University Physics V · Foundations of Quantum Mechanics · 2.9
Quantum States, Dirac Notation & Mixtures
This is the grammar the rest of the course is written in. Practise the moves until they are reflex: normalise, project, insert the identity, drop a global phase and keep a relative one — then meet the density operator, because most of the systems you will ever calculate on do not come with a ket.
Build the model
Connect the measurement to the mechanism.
Quantum mechanics assigns an isolated system a unit vector in a complex separable Hilbert space, then immediately weakens the claim: |ψ⟩ and e(iα)|ψ⟩ give identical predictions for every observable, so the physical state is the ray, not the ket. Everything computable flows through the inner product — probabilities are squared overlaps, and any orthonormal basis carries a resolution of the identity Σ|n⟩⟨n| = 1̂ that can be inserted anywhere to change representation at zero cost. The formalism's power is also its limit: a ket encodes the most that can be known, and the moment the preparation itself is uncertain — a source emitting |0⟩ or |1⟩ on a coin flip — no ket exists for the output.
The honest object is the density operator ρ = Σ pₖ|ψₖ⟩⟨ψₖ|, positive with unit trace, whose off-diagonal coherences are the entire difference between a superposition, which can interfere, and a classical mixture, which cannot. What the ray convention costs is vigilance about phase: a global phase never matters, a relative phase inside a superposition always does, and half the errors in this course come from confusing the two.
- Simple definition
- A quantum state is the ray a unit ket spans in a complex separable Hilbert space; when the preparation is uncertain it is a density operator — a positive, Hermitian, unit-trace operator on that space.
- Example
- (|0⟩+|1⟩)/√2 and the mixture ½|0⟩⟨0| + ½|1⟩⟨1| both give P(0) = P(1) = ½, yet σₓ returns +1 with probability 1 on the first and ½ on the second: the coherence ⟨0|ρ|1⟩ = ½ versus 0 is the entire difference.
|⟨φ|e(iα)ψ⟩|² = |⟨φ|ψ⟩|² for every |φ⟩, so the representative ket is yours to choose — a phase inside a sum is not covered.
α dimensionless (radians); both kets unit-normalised, ⟨ψ|ψ⟩ = 1
Insert it anywhere to change basis or read off a matrix element; Parseval keeps total probability at one.
(|n⟩) any orthonormal basis; cₙ dimensionless amplitudes with Σ|cₙ|² = 1
The state of a system whose preparation is uncertain — the object that exists when no single ket does.
pₖ preparation probabilities; ⟨A⟩ = Tr(ρA), P = Tr(ρP̂) for a projector P̂
One basis-independent number answers 'ket or mixture?' — no diagonalisation, no choice of axes.
d = dim H; the floor 1/d is reached only by the maximally mixed ρ = 1̂/d
Every qubit state is a point in the unit ball: pure states on the sphere, the even mixture at the centre.
coherences: aₓ = 2 Re ρ₀₁, ay = −2 Im ρ₀₁; Tr ρ² = ½(1 + |a|²)
The state is a ray in a separable Hilbert space
The first postulate assigns an isolated system a unit vector |ψ⟩ in a complex separable Hilbert space H — two dimensions for a spin-½, the square-integrable functions L² for a particle on a line. But the assignment is many-to-one: |ψ⟩ and e(iα)|ψ⟩ satisfy |⟨φ|e(iα)ψ⟩|² = |⟨φ|ψ⟩|² for every |φ⟩, so no experiment distinguishes them. The physical state is therefore the ray — the one-dimensional subspace {c|ψ⟩ : c ∈ ℂ} — with the unit ket a convenient representative. Separability earns its place in the sentence: it guarantees a countable orthonormal basis, which is what lets a state be a column of amplitudes and an operator a matrix rather than something worse.
Bras, inner products, and what they compute
For each ket |ψ⟩ the Riesz theorem supplies exactly one dual functional ⟨ψ|, and the pairing ⟨φ|ψ⟩ is the inner product — conjugate-linear in the bra slot in the physics convention, so ⟨φ|ψ⟩ = ⟨ψ|φ⟩*. Everything measurable is built from it: ⟨ψ|ψ⟩ = 1 is the normalisation Born's rule needs, |⟨φ|ψ⟩|² is the probability that a system prepared in |ψ⟩ passes a test for |φ⟩, and ⟨φ|A|ψ⟩ is a matrix element once an operator sits between. Try it on |ψ⟩ = (3|0⟩ + 4i|1⟩)/5: ⟨ψ|ψ⟩ = (9 + 16)/25 = 1 — but only because the bra conjugates 4i to −4i, making the cross term (−4i)(4i) = +16. Drop the conjugate and you get (9 − 16)/25, a negative probability; the star is not decoration.
