A quantum state is a normalized complex description of possible observations

What a qubit is, and what superposition is not

A classical bit has values zero or one. A qubit state is a unit vector in a two-dimensional complex Hilbert space, written α|0⟩ + β|1⟩ where α and β are complex and |α|² + |β|² = 1. The amplitudes are not probabilities. Measurement in the computational basis returns one bit — 0 with probability |α|², 1 with probability |β|² — and destroys the state; you do not get to look at the amplitudes afterward.

Interviewers open here precisely because the weak answers cluster. The two failures they are fishing for:

• "A qubit is 0 and 1 at the same time." Superposition is a statement about the vector description, not about the qubit holding both bit values. Measurement yields exactly one outcome.
• "It's like a coin spinning in the air — 50/50 until you look." A classical hidden-variable story predicts the same computational-basis counts for |+⟩ and for a fair coin, but they are different physical situations: |+⟩ has a relative phase that shows up the moment you measure in a different basis or run it through interference. If you can't say what distinguishes them, you haven't answered the question.

A strong answer names the mechanism: complex amplitudes, squared magnitudes as probabilities, and interference as the observable consequence of amplitude structure. If the interviewer asks "so what is a qubit physically?" — a trapped ion's two internal levels, a superconducting circuit's ground and first excited states — the answer is that the physical system realizes the two-dimensional complex vector space; the math is the same across platforms.

Bra-ket fluency: Dirac notation is shorthand for linear algebra

A ket |ψ⟩ is a column vector; its bra ⟨ψ| is the conjugate transpose (a row vector). ⟨φ|ψ⟩ is the complex inner product, conjugate-linear in the first argument. |⟨φ|ψ⟩|² is the probability of obtaining φ's outcome when measuring ψ against a basis containing φ. Orthonormal basis vectors have unit norm and zero pairwise inner product, which is what makes measurement probabilities sum to one.

The common interview miss here is dropping the conjugation: writing ⟨φ|ψ⟩ as an ordinary dot product fails on complex amplitudes. If asked to compute ⟨+|ψ⟩ for ψ = (α, β), the answer is (α + β)/√2, conjugates and all, and the probability is |α + β|²/2. Dirac notation earns its keep because it makes outer products legible too: |ψ⟩⟨ψ| is a matrix, which is exactly the density operator for a pure state. If you treat the notation as magic syntax rather than linear algebra, the follow-up "derive the measurement rule from the inner product" will expose it.

Global phase versus relative phase

Multiplying every amplitude by the same unit-modulus complex number e^{iφ} changes no observable prediction: global phase is physically irrelevant, and α|0⟩ + β|1⟩ and e^{iφ}(α|0⟩ + β|1⟩) are the same state for all purposes you can measure. The phase of β relative to α is observable through interference. |+⟩ = (|0⟩ + |1⟩)/√2 and |−⟩ = (|0⟩ − |1⟩)/√2 have identical computational-basis probabilities — half and half — but a Hadamard maps |+⟩ to |0⟩ and |−⟩ to |1⟩, deterministically. The sign between the components decided the outcome.

Interviewers probe this with the pair question: "I give you two states that both measure 50/50 in the computational basis. Are they the same state?" The weak answer is yes. The strong answer is that computational-basis counts alone cannot distinguish them, and then names the measurement that can. The follow-up is usually "why is global phase unobservable?" — because every measurement probability is a squared magnitude of an inner product, and the phase cancels in |e^{iφ}c|². In code, this means comparing states up to a fitted global phase or via fidelity, not elementwise equality.

Basis dependence: |0⟩/|1⟩ is a choice

The computational basis is one orthonormal basis among infinitely many. |+⟩ and |−⟩ are the same states you already know, written in the X basis — and conversely, |0⟩ = (|+⟩ + |−⟩)/√2. Expanding a state in a different basis changes its coordinates, not the physical state. Measuring in one basis gives you no information about a state aligned with another: measuring |+⟩ in the computational basis yields a uniformly random bit and tells you nothing about the fact that the state was perfectly definite in the X basis.

This is where the Bloch sphere earns its place: a pure single-qubit state is two angles (θ, φ) after removing normalization and global phase, opposite points are orthogonal, and the choice of measurement basis is the choice of an axis. The interior of the Bloch ball is mixed states in the density-operator picture — not available to pure state vectors — and one Bloch sphere cannot represent the full state of multiple entangled qubits. If an interviewer asks "can you draw this two-qubit state on the Bloch sphere?" for an entangled pair, the correct answer is no, and the reason is that the joint state does not factor.

Composite systems, entanglement, and the exponential

Composite systems use tensor products. Two qubits occupy a four-dimensional space with basis |00⟩, |01⟩, |10⟩, |11⟩; n qubits need 2ⁿ complex amplitudes in a general statevector. A product state factors into one state per subsystem; nonfactorable pure states are entangled. Even for a product state, tensor-factor order matters to array layout, and dimension metadata should preserve subsystem structure instead of treating a length-four vector as ambiguously one four-level system or two qubits.

