1)=1P(0)=\frac{\lvert1+i\rvert^2}{6}=\frac{2}{6}=\frac13,\qquad P(1)=\frac{4}{6}=\frac23,\qquad P(0)+P(1)=1
    Born rule in the computational basis; the probabilities sum to one precisely because we normalised.
  5. Step 5:
    ∣ψ⟩=eiπ/4(13∣0⟩+23 e−iπ/4∣1⟩)\lvert\psi\rangle=e^{i\pi/4}\Big(\frac{1}{\sqrt3}\lvert0\rangle+\sqrt{\tfrac23}\,e^{-i\pi/4}\lvert1\rangle\Big)
    Pull out the phase of the first amplitude, 1+i=2eiπ/41+i=\sqrt2e^{i\pi/4}: up to a global phase the state is cos⁡θ2∣0⟩+eiϕsin⁡θ2∣1⟩\cos\tfrac\theta2\lvert0\rangle+e^{i\phi}\sin\tfrac\theta2\lvert1\rangle with cos⁡θ2=13\cos\tfrac\theta2=\tfrac1{\sqrt3} and relative phase ϕ=−π/4\phi=-\pi/4.

Result

∣ψ⟩=16((1+i)∣0⟩+2∣1⟩),P(0)=13, P(1)=23\boxed{\lvert\psi\rangle=\tfrac{1}{\sqrt6}\big((1+i)\lvert0\rangle+2\lvert1\rangle\big),\quad P(0)=\tfrac13,\ P(1)=\tfrac23}

Superposition is not a mixture. The state ∣+⟩\lvert+\rangle gives 00 or 11 with probability 12\tfrac12 each, like a fair coin, but it is not "a qubit that is secretly ∣0⟩\lvert0\rangle or ∣1⟩\lvert1\rangle with probability 12\tfrac12". The relative-phase derivation shows the difference: after HH, ∣+⟩\lvert+\rangle gives 00 with certainty, while the coin-like mixture would still give 12,12\tfrac12,\tfrac12 (half the time H∣0⟩=∣+⟩H\lvert0\rangle=\lvert+\rangle, half the time H∣1⟩=∣−⟩H\lvert1\rangle=\lvert-\rangle, each yielding 12,12\tfrac12,\tfrac12). Amplitudes add before squaring — 12(1+1)\tfrac12(1+1) versus 12(1−1)\tfrac12(1-1) — and that interference is precisely what probabilities cannot do. Mixtures are handled by density operators in Module 1.4.

Counting parameters. An nn-qubit state is a unit vector in C2n\mathbb C^{2^n}: 2n+12^{n+1} real numbers, minus one for normalisation, minus one for the global phase, leaves 2n+1−22^{n+1}-2 real parameters. For n=1n=1 this is 22 (the Bloch angles); for n=10n=10 it is already 20462046. This is the exponential description cost that the statevector simulator pays and that quantum algorithms try to exploit.

Quick check

Normalise v=∣0⟩+(1−i)∣1⟩+i∣2⟩v=\lvert0\rangle+(1-i)\lvert1\rangle+i\lvert2\rangle in C3\mathbb C^3. What is the probability of outcome 11?

tolerance ±0.000001

References

  1. M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010), §2.2 (the postulates of quantum mechanics; §2.2.1 state space and §2.2.7 global phase) and §1.2 (single qubits and the Bloch sphere).
  2. J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press (2020), §1.2 (kets, bras and operators) and §1.4 (measurements, observables and the uncertainty relations).
  3. J. Preskill, Lecture Notes for Physics 219: Quantum Computation, Caltech, Chapter 2 (foundations I: states and ensembles), §2.1 (axioms of quantum mechanics).theory.caltech.edu

Hands-on labs

  • Prepare exactly |−⟩ (global phase included)

    1 step · auto-graded

    Prepare the exact vector 12(1,−1)T\tfrac{1}{\sqrt2}(1,-1)^T from ∣0⟩\lvert0\rangle. Physically eiγ∣−⟩e^{i\gamma}\lvert-\rangle is the same state, but this lab compares vectors component by component, so the global phase must be γ=0\gamma=0.

    Open lab