Insert the identity to change basis
Any orthonormal basis (|n⟩) resolves the identity: Σ|n⟩⟨n| = 1̂. Insert it before a ket and the expansion falls out, |ψ⟩ = Σ⟨n|ψ⟩|n⟩, each coefficient a projection cₙ = ⟨n|ψ⟩, with Parseval guaranteeing Σ|cₙ|² = 1 — probability survives a change of representation because the insertion is the identity, not an approximation. Concretely: the z-basis ket |0⟩ meets the x basis |±⟩ = (|0⟩ ± |1⟩)/√2 through ⟨+|0⟩ = ⟨−|0⟩ = 1/√2, so |0⟩ = (|+⟩ + |−⟩)/√2 and a σₓ measurement returns ±1 at ½ each. The same insertion between an operator and a ket, ⟨m|A|ψ⟩ = Σₙ ⟨m|A|n⟩⟨n|ψ⟩, is nothing but matrix-times-column — Dirac notation makes the linear algebra self-assembling.
Global phase never, relative phase always
Multiply an entire ket by e(iα) and every probability keeps its value: the phase is global and unphysical. Move the same factor inside a superposition and it becomes a relative phase, which measurement can see. Take |ψφ⟩ = (|0⟩ + e(iφ)|1⟩)/√2. In the z basis both amplitudes have modulus 1/√2 whatever φ is: P(0) = P(1) = ½ and φ is invisible. Project on |+⟩ instead: ⟨+|ψφ⟩ = (1 + e(iφ))/2, so P(+x) = ½(1 + cos φ) — a full interference fringe. At φ = 0 the state is |+⟩ itself and P = 1; at φ = π it is |−⟩ and P = 0; at φ = π/2 it is the σy eigenstate and P = ½. The working rule: a phase you can factor out of every term is gauge; one you cannot is data.
When no ket exists: the density operator
A ket encodes maximal knowledge. A machine that emits |ψₖ⟩ with probability pₖ has handed you less, and no single ket in H reproduces its statistics. The right object is ρ = Σ pₖ|ψₖ⟩⟨ψₖ| — Hermitian, positive, unit trace — with every prediction read off as ⟨A⟩ = Tr(ρA) and P = Tr(ρP̂) for the outcome's projector. A pure state is the special case ρ = |ψ⟩⟨ψ|, a rank-one projector with ρ² = ρ; anything else is mixed, and the basis-free test is purity: Tr ρ² = 1 iff pure, falling to 1/d at the maximally mixed 1̂/d. One warning the notation hides: the decomposition into pₖ and |ψₖ⟩ is not unique. Mixing |0⟩ and |+⟩ at ½ each gives ρ = [[¾, ¼],[¼, ¼]], and mixing that ρ's own eigenstates at 0.854 and 0.146 gives exactly the same matrix — the preparations differ, the physics does not.
Coherences are what a mixture is missing
Write both candidates in the (|0⟩, |1⟩) basis. The superposition (|0⟩+|1⟩)/√2 has ρ = ½[[1,1],[1,1]]; the even mixture has ρ = ½[[1,0],[0,1]]. Their diagonals — the populations — agree, so every z measurement agrees. The whole difference sits in the off-diagonal coherence ⟨0|ρ|1⟩: ½ against 0. The Pauli decomposition ρ = ½(1̂ + a⋅σ) makes this geometric: a = (⟨σₓ⟩, ⟨σy⟩, ⟨σz⟩) is a point in the unit ball, pure states on the surface, the even mixture at dead centre, and the coherences are the equatorial components aₓ = 2 Re ρ₀₁, ay = −2 Im ρ₀₁. A σₓ measurement returns +1 with probability ½(1 + aₓ): 1 for the superposition, ½ for the mixture. Interference is nothing exotic — it is a coherence being read out in a rotated basis, and decoherence is that matrix element draining away.
Change one variable at a time
Make the relationship visible.
Slide V from 1 to 0 at any φ: the arrow retreats to the centre and the fringe flattens onto the dashed P = ½ line — a mixture keeps no coherence for any basis to read. At V = 1 the state is pure and φ alone swings P(+x) through the full range 1 to 0.
P(+x)0.75
COHERENCE |ρ01|0.50
PURITY Tr ρ²1.00
EIGENVALUE λ+1.00
Live interpretationP(+x): 0.75. COHERENCE |ρ01|: 0.50. PURITY Tr ρ²: 1.00. EIGENVALUE λ+: 1.00
Catch the common trap
Explain before calculating.