The exponential is worth pricing rather than describing, because it decides which experiments are possible on the machine in front of you. A dense statevector in complex128 costs 16 bytes per amplitude: 30 qubits is 2³⁰ amplitudes, or 17.2 GB, which fits a large workstation; 34 qubits is 275 GB, which does not. The density-matrix picture squares the object — 4ⁿ entries — so a 17-qubit noisy simulation costs the same 275 GB as a 34-qubit pure-state simulation. That factor-of-two collapse in reachable width is the single most useful number to carry into a design discussion: an algorithm validated on 30 qubits cannot be validated above roughly 15 qubits under a density-matrix noise model on the same hardware, and the honest response is to change the simulation method — stabilizer, tensor-network, or trajectory sampling with a stated bond dimension or trajectory count — rather than quietly drop the noise model and keep the width. Tomography prices the same exponential in shots: reconstructing a 5-qubit density matrix takes 3⁵ = 243 measurement settings for 4⁵ = 1,024 real parameters, and at 4,096 shots per setting that is roughly 1.0 million circuit executions to characterize a state a simulator would hold in 512 bytes.

The standard follow-up is "so does that mean quantum computers are just exponentially parallel classical computers?" The answer is no, and the reason is measurement: the 2ⁿ amplitudes exist in the description, but a single execution yields n classical bits. Useful algorithms work by arranging interference so that amplitudes for wrong answers cancel before measurement — which is exactly why relative phase, not amplitude count, is where the power lives.

Ordering conventions: the failure that looks like success

Ordering conventions cause more wrong answers in practice than any misunderstanding of superposition, because they fail silently and produce plausible output. A framework that draws qubit 0 at the top of a circuit diagram while treating it as the least significant bit of the output integer will report the bit string 100 for the state in which qubit 2 is excited and qubits 0 and 1 are not, so a reader tracing the diagram top to bottom concludes the opposite of what happened. The same convention decides which tensor factor a partial trace removes: tracing out subsystem 0 under the wrong convention returns the reduced state of a different qubit, and since both results are legitimate density matrices, nothing raises an error.

The defence is to test the asymmetric case rather than the symmetric one. Prepare a state that is deliberately not invariant under relabelling — apply X to exactly one qubit of a three-qubit register, or use amplitudes 0.1, 0.2, 0.3 and 0.9 on a two-qubit register — then assert the statevector index, the classical register mapping, the histogram key and the partial trace against values worked out by hand. A Bell state or a uniform superposition passes every ordering convention equally well, which is exactly why it is the wrong fixture for this test.

Mixed states, density matrices, and what measurement can't recover

A statevector describes a pure state. A density matrix ρ describes both pure states and probabilistic ensembles: Hermitian, positive semidefinite, trace one. For a pure state |ψ⟩, ρ = |ψ⟩⟨ψ| and Tr(ρ²) = 1; mixed states have lower purity. The distinction that interviewers actually test: a classical mixture of |0⟩ and |1⟩ with equal probabilities differs from the pure |+⟩ state. Both yield half zero and half one in the computational basis, but only |+⟩ retains off-diagonal coherence and measures deterministically in the plus/minus basis. The off-diagonal terms of ρ are precisely the coherence relative to the chosen basis.

The reduced state of a subsystem comes from the partial trace over the rest. An entangled global pure state can give each local subsystem a mixed reduced state — this does not mean the global state was prepared by a classical lottery; the correlations live in the joint density operator. Expect the follow-up: "if each qubit of a Bell pair is locally maximally mixed, how do the correlations show up?" Answer: in the joint state, not in either marginal — measure both qubits and compare outcomes.

Two limits close the topic. No-cloning: state preparation cannot copy an unknown arbitrary quantum state; a universal cloner would violate linear quantum evolution. Known basis states or classical descriptions can be re-prepared, so no-cloning is not a ban on creating two qubits in the same known state — a distinction interviewers probe by asking whether no-cloning forbids a CNOT. (It doesn't: CNOT copies a basis state, and fails to copy superpositions — apply it to α|0⟩ + β|1⟩ and you get entanglement, not a copy.) Tomography: estimating a density matrix from measurements in multiple settings across many independently prepared copies; finite sampling can make naive linear estimates unphysical, so constrained reconstruction enforces positivity and trace, and tomography does not reveal the pre-measurement state of one individual system — each copy is consumed by its measurement.

What a weak answer sounds like, in one paragraph

The weak answer treats a quantum state as a bag of probabilities with mystical language attached: "both values at once," "collapses when observed," "exponentially parallel." The strong answer is a representation contract: one explicit ordered basis, normalized complex amplitudes, probabilities as squared magnitudes, global phase discarded and relative phase kept, mixedness carried in a density matrix when preparation uncertainty or subsystem reduction matters, and numerical comparisons made up to tolerance and global phase. Simulation code should make that contract explicit — qubit order, basis, dimensions, precision, whether the object is a ket, operator or density matrix, and the simulator version and shot count recorded with the artifact. A statevector simulation validates the ideal circuit, not hardware noise, preparation or readout behavior; those need separately validated tests. Every amplitude, probability, subsystem and displayed bit string should refer to one explicit ordered basis and one normalized physical state model, with no conversion of quantum coherence into a misleading classical story.