A source outputs qubits that give 0 and 1 with probability ½ each when measured in the (|0⟩, |1⟩) basis. Which measurement settles whether it emits the superposition (|0⟩+|1⟩)/√2 or the even mixture ρ = 1̂/2?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA qubit is prepared as the unnormalised ket 3|0⟩ + 4i|1⟩. Normalise it, find the probabilities of the outcomes 0 and 1, then the probability of +1 when σₓ is measured, given |+⟩ = (|0⟩ + |1⟩)/√2.
- ⟨ψ|ψ⟩ = |3|² + |4i|² = 9 + 16 = 25 — the bra conjugates 4i to −4i, so the cross product is (−4i)(4i) = +16, never −16. Divide by √25 = 5: |ψ⟩ = (3|0⟩ + 4i|1⟩)/5.
- Born's rule on the z basis: P(0) = |3/5|² = 0.36 and P(1) = |4i/5|² = 0.64. They sum to 1 because the basis is complete — that is Parseval, not luck.
- Project on |+⟩: ⟨+|ψ⟩ = (3 + 4i)/(5√2), so P(+x) = (3² + 4²)/50 = 25/50 = 0.50.
- The i has not vanished — it moved: ⟨σy⟩ = 2 Im(c₀*c₁) = 2 Im(3/5 × 4i/5) = 24/25 = 0.96. The relative phase hides from σz and σₓ and shows up in σy.
Answer|ψ⟩ = (3|0⟩ + 4i|1⟩)/5; P(0) = 0.36, P(1) = 0.64, P(+x) = 0.50, and the relative phase i surfaces as ⟨σy⟩ = 0.96.
MediumFor |ψφ⟩ = (|0⟩ + e(iφ)|1⟩)/√2, find P(+x) as a function of φ and evaluate it at φ = π/3. Then show that multiplying the whole ket by e(iπ/7) changes nothing, while moving that same factor onto |1⟩ alone changes the answer.
- ⟨+|ψφ⟩ = (1 + e(iφ))/2, so P(+x) = |1 + e(iφ)|²/4 = (2 + 2 cos φ)/4 = ½(1 + cos φ).
- At φ = π/3: cos(π/3) = 0.5, so P(+x) = ½ × 1.5 = 0.75. In the z basis the same state still gives P(0) = P(1) = ½ — the phase is invisible there.
- Global: e(iπ/7)|ψφ⟩ multiplies every overlap by e(iπ/7), and |e(iπ/7)| = 1, so every |⟨φ|ψ⟩|² and every expectation value is untouched. Same ray, same state.
- Relative: (|0⟩ + e(i(π/3 + π/7))|1⟩)/√2 has phase 10π/21, so P(+x) = ½(1 + cos(10π/21)) = ½(1 + 0.0747) ≈ 0.54 — measurably different from 0.75.
AnswerP(+x) = ½(1 + cos φ) = 0.75 at φ = π/3; a global e(iπ/7) leaves every prediction fixed, but the same factor as a relative phase drags P(+x) down to ≈ 0.54.
HardA source emits |0⟩ with probability ½ and |+⟩ = (|0⟩+|1⟩)/√2 with probability ½. Construct ρ in the (|0⟩, |1⟩) basis, find its Bloch vector, purity and eigenvalues, and explain why the ½/½ recipe does not appear in the spectrum.
- Outer products: |0⟩⟨0| = [[1,0],[0,0]] and |+⟩⟨+| = ½[[1,1],[1,1]]. Then ρ = ½|0⟩⟨0| + ½|+⟩⟨+| = [[¾, ¼],[¼, ¼]] — Hermitian, trace 1, positive.
- Bloch components: aₓ = 2 Re ρ₀₁ = ½, ay = −2 Im ρ₀₁ = 0, az = ρ₀₀ − ρ₁₁ = ½. So a = (½, 0, ½) and |a| = √(¼ + ¼) = 1/√2 ≈ 0.707 < 1: the state is mixed.
- Purity: Tr ρ² = ½(1 + |a|²) = ½(1 + ½) = ¾ — strictly below the pure value 1, above the qubit floor ½.
- Eigenvalues: λ± = ½(1 ± |a|) = ½ ± √2/4 ≈ 0.854 and 0.146. Cross-check against the matrix: trace ¾ + ¼ = 1 and det = ¾⋅¼ − ¹⁄₁₆ = ⅛ = λ₊λ₋. Both agree.
- The ½/½ weights are absent because |0⟩ and |+⟩ are not orthogonal — ⟨0|+⟩ = 1/√2 — so they are not eigenvectors of ρ. Mixing ρ's own eigenstates at 0.854/0.146 builds the identical matrix: the decomposition of a density operator is not unique, and only ρ itself is physical.
Answerρ = [[¾, ¼],[¼, ¼]]; Bloch vector (½, 0, ½) with |a| ≈ 0.707; purity Tr ρ² = ¾; eigenvalues 0.854 and 0.146. Any preparation producing this ρ is experimentally identical.